Task TK-940TX-DF Blowing fuse

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Thanks for the confirmation, and additional info

What do you mean by washed fans, though? I hope you didn't run them through running water. Otherwise the iron on the stator core will start rusting over time and might make the fan seize.

Yes, i washed the fans under running water and shook them out and sun dried them - it's quite warm here, so the water will evaporate away to align with the current atmospheric equilibrium.
Wont do that again though.

Runs very quietly
Still does a little squeak when it turns off - which i don't mind, if it's a benign quirk of the PSU

[momaka]

Re: Task TK-940TX-DF Blowing fuse

With only the two fans as a load the -5v and 3v rails are about 0.8V below spec, but they come up to spec when a hard drive is connected.

Sounds normal. Some PSUs do that.

Sleeved a 0.2 ohm current sensing resistor (that should really be 0.1 ohm, that i lost). Spent ages reading and trying to figure out how it's used to sense current and why increasing the value of the resistor would cause the protection to kick in at a lower current. Looking at that schematic that i attached earlier on, increasing the resistor resistance means that more current will be forced through (or maybe more voltage pressure onto) the ISENSE terminal of the chip.

Yes, higher resistance sense resistor means it will have higher voltage across it when current flows through the primary of the transformer and main PS switching MOSFET. The higher voltage across the sense resistor means higher voltage going to the ISENSE pin, which means it will trigger earlier... which may not be a bad thing for this PSU, TBH.

Also, if you find another 0.2 Ohm resistor, you can just run the two in parallel, which will give you the same 0.1 Ohms resistance.

This PSU has traces/pads that come away from the board a lot easier than some other boards

Grrr. I hate those... especially the ones that crackle and pop when I run my soldering iron over them, followed by a blistered trace. Though sometimes, it's just the PCB absorbing moisture over time, so not much can be done for that.

Removed the rust, painted the case, and washed the fans.

Good job, it looks nice!

What do you mean by washed fans, though? I hope you didn't run them through running water. Otherwise the iron on the stator core will start rusting over time and might make the fan seize.
I usually just brush the dust off the fans with a brush, then wipe with moist paper towel (often times with window cleaner.) If the fans are really full of dust, I sometimes take the rotor part off and then clean more in-depth.

Put it back together and the top wouldn't go down due to the fact that i didn't use the original primary caps - they were a bit too long and touching the top fan, so i took it apart and put the original primary caps back on.

Been there, done that.

What I don't understand is why some manufacturers do this and add a second fan on the top that is the same size as the rear exhaust fan. I mean, if there is a rear exhaust fan, then that fan on the top is kind of redundant. Moreover, whenever there is a back exhaust fan, most studies suggest that a front-to-back flow is typically more optimal for cooling. But I guess this was considered OK practice in the early 2000's, which is probably the era this PSU was made in.

But i'll do that at the same time when i come across a 0.1 ohm current sensing "shunt" resistor

What's the power rating of the shunt resistor?
Here is my calculation: since most of the power would be dissipated via the main transformer windings at maximum load wattage of 400 watts.
R=V^2/P = 230^2/400 = 132 ohms effective resistance
I=V/R = 1.74A
Voltage across transformer winding (ignoring the comparatively negligible resistance of the shunt resistor) = 132 X 1.74 = 229.6V
Voltage across shunt resistor = 0.1 X 1.74 = 0.174V
Power dissipated by shunt resistor = V X I = 0.3A
So a half watt resistor would suffice?

Well, those calculations for SMPS don't exactly work like that... though you do have the right idea that if a certain average power flows through the primary side at a certain average voltage, then you can guesstimate more or less what the average current and resistance may be like. And in that context, the above calculations do make sense more or less. It's just that we don't know which component(s) would be the most lossy or cause loss of efficiency the most. So with that said, the above power you calculated should only be used as a very rough guesstimate. Also, I think 300 Watts is a more realistic rating for this PSU. But either way, 1/2 Watt resistor will probably get you by OK. However, at least from what I've seen in most PSUs, 1 Watt is a more typical value.

On that note, also make sure your resistor is flame-proof or at least flame resistant. IIRC, that would be either Metal Oxide or Metal Film resistors... but specifically check the part datasheet to determine that. Avoid carbon film or wirewound. The former is not flame-resistant and the latter may be too inductive.

Also noticed that the printing on the label of this PSU says that it's "approved by Intel and AMD" - so i guess that the presence of balancing resistors are not part of the requirements

I really wouldn't take that text to mean anything useful, nor do I think all PSU manufacturers used it with Intel's or AMD's approval. Some just slapped it on there to give the PSU more "merit". If I remember, that label came around the early Athlon XP and Pentium 4 era to signify that the PSU can handle both older 5V-heavy systems (Pentium 3, Athlon, and early Athlon XP) and newer 12V-heavy systems (Pentium 4 and later Athlon XP.) Prior to that, it was standard for most ATX PSUs to carry only 8 Amps or less on the 12V rail (and hence lack a 4-pin 12V CPU connector), which meant you couldn't use those old PSUs on a new (at the time) Athlon XP or Pentium 4 PC. So I think they came up with that label to make it easier for people to distinguish which PSUs can and cannot be used with a newer PC. But that label really meant nothing in terms of how the PSU would perform when cross-loaded one way or the other... and some didn't do too well.

