Samsung 2693HM - Doesn't turn on (no standby)

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

HOW they get charged? And with what voltage? How you can determ how much voltage is going to be inside one charged capacitor (without reading the markings on the capacitor itself).

Simple.

If you connect a 1.5V battery across a capacitor and then disconnect the battery, the capacitor will have.... guess what? 1.5V across it. If you do the same with a 9V battery, then the capacitor will have 9V across it.

Now imagine you put a resistor between the battery and the capacitor. If you "very quickly" connect and disconnect your capacitor to the battery with the resistor between them, the capacitor will not be at the same voltage as the battery - it will be lower, because the resistor limits the rate of current going into the capacitor, and thus allows it to charge up slower. However, if you leave the capacitor connected to the battery (again with the resistor in between them) "long enough", eventually the capacitor will reach the same voltage as the battery.

In the case of the circuits around IC602 and IC924, think of the SW pin as the battery and capacitors C621/C3045 as the capacitors in the above experiment. The main difference here is that instead of using a resistor like in the above experiment, the circuits use an inductor (L600/L902.) Why an inductor and not a resistor? Because with a resistor, there will be a lot of losses as current passes through it and generates heat. But with an inductor, the losses will be much smaller, because it can "store" some of the current that passes through it (at the peaks of the square wave pulses), and then give it back to the circuit (to capacitors C621/C3045) when there is nothing output between the peaks of the square waves.

I keep getting stuck completely when observing a cirucits and see that one end of capacitor is connected to a 'lines' which goes to ground, but...same lines are connected to other elements which should 'go' to close the circle. So I always keep asking myself 'how that dmn voltage is not lost, since its connected to the ground'.

I know what you mean and I was getting stuck in a similar way when I was a beginner in electronics and didn't understand how circuits work.

The reasons circuits are called "circuits" is because there is always a return path (or multiple return paths eventually all recombining back into a single return path.) Again, going back to batteries, since they are easier to understand... you CANNOT have a current go out of one terminal of the battery and not go back to the other. The battery is a electro-chemical reaction waiting to happen. By drawing current from one terminal of the battery and returning it to the other, you are essentially allowing the chemicals inside the battery to react and become more neutral. While doing so, however, you can use the current to do useful work for you. In the case of connecting a simple light bulb across a battery... as the current passes through the filament of the bulb, it makes it heat up so much that the filament starts glowing, which is what produces light. But the current does NOT get lost in the filament of the bulb. If let's say 0.5 Amps of current goes out of the (+) terminal of the battery and makes it to the bulb, then the same 0.5 Amps will continue to flow through the bulb and out of the bulb, back to the (-) terminal of the battery. So current CANNOT get lost. However, the voltage can... or rather, it gets reduced as it passes through the bulb. This is called a voltage drop.

So in the case of complex circuits, like the ones surrounding IC602 and IC924, you have different components that manipulate the current to produce the desired functions you want the circuit to do. Some components will store current and release it later (like L600 and L902). Others, like the resistive divider formed by R604 and R606 simply "waste" power and turn it into heat... but it's needed so that the ICs can "read/see" the output voltages properly on their FB pins. And moreover, these resistors waste very little power, because they have relatively high resistances given the voltages in the circuit. And capacitors C621 and C3045 may seem like they waste power when they allow spikes of current to pass to ground when they charge up... but remember: as these capacitors charge up, they then can act like little sources/batteries themselves, thus returning the current that they seemed to waste initially.

That said, here are some videos on the matter that you may find useful:
https://www.youtube.com/watch?v=X4EUwTwZ110
https://www.youtube.com/watch?v=rbCXKhhzBN0

Ok....but now comes the stuped question: why the negative leg of capacitor is connected 'further' to other cirucit?

It's not. It's connected to ground... which is basically just a common point that other parts of the circuit use as a "reference" for the voltage levels they see/get.

In other words, if we are both standing next to each other on the 1st floor of a building and look up to a window on the 3rd floor, we will both agree that this window is xx meters above us. But if I'm standing on the 1st floor and you're standing on the 2nd floor and we look at that same window on the 3rd floor again, we will not agree that the distance between us and that window is the same xx meters. You may say it's 3 meters and I may say it's 6, and we may both be right.

So ground is just a common reference point for voltages. In the case of capacitors C621 and C3045, when they charge to 3.3V and 1.2V respectively, other parts in the circuit will see those voltages ONLY if both of these capacitors and the other parts of the circuit are using one same point as a reference. In this case, it's ground (though it doesn't have to be, and this will vary with different circuits.)

And 'when' that cirucuit is 'activated'. And...why it goes to other elements (like that schottky diode) but at the same time it 'goes' to the ground.

The Schottky diodes D600 and D942 play a very important role when there is a current passing through inductors L600 and L902 but when the voltage on the SW pin is 0V between the peaks of the square wave pulses: it provides a return path for the current "stored" in inductors L600/L902 to continue to flow out of them and into capacitors C621/C3045. Without these Schottky diodes, inductors L600/902 will try to do "anything" to keep that current going through them, because... remember, inductors don't like instant changes in current. And that "anything" could mean the inductor could generate 1000's of Volts just so that current can jump through and continue to flow. If you're wondering what the heck I mean, think about a time when you have disconnected an inductive high-current device when it was turned On and drawing current - for example, an electric radiator/heater. If you have, notice sometimes you see blue sparks/lights at the plug when you did that? - That's the inductance of the heating element generating high voltage and jumping the air gap between the disconnected plug prongs and wall outlet to try to maintain the current going through it. Another example would be the sparks you see at the brushes on a motor (e.g. a drill.) The sparks are created from the inductive property of the coils in the rotor of the motor.

So basically, this phenomenon is called inductive kickback.
Diodes D600 and D942 simply provide a return path so that the inductive kickback from L600 and L902 doesn't generate extremely high voltages that can damage IC602 and IC924. Moreover, these diodes allow the inductive kickback from L600 and L902 to supply current into C621 and C3045 between the voltage pulses from the SW pins. So L600/L902 and C621/C3045 are used in tandem as energy storage devices to transform the square wave pulses into a smooth DC voltage and provide power to the rest of the components on the logic board.

Again, look up how a buck regulator circuit works, and it may get more clear.

I have real issue while reading schematics, especially when I observe the lines connected to the ground. Everytime I see a ground sign, I say 'if the 'voltage' touch the ground, everything is wasted and nothing exist. But seeing other elmeents connected to the ground and at the same time being part of the full circuit makes me 'hang' and my brain 'freeze'.

Well, the voltage never really directly "touches" ground. If it did, that would be the same as placing a wire across the (+) and (-) of a battery or in the plug of you wall, and you know what that does, right.

But when you have resistors, capacitors, inductors, and other elements going to ground, that doesn't mean they are uncontrollably passing current to ground. Rather, it's exactly the opposite of that - they pass current in a controlled manner, again with the idea to manipulate it to do various function in the circuit. In the case of caps, they'll store charge. In the case of inductors, they'll maintain current. In the case of resistors... they may waste power, but divide voltages or limit currents going to other components.... and so on.

I'm trying to figureout how you got to some conclusions.

No shame in that. It's what this forum is for, after all: sharing knowledge. I've received a ton of help and knowledge from here, so now it's my turn to return the favor if I can.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Ok.... For last 2 hours I was staring at my monitor datasheet. I was looking at IC602 (and IC924) and the circuits around it, trying to understand how it works after you gave me the answer. Both ICs (IC602 & IC924) are the same. So I was checking dataseet of these 2 ICs (MP1583DN).
...
If I understood... once the EN pin 'sense' that there is a voltage (on EN pin) higher than 2.71, it will give the order to IC to start outputting pulsed voltage at SW pin"? Correct?

Yes.

Now (again, if I understood correctly), pin EN can get its > 2.71v internaly (from the circuit inside IC) or it can get it from some 'outside' source (externaly), at least thats how I figureout what datasheet for this (MP1583DN) IC sais.

Yes.

Basically, EN pin has an internal pull-up resistor. So if you don't connect it to anything, it will bring itself to 2.71V or higher and make the IC start... or you can force it with an external voltage source to either be low (less than 0.9V if you want the IC to be OFF) or high (over 2.71V to turn the IC ON.) Also note that datasheet says the IC has an internal Zener diode between EN pin and GND, so voltage on EN pin should NOT be forced over 7V (or you may damage the IC.)

I'm saying that, because circuits around these 2 ICs (IC602 & IC924) in monitor datasheet are the same except for the fact that IC602 has nothing connected on its pin EN while IC924 has a resistor R3016 connected, which receive a voltage from a 3.3v line.

Right.

The designer probably did this so that IC924, which is the regulator for the +1.2V_LIVE rail, will only turn On *after* IC602 brings up the +3.3V_LIVE rail to its proper voltage level. So this ensures that the 1.2V rail turns On a few moments later than the 3.3V rail. My guess is that this is how IC300 or perhaps some other IC expects to see its power come up. It's possible that the reverse order may produce an erroneous state for some of the ICs, and so this is a way to avoid that. Maybe!

(for IC924 3.3v voltage is 'pulled down' by the resistor R3016, because there will be a voltage drop across that R3016 resistor, making the voltage on EN pin of IC924 somewhere around 2.71v or higher, which would be enough to make IC start outputing pulsed (switching) voltage at its SW pin. Correct?

Correct.

Except R3016 is not pulling the voltage down on the EN pin - it's pulling it up. (I.e. it's a pull-up resistor.) But you get the concept: 3.3V rail goes through R3016 and pulls up pin 7 (EN) on IC924 so that it goes above 2.71V so that it can start up. The reason they put a resistor there and not connect it directly is that in case the voltage on the 3.3V rail goes above 7V for whatever reason (perhaps a malfunction of IC602... or like in your case, an external power spike from the HDMI source), then IC924 pin 7 will be protected by the high resistance of R3016, as it will not allow too much current to pass to that pin and damage the IC.

So...I would say, both ICs are being 'powered' by 5V_MICOM line on pin 'IN' so that ICs can start to work once the 'EN' pin gives it a 'go'.

Yes.

And remember, once the monitor is turned on, the 5V_MICOM rail will be superseded by the +18V_IN line coming through diode D602, so IC924 and IC602 will see/use 18V on their pin 2 (IN) then.

... and IC starts to 'work' by outputting some voltage (I don't know what are the values) on its SW pin.

Yes.

The voltage output at the SW pin on these ICs will be either 5V or 18V, but pulsed square wave (and not DC!), depending if +18V_IN is present at the time or not. If not present, 5V_MICOM is the supply and the ICs will output square wave pulses with peaks of 5V on their SW pins. If +18V_IN is present, then the ICs will output square wave pulses with peaks of 18V on their SW pins, but the width of the square wave pulses will be shorter than with 5V.

At the other hand IC924 gets EN voltage externaly (and I guess thats the voltage created by [color-red]IC924[/color]) and once its EN pin gets that ~2.71v which comes after R3016 resistor, its starting a process of ouputting some (again I don't know the values) switching voltage at its SW pin? Correct?

Yes.

But I think you meant to say IC602 there above where I highlithed the "IC924" text in red color.

The pulsed (square) voltage is then smoothed by inductors L600 (for IC602) and L902 (for IC924)?

Yes... though the word "smoothed" here is a very crude description.

You'll need to know some basic circuit theory behind inductors to understand this... but basically, inductors won't allow "instant" changes in current across their pins but will allow instant changes in voltage across their pins (which is the exact opposite of capacitors.) In short, all you need to know is that inductors are analogous to what capacitors do - they both are a type of passive energy storage device. In the case of capacitors, they charge up to some voltage and then maintain that voltage like a tiny small battery. With inductors, they do the same, but with current instead of voltage: they start passing current through them and then try to maintain that current until their magnetic field in their core is "depleted".

So the importance of all of the above is that if inductors L600 and L902 are NOT in the circuit, then those square wave voltages output by the SW pin will drive huge/massive spikes of current through capacitors C621 and C3045 as C621 and C3045 try to charge "instantly" from the square waves. But because of inductors L600 and L902, this won't happen, because the current going to capacitors C621 and C3045 will ramp up and down much more slowly. Thus L600 and C621 work together to transform the 5V/18V square wave pulses down to 3.3V. Same goes for L902 and C3045 to make 1.2V.

Look up how switching "buck regulators" work for more info.

But then, I used the same formula to calculate FB voltage for IC602 (which creates 3.3v output voltage). And the FB voltage for that IC is 1.223v. So....no matter what is the output voltage, FB pin require to always 'sense' ~ 1.2v to keep things 'constant'. Once that voltage starts to oscilate (I guess drop), it will (as you said) send the impuls to IC to 'increase' the voltage that comes out at SW pin to make the output voltage intact. Correct?

Yes, I think you're beginning to understand the concept here.

Basically, think first when the monitor was unplugged and there was no power anywhere on anything. Then you plug it in and PSU starts giving 5V to logic board on 5V_MICOM pin. IC602 sees 5V on its IN pin and waits until voltage on its EN pin goes 2.71V so that it can start switching. At this point, there is still no voltage on the SW pin and thus C621 is also empty/discharged (0V or thereabouts.) Now let's say EN pin has reached 2.71V or higher, thus enabling IC602. At this point, IC602 reads the voltage on its FB pin: it's 0V, because C621 is discharged (the resistive divider formed by R604 and R606 divides the voltage from C621 and feeds it into the FB pin, but because the voltage at C621 is 0V, then diving 0 by any number still gives you 0.) But IC602 expects to see about 1.22V on that pin. To make that happen, it starts outputting 5V square waves on its SW pin. These square waves go through L600 and get smoothed by C621, thus increasing the voltage on C621. Once the voltage on C621 reaches 3.3V, the resistive divider created by R604 and R606 ensures that there will be about 1.22V on the FB pin. If the voltage goes higher than 3.3V on C621, then the voltage on FB pin will go slightly higher than 1.22V, and IC602 will reduce the width of pulses on the SW pin until the voltage on C621 decreases to 3.3V (or whatever voltage that makes the FB pin see 1.22V.)

What I don't know is: what is the voltage that comes out at SW pin? Its pulsed, but... is it 3.3v (for IC602) or 1.2v (for IC924)?

Its 5V or 18V, as mentioned above - same voltage as whatever is found on the IN pin.

For example, lets say that the voltage which comes to pin 'IN' is 18v. What are the voltage values at pin SW? Are they...'random'? Or pulsed with 'peak' being 18v? Or its much lower?

Square wave pulse with 18V peak.

And one thing that bothers me so much for a very long time....

Capacitors. Or in this case 'filter capacitors'. How they exactly work? I know they are getting charged, but then what? They must discharge themselfs in some momment? (or not)?

They do. As IC300 and other ICs on your monitor logic board draw current from the 3.3V and 1.2V lines, they will discharge C621 and C3045 down to nothing if IC602 and IC924 don't replenish them. But you do need C621 and C3045 in there, because without them, the ICs on your monitor will be getting 5V or 18V square wave spikes... and that's not good!

10k char limit. Post #2 coming below.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

First, L307 is connected as you said to AVDD_DVI line. What I don't understand is: why its called '_DVI' line? I see the marking, but what confusing me is that (what I suspect all this time) a higher voltage came through HDMI cable from a cable TV device. At page 79 (service manual) there is 'HDMI_INPUT' and 'DVI_INPUT'. So... (I'm probablby wrong)...but why that line is called '_DVI' and not 'HDMI' if we expect that higher voltage came through HDMI port? Maybe DVI and HDMI is the 'same thing' just port has a different 'shape'?