[socketa]

Re: Task TK-940TX-DF Blowing fuse

With only the two fans as a load the -5v and 3v rails are about 0.8V below spec, but they come up to spec when a hard drive is connected.
Added two 330k balancing resistors in unused holes next to one of the primary caps.
Sleeved a 0.2 ohm current sensing resistor (that should really be 0.1 ohm, that i lost). Spent ages reading and trying to figure out how it's used to sense current and why increasing the value of the resistor would cause the protection to kick in at a lower current. Looking at that schematic that i attached earlier on, increasing the resistor resistance means that more current will be forced through (or maybe more voltage pressure onto) the ISENSE terminal of the chip.
This PSU has traces/pads that come away from the board a lot easier than some other boards
I wrongly threw away a solder pad that i lifted, so used a small section of wire instead - Use an over-length wire, and then cut it off after soldering; otherwise it can easily be lost if you drop it as a pre-cut length.
Removed the rust, painted the case, and washed the fans. Put it back together and the top wouldn't go down due to the fact that i didn't use the original primary caps - they were a bit too long and touching the top fan, so i took it apart and put the original primary caps back on.
After fiddling around with the hot soldering iron inside the case, in hindsight, i realized that it would have being so much easier to pull out the on/off switch and then desolder the wire.
Then looked at the photos and realized that i should have filled in the empty holes in the traces, so that current can flow better (rather than around the holes)
But i'll do that at the same time when i come across a 0.1 ohm current sensing "shunt" resistor

What's the power rating of the shunt resistor?
Here is my calculation: since most of the power would be dissipated via the main transformer windings at maximum load wattage of 400 watts.
R=V^2/P = 230^2/400 = 132 ohms effective resistance
I=V/R = 1.74A
Voltage across transformer winding (ignoring the comparatively negligible resistance of the shunt resistor) = 132 X 1.74 = 229.6V
Voltage across shunt resistor = 0.1 X 1.74 = 0.174V
Power dissipated by shunt resistor = V X I = 0.3A
So a half watt resistor would suffice?

Also noticed that the printing on the label of this PSU says that it's "approved by Intel and AMD" - so i guess that the presence of balancing resistors are not part of the requirements

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Thanks - that tidy's this up nicely enough
Was an interesting, and sometimes arduous and gruelling, journey.
But well worth the persistence.
Will find a way to put the balancing resistors on the surface, and also reload the heavy traces with solder, and put some glue around the primary caps bases (now i know why the glue was there, as they did tend to loosen as the unit was being handled)

i was trying to understand voltage drop across resistors and it occurred to me to mentally collapse them into wires, and to turn the voltage into push/pull, then things became very clear.
There are parallels with other stuff in life: The more that you mindfully repeat an action or an investigative thought process the easier, or more natural (intuitive), it becomes - and sometimes it's a good idea to apply some "thinkering" (i like that) before the physical action.
i also resolved current into two components: electron flow and electron orbital state, and that cleared up the fogginess of why a load would actively decrease it's resistance, and why a lower resistance equals greater dissipation (power/work)
We create a lot of needless resistance (often as a consequence of certain mental or 'physical' habits), and that's why we often lack in power from infinite potential transcendent source.
Also, it's a very unwise idea to open oneself up to more voltage than the rated wattage of one's current (pun not intended) resistorance.

[momaka]

Re: Task TK-940TX-DF Blowing fuse

Thanks for your clarity of input and assistance - so congrad's to you as well.
During the course of this, i've done a lot of thinking, and learned a lot, in more ways than one

No problems. Glad that's the case.

Was the creation of the bulb trick the result of your own thinking/reasoning?

The bulb trick is pretty old concept (dating back many decades ago.)

I first read about it here on badcaps.net and started trying it / using it when repairing amplifiers.

As far as the suggestions I made with adding more light bulbs in parallel or using a heating element - that is indeed my own thinkering with the bulb trick. It stemmed from the first time I encountered a problem with using the bulb trick on a beefy workstation ATX PSU with APFC circuitry. In particular, the APFC made the bulb flash On and Off very erratically for the first 2-4 seconds the PSU was forced to turn On. This prompted me to think how the APFC circuit works, what was causing the bulb flashing phenomenon, and if there was a way to avoid it.

Any thoughts as to which primary capacitor balancing resistors to use, in respect to resistance values and wattage?
I've looked at quite few PSU schematics, and noted that values range from 100k to 330K, with 150k and 330k being the most common.

Both 150k and 330k will be OK. The lower the resistances of these two resistors, the better the voltage distribution will be between the caps and the quicker they will discharge.

As for the power rating (Watts) of these resistors, you can use P = (V^2)/R to determine how much the resistors will dissipate, then multiply that x2 to get the power rating you need (it's a good "rule of thumb" to size resistor power rating to be able to handle 1.5x to 2x the anticipated nominal heat generation / power dissipation). V is going to be about 170 Volts DC across each cap in most cases.

So with 330 KOhm resistors, you'll get:
P = 170^2 / 330 000 = 0.087575 Watts
Times that 1.5 and you have 0.13136 Watts
A 1/8-Watt resistor can do 0.128 Watts, whereas a 1/4-Watt resistor can do 0.25 Watts.
For this case, probably even a 1/8 Watt resistor will work fine above, despite not quite sticking to the "rule of thumb". But it's close enough. If you want a cool-running resistor, of course, then use 1/4 Watt.

On the other hand, if you run the same calculation for the 150 KOhm resistors, you'll see that they should dissipate around 0.193 Watts each. 1/8 Watt rating will not be enough to cover that. So if you use 150 KOhm resistors, you'll need 1/4 Watt or higher-rated resistors.

Do i have to replace the damaged NTC thermistor with one of the same value, or is there some wiggle room here?

There's plenty of wiggle room there, especially for this PSU, since it only appears to be capable of 250-300 Watts output.

SCK-054 is a common part used on 200-350 Watt PSUs. Since you live in 230V country, probably even an SCK-083 will work in your case. (Hint: look up "SCK 10 series" to get some PDF datasheets that show other parts numbers.) 15SP 2R5 and SCK 2R58 are two NTC thermistors I've seen used a few times as well. Or if you have another junk / parts PSU rated for a similar power range, you could reuse the thermistor from that too.