Exactly.
You answered yourself with that last question. HDMI and DVI use the same type of signaling. DVI is older, though... I think. So for monitor ICs with a single digital input, I wouldn't be surprised if they refer to this signaling as DVI.

Now as far as the HDMI and DVI connectors on your monitor: look at page 79 of the SM again. IC11 is your DVI/HDMI switch IC - that is, it picks whether the signals from the HDMI or the DVI connector go through to IC300. And you can see that on the left side of IC11, you have HDMI lines and DVI lines, while on the right you have the RX- lines that go to IC300.

The surprising part is that IC11 didn't get damaged from the incident.... or at least it doesn't look like it. Perhaps if you connect BD1005 back and the +3.3V_LIVE doesn't want to come up after that or you get a low resistance across C621 again, then IC11 could be bad too. But by the looks of it, I think it might have survived. Still, DO NOT solder BD1005 back in - for now.

Another thing that just came through my mind in last couple of days...probably a crazy one again: Could it be possible, hypoteticly, that my graphic card 'somehow' sent 'some higher voltage' through DVI cable to my monitor and (I know this sounds stuped), my monitor then sent higher voltage back through HDMI cable to cable company TV device?

No.

Your video card has the same kind of protection diodes as the monitor, so that should limit its output from damaging devices through its signal ports. Of course, there may be a slight possibility for that to happen if you used a really cheap no-name gutless wonder PSU that can damage stuff when it blows up. But this is much more rare.

I suspect your cable co. TV box burned your monitor because of improper grounding - either on the coax or your house wiring (and I know that in a lot of former Soviet-block countries, it was pretty much standard for many old apartments and houses to use AC wiring with no grounding, so that potentially compounds the problem, particularly when some electricians at the time would ground an outlet by tieing ground to Neutral at each outlet.)

One more thing about that AVDD_DVI line... on page 79, when I observe lines for all L300-307, just above that, there is a 'small part of circuit' (if I can call it that) that shows 2 ceramic capacitors connected to AVDD_DVI line. There other 'pad' is (ofc) connected to ground. Could it be...in theory... that they could be faulty ones can cause of all this trouble? I'm reffering to C306 & C307 (both 1uF). I'll upload a picture so you can see what I mean.

Yes, you were actually reading my mind here.
SMD ceramic caps have a tendency to short-circuit when exposed to over-voltage. So it is indeed quite possible that one (or both) of these shorted... and that would be wonderful if they did, because it means they may save IC300 from over-voltage.

So for C306 and C307, you can try removing them to see if the resistance to ground of the AVDD_DVI line increases. On that note, same applies to C370, since it is in parallel with C373, which you measured as 1 Ohms, and which is connected to pin 69 (AUVREF).

I connected both power board and logic board together, and then connected a power board to a wall socket through a cable. Then I measured voltages on capacitors that you asked me to check.

C621 = 3.297v
C3045 = 1.224v

Well, I guess thats the end of the road for my logic board IC300.

Not yet!

Check/remove the above ceramic capacitors mentioned. If they don't measure short-circuit, then they are probably OK. However, that could imply IC300 is bad. Before we conclude that, though, if you can / have the nerves, try measuring across all ceramic capacitors to see if any more measure 2 Ohms or less. Once you find any that do, just note them down for now.

The thing that confuse me here is... why resistance on C3045 is only 3.6 Ohm, but once I checked the voltage on C3045 it shows voltage of 1.224v?

Don't know.
It could be normal, but it could also be something partially-shorted inside IC300, drawing more current but still not so much to the point where it trips IC924.

On that note, if you do perform the power-up experiment again with the main/logic board connected to the PSU board, does anything on the main/logic board get very hot, like IC300 or any other chip?

Couuld it be that these 2 caps are faulty? I'm thinking...I remember how I felt a 'hard strong' smell when I came back into my room. Like some burned electronics. Could that smell be from something like these caps? (though I don't see any physical damage when I look at them).

Electrolytic cap C373 is rated for 35V, so it is unlikely to get damaged before IC300. Same goes for C367, which is rated for 16V - a lot higher than what IC300 could withstand, so it's unlikely that either of these electrolytic caps got damaged. But the ceramic caps across them, C370 and C358, respectively, could have taken a hit.

With that said, I think the next thing for you to try is to solder back L302, L303, and L304, since you measured fairly high resistance on their "right" pads on the board.

Also, what resistance (to ground) did the other of these inductors measure on their "right" pads - i.e. L300, L301, and L305. We already know AVDD_DVI line (L307) is no good, so DO NOT solder that one back in. But if the others show high enough resistance (over 50 Ohms minimum, I'd say), then you can solder back their corresponding inductors too. That way, maybe we can get IC300 to work partially without L307 soldered back... which with a bit of hope, maybe you just loose the DVI/HDMI input but could still use the VGA input? IDK, we will see.

Once you solder these back, power on the monitor again. You may also want to connect the pannel / front button board to the logic board to see if your power LED works and if you can turn the monitor On and Off. If you can't, check again if the +3.3V_LIVE and +1.2V_LIFE are still OK and report back what you saw/measured.

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Hey momaka! Thanks for another reply!

So now, just before I did everything, I measured resistance of C621, just to confirm its still 1.2 Ohm. And...it is (was) still 1.2 Ohms.

Then I:

removed L307 -> resistance of C621 is 245.7 Ohm

Ouch! That's not a very good thing. Well, the 245.odd Ohms of resistance is probably fine - it actually means we likely have eliminated the short-circuit on the +3.3V_LIVE rail. Unfortunately, the short-circuit seems to be on the AVDD_DVI line that goes to IC300... and AVDD_DVI is essentially the same thing as the +3.3V_LIVE. L307 is basically what separates the two.

I was thinking (kinda) in the same direction when I removed that L307 and observed monitor service manual at page 79. But... there are things that I coudn't figureout there.

First, L307 is connected as you said to AVDD_DVI line. What I don't understand is: why its called '_DVI' line? I see the marking, but what confusing me is that (what I suspect all this time) a higher voltage came through HDMI cable from a cable TV device. At page 79 (service manual) there is 'HDMI_INPUT' and 'DVI_INPUT'. So... (I'm probablby wrong)...but why that line is called '_DVI' and not 'HDMI' if we expect that higher voltage came through HDMI port? Maybe DVI and HDMI is the 'same thing' just port has a different 'shape'?
Another thing that just came through my mind in last couple of days...probably a crazy one again: Could it be possible, hypoteticly, that my graphic card 'somehow' sent 'some higher voltage' through DVI cable to my monitor and (I know this sounds stuped), my monitor then sent higher voltage back through HDMI cable to cable company TV device? I would say thats not possible but... I'm a newb and I keep allowing myself to have some crazy/stupid thoughts. My graphic card works. I bought a new monitor and connected it through DP port, since my new monitor only has DP and HDMI ports. It doesn't have (like Samsung monitor did) also DVI and VGA ports, so I can't currently test if DVI port on my GPU works. I would say its ok, otherwise GPU would be faulty, but... as I said... even if I know it sounds (and for sure, is) stupid, I had to put this in my message, just to be out in the open, for you to see what also went through my mind.


One more thing about that AVDD_DVI line... on page 79, when I observe lines for all L300-307, just above that, there is a 'small part of circuit' (if I can call it that) that shows 2 ceramic capacitors connected to AVDD_DVI line. There other 'pad' is (ofc) connected to ground. Could it be...in theory... that they could be faulty ones can cause of all this trouble? I'm reffering to C306 & C307 (both 1uF). I'll upload a picture so you can see what I mean.

So forget about removing FT300 and D600 for now. We are going to go back to testing if the +3.3V_LIVE rail will turn On. It is OK that L300-L307 and other ferrite beads are removed. DO NOT put them back in the monitor (we will do that sometime later only once we find that the +3.3V_LIVE is working.) So connect the PSU board to the main/logic board and plug it in the wall. You probably won't need to connect the monitor bezel / front buttons for this test, but you can if you like.

When you plug the monitor in, measure the voltage across capacitor C621. Does it measure 3.3V or close? If yes, check the voltage across capacitor C3045. Does it measure 1.2V or close? Please report back what voltages you get.

I connected both power board and logic board together, and then connected a power board to a wall socket through a cable. Then I measured voltages on capacitors that you asked me to check.

C621 = 3.297v
C3045 = 1.224v

If both of these voltages come up, I suspect that would mean bad news for IC300. But let's wait and see what voltages you come back with.

Well, I guess thats the end of the road for my logic board IC300.

If there is still a chance that something else could be faulty (and not IC300), I'm ready to do further testings (or solder/desolder) of the components.

Last thing that I did was to measure resistance on other polarized caps close to C621. Most of them have resistance in MegaOhms, except that:

resistance on C3045 is 3.6 Ohm

Another ouch!
I suspect that may be another unhealthy low resistance, indicating the +1.2V_LIVE rail may also be shorted... and given that IC300 has direct connections to it, things are not looking good for IC300. But I guess we will see what happens when you try to measure voltages above.

On that note, if neither +3.3V_LIVE nor +1.2V_LIVE show any voltage, check the voltage on capacitor C619 (which is the input 5V/18V for IC602 and IC924 that generate the +3.3V_LIVE nor +1.2V_LIVE rails, respectively.)

Well, since (unfortunately) both caps (C621 & C3045) hand their expected voltages, I didn't measured voltages on other caps (like C619).
The thing that confuse me here is... why resistance on C3045 is only 3.6 Ohm, but once I checked the voltage on C3045 it shows voltage of 1.224v?

The resistances on C373 and C367 are quite worrisome.
C373 appears to have the same resistance as AVDD_DVI to ground, which was a similar resistance we had on +3.3V_LIVE before removing L307, implying AVDD_DVI was the likely culprit.
And C367 appears to have a similar resistance to ground as the 1.2V_LIVE rail... so that's something to keep in the back of our mind.

Couuld it be that these 2 caps are faulty? I'm thinking...I remember how I felt a 'hard strong' smell when I came back into my room. Like some burned electronics. Could that smell be from something like these caps? (though I don't see any physical damage when I look at them).

Well, thanks for the resistance measurements. They really do help and give a better clue as to what may be going on. So I appreciate the thorough work you did here.

Don't mention it! All I'm doing as a newb is trying to think of what measurement I can do in order to help you to help me.

BTW, I'll answer some of the other questions you had in post #55 in another post either later tonight or tomorrow.

No rush! (especially about my rant/cry parts of posts). You can answer them (if you decide to) when you want to.

Thanks again for your reply!

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

So, I guess same would be for reading/writting the firmware for that Samsung monitor. The only difference is: I pulled out the bios IC from its socket on Asus motherboards, while in the case of monitor firmware IC, I would have to desolder it first.

Well, if you do obtain a 2493HM board, you might as well just swap EEPROM ICs between the two boards rather than messing with your programmer. It would just be another step in the process that is not needed.

But, before I start removing things, I wanted to first finish removing the last ferrite bead (L307) which I didn't removed last time, since after removing L305, resistance on C621 jumped to 1.2 Ohms. I was thinking thats that, so I stopped and didn't removed L307 as I mantioned in previous post.

Ah OK, there was a misunderstanding on my part then. I thought you already had removed L300-L307 already and the resistance was at 1.2 Ohms.

So now, just before I did everything, I measured resistance of C621, just to confirm its still 1.2 Ohm. And...it is (was) still 1.2 Ohms.

Then I:

removed L307 -> resistance of C621 is 245.7 Ohm

Ouch! That's not a very good thing. Well, the 245.odd Ohms of resistance is probably fine - it actually means we likely have eliminated the short-circuit on the +3.3V_LIVE rail. Unfortunately, the short-circuit seems to be on the AVDD_DVI line that goes to IC300... and AVDD_DVI is essentially the same thing as the +3.3V_LIVE. L307 is basically what separates the two.

So forget about removing FT300 and D600 for now. We are going to go back to testing if the +3.3V_LIVE rail will turn On. It is OK that L300-L307 and other ferrite beads are removed. DO NOT put them back in the monitor (we will do that sometime later only once we find that the +3.3V_LIVE is working.) So connect the PSU board to the main/logic board and plug it in the wall. You probably won't need to connect the monitor bezel / front buttons for this test, but you can if you like.

When you plug the monitor in, measure the voltage across capacitor C621. Does it measure 3.3V or close? If yes, check the voltage across capacitor C3045. Does it measure 1.2V or close? Please report back what voltages you get.

If both of these voltages come up, I suspect that would mean bad news for IC300. But let's wait and see what voltages you come back with. Also, if you did remove D600, please solder it back in before performing the above voltage tests. Without D600 in the circuit, IC602 may get damaged. Same applies with diode D942. So DO NOT plug in the monitor if D600/D942 are removed. The switching regulators IC602 and IC924 cannot function properly without them.

Now, without removing any other element that you asked me to remove (like FT300 or D600), I measured resistance to ground on every pin of IC113:

IC113 - (its the 2.5V regulator for the RAM chip, IC301)
==================================
pin 1 resistance to ground: 245.7 Ohm
pin 2 resistance to ground: 0.1 Ohm
pin 3 resistance to ground: 147.7 Ohm

OK, those appear normal.
Pin 1 is the input of IC113 and connected to the +3.3V_LIVE, so it's normal that it measure the same resistance as C621 to ground.
Pin 2 is ground... so that's normal.
Pin 3 is the output +2.5V_M rail going to the memory, which is also a pretty normal resistance for a RAM chip.

Last thing that I did was to measure resistance on other polarized caps close to C621. Most of them have resistance in MegaOhms, except that:

resistance on C3045 is 3.6 Ohm

Another ouch!
I suspect that may be another unhealthy low resistance, indicating the +1.2V_LIVE rail may also be shorted... and given that IC300 has direct connections to it, things are not looking good for IC300. But I guess we will see what happens when you try to measure voltages above.

On that note, if neither +3.3V_LIVE nor +1.2V_LIVE show any voltage, check the voltage on capacitor C619 (which is the input 5V/18V for IC602 and IC924 that generate the +3.3V_LIVE nor +1.2V_LIVE rails, respectively.)

Then, (just for 'fun') I measured resistance of schottky diodes close to IC602 and IC942. I checked twice, by switching position of the multimeter probes. I checked directly on the diodes. So....

D600 -> resistance: 241.8 Ohm , 245.3 Ohm (reversed)
D942 -> resistance: 3.8 Ohm , 3.8 Ohm (reversed)
D601 -> resistance: 1.857 KOhm , 1.726 MOhm and dropping (reversed)
D602 -> resistance: 1.833 KOhm, 14.03 MOhm and dropping (reversed)

Those appear normal.
D600 is connected to the +3.3V_LIVE on one side and ground on the other, so it's normal it will read the same resistance as C621.
Likewise, D942 is connected to the +1.2V_LIVE on one side and ground on the other, so it's normal it will read the same resistance as C3045.

As for D601 and D602... nothing shorted on their end, so that's good (and I expected that anyways, since we did get 5V/18V on C619 before.)

After this, since that C3045 capacitor has small resistance (of 3.6 Ohm), I decided to check resistances of every electrolytic capacitor on the whole main/logic board. And, while most of them had their resistances in a KOhms or MOhms range, some had resistance in 'Ohm' range. These 'few' are:

C3045 resistance: 3.7 Ohm (^ as I said above already)
C650 resistance: 148.5 Ohm
C373 resistance: 1 Ohm (one Ohm)
C367 resistance: 3.7 Ohm
C649 resistance: 244.9 Ohm

The resistances on C373 and C367 are quite worrisome.
C373 appears to have the same resistance as AVDD_DVI to ground, which was a similar resistance we had on +3.3V_LIVE before removing L307, implying AVDD_DVI was the likely culprit.
And C367 appears to have a similar resistance to ground as the 1.2V_LIVE rail... so that's something to keep in the back of our mind.