Just for reference, SCK-054 is rated for 5 Ohms resistance when there is no current going through it and can do a maximum of 4 Amps.

Since there is no balancing resistors in this PSU, but the PSU does discharge (you could measure and hear the point when it's discharged - it aligned with the squeak sound), how do you think that could be discharging?

5VSB circuit.
It's always connected and the minimum load resistor and feedback resistors on the 5VSB will discharge the primary caps relatively quickly (usually under 30 seconds on most PSUs).

On your particular PSU with it's 2-transistor 5VSB design, the caps can technically discharge down to zero because of that 2 MOhm resistor connected between (+) VDC on the primary and the 5VSB MOSFET.

What was their function?

Referring to the "102" ceramic caps by the bridge rectifier... those are just for extra "extra" EMI/RFI filtering. But they are rather redundant, because your PSU already has proper Y2 (blue disc) and X2 (square box) -class caps on the input. Most PSUs (even the really good quality ones) never bother putting such small ceramic caps across the bridge rectifier because of that. So just leave them out - extra parts for your spare parts bin.

Yes it is, if the bridge rectifier is installed the wrong way around.

Well, now, I wasn't expecting that kind of user error. (Though at some point it did cross my mind something like that must be going on.)

Would the PSU have a 5VSB critical cap, and, if so, what circuity configuration would i be looking for, in order to find it?
(maybe there's one in the attachment that i've included)

Yeah, on the attached shcematic photo, capacitor C14 (50V, 47 uF) in the lower-left corner by optocoupler IC4 would be considered the "startup" / "critical" cap.
However, looking at this picture on the left side:
https://www.badcaps.net/forum/attach...1&d=1587365635
... that would correspond to capacitor C15 on your PSU (and you can see that cap is located right by the 5VSB transformer and its optocoupler, so that's one way to tell which one is the critical cap on a 2-transistor 5VSB circuit.)
Make sure you use a good reliable low-ESR cap there, like Nichicon PW/PM/PS, Panasonic FC and similar.

And since the 5VSB voltage is not coming from any chip that needs to be protected, then drive voltage is not a concern, right?

Correct.
The Zener in that circuit is just to protect the 5VSB MOSFET Gate, in case higher voltage goes through the 2 MOhm resistor connected to (+) VDC.

So it seems to be more ideal to find another zener that's closer to the maximum MOSFET gate voltage of 30W.
Or, realistically, what's the likelihood of the MOSFET gate-source voltage going above 18V, since i measure only 1V across the zener?

Not likely, so you can leave the original 18V Zener in there.

P.S there is another 12V fan on the other half of the case and the PSU starts when that's plugged in - so no extra load is required for this PSU to start

LOL.
I guess that load is just borderline close enough to keep it running.

i found both of zeners in a small plastic bag, with (post)#24 written on it.
Both of them have "18 2" printed on them.
They didn't do a good job of protecting the 5VSB MOSFET, or one of the main PSU MOSFETs
Neither of the MOSFETS should have being short-damaged; as both of the zeners would have conducted at 18V, which is well below the rated MOSFET gate voltages of 30V

Well, in the case of the 5VSB MOSFET... the MOSFET probably shorted because the output caps on the 5VSB were bad, causing the 5VSB regulation to go haywire, which often puts a lot more stress on the driving MOSFET. This applies to both 2-transistor and IC-based 5VSB circuit. Bad caps on the output are simply not good for it.

And in the case of the main PS MOSFET... probably the same thing happened: MOSFET was overloaded due to regulation going crazy with the bad caps on the output. With that said, the Zener diode there did DO its job - it protected your UC3843 PWM controller from going bad. Otherwise, you'd have to had that replaced too by now.

[socketa]

Re: Task TK-940TX-DF Blowing fuse

i found both of zeners in a small plastic bag, with (post)#24 written on it.
Both of them have "18 2" printed on them.
They didn't do a good job of protecting the 5VSB MOSFET, or one of the main PSU MOSFETs
Neither of the MOSFETS should have being short-damaged; as both of the zeners would have conducted at 18V, which is well below the rated MOSFET gate voltages of 30V

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Whoops! You already explained that the zener voltage must be close to, but less than, the maximum MOSFET gate voltage
And since the 5VSB voltage is not coming from any chip that needs to be protected, then drive voltage is not a concern, right?
I removed the zener, and it has "18" with "2" underneath that's printed it - so that means 18V 2W right? - So it seems to be more ideal to find another zener that's closer to the maximum MOSFET gate voltage of 30W.
Or, realistically, what's the likelihood of the MOSFET gate-source voltage going above 18V, since i measure only 1V across the zener?

P.S there is another 12V fan on the other half of the case and the PSU starts when that's plugged in - so no extra load is required for this PSU to start

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Thanks for your clarity of input and assistance - so congrad's to you as well.
During the course of this, i've done a lot of thinking, and learned a lot, in more ways than one

Was the creation of the bulb trick the result of your own thinking/reasoning?

I'm not comfortable about ZD8 acting as protection for the 5VSB MOSFET though - should i remove and check to see if it's appropriate?
The zener voltage might be too high, and if the voltage spikes at the drain, the MOSFET could be damaged.
Any idea as to what voltage it should be, or how to calculate it?
I think that it's a similar setup as in the attachment that i've included - My PSU has the same 2Mohm resistor that is between the +ve DC and the zener

Any thoughts as to which primary capacitor balancing resistors to use, in respect to resistance values and wattage?
I've looked at quite few PSU schematics, and noted that values range from 100k to 330K, with 150k and 330k being the most common.