At last, since both IC602 and IC924 are the same ICs (same model) I decided to check resistance of every pin (on both ICs) to ground. While comparing measured values of the same pins I noticed that the same pins had very large difference of resistance.

That's OK, because IC602 and IC924 generated different voltages and are thus connected to different ICs and parts of the logic board. So I wouldn't worry about this.

You know those ferrite beads (inductors) L300-L307 that I removed? Well, after I removed them there are two pads (per inductor place) which are exposed now. I mean pads where inductors were being soldered. Since now they are desoldered, I decided to measure resistance to ground of all pads of L300-L307.

Well, I put a black probe on metal part of HDMI connector (cause it was easy for me to 'lock it in place'), and then moved the red probe from one pad to other pad (of the same side) of every inductor place. I ended up measuring around 244.5 Ohms for all pads (that were on the same 'side', lets call it 'left' side). I would say: ok, those are positive pads. But then, I started moving red probe (while still holding black probe on HDMI metal plate) over oposite pads (lets call it 'right pads'). And while some measures went from 0.3 Ohms to 1.2 Ohms and 2.2 Ohms, there are 3 pads that measure much more resistance (2 even measure in MOhms).


L302 > 193.1 Ohm
L303 > 13 MOhm and increasing to 55MOhm and then reach OL state
L304 > 0.745 MOhm (at first unstable but finally stay calm at this value)

Well, if anything, those readings reaffirm that AVDD_DVI (connected to L307) and some of the other lines with low resistances may be shorted, while the "right" pads for L302, L303, and L304 show resistances that appear a lot more normal... or at least not shorted. (I don't actually really know what resistances any of these pads should read... and there may not even be a "stable" / solid value that they should have anyways. But generally, whenever I see a resistance under 10 Ohms on a supply rail that isn't know to be super-heavy on current draw, I can't help but suspect it. And those low-Ohm resistances, especially on L307 "right" pad, suggest that IC300 may have taken a hit when the whole accident happened. However, note that this does not mean that IC300 is dead. If we are lucky, maybe the voltage spike only took out the circuit parts inside IC300 that are for the DVI/HDMI input. So that still may leave a possibility of IC300 with VGA input only. But I don't know, we will see when you get the above voltages of +3.3V and +1.2V_LIVE.

At the 2nd thought I said: wait, maybe these 3 (L302, L303, L304) are actually goood. ...'cause somewhere (inside big IC300) connections to ground are currently 'broken'. Maybe the other ones (L300, L301, L305 & L307) are not good?

Yup, quite likely L300, L301, L305, and/or L307 maybe have transistors inside IC300 that are shorted to ground - hence the low resistance to ground.

As far as the actual inductor components L300-L307 you removed, though - those will still be good to re-use.

And then I said, lets tell all this to momaka, for sure he will have an idea what all of this means (if it means anything at all).

Well, thanks for the resistance measurements. They really do help and give a better clue as to what may be going on. So I appreciate the thorough work you did here.

NOTE: I'll upload again image of the main/logic board, in case you would have to look for the post where I shared it. Also datasheet for ICs IC602 & IC924 is re-shared again.

Thanks. I still remember where the stuff is on page 2, though. And I have the datahsheets for these ICs and the service manual downloaded to my PC so I can access them quicker.

BTW, I'll answer some of the other questions you had in post #55 in another post either later tonight or tomorrow.

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Most likely, that would be IC303 (MX25L4005M2C-15G), which is a 4-Mbit CMOS flash... or it may also be IC302 (S-24CS08AFJ-TB), which is an 8 Kb EEPROM... but I doubt it (most likely it's IC303.)

But in any case, most likely you can try plugging a 2493HM logic board in your 2693HM as-is without reprogramming. At worst, the backlight dimming might not work correctly if the 2493HM logic board isn't sending a "proper" signal to your 2693HM inverter board. But nothing should be damaged by that.

Yes, uploading the proper firmware... or even just taking the actual chip from your old 2693HM board and swapping it on the 2493HM board should do the trick (if you want to avoid doing any programming, like I do, since I'm not well-versed in that area and not too interested either - I like to stick to hardware whenever possible and leave the software/firmware stuff for someone else that can't or doesn't want to do the hardware stuff)

Ok, thanks for this reply. Couple years ago I bought a MiniPro TL866A programmer, and I already read and write few bios firmwares from PC motherboards. So, I guess same would be for reading/writting the firmware for that Samsung monitor. The only difference is: I pulled out the bios IC from its socket on Asus motherboards, while in the case of monitor firmware IC, I would have to desolder it first. I beleive I already ahve 'sockets' (or whats they called?) for putting a desoldered IC into it, and further connecting it to a programmer board. Thats why I asked if the 'information' for model 2493HM is writting somehwere which operating system use to 'read' what model is plugged in.And since such information is written (loaded) inside certain IC, I could re-write it if that is whats required to make OS correctly 'read' model of my monitor.





Ok, now back to the most important part: things that you requested to be done.


So.... I was ready to remove FT300, D600, and then at last measure resistance to ground of all 3 pins on IC113. Ofcorse I planned to check resistanced on C621 everytime I removed some element (in this case, efter removing FT300, and also after removing D600).
But, before I start removing things, I wanted to first finish removing the last ferrite bead (L307) which I didn't removed last time, since after removing L305, resistance on C621 jumped to 1.2 Ohms. I was thinking thats that, so I stopped and didn't removed L307 as I mantioned in previous post.

So now, just before I did everything, I measured resistance of C621, just to confirm its still 1.2 Ohm. And...it is (was) still 1.2 Ohms.

Then I:

removed L307 -> resistance of C621 is 245.7 Ohm

Huh!

Now, without removing any other element that you asked me to remove (like FT300 or D600), I measured resistance to ground on every pin of IC113:



IC113 - (its the 2.5V regulator for the RAM chip, IC301)
==================================
pin 1 resistance to ground: 245.7 Ohm
pin 2 resistance to ground: 0.1 Ohm
pin 3 resistance to ground: 147.7 Ohm



Last thing that I did was to measure resistance on other polarized caps close to C621. Most of them have resistance in MegaOhms, except that:

resistance on C3045 is 3.6 Ohm


Then, (just for 'fun') I measured resistance of schottky diodes close to IC602 and IC942. I checked twice, by switching position of the multimeter probes. I checked directly on the diodes. So....

D600 -> resistance: 241.8 Ohm , 245.3 Ohm (reversed)
D942 -> resistance: 3.8 Ohm , 3.8 Ohm (reversed)
D601 -> resistance: 1.857 KOhm , 1.726 MOhm and dropping (reversed)
D602 -> resistance: 1.833 KOhm, 14.03 MOhm and dropping (reversed)


After this, since that C3045 capacitor has small resistance (of 3.6 Ohm), I decided to check resistances of every electrolytic capacitor on the whole main/logic board. And, while most of them had their resistances in a KOhms or MOhms range, some had resistance in 'Ohm' range. These 'few' are:

C3045 resistance: 3.7 Ohm (^ as I said above already)
C650 resistance: 148.5 Ohm
C373 resistance: 1 Ohm (one Ohm)
C367 resistance: 3.7 Ohm
C649 resistance: 244.9 Ohm


At last, since both IC602 and IC924 are the same ICs (same model) I decided to check resistance of every pin (on both ICs) to ground. While comparing measured values of the same pins I noticed that the same pins had very large difference of resistance.

These two ICs are MP1583DN. Here are their pins:

Pin 1: BS
Pin 2: IN
Pin 3: SW
Pin 4: GND
Pin 5: FB
Pin 6: COMP
Pin 7: EN
Pin 8: SS


While keeping black probe on the screw hole of the main/logic board, I moved red probe from pin to pin, and measured their resistances. Here it is:


IC 602
=======
Pin 1: 19.04 MOhm
Pin 2: 4.1 MOhm and decreasing
Pin 3: 244.9 Ohm
Pin 4: 0.1 Ohm
Pin 5: 2.127 KOhm
Pin 6: 11.96 MOhm
Pin 7: OL
Pin 8: 85.6 KOhm



IC 924
=======
Pin 1: 22.4 MOhm
Pin 2: 4.7 MOhm and decreasing
Pin 3: 3.7 Ohm
Pin 4: 0.1 Ohm
Pin 5: 3.7 Ohm
Pin 6: 11.96 MOhm
Pin 7: 10.4 KOhm
Pin 8: 81.1 KOhm


^ For both ICs, Pin 4 resistance is 0.1 Ohm (because its a GND pin), so thats good.
For IC 602, pin 7 is OL, and thats because monitor schematics shows that its not connected to anything, so thats good as well.

There are values on certain pins which made me think that something is not right, but now when I observed the monitor schematic while checking measured values, I think that both ICs have correct resistance values on their pins, or at least when they are observed 'together'. Maybe if every IC is observed as a single entity, then maybe something is wrong with them.







And (now for real) finally, I did one more thing:

You know those ferrite beads (inductors) L300-L307 that I removed? Well, after I removed them there are two pads (per inductor place) which are exposed now. I mean pads where inductors were being soldered. Since now they are desoldered, I decided to measure resistance to ground of all pads of L300-L307.

Well, I put a black probe on metal part of HDMI connector (cause it was easy for me to 'lock it in place'), and then moved the red probe from one pad to other pad (of the same side) of every inductor place. I ended up measuring around 244.5 Ohms for all pads (that were on the same 'side', lets call it 'left' side). I would say: ok, those are positive pads. But then, I started moving red probe (while still holding black probe on HDMI metal plate) over oposite pads (lets call it 'right pads'). And while some measures went from 0.3 Ohms to 1.2 Ohms and 2.2 Ohms, there are 3 pads that measure much more resistance (2 even measure in MOhms).


L302 > 193.1 Ohm
L303 > 13 MOhm and increasing to 55MOhm and then reach OL state
L304 > 0.745 MOhm (at first unstable but finally stay calm at this value)

I also measured resistance on both pads of BD1005:

1st pad resistance is: 3.519 KOhm
2nd pad resistance is: 205.5 Ohm

I tried to figureout what that means, but the more I think, I'm getting more confused.
AT first I thought: shouldn't all the pads (of that side) be connected to ground. Then I remember we desolder some elements to 'brake' the line that connect the ground. But...thats the thing. We desoldered only L300-L307 and BD1005, nothing else.

At the 2nd thought I said: wait, maybe these 3 (L302, L303, L304) are actually goood. ...'cause somewhere (inside big IC300) connections to ground are currently 'broken'. Maybe the other ones (L300, L301, L305 & L307) are not good?

And then I said, lets tell all this to momaka, for sure he will have an idea what all of this means (if it means anything at all).


^ This was 'me' trying to do 'detective' work (for sure I failed, but I gave it a try)



If you still are not tired of my posts, let me know what I could try next.

Thanks again for your reply, momaka!

NOTE: I'll upload again image of the main/logic board, in case you would have to look for the post where I shared it. Also datasheet for ICs IC602 & IC924 is re-shared again.

p.s. My PM box is currently 100% full, but I'll try to make more space soon.

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Hey momanka! Thanks again for your reply!

I'll first comment on the subject revolving around that IC602 and IC924 circuit. I wont quote that part of your quote because I will probably again exceed the 10K character limit.


Ok.... For last 2 hours I was staring at my monitor datasheet. I was looking at IC602 (and IC924) and the circuits around it, trying to understand how it works after you gave me the answer. Both ICs (IC602 & IC924) are the same. So I was checking dataseet of these 2 ICs (MP1583DN).

I need to write this, and ask you to confirm that I understood it (or not).

Datasheet for MP1583DN (IC602 & IC 924) sais for EN pin:

Enable/UVLO. A voltage greater than 2.71V enables operation. For complete low
current shutdown the EN pin voltage needs to be at less than 900mV. When the
voltage on EN exceeds 1.2V, the internal regulator will be enabled and the soft-start
capacitor will begin to charge. The MP1583 will start switching after the EN pin
voltage reaches 2.71V. There is 7V zener connected between EN and GND. If EN is
driven by external signal, the voltage should never exceed 7V.

If I understood... once the EN pin 'sense' that there is a voltage (on EN pin) higher than 2.71, it will give the order to IC to start outputting pulsed voltage at SW pin"? Correct?

Now (again, if I understood correctly), pin EN can get its > 2.71v internaly (from the circuit inside IC) or it can get it from some 'outside' source (externaly), at least thats how I figureout what datasheet for this (MP1583DN) IC sais.

I'm saying that, because circuits around these 2 ICs (IC602 & IC924) in monitor datasheet are the same except for the fact that IC602 has nothing connected on its pin EN while IC924 has a resistor R3016 connected, which receive a voltage from a 3.3v line.

(for IC924 3.3v voltage is 'pulled down' by the resistor R3016, because there will be a voltage drop across that R3016 resistor, making the voltage on EN pin of IC924 somewhere around 2.71v or higher, which would be enough to make IC start outputing pulsed (switching) voltage at its SW pin. Correct?

So...I would say, both ICs are being 'powered' by 5V_MICOM line on pin 'IN' so that ICs can start to work once the 'EN' pin gives it a 'go'.
IC602 created (internaly) voltage higher than 2.71v on its EN pin, and IC starts to 'work' by outputting some voltage (I don't know what are the values) on its SW pin. That way, (I think) IC602 is the IC which creates 3.3v line to be used 'somewhere else on the main/logic board).

At the other hand IC924 gets EN voltage externaly (and I guess thats the voltage created by IC924) and once its EN pin gets that ~2.71v which comes after R3016 resistor, its starting a process of ouputting some (again I don't know the values) switching voltage at its SW pin? Correct?

The pulsed (square) voltage is then smoothed by inductors L600 (for IC602) and L902 (for IC924)? I don't know what is the value of voltages that comes out at SW pin, but before inductors (L600 and L902).


FB pin reads whats the voltage of 'voltage divider cirucuit). I used the calculation and for IC924 (which consist of one 'zero' ohm reisttor), the FB pin voltage will be the same as the output voltage (as you said). But then, I used the same formula to calculate FB voltage for IC602 (which creates 3.3v output voltage). And the FB voltage for that IC is 1.223v. So....no matter what is the output voltage, FB pin require to always 'sense' ~ 1.2v to keep things 'constant'. Once that voltage starts to oscilate (I guess drop), it will (as you said) send the impuls to IC to 'increase' the voltage that comes out at SW pin to make the output voltage intact. Correct?

I hope I got it right this time.


There are some things (but ofcorse) which I don't know and one which I can't figureout.


What I don't know is: what is the voltage that comes out at SW pin? Its pulsed, but... is it 3.3v (for IC602) or 1.2v (for IC924)? I know that these two inductors (L600 & L902) makes it smoothed, but...what kind of voltage (ranges) these two inductors 'smooth out'?
For example, lets say that the voltage which comes to pin 'IN' is 18v. What are the voltage values at pin SW? Are they...'random'? Or pulsed with 'peak' being 18v? Or its much lower?



And one thing that bothers me so much for a very long time....

It just filters the voltage and makes it smoother. Think of it like a small battery in the circuit - it charges when IC602/924 is outputting a voltage and then maintains that voltage when the IC stops outputting a voltage. The reason IC602/924 output a square wave voltage on pin 3 (SW) with varying width is because this is how you regulate a high voltage input (in this case 5V or 18V) into a much lower output (3.3V or 1.2V) with very little losses. The varying of the width of the square waves is called Pulse Width Modulation... or PWM for short.