Do i have to replace the damaged NTC thermistor with one of the same value, or is there some wiggle room here?

until the discharge/balance resistors across the caps discharge them

Since there is no balancing resistors in this PSU, but the PSU does discharge (you could measure and hear the point when it's discharged - it aligned with the squeak sound), how do you think that could be discharging?

those orange "102" ceramic caps by the bridge rectifier (C43 and...?) Remove them permanently and don't put them back on. They aren't really needed

What was their function?

Have removed all of the primary DC components and thoroughly cleaned the backside of the board - but the bulb still glows (no surprise there)

That is not possible.

Yes it is, if the bridge rectifier is installed the wrong way around.


Would the PSU have a 5VSB critical cap, and, if so, what circuity configuration would i be looking for, in order to find it?
(maybe there's one in the attachment that i've included)

[momaka]

Re: Task TK-940TX-DF Blowing fuse

The Task and the Bestec seem to be significantly inefficient, compared to the Aywun - but maybe that is because the Aywun is gutless?

Yup.

Generally, the bigger / more powerful a PSU is, the more inefficient it becomes at very low loads. It's actually not uncommon even for very expensive and high-end PSUs to get terrible efficiency at very low loads (like for example, a PC that only draws 30-50 Watts, and the PSU being rated for 500 Watts or more.)

With the Aywun being a gutless wonder, it will get better efficiency at low loads. But try all of those PSUs with something like a PC that draws around 120-150 Watts. The Aywun efficiency will likely not be stellar... if not outright appalling (lower end of 70%, I'd guess.) Meanwhile, the Bestec and this PSU might do something like 75 to almost 80% with the same load, despite appearing to be more inefficient at low loads.

In general, PSU efficiency is a bit like a bell curve.
- At very low loads, efficiency is low
- At very high loads (usually near the PSU's power rating, or slightly above it, if it is build well), the efficiency will also drop sharply past a certain point (and for PSUs with questionable protections, this is a sign to STOP loading further, or the PSU will likely get damaged / blow something.)
- At approx. 1/3 to 2/3 of the power rating on the label, the PSU will have optimal efficiency... provided the PSU label is accurate, of course.

Also, bearing in mind that the big pink 0.1 ohm resistor that i lost, has being replaced with one that is 0.2 ohms

That's fine.
This will just lower the power limit where over-power protection (OPP) kicks in on the primary side... which may not be a bad thing for this PSU, all things considered.
Just keep an eye on it after using it with a real PC (or similar) load for the first time. If it seems to have ran extremely hot and discolored the PCB, you may have to find a 0.1 Ohm resistor (or just another 0.2 Ohm resistor to put in parallel )... but this is very unlikely to be the case.

Anyways, congrats on getting it working!

[momaka]

Re: Task TK-940TX-DF Blowing fuse

Why does this PSU have a 400W rating?, when the info in the label says
+5V = 30A
+3.3V = 28A
+12V = 24A
150W + 288W + 92W = 530W, which exceeds the 400W max for this PSU

Ah, that's a classic "play on the numbers" by manufacturers.
This doesn't mean the above numbers are bogus or made up. But I didn't see you post a picture of the label of this PSU, so I can't say for sure if they are or aren't.

With that said, reputable PSU manufacturers will usually specify not only the maximum power the PSU can provide, but also the maximum power available from each rail and all of the rails combined.

Nowadays, most good quality modern PSUs typically have just a single 12V rail output, and they derive the 5V and 3.3V rails from it with switching / buck converters. But older PSUs, like the one in this thread (or modern cheaper ones) often use a "group regulated" design, where the main transformer outputs a 5V and a 12V rail (and sometimes, but not always, a 3.3V rail). In many cases, regardless if there is a 3.3V rail tap on the transformer or not, the 3.3V and 5V rails usually share turns on the same winding on the main transformer. For this reason, group regulated PSUs will usually specify a "3.3V + 5V combined power" rating. Here is a good example of that:
https://www.badcaps.net/forum/attach...1&d=1425959741

... which is the PSU found here:
https://www.badcaps.net/forum/showpo...postcount=2230
The Inno Power PSU in the above link has a 30 Amp rectifier for the 5V rail and a 30 Amp rectifier for the 3.3V rail as well. While the 5V rail is indeed specified for up to 30 Amps, the 3.3V rail is specified for only 14 Amps max. Why? Because the 3.3V rail uses a mag-amp circuit, so that's only how much it can do.

Furthermore, you see that 5V * 30A = 150 Watts and 3.3V * 14 Amps = 46.2 Watts. Thus if you add the 3.3V and 5V rail powers you get 196.2 Watts. The label, however, says you can get only 160 Watts max from the 3.3V and 5V combined, and that's because of that shared transformer winding tap for the two rails.

On the other hand, some newer group regulated PSUs have a "7V rail" tap on the main transformer instead of a 12V rail, and the 12V rail is derived by adding the 7V tap on top of the rectified (DC) 5V rail for a total of 12V. This means that any current pulled on the 12V rail must also go through the 5V rail rectifier. For PSUs like that, you may also see a 5V + 12V rail combined ratings.

And last but not least, the power/current numbers on the label are irrelevant if the rectifiers are not rated adequately to support what the label claims for each rail. Also, the primary side is another limiting factor to how much power the PSU can do.

For the Taks PSU you posted in this thread, it is a single-transistor forward converter topology (STF for short.) Generally, this topology/design is limited to about 350 Watts continuous. It may do 400W "peak", but certainly not for very long. To me, this PSU looks more what I would expect to see in a 250-300 Watt build. So the values on your label may be slightly fictional.

Anyway, i was feeling confident about this being one of those PSU's that requires an external load to operate, and whatever it is about the input impedance sensitivity that you mentioned, so connected a hard drive, and used just the 5A fuse wire on the AC Live wire.