Capacitors. Or in this case 'filter capacitors'. How they exactly work? I know they are getting charged, but then what? They must discharge themselfs in some momment? (or not)?
HOW they get charged? And with what voltage? How you can determ how much voltage is going to be inside one charged capacitor (without reading the markings on the capacitor itself).
When you say 'filter' capacitor...how he 'filter' it? And what he filters? Does it 'take' the 'extra' voltage out of the circuit and hold it (of there was more voltage than whats required at certain end of the cirucuit)? What does that capacitor do with its charging?
I keep getting stuck completely when observing a cirucits and see that one end of capacitor is connected to a 'lines' which goes to ground, but...same lines are connected to other elements which should 'go' to close the circle. So I always keep asking myself 'how that dmn voltage is not lost, since its connected to the ground'.
I know that a lot of people would say 'this guy...'. BUT no matter how capacitors may be 'simple' parts of the circuit, the more I think and try to understand them completely, the more I'm getting confused.


I know that you said that filter capacitor acts like a battery and that it charges when ICs are outputting the voltages and then maintains that voltage when ICs stops outputting the voltage. Ok, but confusing is in this: the way that capacitor 'received' its chages was through its positive leg. Then... the cap 'waits' untill IC stops outputtting the voltage, and then capacitor....'return' that voltage charge onto the 'output' end of the cirucuit and keep that output voltage 'constant'. So I would conclude voltage over that 'filter capacitor' is the same as the output voltage (3.3v for IC602, and 1.2v for IC924).
Ok....but now comes the stuped question: why the negative leg of capacitor is connected 'further' to other cirucit? And 'when' that cirucuit is 'activated'. And...why it goes to other elements (like that schottky diode) but at the same time it 'goes' to the ground.
I have real issue while reading schematics, especially when I observe the lines connected to the ground. Everytime I see a ground sign, I say 'if the 'voltage' touch the ground, everything is wasted and nothing exist. But seeing other elmeents connected to the ground and at the same time being part of the full circuit makes me 'hang' and my brain 'freeze'.

Seams like electronics is not for everyone. ...but I have interest in it. What makes me so much frustrated is that I put an effort to understand things, and the more I spend time on the subject, I only 'understand' that I know less than I thought I do. Its like I didn't made my brain to understand some new things, but just the opposite. I only understand that my ignorance is even larger. Thats how I feel...

I understand that I keep asking you questions like this, and thats probably over a line of decency. But what I'm trying to do everytime you try to explain to me why we have to desolder certain elements, is to look at the circuit and understand the things by myself before I do what you asked me to do. In other words, I'm doing my best to keep up with you and not just blindly fullfill your requests. I'm trying to figureout how you got to some conclusions.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

I wont care much if my 2693HM monitor is recognized as 2493Hm monitor. But still.... that 'information' must be 'written' somewhere (inside some cheap or something).

Yes.
Most likely, that would be IC303 (MX25L4005M2C-15G), which is a 4-Mbit CMOS flash... or it may also be IC302 (S-24CS08AFJ-TB), which is an 8 Kb EEPROM... but I doubt it (most likely it's IC303.)

but for instance... 2493HM has less CCFL lamps than 2693HM (at least thats what I concluded by observing different inverter boards between 2493HM and 2693HM.) So I was thinking.... 'maybe' different kind of 'information' needs to be sent to 2493HM inverter board from IC chip located at main/logic board).

Generally, no, since most monitor inverters have two inputs only: backlight ON (typically called BL_ON or BL_EN or similar) and backlight dimming (typically called BDIM or DIM). The first just tells whether the inverter should turn On or not. The latter is usually a 0 to 3.3V analog or PWM signal to control the light intensity (brightness) of the backlights.

However, this monitor might be a slightly more special case, as I see 3 other control signals going to the inverter board, besides BL_ON/OFF, which are listed on page 81 of the SM. These are labeled as SOS, DIM, and STATUS on connector CN601, which translate to signal lines A_DIM, BRT_CTL, and PWM_DIM. From what I can understand, PWM_DIM is used to tell the inverter whether to use PWM method for the dimming or analog voltage range (0 to 3.3V). And with that, BRT_CTL and A_DIM seem to be used for the dimming (the former being PWM control and the latter analog dimming) - at least for the 2693HM model. But why have two dimming methods, that I can't answer. Perhaps these monitors sometimes come with different inverter boards... or like you said, the 2493HM perhaps having less CCFLs and a different inverter, maybe that's where the difference comes.

But in any case, most likely you can try plugging a 2493HM logic board in your 2693HM as-is without reprogramming. At worst, the backlight dimming might not work correctly if the 2493HM logic board isn't sending a "proper" signal to your 2693HM inverter board. But nothing should be damaged by that.

At the end, if the only difference that I will see if I use main/logic board from 2493 instead of my faulty 2693 main/logic board, when I load my OS, and it shows its a 2493HM monitor connected (instead 2963HM), I wont care much about it. But I just want to know if anything can be done about it and 'what'? (I mean would uploading firmware or whatever, would 'fix that').

Yes, uploading the proper firmware... or even just taking the actual chip from your old 2693HM board and swapping it on the 2493HM board should do the trick (if you want to avoid doing any programming, like I do, since I'm not well-versed in that area and not too interested either - I like to stick to hardware whenever possible and leave the software/firmware stuff for someone else that can't or doesn't want to do the hardware stuff .)

Ok so, every resistor with '0' (zero) stamp on it is a 0 Ohm (link resistor)?

Yes.

btw. whats 'link resitor'? Like.... wire? If 'yes' than why not just put wire there? Wire has some small reistance (0.2 or 0.3 Ohm, right)?

Well, technically, a 0-Ohm "resistor" is indeed just a piece of wire inside.

The reason it is labeled as a resistor on the board is that let's say there are several different monitor models that use this logic board, and maybe one of them uses a different chip with a different voltage. All the manufacturer would have to do then is swap that 0-Ohm resistor with another SMD resistor during production on the assembly line. So in other words, it makes it easier to change/modify the circuit in production as demand for the two models varies.

So if this 'link reistor' has zero Ohms... how can circuit even consider it as a resistor and 'devide' the voltage?

It's not. That's why the output voltage will practically equal the reference voltage, V_FB.

Ok...you might not believe me, but I spent last 4 hours trying to understand dmn voltage divider. I watched few youtube clilps. They way they draw the circle there is with 'battery source'. So circuit goes from positive to first resistor, then 2nd reistor (in series with first reistor), then goes to negative 'pin' on the battery. Divider there is what you get 'after' first resistor. And I (think) I understand that.

Correct.

BUT, when I check how circuit looks like around that IC924 in the service manual, it doesn't look like those 'voltage divider circuits' from youtube that I was watching.

I can see why you're getting confused. To be honest, I myself had to look aat IC602 and IC924 several times to try to understand which pins are power input and which are power output.

The way it goes...

You have 5V or 18V (depending if monitor is in standby or turned On) coming into pin 2 (IN) of IC602/924. Capacitors C619 (electrolytic) and C3050 (SMD ceramic) just filter (smooth out) the voltage on this 5V/18V input pin.

Now, once power goes into pin 2 (IN), IC602/924 outputs voltage on pin 3 (SW) via an internal MOSFET inside the IC. This output voltage on pin 3 is a bunch of square wave pulses with amplitude that is either 5V or 18V - in other words, it is NOT a smooth DC voltage (it's pulsed DC.) Anyways, these square wave pulses are turned into a (fairly) smooth DC voltage by inductor L600/L902, capacitor C621/C3045, and diode D600/D942.

To get the proper output voltage, though (i.e. 3.3V or 1.2V), the smoothed DC output (from inductor L600/L902, capacitor C621/C3045, and diode D600/D942) is then fed to a voltage divider circuit consisting of R604 and R606 (for IC602) or R3027 and R3050 (for IC924). The output of the voltage divider (that is, the voltage after the "first" resistor in the voltage divider circuit - i.e. either R604 or R3027) is fed in to pin 5 (FB - short for "feedback") so that IC602/924 can "see" what voltage it is actually outputting and whether it should change the pulse width of the square wave pulses that come out of pin 3 (SW). In other words, pin 5 (FB) is used for monitoring the voltage output on the +3.3V_LIVE/+1.2V_LIVE rails and telling the IC when the voltage dips so that the IC can increase the width of the square wave pulses to make up for the voltage dip, or do the reverse if the voltage on these rails goes higher than it should.

Also I don't understand whats the purpose of capacitor C3045 there (in service manual)? Its polarized capacitor. Is it there to... make current flow in certain direction?

It just filters the voltage and makes it smoother. Think of it like a small battery in the circuit - it charges when IC602/924 is outputting a voltage and then maintains that voltage when the IC stops outputting a voltage. The reason IC602/924 output a square wave voltage on pin 3 (SW) with varying width is because this is how you regulate a high voltage input (in this case 5V or 18V) into a much lower output (3.3V or 1.2V) with very little losses. The varying of the width of the square waves is called Pulse Width Modulation... or PWM for short.

And the line that comes out of FB pin of IC924 is spliting in two directions just there (one to R3024 and one to R3050).

Yeah, that's probably the part that had you tripping.
Pin 5 (FB) is NOT outputting a voltage. It is inputting / "reading" a voltage from the voltage divider formed by resistors R3024 and R3050 and comparing it to an internal feedback voltage, V_FB, that is equal to 1.22V (typical.)

Basically, IC924 will change the width of the output square waves on pin 3 (SW) in accordance with what it "sees" back on pin 5 (FB). If the output of the +1.2V_LIVE rail lowers (perhaps due to a higher demand of power from some IC somewhere on the logic board), the voltage on pin 5 (FB) will also become lower due to the resistor "divider" circuit formed by R3024 and R3050. When IC924 sees that its voltage on pin 5 (FB) is running lower than the 1.22V internal reference voltage (V_FB), it will increase the pulse width of the square waves coming out of pin 3 (SW), which will then increase the voltage on the +1.2V_LIVE rail to try to bring it back to 1.2V, or as close as possible.

^ Also, one extra thing that bothers me: While you tried to give me example how we can get specific output voltage (2.44), I don't know how you came up with these voltages (2.44) or 1.22v? Where did you found 1.22v ? The schematic show the output voltage is 1.2. So if the input voltage (feedback voltage at pin 5 of IC924) is the same as output voltage (after is divided) is 1.2v .... then where did you found 1.22v ? (maybe you meant 1.2v)

Page 3, row 3 of the MP1583DN (IC924) datasheet tells you what its internal reference voltage, V_FB is... which is normally 1.22V... but it could be as low as 1.194V or as high as 1.25V (it can vary with temperature, load, and other factors like that.)

And at last: why use voltage divider if the voltage is the same from pin 5 to the 1.2 output pin? Why not linera voltage regulator instead?

Well, I think I have that answered now with all of the above info.
Since now you know pin 5 is NOT an output, hopefully you can see why this is the case. On that note, the internal reference voltage (V_FB) of MP1583DN is too weak to be used as an output voltage to power other chips. It is only used for signal purposes inside the IC.

Hope that answers some of these questions.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

OK, time to get back at it.
Sorry for the delay, I've been away from BCN for a few days.

And since you said that if the resistance changed during L300-L307 desoldering then 'very likely Ic300 was damaged', I guess thats that.

Correct?

Nope.
1.2 Ohms is still too suspiciously low to be a normal resistance on the +3.3V_LIVE. So I think we still haven't found the source of the short-circuit yet. The reason the resistance changed (slightly) is because IC300 also has some resistance to ground... just not that low.

Anyways, there are a few more suspect components we need to check. Reason I stopped above at L307 is to be able to keep track of what was tested and what wasn't. It appears there are no other connections between +3.3V_LIVE and IC300, so if anything, the fact that the low resistance remains while we have IC300 pretty much disconnected from +3.3V_LIVE means IC300 is likely OK.

So, try removing the following components below (one at a time), and then repeat the same procedure as before with the other ferrite beads: measure resistance to ground at positive (+) lead of C621 and report back the results.

Components to remove:
FT300 - this is an SMD inductor above IC300 and below IC303.
D600 - this is an SMD Schottky diode next to IC602 - the +3.3V_LIVE switching regulator IC.

If you still have a low resistance after removing these two components, things are going to be a little more complicated. First, find IC113 - it's located to the lower right side of the big main chip, IC300. IC113 is the 2.5V regulator for the RAM chip, IC301. Check resistance to ground on all 3 pins of IC113 (one at a time) and report back what you get. For the purpose of repeatability, reference pins 1, 2, and 3 the same way as the pins on IC304, like you did in this picture:
https://www.badcaps.net/forum/attach...6&d=1604372177

Now, pin 2 (the big tab) should obviously read short-circuit (0.5 Ohms or less) to ground, since that's how it's supposed to be connected. But I'm interested in the reading you get on pin 3, which is the +2.5V_M output. So check that resistance carefully and report back what you get.

That said, I doubt IC113 is going to be bad. The way things are looking, most likely either IC602 may be faulty or one (or more?) of the protection diodes (D105-D112 and D121-D128) on the DVI or HDMI connector are short-circuited. There is also a chance that a small SMD ceramic cap could be shorted somewhere that is directly connected on the +3.3V_LIVE rail, like C648 (next to IC113), or C109 & C110 (next to D-sub / VGA connector). If it gets to that point, you will likely either have to remove all of the protections diodes and ceramic caps mentioned above along with IC602... one by one... until the short-circuit low resistance disappears. Alternatively, we can also try forcing 3.3V externally from a PC ATX PSU (preferably with current limiting, like a 20-50 Watt, 12V incandescent bulb in series) into the +3.3V_LIVE and hope that if any of these components above are faulty, it will eat all the power and heat up or burn itself out.

The desoldering of components one by one should allow us to track better what was shorted... but it's A LOT more work, as you might imagine removing all of these tiny 3-pin SMD diodes. On the other hand, forcing 3.3V externally may be quicker at finding the fault... but if the faulty component is "hard" short-circuited, it may not heat up at all, even with very large currents flowing through it - and that can end up damaging traces on the board (hence the suggestion to use current limiting with a 20-50 Watt, 12V incandescent light bulb, like from a card headlight.)

But let's see first what results you get after removing FT300 and D600, and then measuring the 3 pins of IC113. If nothing looks to be changing, then you can either desolder the other components, or I can give you instructions on how to feed 3.3V into the +3.3V_LIVE with a PC ATX PSU safely (to the board.)

Anyways, that is all I have for you for now. I'll answer all the other questions above in another post, shortly... or I'll edit this one if time still allows.

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Resistance of capacitor C621 is: 0.7 Ohm (black probe on negative, red probe on positive leg)

That's NOT good.

It means the +3.3V_LIVE rail has a short-circuit across it somewhere.

I suppose let the hunt begin.

With that said, let's get to troubleshooting the short-circuit (0.5 Ohms) on the +3.3V_LIVE rail first.

to make things a bit easier, let's try removing components that are more "friendly" to a soldering iron, one at a time, which I think you should be able to do with a wide(r) chisel or flat tip on your soldering iron (or lots of solder on the tip if it's conical type.) Start with SMD ferrite bead BD1005. Once you remove it, check resistance to ground again at capacitor C621 positive (+) pin. If the resistance is still the same more or less (i.e. around 0.5 Ohms) on the +3.3V_LIVE, continue reading below. If not, stop and report what resistance you get now.
.
.
If you continued due to resistance still being 0.5 Ohms on the 3.3V rail... now remove SMD inductor / ferrite bead L300. Inductor L300 is a black / dark gray SMD component that can be found close to the lower-right corner of IC300. Check resistance to ground after that same way as when you removed BD1005 and either report back what you get (if resistance has changed), or continue below.
.
.
If you continued due to resistance still being 0.5 Ohms after removing L300, keep going at it with L301, then L302, then L303... and so, until you get to L307. These inductors / ferrite beads supply +3.3V_LIVE to IC300, so if the 0.5 Ohms resistance on the 3.3V rail disappears at some point while removing there, then very likely IC300 was damaged (though I doubt IC300 is damaged.)

Ok... I fear that last sentance between the brackets described it perfectly.


So here it is:

I desoldered BD1005 ferrite bead and measured resistance on the C621 capacitor (black on negative leg, red on positive let).
Resistance was the same as before: 0.7 Ohm

Then I started with:

removed L300 -> resistance of C621 was 0.7 Ohm
removed L301 -> resistance of C621 was 0.7 Ohm
removed L302 -> resistance of C621 was 0.7 Ohm
removed L303 -> resistance of C621 was 0.7 Ohm
removed L304 -> resistance of C621 was 0.7 Ohm
removed L305 -> resistance of C621 was 1.2 Ohm

This is where I stopped desoldering the rest of ferrite beads (only L307 was left, L306 doesn't exist on the board or in the service manual)


L304 was physically located on the board between L305 and L307. Yet, I was removing them by their number order ^ like I just wrote above.

And since you said that if the resistance changed during L300-L307 desoldering then 'very likely Ic300 was damaged', I guess thats that.

Correct?

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Samsung 2493HM monitor has the same main/logic board as Samsung 2693HM. They have exactly the same mark: BN41-00962A
(and I was thinking, IF all this fails, I could try to buy used main/logic board from model 2493HM, since they do look the same, and has the same makings: BN41-00962A).
But now, when you said that because for model 2493HM Service manual has some cirucits marked as 'delete', that such circuits were not included on main/logic board for model 2493HM. If thats true, then main/logic board model mark wont be the same as it is for Samsung 2693HM (BN41-00962A). OR, 2493HM and 2693HM models use the same main/logic board, but the Service Manual was initially made for 2493HM (before Samsung used the same schematics for 2693HM) so the markings inside Service Manual that was made for 2493HM stayed the same for 2693HM (they just didn't put a note '2693 'Delete, but 2493HM 'delete', like they were 'lazy' to chance the note from 2493HM to 2693HM). What do you think? Could I (if all this fails) buy the same main/logic board model type from 2493HM monitor (BN41-00962A) and use it on my 2693HM monitor? Would I need to 'upoload something' to an IC chip by using a programmer or I should not 'change' anything, just attach the board (from 2493HM model). Both monitors have 19200x1200 resolution.\

If both monitors have the same native resolution, I expect the boards to be cross-compatible without an issue. No, you shouldn't have to do any kind of firmware programming or anything like that. However, if you do go with that route, connecting the monitor to your PC will be recognized as a 2493HM now instead of 2693HM.

I wont care much if my 2693HM monitor is recognized as 2493Hm monitor. But still.... that 'information' must be 'written' somewhere (inside some cheap or something). Thats why I asked if I should use programmer or something to 'upload' the information into an IC if such information is required. I know it might be supid of me to think that (I'm a newb ), but for instance... 2493HM has less CCFL lamps than 2693HM (at least thats what I concluded by observing different inverter boards between 2493HM and 2693HM.) So I was thinking.... 'maybe' different kind of 'information' needs to be sent to 2493HM inverter board from IC chip located at main/logic board).

At the end, if the only difference that I will see if I use main/logic board from 2493 instead of my faulty 2693 main/logic board, when I load my OS, and it shows its a 2493HM monitor connected (instead 2963HM), I wont care much about it. But I just want to know if anything can be done about it and 'what'? (I mean would uploading firmware or whatever, would 'fix that').

Lets for a second imagine that IC924 reaceive regular 3.3v Enable signal. If we want to check the Feedback (FB) signal of IC924 which comes out on pin 5, would be best way to test it , to put black probe on 'screw-hole' and positive probe on pin 5 of IC924 and check if there is a voltage there in range between 1.194v and 1.25v ?

Yes, that's how I'd measure it too.

That said, pin 5 voltage of IC924 should be exactly the same as the output +1.2V_LIVE output, since resistor R3027 is a "0"-Ohm link resistor, and you can see that on your board too if you zoom in:
https://www.badcaps.net/forum/attach...8&d=1604050269
(It's the 2nd SMD resistor to the right of the lower-right corner of L902.)

Ok so, every resistor with '0' (zero) stamp on it is a 0 Ohm (link resistor)? btw. whats 'link resitor'? Like.... wire? If 'yes' than why not just put wire there? Wire has some small reistance (0.2 or 0.3 Ohm, right)? So if this 'link reistor' has zero Ohms... how can circuit even consider it as a resistor and 'devide' the voltage?

In case there is a correct voltage on pin 5, could that resistor R3050 (15KOhms) lower the voltage on the capacitor (C3045) ?

No.
Resistor R3050 and R3027 are used as a voltage divider in a circuit like this. So if you want a voltage output higher than V_FB of the IC, you change the values of resistors R3050 and R3027 accordingly. For example, if you want 2.44V to be the output, then one way to do it is to make R3027 a 15-KOhm resistor (i.e. same as R3050 right now.) Thus, when 2.44V is output on capacitor C3045, since R3050 and R3027 have equal values, that 2.44V will be split exactly in half to 1.22V in the middle, which is what V_FB is expecting to see. Or in other words, IC924 will crank the voltage on its output (i.e. at C3045) either up or down in order to keep V_FB at 1.22V. So because of that property, the values of R3050 and R3027 can be changed to make IC924 output any desired voltage of value between V_FB and the input voltage at pin 2 (IN)... minus one or two diode voltage drops (so probably anywhere from 1.22V to 4.3V for a 5V input... more or less.)

Ok...you might not believe me, but I spent last 4 hours trying to understand dmn voltage divider. I watched few youtube clilps. They way they draw the circle there is with 'battery source'. So circuit goes from positive to first resistor, then 2nd reistor (in series with first reistor), then goes to negative 'pin' on the battery. Divider there is what you get 'after' first resistor. And I (think) I understand that.


BUT, when I check how circuit looks like around that IC924 in the service manual, it doesn't look like those 'voltage divider circuits' from youtube that I was watching. What confusing me is that (the one around IC924 in service manual) has second resitor 'after' it touches ground. Its just that my mind is stuck there and for hours I was trying to use online circuit simulators to see how it works. And even though I managed to get output voltage (once I put other values on reistors), when I observe current flow its like...I don't know... it makes no sense. Also I don't understand whats the purpose of capacitor C3045 there (in service manual)? Its polarized capacitor. Is it there to... make current flow in certain direction?

For last 4 hours I got the feeling that there are 2 types of circuit schematics which I keep mixing in my head and until I figureout that, I think I'll be stuck in this loop forever trying to understand this.

The way the schematic is around IC924 in service manual, I would say courent will go from capacitor directly to ground. It wont 'to to' R3050'. And the line that comes out of FB pin of IC924 is spliting in two directions just there (one to R3024 and one to R3050). Thats not how every other dmn youtube clip shows voltage divider schematics. Can't be that every 'expert' at youtube just decided to pick the same schematics to show how they're good at understanding electronics.

Can you help me here to understand this? I understand the principle and the schematics from youtube is what I understand. But its not the same compared to Service manual schematics around that IC924.

I watched these clips (for intsance):
How Voltage Dividers Work
Voltage Dividers - Electronics Basics 12


^ Also, one extra thing that bothers me: While you tried to give me example how we can get specific output voltage (2.44), I don't know how you came up with these voltages (2.44) or 1.22v? Where did you found 1.22v ? The schematic show the output voltage is 1.2. So if the input voltage (feedback voltage at pin 5 of IC924) is the same as output voltage (after is divided) is 1.2v .... then where did you found 1.22v ? (maybe you meant 1.2v)



And at last: why use voltage divider if the voltage is the same from pin 5 to the 1.2 output pin? Why not linera voltage regulator instead? Its voltage wont change a lot from input to output and since there is no much difference between 'in' and 'out' voltage, then linear voltage regulator wont get hot, so....in my newbish mind... I would say they could've used that instead of voltage regulator? ....or not?

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Ok, connected ONLY Power (PSU) board through cable to a wall socket rail (230v). Here are the results:

All 3 pins for 5.3v rail show 0v (zero volts)
All 3 pins for 18v rail show 0v (zero volts)
Power On/oFF pin show 0v (zero volts)
Standby (ST_BY) pin show 5.24v

Good.

That shows the PSU board is powering on correctly (i.e only when it receives a voltage above 0V on PS_ON/OFF pin.)

I measured resistance between each pair of pins (each end of the 'wire') and its everywhere 0.4 Ω. But then, I thought, lets measure resistance between 2 'sides' (between 2 different 'wires'). I measured resistance between 'bottom-left' pin and 'top-right' pin. I also checked the measured values against the nubmers I got when I put probes on 'top-left' pin and 'bottom-right' pin, and they were in both scenarious almost the same, so I wrote down only one final value 'between wires'. (please check the attached picture).

So, resistance between two different 'wires' of those Ferrite beads is:

FT101 = 121.3 Ω
FT102 = OL
FT103 = 479.2 Ω
FT104 = 184.4 Ω


^ FT102 is a surprise for me. Or, maybe its resistance is the best among all 4 Ferrite beads? Can you tell me what you think about it, please?

The 0.4 Ohm resistance between each wire pair shows these ferrite beads are good.

As for the resistances between different wire pairs... hard to say at this point if that means anything, since those wires go to IC11 (the HDMI-DVI switch chip) and also connected to diodes D105-D112. Thus, if either IC11 is bad (due to getting "zapped" by the accident with the cable box) or diodes D105-D112 are bad, those resistances could be coming from there.

If anything, I expected most resistances to appear like the one on FT102.

That said, I think we can give FT101-FT104 a "pass" - i.e. they should let a signal through just fine, as evidenced by the 0.4 Ohms across each wire pair. So we can leave these alone.

Now, an extra question: You said that at high voltage output, 5-10 mA current can be lethal. I thought (I know that sound stuped but...) that 5-10 mA is a very 'low' current.

Well, above 10 mA is considered dangerous, because it has the potential to damage human nerves. So relatively speaking, that current is considered "high" to human nerves.

Of course, for that kind of current to go through your body, you also need a high enough voltage to push it through, due to resistance of your body. And with that, it will matter whether the current is DC or AC (AC tends to break through the skin easier.) This is why even though a 1.5V AA/AAA or 9V battery can source above 200-300 mA of current, you don't get a shock from it, because the voltage is too low to break through your skin.

But that's not all. Even if the voltage is high enough and the source is capable of supplying lethal currents, you also have to consider the exposure time. This is analogous to running a very high current through a very thin wire: if you do that, the wire will likely burn and melt if left long enough. However, if you just "pulse" that thin wire very quickly for a very short amount of time, the wire may not melt or get damaged. So the same principle applies to your body nerves too. This actually happens when you get zapped by static electricity from, say, walking on carpet with wool socks, for example, (which can generate 1000's of Volts, if not 10's of KiloVolts on your body.) The current when you get that "zap" can also be very high. But because the exposure ("pulse") from the shock is extremely short/narrow, the energy delivered from it is also relatively small and usually not enough to cause damage to your nerves.

And finally, if you're getting shocked from AC electricity, the frequency of the AC also matters. Relatively high frequency (to your body) up in the KHz or 10's of KHz range (or more) tends to flow only on or very near the surface of the conductor. The higher the frequency, the closer to the surface the flow of electrons becomes. This is called the "skin effect". Thus very high frequency is unlikely to cause nerve damage. In fact, very very high frequency (to your body) you may not even feel. But at sufficiently large currents, it will just cause skin burns instead... which could be fatal, after all, from the wounds you get from that.

So in short, there are a lot of factors to consider here besides just the current.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Ok, after you made this post, I created one extra post asking you about correct pin location (because their position was not the same when compared between Samsung Service Manual and diode datasheet). You said, thats not important right now, and we are only seeking to find resistances between them.

So, I'll use (as you suggested) Service manual as a refference. That means, when I observe the diode (on main/logic board), at one said it has 2 pins, and on the oposite side there is only 1 pin. The side with 2 pins I would reffer as 'TOP' side. So, while observing Service Manual, I would say that pin 1 is top-RIGHT, and pin 2 is top-LEFT. (pin 3 is on the oposite side, all alone by itself). So:

Pin 1 --> top-RIGHT
Pin 2 --> top-LEFT
Pin 3 --> at the bottom


Now, in the measurement results below I'll write something like this:

R1-B3 0.2 Ohm

^ That means that RED probe was touching pin 1 (R1) and that the black probe was touching pin 3 (B3), and the rest is the measured resistance between these two pins.

You asked me to meaure resistance between pins 1 & 3, and 3 & 2 (as a 2nd test). But I'll also include a 'Test 3' which is measured resistance between pins 1 & 2.

Perfect!

That really is the easiest way to get the information across without getting hung up on component datasheets.

Going by your resistances, this is what I'm thinking so far (note: I say thinking, because I'm still trying to make sense of some of the numbers.)...

D121 through D128 appear to show normal resistances, except for test #3, but that's because the +3.3V_LIVE rail has a short-circuit to ground... and essentially, test #3 is the same as measuring across capacitor C621. Since all of the resistances for tests #1 and #2 on these diodes are in the Mega-Ohm range, I think that at least for now, we can probably conclude that all of these diodes are OK.

Next, D105 through D112...
Again, test #3 confirms the same short-circuit on the +3.3V_LIVE rail, so we can't really use that to tell if any of these diodes are bad or not.
Tests #1 and #2 however, appear to show some really low resistances that I can't tell for sure if they are bad or not.

In particular, the resistances of D105, D109, and D112 do look alarmingly low. It's hard to say at this point, though, if those low resistances are the cause of the short-circuit on +3.3V_LIVE. I think NOT for the time being. Also, the low resistances could be coming from IC11 (the HDMI-DVI switch.) So I'm neither giving a "pass" nor a "fail" on D105, D109, and D112 for now. On that note, same goes for D106, D110, and D111, due to resistances still being in the 100's of Ohms range. In any case, you do NOT have to remove these diodes from the circuit at this point. We will try some other stuff first.

Meanwhile, the physically damaged D108 with it's open-circuit resistances also throws a red flag that its internal connection at pin 3 may have become disconnected (open-circuit) from the two diode junctions. So with that said, I think D108 will very likely need to be replaced. However, I don't think D108 is the cause of the short-circuit we see on +3.3V_LIVE, so you can leave it alone for now.

Really only D107 appears to show normal resistances for tests #1 and #2, if we are to judge by the results from diodes D121-D128. So at least D107 is also likely fine.

Ok, I checked resistance between my red and black multimeter probes, and it shows most of the time 0.2 Ohm (sometimes it shows 0.1 Ohm).

For measuring I put black multimeter probe on a screw hole of the main/logic board and kept moving red probe where it needed to be. I also waited few sec before I wrote measured values.

So, here it is:


D105
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 48.2 Ω
.
.
.
.

OK, so these tests basically show that pin 1 on each diode shows the same resistance as when you short the leads on your multimeter together. Thus, you can conclude that pin 1 on each of these diodes is connected to ground, which is how the service manual shows it is anyways.

Meanwhile, on pin 2 you get the same ~0.5 Ohms resistance as when you had your red multimeter probe on the positive lead of cap C621, which is also the +3.3V_LIVE output rail. This shows that pin 2 is connected to +3.3V_LIVE, which is how the SM shows it too.