Turned it on and it went good.
Turned it off after a couple of minutes, being grateful that nothing went "pop" or "bang"
The voltages fluctuate plus or minus 0.03V

So it's working, finally!
This is why I suggested the heating element in place of the bulbs, as you would have likely seen the PSU work with that too. But since you "jumped the gun" and tried the PSU with only a fuse and it worked... no need for the series bulbs or heating element anymore.

This is the first PSU that i've had that appears to require an external load to function

Unfortunately, it's not my first one.
There was another thread here, where a chap was troubleshooting his PSU and many people offered suggestions what to check/measure, myself included. Finally, it dawned on me (having read this somewhere else, again on badcaps a very very long time ago) that the PSU may be misbehaving due to low loading, I suggested that, the chap tried it, and his PSU worked.

So I always keep it in mind when testing PSU unloaded that they may not be working simply because they need a bigger load... which is a shame, because when using a series bulb / heating element, you can't have a large power draw from the PSU. That makes it hard or impossible to use the series bulb trick on such PSUs, unless you use something like a large heating element.

Can you confirm that this PSU has no balancing resistors, because, unless i'm delusional, this PSU has no resistors attached to the middle terminals trace of the primary caps.

No, fortunately you're not delusional.
UNfortunately, this PSU does not seem to have the balancing resistors across the input caps. Usually, they are almost always found next to the primary caps or near the bridge rectifier. The picture you put in post #5 shows no space for such resistors, though:
https://www.badcaps.net/forum/attach...6&d=1586164479

... so that's poor design on the manufacturer's part.

Would it be a good idea to add the resistors on the underside of the board?

Yes, you can add them to the underside of the board if you don't want to drill holes through the PCB to install them there. (Though generally installing through-hole components directly on a bottom of a PCB is considered "cheap" and often frowned upon... but if done well, it can be done safely without issues.) Just make sure to put a layer or two of heatshrink wrap around these resistors if you install them on the bottom/solder side of the PCB... or wrap with electrical tape. Don't leave them exposed, even if the PSU case has a plastic insulation sheet on the bottom.

[momaka]

Re: Task TK-940TX-DF Blowing fuse

Now, you were wondering when you had the 1x 60W bulb + 1x 70W bulb, why did the PSU not work, since the total power of both bulbs combined is 130 Watts - i.e. much higher than what the PSU was drawing.

To answer that, consider this much simpler example (and you are welcome to try and recreate it if you like):
Two resistors in series, R1 and R2, connected across a power supply Vs.
. ___________
.|...I -->.......|
.|...............[R2]
(Vs)..............|
.|.................|
.|................[R1]
.|___________|

(^ please ignore the "...." dots above - they're just for spacing my beautiful text art )
Imagine R1 represents the resistance of the PSU at any given moment, and R2 representing the resistance of the incandescent bulb / heating element.

To make math simple, let's have Vs = 6V and R2 (bulb) = 1 KOhms.

Now imagine the first case is where the PSU is short-circuited (i.e. R1 = 0 Ohms.) Then the total resistance is R1 + R2 = 1 KOhms and the current, I, is Vs / (R1 + R2) = 6 / 1K = 6 mA.
Using V=IR again, you can see that the voltage across R1 is V1 = I * R1 = 6 mA * 0 = 0V
and the voltage across R2 is V2 = I * R2 = 6 mA * 1k = 6V
... which makes sense, because V1 + V2 must always equal to Vs
and power draw of the PSU is P_psu = V1 * I = 0V * 6 mA = 0 mW

Next, let's repeat that with the PSU trying to work "hard" and having R1 = 0.5 KOhms. Then the total resistance is R1 + R2 = 1.5 KOhms and the current, I, is Vs / (R1 + R2) = 6 / 1.5K = 4 mA.
Using V=IR again, you can see that the voltage across R1 is V1 = I * R1 = 4 mA * 0.5k = 2V
and the voltage across R2 is V2 = I * R2 = 4 mA * 1k = 4V
as a check..... V1 + V2 = 2V + 4V = 6V, which is equal to Vs.
and power draw of the PSU is P_psu = V1 * I = 2V * 4 mA = 8 mW

Now let's try again with the PSU resistance being the same as the bulb's - i.e. R1 = R2 = 1 KOhms. Then the total resistance is R1 + R2 = 2 KOhms and the current, I, is Vs / (R1 + R2) = 6 / 2K = 3 mA.
Using V=IR, the voltage across R1 is V1 = I * R1 = 3 mA * 1k = 3V
and the voltage across R2 is V2 = I * R2 = 3 mA * 1k = 3V
as a check..... V1 + V2 = 3V + 3V = 6V, which is equal to Vs.
and power draw of the PSU is P_psu = V1 * I = 3V * 3 mA = 9 mW

We can continue this "trend" with increasing PSU resistance R1. Let's try 2 KOhms. Then the total resistance is R1 + R2 = 3 KOhms and the current, I, is Vs / (R1 + R2) = 6 / 3K = 2 mA.
Using V=IR, the voltage across R1 is V1 = I * R1 = 2 mA * 2k = 4V
and the voltage across R2 is V2 = I * R2 = 2 mA * 1k = 2V
as a check..... V1 + V2 = 4V + 2V = 6V, which is equal to Vs.
and power draw of the PSU is P_psu = V1 * I = 4V * 2 mA = 8 mW

Once more... let's set R1 = 5 KOhms. Then the total resistance is R1 + R2 = 6 KOhms and the current, I, is Vs / (R1 + R2) = 6 / 6K = 1 mA.
Using V=IR, the voltage across R1 is V1 = I * R1 = 1 mA * 5k = 5V
and the voltage across R2 is V2 = I * R2 = 1 mA * 1k = 1V
as a check..... V1 + V2 = 5V + 1V = 6V, which is equal to Vs.
and power draw of the PSU is P_psu = V1 * I = 5V * 1 mA = 5 mW