And that's what I meant above that we can't really use test #3 to conclude anything about the diode resistances above, since it measures between the +3.3V_LIVE rail and ground every time.

That aside... the resistance tests you did to ground on each pin of these diodes confirms that D105-D112 are all possible suspects, with the exception of D107, which shows good resistance... and D108, which appears to have an open connection at pin 3 from the physical damage.

With that said, let's get to troubleshooting the short-circuit (0.5 Ohms) on the +3.3V_LIVE rail first.

This short-circuit could be caused by any of the D105-D112 diodes that I marked as suspicious. However, the short-circuit could also be caused by IC11 (the HDMI-DVI switch IC), because if you look at page 78 schematic on the service manual (all the way on the right side), you can see that IC11 has a direct connection to the +3.3V_LIVE rail on its pins labeled with "VDD33". So if IC11 was "zapped" by the accident with the cable box, it could be why it's pulling the 3.3V rail low.

Finally, because diodes D105-D112 and diodes D121-D128 have both connections to ground and the +3.3V_LIVE rail, it's possible that the accident from the cable box forced a high-voltage back into the +3.3V_LIVE rail, which means any IC with pins directly connected to +3.3V_LIVE could have been damaged, including the big chip (IC300), IC11 (HDMI-DVI switch), and even the 3.3V regulator IC - i.e. IC602 and its Schottky diode, D600.

With that said, it now basically down to removing components to see what causes that low / shot-circuit resistance on the +3.3V_LVIE rail to disappear. Surely removing either IC11 or IC300 would be the quickest way to tell, in theory... but that would assume you have a hot air station and the skills to remove those chips easily... which I don't assume is the case (remember what my friend told me about ASSUMING! )

So to make things a bit easier, let's try removing components that are more "friendly" to a soldering iron, one at a time, which I think you should be able to do with a wide(r) chisel or flat tip on your soldering iron (or lots of solder on the tip if it's conical type.) Start with SMD ferrite bead BD1005. This component is located to the left of IC11. It looks like an SMD ceramic capacitor, except that it's gray in color. Once you remove it, check resistance to ground again at capacitor C621 positive (+) pin (or alternatively, to pin 2 on any diode from D105-D112 and D121-D128... just whichever you prefer - we only need one measurement.) If the resistance is still the same more or less (i.e. around 0.5 Ohms) on the +3.3V_LIVE, continue reading below. If not, stop and report what resistance you get now.
.
.
If you continued due to resistance still being 0.5 Ohms on the 3.3V rail... now remove SMD inductor / ferrite bead L300. Inductor L300 is a black / dark gray SMD component that can be found close to the lower-right corner of IC300. Check resistance to ground after that same way as when you removed BD1005 and either report back what you get (if resistance has changed), or continue below.
.
.
If you continued due to resistance still being 0.5 Ohms after removing L300, keep going at it with L301, then L302, then L303... and so, until you get to L307. These inductors / ferrite beads supply +3.3V_LIVE to IC300, so if the 0.5 Ohms resistance on the 3.3V rail disappears at some point while removing there, then very likely IC300 was damaged (though I doubt IC300 is damaged.) With that said, try to NOT damage these inductors / ferrite beads (L300-L307 and BD1005) when desoldering them, if you can. And even if it looks like you did, still save them - in most cases, they can still be re-used.
.
.
If you had removed all of the above components and you still have 0.5 Ohms on the +3.3V_LIVE, just stop and report back.

That's all I have for you now. Let me know if any step needs further explanation.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

You see, Samsung 2493HM monitor has the same main/logic board as Samsung 2693HM. They have exactly the same mark: BN41-00962A
(and I was thinking, IF all this fails, I could try to buy used main/logic board from model 2493HM, since they do look the same, and has the same makings: BN41-00962A).
But now, when you said that because for model 2493HM Service manual has some cirucits marked as 'delete', that such circuits were not included on main/logic board for model 2493HM. If thats true, then main/logic board model mark wont be the same as it is for Samsung 2693HM (BN41-00962A).

Well, it could be the same board, but just with the parts for those circuits marked as "DELETE" not installed on the board. Technically speaking though, you are correct that the 2493HM main board should have a different part number then, despite being nearly the same thing. But I don't know 100%, TBH. That's just something only Samsung may be able to answer (though I doubt if you send them an email they'd reply with that info... but it's an option.)

What do you think? Could I (if all this fails) buy the same main/logic board model type from 2493HM monitor (BN41-00962A) and use it on my 2693HM monitor? Would I need to 'upoload something' to an IC chip by using a programmer or I should not 'change' anything, just attach the board (from 2493HM model). Both monitors have 19200x1200 resolution.

If both monitors have the same native resolution, I expect the boards to be cross-compatible without an issue. No, you shouldn't have to do any kind of firmware programming or anything like that. However, if you do go with that route, connecting the monitor to your PC will be recognized as a 2493HM now instead of 2693HM.

Lets for a second imagine that IC924 reaceive regular 3.3v Enable signal. If we want to check the Feedback (FB) signal of IC924 which comes out on pin 5, would be best way to test it , to put black probe on 'screw-hole' and positive probe on pin 5 of IC924 and check if there is a voltage there in range between 1.194v and 1.25v ?

Yes, that's how I'd measure it too.

That said, pin 5 voltage of IC924 should be exactly the same as the output +1.2V_LIVE output, since resistor R3027 is a "0"-Ohm link resistor, and you can see that on your board too if you zoom in:
https://www.badcaps.net/forum/attach...8&d=1604050269
(It's the 2nd SMD resistor to the right of the lower-right corner of L902.)

In case there is a correct voltage on pin 5, could that resistor R3050 (15KOhms) lower the voltage on the capacitor (C3045) ?

No.
Resistor R3050 and R3027 are used as a voltage divider in a circuit like this. So if you want a voltage output higher than V_FB of the IC, you change the values of resistors R3050 and R3027 accordingly. For example, if you want 2.44V to be the output, then one way to do it is to make R3027 a 15-KOhm resistor (i.e. same as R3050 right now.) Thus, when 2.44V is output on capacitor C3045, since R3050 and R3027 have equal values, that 2.44V will be split exactly in half to 1.22V in the middle, which is what V_FB is expecting to see. Or in other words, IC924 will crank the voltage on its output (i.e. at C3045) either up or down in order to keep V_FB at 1.22V. So because of that property, the values of R3050 and R3027 can be changed to make IC924 output any desired voltage of value between V_FB and the input voltage at pin 2 (IN)... minus one or two diode voltage drops (so probably anywhere from 1.22V to 4.3V for a 5V input... more or less.)

With that said, I suggest you do NOT play with the values of R3050 and R3027 at this point. Reason why is because there is also this thing called regulation compensation, which is done by the resistor and capacitors connected to pin 6 (COMP) of IC924. These values are carefully calculated so that IC924 outputs a steady voltage over a given expected load. This is one area where I never really learned much about, so how the calculations are done exactly, I cannot explain or do myself. Generally, you can make small changes to either R3050 or R3027 to "tweak" the voltage slightly. But any large changes (probably over 20-30% difference) will likely require the compensation circuit to also be re-designed/changed too.

btw. Can I measure voltages across any resistors and capacitors while the board is powered ON?

Yes, absolutely.

And... does it metter where black and red probes goes on those small (rice-size) capacitors? Or resistors?

Yes, but only if you care about the sign of the voltage - i.e. whether it's negative or positive.

For example, take a 1.5V AA/AAA or 9V battery. If you put red probe on (+) and black probe on (-) of the battery, you will get either 1.5V or 9V (depending if you did this experiment with a 1.5V or 9V battery.)
Now try reversing the probes. You will get the same voltage (1.5V or 9V), but negative... i.e. the meter will read either -1.5V or -9V.

So with that said, if you arbitrarily pick which way the probes go across any component (be it a cap, resistor, or whatever), the only difference will be that the voltage measured will be either positive or negative - at least when you measure DC voltages. With AC, it won't matter.

I think for resistors it doens't matter, but I 'think' that I heard for small SMD caps, they have no 'negative' marked ends, so for them you can solder them however you want. And because of that, I would conclude for SMD capacitors, I can measure the voltage across them with black and red probes in both positions. Correct?

You are correct that SMD ceramic caps are not polarized (i.e. have no negative marked ends.) In fact, ceramic and film capacitors in general are not polarized.

However, that doesn't mean that if you reverse the probes when measuring DC voltages that you will see the same value. If anything, it would be just like the experiment with the 1.5V or 9V battery. That's because while ceramic and film caps are not polarized doesn't mean the circuit they are installed in will not be polarizing them.

Take for example SMD ceramic capacitors C618 or C3050, which is installed between the input and ground of IC602 and IC924. Although these caps are not polarized, one of their ends is soldered to ground on the board and the other end is soldered to the voltage input pin (which is either +5V_MICOM or +18V_IN). So if you put the red probe on the grounded end instead of the black probe on these capacitors (and consiquently the black probe on the + end), then your multimeter will read either -5V or -18V, depending on if +5V_MICOM or +18V_IN is supplying the voltage. But if you place the probes the "correct" way (I put that in quotes, since it's not necessarily the correct way if you don't care about the numeric sign in front of the voltage), then you will get either +5V or +18V.

So in short, the sign of all DC voltage measurements to your multimeter are relative, depending on where you chose to place the probes.

I enjoy reading your detective way of thinking.

Thanks.
And I enjoy trying to "solve" this puzzle too (at least to my mind, this is one.)

I would love if my mind can work like that. I was checking the Service manual, before I uploaded it, and I didn't concluded what you did. You figureout that its IC602 which create 3.3v rail. I thought its IC602 that is also reaceiving 3.3v from... 'somewhere'.

Well, those particular ICs are a bit tricky, because it's hard to tell where power is coming IN and where it is going OUT, unless you checked the IC datasheet.

Honestly, I wasn't 100% sure what these ICs were at first either. However, whenever I see an inductor in a circuit like this, followed by a set of transistors (or a transistor and diode... or in the case of this monitor, just a diode and a switching regulator IC), chances are we are dealing with a switching regulator circuit, and my guess was that these ICs were the switching regulators. Note that some switching regulator ICs may still use discrete MOSFETs besides them. This is the case with computer motherboards, for example, where you can see this next to the CPU that there are a bunch of MOSFETs, inductors (usually toroids on older boards), and capacitors, and they are all controlled by one single PWM IC. The whole circuit forms a switching regulator, though.

This is different from linear regulators, where those circuit are typically made of only a single linear regulator or a control IC and a transistor of some sort... and output cap(s), of course. The difference is that on a linear regulator circuit, you will NOT see an inductor anywhere nearby (or if you do, it won't be part of the linear regulator circuit.)

Resistance of capacitor C621 is: 0.7 Ohm (black probe on negative, red probe on positive leg)

That's NOT good.

It means the +3.3V_LIVE rail has a short-circuit across it somewhere.

I suppose let the hunt begin. I'll answer your other post's questions before returning back to this and what to try to locate the short-circuit.

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

On that note, you said your monitor died when cable co. came in and installed new tuner box and had said box connected to monitor. If that really was the case and not a coincidence (after all, it could be, despite the guy saying this happens all the time - we still don't know the problem, so we have to keep open mind about any possible scenario), then I suspect there may also be shorted protection diode(s) on the HDMI or DVI-I (or whichever input you used) that's pulling the 3.3V rail down. In particular, I was looking at the monitor SM and on page 79, look at the schematic of the HDMI and DVI_I inputs. For the HDMI connector, diodes D105 through D112 provide anti-static / surge protection on the signal lines by clamping high voltages to the 3.3V rail and negative voltages to ground. Likewise, diodes D129 and D130 provide clamping to 5V_DDC and 5V_DDCH rails. On the DVI_I connector, it's the same scheme with the surge protection - you have diodes D121 through D128 for clamping signal lines to 3.3V or ground and diode D133 providing protection on SCL and SDA lines.

Thus, since the 3.3V line is low, it is possible that one of these diodes (in particular diodes D105-D112 and D121-D128) to be shorted. Again, with the power OFF and mainboard disconnected from all other boards, you can check these diodes either on RESISTANCE setting or in DIODE MODE. Just for consistency, let's use RESISTANCE mode. These are the tests to carry out:

Test 1:
Place red MM probe on pin #1 of diode (see page 79 of SM - it shows pin numbers for these diodes and how they are wired) and black MM probe on pin #3. You may or may not see a reading on your multimeter, but if you do, it should NOT be lower than probably 200-400 Ohms (if it is, report what you measured.) Now reverse the positions of the red and black MM probes on pins #1 and 3. Your MM should show very high resistance or OL. If not, report what you get.

Test 2:
Place red MM probe on pin #3 of diode and black MM probe on pin #2. You may or may not see a reading on your multimeter, but if you do, it should NOT be lower than probably 200-400 Ohms (if it is, report what you measured.) Now reverse the positions of the red and black MM probes on pins #3 and 2. Your MM should show very high resistance or OL. If not, report what you get.

Repeat the above tests for all protection diodes (D105-D112 and D121-D128).
You don't need to report all resistances back here. But if you do, perhaps do it in this format so it's easier to read.
Dxxx
Test 1: resistance A, resistance B (reversed probes)
Test 2: resistance A, resistance B (reversed probes)
.
.
and etc.

Ok, after you made this post, I created one extra post asking you about correct pin location (because their position was not the same when compared between Samsung Service Manual and diode datasheet). You said, thats not important right now, and we are only seeking to find resistances between them.

So, I'll use (as you suggested) Service manual as a refference. That means, when I observe the diode (on main/logic board), at one said it has 2 pins, and on the oposite side there is only 1 pin. The side with 2 pins I would reffer as 'TOP' side. So, while observing Service Manual, I would say that pin 1 is top-RIGHT, and pin 2 is top-LEFT. (pin 3 is on the oposite side, all alone by itself). So:

Pin 1 --> top-RIGHT
Pin 2 --> top-LEFT
Pin 3 --> at the bottom


Now, in the measurement results below I'll write something like this:

R1-B3 0.2 Ohm

^ That means that RED probe was touching pin 1 (R1) and that the black probe was touching pin 3 (B3), and the rest is the measured resistance between these two pins.

You asked me to meaure resistance between pins 1 & 3, and 3 & 2 (as a 2nd test). But I'll also include a 'Test 3' which is measured resistance between pins 1 & 2.