And let's say the PSU is very "lightly" loaded... perhaps R1 = 11 KOhms. Then the total resistance is R1 + R2 = 12 KOhms and the current, I, is Vs / (R1 + R2) = 6 / 12K = 0.5 mA.
Using V=IR, the voltage across R1 is V1 = I * R1 = 0.5 mA * 11k = 5.5V
and the voltage across R2 is V2 = I * R2 = 0.5 mA * 1k = 0.5V
as a check..... V1 + V2 = 5.5V + 0.5V = 6V, which is equal to Vs.
and power draw of the PSU is P_psu = V1 * I = 5.5V * 0.5 mA = 2.75 mW

If you look at the pattern of the values in this example, you can see that as the PSU power draw decreases, the voltage across it increases, whereas the voltage across the bulb decreases. Not only that, but you can see that the maximum power draw from the PSU occurs when the PSU resistance matches that of the light bulb. However, do note that the voltage across the two is equally split in half. In the real world, this would correspond to the PSU getting only 115V when set to operate on 230V AC line and only 57.5V when set to operate on 115V AC line. Clearly, this isn't going to work!

So as you can see, the PSU must be very lightly loaded, if at all, to use the series bulb trick. Alternatively, you must use higher-power bulbs / element, if you want the PSU to draw more power without the AC across it dropping out of spec.

[momaka]

Re: Task TK-940TX-DF Blowing fuse

From my understanding, the reason for the bulbs or the element on the AC Live input, is to restrict the current so that you don't keep wasting fuses

Yes, but thinking that the dim bulb device is for limiting current is not quite the correct term here. Rather, the bulb(s) limit the amount of power to the DUT (device under test)... and with that, of course, the current also becomes limited. Keep this small distinction in mind for the discussion that is to follow.

The three parallel bulbs is 194W, which is 0.84A, which is nicer working result than an issue that would necessitate usage of the 4-5A fuse that you previously mentioned.
But what is the reason for using a 1500W heating element, (for which current would be 6.5A in the event of a completely shorted PSU?

So here is why I mentioned the distinction above: the series bulb(s) limit power, whereas a fuse limits (interrupts) current once it goes past a certain threshold.

Not only that, but the fuse acts pretty much like a short-circuit (extremely low impedance device) right up until the maximum current is exceeded. And on that note, most fuses (even fast blow ones) will usually take a long time (could be up to several seconds) to pop when running at or near their maximum current. It's only when you go significantly above their rating that they will act fast and blow right away. Meanwhile, the voltage going to the device is mostly unaffected (that is, there's no significant voltage drop over the fuse). And because of that, the power going to the device is mostly still unlimited and the device can draw as much as it pleases... or at least until the fuse blows. So in the relatively long time it takes for the fuse to act / blow, you can have a large spike of surge current pass through it for a few moments, and thus also have a large surge / spike in power draw. These surges in power and current can be so high that other parts downstream may get damaged.

With that said, fuses are not necessarily meant to protect parts from blowing up. Rather, they are a "last resort" device so that components don't overload uncontrollably for an extended period of time, leaving the possibility of starting a fire.

On the other hand, an incandescent bulb or heating element has a fairly defined and relatively high resistance (at least compared to a fuse.) Because of that, the moment your device tries to draw more power (and more current) through the bulb / heating element, the voltage drop across the bulb / element will increase proportionally with the increase in current draw (because of Ohm's law), and so will the bulb's / heating element's power. So in other words, the bulb / heating element imposes a power limit and this power can only be shifted between the DUT and the bulb / heating element. Moreover, everything happens instantly and there is no delay or surge in current (and power) like there can be with the fuse. As such, there is much less chance of blowing components... though if the current is too high for some components, they can overheat and even burn / start a fire if there is no fuse to interrupt the current. This is why I suggested to use the 4-5 Amp fuse if testing with the 1.5 KW heating element - that way, you have a current and power limiting.

And why would input impedance of the bulbs affect the operation of the PSU?

Right.

So try to keep in mind everything mentioned in the above discussion.

Before I continued with more on that topic, though, first consider this:
most devices and electrical appliances are rated for a specific voltage input and frequency. In the case of *most* SMPS's, the input frequency isn't really that important (the exceptions, of course, being ones with APFC and large PPFC chokes.) However, the input/line voltage is very important.

While APFC and many flyback PSUs are typically specified to work over a "wide range input" - i.e. a line voltage anywhere in the range of 90 to 240V AC... older PSUs with a voltage doubler circuit (like yours in this thread) are typically specified to work either with 115V (110-120V AC) or 230V (220-240V AC), +/- 6%. Going below 110V (with the voltage selector switch set to 115V) or 220V (with voltage selector switch set to 230V) may render the PSU inoperable, just like yours did with the bulbs.

How low can you go on the line voltage before the PSU stops working?
- That tends to vary between different PSUs with different designs. In my experience, 100-105V is about as low as I could go on most PSUs (with their switch on the "115V" position, since I live in the USA and have 120V AC lines.) But it's not a set number. I've seen some very old half-bridge PSUs work on as low as 90V AC and also saw one that stopped below 108V AC.

Now, the reason I brought this discussion up is because with the incandescent bulb(s) or heating element in series with the PSU, you're going to see a voltage "drop" across the bulb(s) / element as the PSU starts to draw current. An increase in the current draw by the PSU will cause an increase in current through the bulb(s) / heating element. Because the bulb(s) / element have a significant resistance/impedance (unlike a fuse), Ohm's law dictates that the voltage across the bulb(s) / element will also increase. This means the voltage across the PSU will actually decrease. In other words, the AC voltage across the PSU (let's call it V_psu_ac) plus the voltage across the bulb / heating element (let's call that V_bulb_ac) is always equal to the line voltage (230/240V in your case), or:
V_psu_ac + V_bulb_ac = 230V

When you have less bulbs in series with the PSU, the resistance/impedance of the bulbs is higher, and so with the same amount of current drawn by the PSU, the voltage drop across the bulbs will be higher with less bulbs, and that means PSU getting lower voltage input. Thus, it's only a matter of how much current the PSU draws before the voltage across it decreases too much and falls out of the specified working range.
.
... and I'm splitting posts here, because I got hit with 10k char limit. lol.

[socketa]

Re: Task TK-940TX-DF Blowing fuse

The ratings (current draw?) of the hard drive is 5V 0.85A, 12V 0.75A (= 13.2W)
The current drawn by the Task PSU + Hard drive = 22.5W
So tried a Bestec PSU + Hard drive, and got 20.3W
And then tried an Aywun (that couldn't fully power a particular non-faulty optical mouse (was only recognized by windows when i changed the PSU) when plugged into the GA-78LMT-USB3 that you helped me get working) + Hard drive, and got 13.3W

The Task and the Bestec seem to be significantly inefficient, compared to the Aywun - but maybe that is because the Aywun is gutless?

Also, bearing in mind that the big pink 0.1 ohm resistor that i lost, has being replaced with one that is 0.2 ohms

[socketa]

Re: Task TK-940TX-DF Blowing fuse

From my understanding, the reason for the bulbs or the element on the AC Live input, is to restrict the current so that you don't keep wasting fuses

The three parallel bulbs is 194W, which is 0.84A, which is nicer working result than an issue that would necessitate usage of the 4-5A fuse that you previously mentioned.
But what is the reason for using a 1500W heating element, (for which current would be 6.5A in the event of a completely shorted PSU?

So i used 5A fuse wire, which can handle (5A X 230V) 1150W (yes, i know that it was risky since the PSU is only rated for 400W),
1.5A fuse wire would have being ideal, since that would melt at 345W

Why does this PSU have a 400W rating?, when the info in the label says
+5V = 30A
+3.3V = 28A
+12V = 24A
150W + 288W + 92W = 530W, which exceeds the 400W max for this PSU

Anyway, i was feeling confident about this being one of those PSU's that requires an external load to operate, and whatever it is about the input impedance sensitivity that you mentioned, so connected a hard drive, and used just the 5A fuse wire on the AC Live wire.

Turned it on and it went good.
Turned it off after a couple of minutes, being grateful that nothing went "pop" or "bang"
The voltages fluctuate plus or minus 0.03V

This is the first PSU that i've had that appears to require an external load to function

Can you confirm that this PSU has no balancing resistors, because, unless i'm delusional, this PSU has no resistors attached to the middle terminals trace of the primary caps.
So maybe it's the first PSU that i've had that doesn't have balancing resistors.
Would it be a good idea to add the resistors on the underside of the board?

And why would input impedance of the bulbs affect the operation of the PSU?

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Check your connections on the 12V bulbs. They should all light up or at least glow a little.
Otherwise this result suggests the PSU may be OK.

The connections are good, they do glow, except, or course, for when the PSU shuts itself down after swapping the 12V bulbs around (actually, just swapping the wires that are inserted into the 5V and 12V molex connector holes), and there is no voltage on the 3.3V, 5V and 12V rails.

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Before i get immersed in all of that,

After thinking about this, if i use only a 60W and a 75W bulb in parellel, in which 1/R = 1/(0.00113 + 0.00142) = 392 ohms > 134W, instead of the three bulbs, and the PSU malfunctions, then that means that there is still a problem with the PSU, because the amount of bulbs that are now present, which is still well able to handle the required current, should not be relevant to the functioning of the PSU.

The actual metered Watts are 24W with three bulbs (so no surprise that that three bulbs dont glow?)

After doing some ohms law calcs etc, and Watt Meter measurements, i can see that the load of the PSU is the determining factor for current - not any bulbs that are inserted in the live AC line - thus, considering only that fact, removing one bulb shouldn't make any difference to the well-functioning "three bulbed" PSU,


i.e., what does that result narrow the cause down to?
The PSU seems to be aware that a bulb on the AC live has being removed - could it be the input impedance sensitivity that you previously mentioned?
Why would a PSU care about the impedance of any bulbs on the primary filtering section?
Would it cause some imballance in the two primary input coils?
Another thought is that removing one parallel bulb, would increase the input impedance - so that correlates with the test result and the theory of failure due to increased input impedance.
According to the calcs that i did, the Wattage of the combined three parallel bulbs will equal 194W, and after removing a 60W bulb it equals 134W - so maybe that is the reason why two bulbs both light up (and the PSU goes AWOL), while three bulbs don't.
When i remove a 60W bulb and power on the PSU (fan doesn't spin with two bulbs now), the metered wattage increases to 127W (was 24W with the three bulbs)
127W is pretty close to what the two parallel bulbs would naturally draw (134W)

With just the 60W and 75W bulbs connected, and the 3.4W bulb on the 12V rail, and 8W bulb on the 12V rail, all four bulbs glow, and the rail voltages are not high enough
and if i switch the small bulbs around, the PSU squeaks for half a second then shuts down with only the two 60 and 70W bulbs glowing.

---

My calcs are as follows:
The 3 parallel bulbs that are connected in series with the AC live wire briefly glow (due to in-rush current?)

so the max combined output resistance of the PSU with the two 12V bulbs, in post #89 was:
using P=V²/R
15 ohms resistor + (43 ohms resistor + 5V^2/8W) + (233ohms resistor + 12V^2/3.4W)
= 15 ohms on 3.3 rail + 2.91 ohms on 5V rail + 42.36 ohms on the 12V rail
Converting to watts: 3.3^2/15 + 5^2/2.91 + 12^2/42.36
= 0.72W + 14W + 3.4
So that means that the PSU should draw close to 19W with those two 12V bulbs

But the wall power meter says that it's drawing 24W - which is only 5W difference, which could be attributable to heating of the PSU resistors, coils, and other components.