D105
Test 1 (R1-B3): 48.2 Ω , 48.2 Ω (reversed probes)
Test 2 (R3-B2): 47.8 Ω , 47.8 Ω (reversed probes)
Test 3 (R1-B2): 0.5 Ω , 0.5 Ω (reversed probes)

D106
Test 1 (R1-B3): 73.2 Ω , 73.2 Ω (reversed probes)
Test 2 (R3-B2): 73.4 Ω , 73.4 Ω (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D107
Test 1 (R1-B3): 2.056 MΩ , 2.269 MΩ (reversed probes)
Test 2 (R3-B2): 2.256 MΩ , 2.090 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D108 - phisically damaged diode
Test 1 (R1-B3): OL , OL (reversed probes)
Test 2 (R3-B2): OL , OL (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D109
Test 1 (R1-B3): 32.1 Ω , 32.1 Ω (reversed probes)
Test 2 (R3-B2): 31.7 Ω , 31.7 Ω (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D110
Test 1 (R1-B3): 440.5 Ω , 445.8 Ω (reversed probes)
Test 2 (R3-B2): 444.6 Ω , 440.1 Ω (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D111
Test 1 (R1-B3): 163.4 Ω , 163.5 Ω (reversed probes)
Test 2 (R3-B2): 163.1 Ω , 163.1 Ω (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D112
Test 1 (R1-B3): 21.4 Ω , 21.4 Ω (reversed probes)
Test 2 (R3-B2): 21.1 Ω , 21.1 Ω (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)




D121
Test 1 (R1-B3): 1.795 MΩ , 1.922 MΩ (reversed probes)
Test 2 (R3-B2): 1.969 MΩ , 1.832 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D122
Test 1 (R1-B3): 1.776 MΩ , 1.869 MΩ (reversed probes)
Test 2 (R3-B2): 1.883 MΩ , 1.795 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D123
Test 1 (R1-B3): 1.716 MΩ , 1.859 MΩ (reversed probes)
Test 2 (R3-B2): 1.892 MΩ , 1.807 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D124
Test 1 (R1-B3): 1.841 MΩ , 1.977 MΩ (reversed probes)
Test 2 (R3-B2): 1.984 MΩ , 1.862 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D125
Test 1 (R1-B3): 1.847 MΩ , 1.990 MΩ (reversed probes)
Test 2 (R3-B2): 1.926 MΩ , 1.721 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D126
Test 1 (R1-B3): 2.117 MΩ , 1.975 MΩ (reversed probes)
Test 2 (R3-B2): 1.934 MΩ , 2.084 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D127
Test 1 (R1-B3): 1.644 MΩ , 2.197 MΩ (reversed probes)
Test 2 (R3-B2): 2.170 MΩ , 1.490 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)


D128
Test 1 (R1-B3): 1.624 MΩ , 2.109 MΩ (reversed probes)
Test 2 (R3-B2): 2.107 MΩ , 1.623 MΩ (reversed probes)
Test 3 (R1-B2): 0.6 Ω , 0.6 Ω (reversed probes)








Ok, here comes the messy part. Although you did answer my question, and you said what you said (I'll quote that below), the numbers you see that I measured do not make any sense to me. And that's that makes me frustrated so much, because I can't figureout why those numbers are like that (you'll see them below too). I hope you can figureout whats wrong and let me know, so that my burned out newbish brain can try to understand.

You just need to know which pin is connected to ground and which pin is connected to +3.3V_LIVE rail. You can do that with a simple resistance check. First, select lowest resistance scale on your multimeter. Then, touch both probes together to each other and nothing else. The screen should show very low resistance - probably less than 2 Ohms and preferably under 1 Ohm (the value will vary here, based on the quality and calibration of your multimeter.) Note the resistance, as this is your short-circuit "resistance" (i.e. the lowest the meter can measure.)

Now, put black (-) MM probe on a monitor/logic board ground and red (+) MM probe on... let's use D105 as an example here... pin 1, based on the schematic shown in the service manual. Do you get that same short-circuit resistance (or nearly) when you touched the probes together? If yes, then pin 1 IS likely connected to ground. But now we verify that, just in case: move red (+) MM probe to pin 2 (while black MM probe is still connected to ground), which according to the SM should connect to +3.3V_LIVE. If that's the case, then you shouldn't see a short-circuit resistance.

Ok, I checked resistance between my red and black multimeter probes, and it shows most of the time 0.2 Ohm (sometimes it shows 0.1 Ohm).

For measuring I put black multimeter probe on a screw hole of the main/logic board and kept moving red probe where it needed to be. I also waited few sec before I wrote measured values.

So, here it is:


D105
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 48.2 Ω

D106
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 73.2 Ω


D107
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 2.286 MΩ


D108 - phisically damaged diode
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: OL

^ its like pin 3 doesn't exist. If I touch an inside contact of the diode (because its 'inside content' is revealed) resistance jumps to 3-6 MΩ , then when I try to measure again it jumps to 12-13 MΩ, then next time 30-40 MΩ... all crazy numbers. I tried to press a bit more or 'twist' the probe on Pin 3 but the probe finds nothing on pin 3 (except) when i try to touch the opened content of the inside of diode D108.


D109
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 32.1 Ω


D110
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 536.6 Ω


D111
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 163.5 Ω


D112
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 21.4 Ω




D121
Pin 1: 0.2 Ω
Pin 2: 0.6 Ω
Pin 3: 1.974 MΩ


D122
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 1.891 MΩ


D123
Pin 1: 0.2 Ω
Pin 2: 0.5 Ω
Pin 3: 1.906 MΩ


D124
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 1.987 MΩ


D125
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 1.971 MΩ


D126
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 1.975 MΩ


D127
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 2.221 MΩ


D128
Pin 1: 0.1 Ω
Pin 2: 0.5 Ω
Pin 3: 2.128 MΩ

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

One extra (but) offtopic question: About 6-7 months ago, I disassembled one old 17" LG monitor. I wanted to see if there is maybe some cap leaked from preventing it to power up the monitor. And yes, I found the bulged cap. BUT, in the process of disassembling that monitor, I cut out one of the inverter connectors. There were 2 'holes' hard to access. From both holes there were 2x2 wires with 2 connectors. So it had 4 connectors total.
I don't want to bother you with this now, but can that connector be... 'patched' somehow, since wires are so small and can easily brake.

Yeah, you can patch it. Just make sure to isolate it well with lots of electrical tape afterwards... or preferably, heatshrink tubing. Really, each CCFL has only one wire that is considered high-voltage (the thicker one.) The thinner wire is the "return" path that is close to ground value and unlikely to jump gaps. So really you only need to make sure to isolate the thick wire well, if that's the one you cut. The thin one can just do with a layer of tape or two fine.

Ok, thanks again for this information. I'll leave this case for some other time. I think I started way too much to flood this topic with my offtopic questions. So... once when we're done with this (Samsung 2693HM) monitor, I might ask more questions regarding that 'other' monitor (in maybe some other topic).

OK, so I checked the voltage on PWR_ON/OF (PS_ON) pin on the power board.
...
So to sum things up:

PWR_ON/OFF voltage on power board CNM803 connector: 0v (zero volts)
ST_BY (standby) voltage on power board CNM803 connector: 5.24v

Thanks. So looks like PWR_ON/OFF is 0V when logic board is disconnected.

That said, I forgot to ask... what voltage do you get on the 18V rail when PSU board is plugged in the wall but logic board DISCONNECTED from it (i.e. only have PSU board plugged in the wall and check voltage on 18V rail.) I'm just curious if the PSU board self-turns On or not in the lack of signal from the logic board.

Ok, connected ONLY Power (PSU) board through cable to a wall socket rail (230v). Here are the results:

All 3 pins for 5.3v rail show 0v (zero volts)
All 3 pins for 18v rail show 0v (zero volts)
Power On/oFF pin show 0v (zero volts)
Standby (ST_BY) pin show 5.24v

And finally, there are 4 elements (close to diodes D105-D112) which are marked as FT101-FT104). I took a picture of these 4 elements. Hmm, one (or two) seams like it has some... 'black dots' on top of it. Kinda like its...damaged (or burned). So, can you check it out and tell me what you think, please?
Also, I coudn't find these FT101-FT104 elements listed in Service manual so I have no idea what they are. Can you tell me what are they?

They should be just Ferrite beads - i.e. small inductors... i.e. a piece of wire wound around a Ferrite core.

Thus, you should be able to read a short-circuit across each pair of pins - that is, the top pins on each Ferrite bead should show short-circuit to each other, and the bottom two pins should also show short-circuit to each other.

I would be surprised if any of them have burned out, though.

Ok, thanks for this information.

I measured resistance between each pair of pins (each end of the 'wire') and its everywhere 0.4 Ω. But then, I thought, lets measure resistance between 2 'sides' (between 2 different 'wires'). I measured resistance between 'bottom-left' pin and 'top-right' pin. I also checked the measured values against the nubmers I got when I put probes on 'top-left' pin and 'bottom-right' pin, and they were in both scenarious almost the same, so I wrote down only one final value 'between wires'. (please check the attached picture).

So, resistance between two different 'wires' of those Ferrite beads is:

FT101 = 121.3 Ω
FT102 = OL
FT103 = 479.2 Ω
FT104 = 184.4 Ω


^ FT102 is a surprise for me. Or, maybe its resistance is the best among all 4 Ferrite beads? Can you tell me what you think about it, please?

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Thanks again momaka for your answer(s)!

I'll try to answer on most questions. First I'll try to post answers (or ask question) regarding previous posts, and then I'll post some measurements that you asked me to do

Well, good thing you provided the SM (service manual) and IC924 datasheet.
From the MPS MP1583DN (IC924) datasheet, the reference/feedback voltage V_FB of the IC is typically 1.222V (see page 3, parameter "V_FB"). Resistors R1 and R2 in the app note in the datasheet set the output voltage. In the SM of the monitor, we can see these resistors correspond to R3027 and R3050. From the picture of the mainboard you provided, we can see that R3027 has a big zero (0) written on it, so it's a zero Ohm resistor (i.e. a jumper)... which means the output of IC924 should equal to the reference voltage, V_FB... so that means 1.2V is the correct voltage output and 1.8V is not. I think also any bold text in the SM is basically to indicate a correction or later revision. (And for example, you can see that for the 2493HM model of the monitor, there are certain parts that say "DELETE", which probably means those parts of the circuit are not installed. So in other words, the bold text supersedes anything else written on the schematic.)

Well... thats the thing thats confusing me a bit, now. You see, Samsung 2493HM monitor has the same main/logic board as Samsung 2693HM. They have exactly the same mark: BN41-00962A
(and I was thinking, IF all this fails, I could try to buy used main/logic board from model 2493HM, since they do look the same, and has the same makings: BN41-00962A).
But now, when you said that because for model 2493HM Service manual has some cirucits marked as 'delete', that such circuits were not included on main/logic board for model 2493HM. If thats true, then main/logic board model mark wont be the same as it is for Samsung 2693HM (BN41-00962A). OR, 2493HM and 2693HM models use the same main/logic board, but the Service Manual was initially made for 2493HM (before Samsung used the same schematics for 2693HM) so the markings inside Service Manual that was made for 2493HM stayed the same for 2693HM (they just didn't put a note '2693 'Delete, but 2493HM 'delete', like they were 'lazy' to chance the note from 2493HM to 2693HM). What do you think? Could I (if all this fails) buy the same main/logic board model type from 2493HM monitor (BN41-00962A) and use it on my 2693HM monitor? Would I need to 'upoload something' to an IC chip by using a programmer or I should not 'change' anything, just attach the board (from 2493HM model). Both monitors have 19200x1200 resolution.

With that said, that 1.2V rail is still a bit too low. According to MP1583DN datasheet, V_FB can be as low as 1.194V and as high as 1.25V... so even if your multimeter is slightly off, it should still measure close to those values if IC924 is working correctly. And on that note, I think IC924 IS working fine. However, the monitor SM says that IC924 gets its enable signal from the output of 3.3V_LIVE (i.e. the voltage generated by IC602.)

Lets for a second imagine that IC924 reaceive regular 3.3v Enable signal. If we want to check the Feedback (FB) signal of IC924 which comes out on pin 5, would be best way to test it , to put black probe on 'screw-hole' and positive probe on pin 5 of IC924 and check if there is a voltage there in range between 1.194v and 1.25v ?
In case there is a correct voltage on pin 5, could that resistor R3050 (15KOhms) lower the voltage on the capacitor (C3045) ?


btw. Can I measure voltages across any resistors and capacitors while the board is powered ON? And... does it metter where black and red probes goes on those small (rice-size) capacitors? Or resistors? I think for resistors it doens't matter, but I 'think' that I heard for small SMD caps, they have no 'negative' marked ends, so for them you can solder them however you want. And because of that, I would conclude for SMD capacitors, I can measure the voltage across them with black and red probes in both positions. Correct?

Now, in my last post, I mistakenly assumed that since the voltage on C621 is close to 1.6V, then that must be what the IC is supposed to output. However, that is INCORRECT. The fact that you are getting only about 1.6V on the output of the 3.3V rail when both 5V_MICOM / ST_5.2V and +18V_IN are more or less normal, then that means there is something wrong with the 3.3V rail and this is where we should start digging deeper.

I enjoy reading your detective way of thinking. I would love if my mind can work like that. I was checking the Service manual, before I uploaded it, and I didn't concluded what you did. You figureout that its IC602 which create 3.3v rail. I thought its IC602 that is also reaceiving 3.3v from... 'somewhere'. I need a LOT to learn things that you know, and I envy you on your knowledge.

With the power turned OFF and logic/mainboard disconnected from PSU board and all other boards, measure the RESISTANCE between a ground (screw hole metal ring or C621 negative '-' terminal, whichever you prefer) and positive (+) terminal of C621. Keep the multimeter connected for 2-4 seconds and tell me what you see. Stable resistance? Or resistance climbing? (And approximate #'s, if value is climbing - this doesn't need to be exact.) Post back what results you get here

Resistance of capacitor C621 is: 0.7 Ohm (black probe on negative, red probe on positive leg)

And... what about inverter board? I has those small tranformers. And you said it has 700-1000v. That scares me so much. You said don't touch tranformers there. But...maybe I check voltages (for instance) on on capacitors (on inverter board) but touching screw-hole rings with negative probe, while using positive (red) probe to touch positive leg of a capacitor on inverter board)? Or I MUST touch negative capacitor leg directly with multimeter black probe? I'm asking because of those 700-1000v that you mantioned.

Inverter board ground is same as for other secondary side voltages, so same rule applies: use either metal chassis, metal ring around screw holes, or electrolytic cap negative lead (on the inverter board.)

Just avoid touching the transformers on the inverter board when it's powered, as that's when they may be outputting high voltage AC for the CCFLs. The good thing is that the moment you unplug the monitor, inverter CCFL voltages will are gone. There's no cap to hold the high voltages there, since it's all AC.

You said AC two times, so I guess you didn't made a typo. Ok, here comes another of my many stupid questions: How can it be AC there on the inverter board, since its gets voltages from secondary tranformer side (located on Power PSU board)? I thought there are no AC voltages anywhere else except for the power board (transfomer primary area marked with white lines)

It's a different type of AC, generated from the DC voltage on the secondary.

From your monitor, I think it's the 24V DC rail that goes through a set of MOSFETs that then pulse (or "chop", if you like) On and Off that 24V rail into a transformer. The pulsed output of these MOSFETs is basically a square wave with peaks up to 24V... which is a type of AC voltage. Once this 24V "chopped" AC goes through the inverter transformer (because transformers can only work with AC, regardless if it's nice and sinusoidal like the AC at your wall or square pulse, like from the MOSFETs), the output of the inverter transformer is wound to make roughly 700-1200V AC (which will also be a square wave, BTW.) This high-voltage AC then goes directly to the CCFLs through a set of small coupling capacitors. And that's all there is to it, really.

So essentially, the inverter takes low voltage and high current to produce high voltage and low current needed for the CCFLs. Although I say "low current" here, the current can be 5-10 mA at the high voltage output, which is enough to be lethal. On the positive side, since the inverter switches at high frequency (typically 50 KHz or more), this high voltage is also high frequency... and high frequency current tends to flow more on the surface of the skin rather than go deeper through nerves and muscles. So that makes it slightly less likely to be lethal... but still, DON'T TOUCH THE INVERTER WHEN IT'S OPERATING.

Ok, thanks momaka for this explanation. And sorry, but I thought its important to quote-quote-quoted-quotes, so if I decide to read it again (or someone else) it would be clear what we were talking about.

Now, an extra question: You said that at high voltage output, 5-10 mA current can be lethal. I thought (I know that sound stuped but...) that 5-10 mA is a very 'low' current. Seams like its not if the there are high voltage output. Ok, but... at what voltage output such current (5-10mA) or... even 250 mA is NOT leathal?

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Also would probably need to test that D018 diode extensively, as a broken case like that may eventually compromise the diode over time, even if it may still be working now (though I doubt it.)

Also diodes D121-D128 (Service manual page 79, DVI_INPUT info at the bottom) are different type. According to Service manual, they are 'marked' as 'MMBD4148SE', I googled it and got the datasheet. There everything (at least pin location) seams in order (compared to Service manual pin markings). But now, there is another thing here: these diodes (D121-D128) have 'top mark' (on themself) 'KJE V5'. I googled that one and found a datasheet for diodes that has mark 'KJE' and 'V5' represent that they are made in May 2008. These datasheets sais that they are 'spice model: BAV99', whatever that means)

Yeah, again, don't worry about the datasheet.
We just need to see if the diodes are bad or not - i.e. any of them showing abnormally low resistance between any pins.

And finally, there are 4 elements (close to diodes D105-D112) which are marked as FT101-FT104). I took a picture of these 4 elements. Hmm, one (or two) seamsm like it has some... 'black dots' on top of it. Kinda like its...damaged (or burned). So, can you check it out and tell me what you think, please?
Also, I coudn't find these FT101-FT104 elements listed in Service manual so I have no idea what they are. Can you tell me what are they?

They should be just Ferrite beads - i.e. small inductors... i.e. a piece of wire wound around a Ferrite core.

Thus, you should be able to read a short-circuit across each pair of pins - that is, the top pins on each Ferrite bead should show short-circuit to each other, and the bottom two pins should also show short-circuit to each other... using this picture as a reference, that is:
https://www.badcaps.net/forum/attach...1&d=1605276370

I would be surprised if any of them have burned out, though. The one with the black dots might be just some kind of leftover residue (flux?) from manufacturing. If they have been subjected to enough overload to burn, usually you'll see the PCB darkened around them as well.

I'll try to make a shorter post this time (but for sure I'll fail again)

I think it's moot at this point - might as well do things the way it makes sense to us. And if that means making long posts, then so be it. I had to split my reply in two again, because it wouldn't fit in the 10k char limit, lol.

On a side note, probably both of us could surely have made fine careers as typists.

[momaka]

Re: Samsung 2693HM - Doesn't turn on (no standby)

You asked me to measure resistance between pins of the diodes. I checked the Service manual (page 79) to see which pin is pin 1, and which is pin 2 (and pin 3). From the service manual it looked like top-right pin is pin 1, and top-left pin is pin 2.

Then I went to make zoomed-in image of the diode from a main/logic board, so I can read its markings and then try to find a datasheet. Markings on the diode is 'BA 16'. Since google didn't gave much of results, I checked the service manual by searching for 'D105' keyword. I got to page 37 of the Service manual where specs of D105 diode are listed. It sais:

DIODE-SWITCHING;DAN217,80V,100MA,SOT-23

So I was seaching for that, especially for DAN217 keyword. Looks like that DAN217 diode has couple more variations, but I found the one which has a note that its markings are 'BA' (which is whats written on top of the diode from main/logic board), 'BA 16' (I guess 16 reffers to maybe week of the year when it was produced or some other 'manufacture day' code). Ok so, if thats the correct one, then its a DAN217T146.

In all honesty, I wouldn't worry myself with finding the datasheet for these protection diodes. You just need to know which pin is connected to ground and which pin is connected to +3.3V_LIVE rail. You can do that with a simple resistance check. First, select lowest resistance scale on your multimeter (if it is a manual one... which IIRC, it wasn't, but I always state this for clarity anyways.) Then, touch both probes together to each other and nothing else. The screen should show very low resistance - probably less than 2 Ohms and preferably under 1 Ohm (the value will vary here, based on the quality and calibration of your multimeter.) Note the resistance, as this is your short-circuit "resistance" (i.e. the lowest the meter can measure.)

Now, put black (-) MM probe on a monitor/logic board ground and red (+) MM probe on... let's use D105 as an example here... pin 1, based on the schematic shown in the service manual. Do you get that same short-circuit resistance (or nearly) when you touched the probes together? If yes, then pin 1 IS likely connected to ground. But now we verify that, just in case: move red (+) MM probe to pin 2 (while black MM probe is still connected to ground), which according to the SM should connect to +3.3V_LIVE. If that's the case, then you shouldn't see a short-circuit resistance.

Also here's a hint: when in doubt, cross-verify with the other diodes of the same type - i.e. D106, D107, since they should be connected the same way. Basically, all of these protection diodes should have 1 pin tied to ground and another pin tied to +3.3V_LIVE. Even if the service manual switched the picture of the diode or mislabeled their pins, we don't care, because we should be able to clearly see which pin on each diode is connected to ground and which isn't, based on the test above. Either pin 1 or pin 2 should show a very low-resistance to ground. Whichever one it is, that's the one that is connected to the Anode of one of the MOSFETs in that package.

That said, it only gets tricky if one of these diodes has gone completely short-circuit for both of its diodes. In such case, you may get short-circuit resistance on both pins 1 and 2. But if it comes to that, then we know probably one of these protection diodes is bad... and because they are all connected in the same way, it will appear they are all bad. But then that's why I have you do the RESISTANCE tests in post #40 on these diodes, as that should tell us which diode is bad / short-circuited (if any.)

Anyway, I checked the datasheed and there, the pin 1 and pin 2 had 'switched' places (compared to their position in Service manual). I was like 'fck, why can't it be easy for once?'.
Ok, so I was thinking how to be sure which pin is which, and I thought, ok I'll put the multimeter in 'diode mode'.

No, please use RESISTANCE mode for now, as I think it's less likely to cause confusion at this point.

Since this 'diode' consist of 2 diodes (btw. isn't that a transistor if there are two diodes connected, and not a 'diode'?).

No, two diodes are two diodes.

I do understand why you ask that, though. BJTs (Bi-polar Junction Transistors) are -sort of- like two diodes and they certainly measure like that on a multimeter on DIODE TEST. However, function-wise, you can never have two diodes connected like a BJT internally to work like a BJT.

Anyway, I thought, once I put red probe on (lets say) top-right pin, and one probe to 'bottom' pin (I guess pin 3), I would get 'OL', or something like 0.3-0.7v. And when multimeter probes switch places I would get 0.3-0.7v or 'OL' (depends what I had in first checking). In other words, I thought I can't get measurement in both prositions. But noooo, I got measuremants in both positions.

Yeah, it's possible, because those HDMI and DVI-D lines go to a chip, which too, could have internal diodes and other semiconductor device inside, which can show the double-reading you saw.

Now, if that diode was tested out of circuit, you would indeed see a diode reading one way and not the other when you reverse the probes, like you anticipated.

And that's the problem with measuring components in circuit - there can be other components that impact the readings. That's not to say you have to remove those diodes. It may be able to test them in the circuit. You just have to keep in mind that if they don't test as they should, then there may be something else in the circuit affecting your readings. For this reason, I suggest you use the resistance setting to test these diodes, since right now, we really are only looking if any of them are shorted or not.

In one it goes like 0.6**v and in different position it goes around 0.4**v. Maybe this has to do cause they are (maybe) 'switching diodes' that 'switch directions'???

No, it just has to do with other components in the circuit affecting those values.

The term "switching diodes" is just to signify the diodes are suitable for high frequency use.

Then I checked measuremant between pin 1 and pin 2, and I got some readings, then in different position it was different. Then I checked the same diode type and I got different measuremants. Also resistance between pin 1 and pin 2 on some diodes was 'OL' and between some other diodes, resistance for pin1-pin2 was like 0.0002 Ohm. I was like WTF?!

Well, in that case, you may want to note down which diodes showed 0.xxx Ohms.

Also, make sure to keep the probes connected for at least 2-3 seconds on each measurement. This is because large value electrolytic capacitors can take time to charge when you try to measure the diodes one way, and then try to discharge when you reverse the probes.

So now, I'm trying to calm down, but I'm pizzed pretty much because I can't give you the answers you asked, and I'm forced to make this in-between post, before I can answer properly to your questions.

No problems, take it easy.

Worst case, just record all (stable) readings you see for each of these diodes (using the RESISTANCE setting on your MM.) And then maybe we can try to figure any sense out of it.

My question is: how to check and findout which pin is pin 1 and which pin is pin 2?

As explained above, just use resistance test to see which pin shows 0.x Ohm (or whatever lowest resistance your MM can measure) to ground. If both pin 1 and pin 2 show that resistance, note down on which diode you saw that. (Really, it should be on all of them ,since they are connected all in the same manner... but I'm just verifying.)

Also, hint you may want to keep in mind: sometimes there is leftover flux residue on the board, including on all solder joints. This flux residue is non-conductive and sometimes act's as a hard "skin" on top of the solder. So in some cases, it's possible not to get a reading or get a partial reading only, due to the flux preventing your multimeter probes to make good electrical contact with the solder joint. In cases like that, I just slightly move, twist, and apply slightly more pressure with the MM probe on the test point I'm trying to measure.

When I wanted to make a picture of top diode markings so I can seach for datasheet, I noticed something. One freaking diode (D108) was destroyed (split in half). I thought maybe I damaged it because I moved my finger across diodes in order to clean their markings a bit, so that the mobile phone can make a clear zoomed picture of the markings.

Then I went to my PC and checked the image of the main/logic board that I shared in my prevous posts. Freaking damaged diode was there all the time, in front of our nose (my nose). I'll upload a picture, so please check it out.

Oh wow, I guess I missed that too, even though I could swear I carefully tried to look over each component of your logic board.

I don't think you damaged that diode with your finger. Their case is hard epoxy, and usually it takes something with a striking force (or a bad short-circuit) to blow them apart like that.

It may be related to the problem or it may be just damage that happened when you were disassembling the monitor.

Could it be... that because of that didoe (D108) the other diods (D105-D112) give different and messed up values that doesn't make sense (at least to me)?

Yes, maybe.

[twist3r]

Re: Samsung 2693HM - Doesn't turn on (no standby)

Thanks momaka for your answer!

First, one thing:
I've tested my multimeter by checking voltage across 2 AAA unused (not alcaline) batteries. Both bateries were measured as 1.569v

Then I pulled out 2 AAA batteries that were in my multimeter and with new batteries 'in', I checked the 'old' ones. Old ones had 1.488v and 1.489v.

I put back old batteries in my multimeter.




Ok... well this was not the kind of post that I wanted to create. For last two days I'm trying to make a 'nice long post(s)' but there are things that stops me from doing it. So, now I'm forced to make 'in-between' post before I can make a proper respond to your previous posts.

I'll try to make a shorter post this time (but for sure I'll fail again):

You asked me to measure resistance between pins of the diodes. I checked the Service manual (page 79) to see which pin is pin 1, and which is pin 2 (and pin 3). From the service manual it looked like top-right pin is pin 1, and top-left pin is pin 2.

Then I went to make zoomed-in image of the diode from a main/logic board, so I can read its markings and then try to find a datasheet. Markings on the diode is 'BA 16'. Since google didn't gave much of results, I checked the service manual by searching for 'D105' keyword. I got to page 37 of the Service manual where specs of D105 diode are listed. It sais:

DIODE-SWITCHING;DAN217,80V,100MA,SOT-23

So I was seaching for that, especially for DAN217 keyword. Looks like that DAN217 diode has couple more variations, but I found the one which has a note that its markings are 'BA' (which is whats written on top of the diode from main/logic board), 'BA 16' (I guess 16 reffers to maybe week of the year when it was produced or some other 'manufacture day' code). Ok so, if thats the correct one, then its a DAN217T146.
Anyway, I checked the datasheed and there, the pin 1 and pin 2 had 'switched' places (compared to their position in Service manual). I was like 'fck, why can't it be easy for once?'.
Ok, so I was thinking how to be sure which pin is which, and I thought, ok I'll put the multimeter in 'diode mode'. Since this 'diode' consist of 2 diodes (btw. isn't that a transistor if there are two diodes connected, and not a 'diode'?). Anyway, I thought, once I put red probe on (lets say) top-right pin, and one probe to 'bottom' pin (I guess pin 3), I would get 'OL', or something like 0.3-0.7v. And when multimeter probes switch places I would get 0.3-0.7v or 'OL' (depends what I had in first checking). In other words, I thought I can't get measurement in both prositions. But noooo, I got measuremants in both positions.
In one it goes like 0.6**v and in different position it goes around 0.4**v. Maybe this has to do cause they are (maybe) 'switching diodes' that 'switch directions'???
Then I checked measuremant between pin 1 and pin 2, and I got some readings, then in different position it was different. Then I checked the same diode type and I got different measuremants. Also resistance between pin 1 and pin 2 on some diodes was 'OL' and between some other diodes, resistance for pin1-pin2 was like 0.0002 Ohm. I was like WTF?!

So now, I'm trying to calm down, but I'm pizzed pretty much because I can't give you the answers you asked, and I'm forced to make this in-between post, before I can answer properly to your questions.

My question is: how to check and findout which pin is pin 1 and which pin is pin 2?
I thought once I put red probe on top-right pin and black to bottom (pin 3) I would get some reading between 0.3 and 0.7v, but when I reverse probes I would bet 'OL', but no. Some diodes gave also some values (lower values but still 'values) in reversed position.
I don't know, maybe thats how they should behave (since their name is 'switching, so maybe they 'switch' the 'flow' in some momment). I'm sure I'm making no sense here but... I'm trying to figureout whats happening (and for what reason), but since I lack the knowledge I can't.


So, can you help me find out where are the correct pin 1-2-3 positions?


And now one thing that I found out and I wanted to keep as a nice thing for my post, but now I'm forced to share it in this post:

When I wanted to make a picture of top diode markings so I can seach for datasheet, I noticed something. One freaking diode (D108) was destroyed (split in half). I thought maybe I damaged it because I moved my finger across diodes in order to clean their markings a bit, so that the mobile phone can make a clear zoomed picture of the markings.

Then I went to my PC and checked the image of the main/logic board that I shared in my prevous posts. Freaking damaged diode was there all the time, in front of our nose (my nose). I'll upload a picture, so please check it out.

Now, because of that, I'm asking this:

Could it be... that because of that didoe (D108) the other diods (D105-D112) give different and messed up values that doesn't make sense (at least to me)?




Also diodes D121-D128 (Service manual page 79, DVI_INPUT info at the bottom) are different type. According to Service manual, they are 'marked' as 'MMBD4148SE', I googled it and got the datasheet. There everything (at least pin location) seams in order (compared to Service manual pin markings). But now, there is another thing here: these diodes (D121-D128) have 'top mark' (on themself) 'KJE V5'. I googled that one and found a datasheet for diodes that has mark 'KJE' and 'V5' represent that they are made in May 2008. These datasheets sais that they are 'spice model: BAV99', whatever that means)

So now, even for these diodes I can't be sure which is which, because for 'KJE V5' datasheet there is no info of correct position of pins 1-2-3. I'll also attach (all) the datasheets (that I've found) for D121-D128 diodes.



And finally, there are 4 elements (close to diodes D105-D112) which are marked as FT101-FT104). I took a picture of these 4 elements. Hmm, one (or two) seamsm like it has some... 'black dots' on top of it. Kinda like its...damaged (or burned). So, can you check it out and tell me what you think, please?
Also, I coudn't find these FT101-FT104 elements listed in Service manual so I have no idea what they are. Can you tell me what are they?


There...the post I didn't wanted to make, but was forced to. Sorry momaka that I didn't gave you the proper answer to your requested testings. I hope to do it once I learn the correct positions of diode pins. I hope you can help me with that.