[momaka]

Re: Task TK-940TX-DF Blowing fuse

Good thinking, Batman! (0% alcohol version)
Even though it has "minimum load"(?) resistors on the three rails?

Never assume the designer necessarily did their job properly - at least with lesser known PSU brands.

Sirtec PSUs, for example, are known to oscillate badly with a heavy 12V load and light 5V load, despite having a 12V rail strong enough for a 12V-based system and the label claiming them being suitable for "Pentium 4" (goes to show how old some of these are )

OK, i put the 15-Ohm minimum load resistor back on, and carried out the above test, with the following results:

I used two 60W bulbs and a 70W bulb,
and connected the 12V bulbs as you suggested, turned on the PSU, and got good voltages on all three rails.
The PSU didn't shut down.
Only the 12V bulbs glowed.

Check your connections on the 12V bulbs. They should all light up or at least glow a little.
Otherwise this result suggests the PSU may be OK.

Try running the PSU longer with this test - something like 10-15 minutes to see if anything overheats. Run the test in 5 minute intervals while checking heatsink temperatures - and on that note, unplug PSU while it is still running to discharge primary caps so that the primary heatsink is not live or at a dangerous voltage when you try to check its temperature.

If all is good, try putting the 23W, 12V bulb on the 12V rail and the 8W, 12V bulb on the 5V rail. Then switch the three parallel-series bulbs on the input of the PSU with that 1.5 KW heating element you had before (and make sure you have a 4-5 Amp fuse in the PSU, in case something shorts hard so the element will disconnects.) Now see if the PSU runs and if anything gets hot. Note: you may or may not have to switch the 23W and the 8W bulbs between the 12V and 5V rails, depending if this PSU prefers a more 12V-heavy or 5V-heavy load.

Then i removed one 60W bulb and turned on the PSU, and one of two thing happened:

1. All bulbs glowed and the PSU made a high pitched squealing noise (what causes that?) with poor voltages (3.3V rail=2.3V, 5.5V rail = 2.4V , 12V rail = 6V) The PSU didn't shut down. The fan may or may not start, but turns at about 1/2 speed if i give it a push with my hand

Ouch!
Whatever protections there are in this PSU, they don't seem to be engaging. Makes me wonder if the supervisory IC on the secondary side is still good or not (maybe it got taken out from all of the ripple of the previously bad caps or when the 5VSB went bad.)

The squealing is the PSU trying to keep up with the "load"... whether that means it's loaded too hard (given the three series bulbs on the input possibly lowering the input AC voltage too much - more on that in a bit), or loaded too lightly (unlikely - you'd see at least one rail with a much higher voltage then), or loaded unevenly (unlikely - same as loaded too lightly)... it will just have to be determined by experiments.

On that note, while trying the above tests with the three series bulbs or heating element on the input...
1) Check your line AC voltage at a plug to get a reference of what it is. Post the value here.
2) When the PSU turns On with stable voltages, check the line AC voltage that the PSU is getting _after_ the series bulbs / heating element. Note the value here as well. Some PSUs don't work well when the AC voltage drops too much. I'm curious if that is the case here...

Hope I'm not annoying you to death with these tests.

2. All bulbs glowed. No high pitched squeal.The fan spins at about half-speed, (3.3V rail = 3V, 5.5V rail = 3.1V , 12V rail = 7.5V).

Yeah, something is not right with this thing. If you can't get proper output voltages with the 1.5 KW heating element as the series current limit, then it might be time to change the supervisory IC... and possibly PWM controller too. While at it, grab a few spare 431 shunt regulators as well.

If this doesn't revive the PSU, I really don't know what else can... at least for the time being (it's going to be at least a few more bathroom sessions with me sitting on the toilet and pondering about circuits stuff )

Was the reason for the higher wattage bulb across the lower voltage rail, because this PSU is one that "5V heavy"?

Yeah, it could be.
If you can determine with which bulbs on what rails the PSU can run and output normal voltages and with which it can't, then that should give you a clue whether it's a 5V-heavy or 12V-heavy design... or just a heavy oversized paperweight.

Just out of interest, is the 3.3V derived via a mag-amp in this schematic,
http://danyk.cz/s_atx01u.png,
and if so, where is the mag-amp?

Yes, it is.

In that circuit...
SD1 is the 3.3V rail rectifier
L4 and L(?) are the 3.3V rail filter coils
C18 and C19 are the output filter caps
R28 is the minimum load resistor

And the mag-amp circuit components...
L2 is the saturation coil/toroid (the white one next to the main transformer in your PSU - looks like L7, I think.)
Q5 is the mag-amp switching transistor
Q6 (TL431 - it should really be labeled as "IC" and not "Q" in the schematic) is the current shunt regulator for the 3.3V rail.
D8 and D9 are the diodes for the mag-amp coil
R26 and R27 (1k VR) are the resistors forming a resistive divider to set the 3.3V rail voltage to 3.3 Volts.

But really, the easiest way to recognized when a PSU has a mag-amp circuit for the 3.3V rail is that you will see either one or two saturation coils in the proximity of the main transformer (sometimes can be behind the secondary heatsink) in addition to a main toroid filter inductor and a separate (and slightly smaller) toroid filter inductor for the 3.3V rail - both located behind the secondary heatsink.

During the course of all this soldering/desoldering, i found one tip that someone may find useful:
Cleaning off flux with isopropyl alcohol doesn't do a very good job with "kingbo" flux; so i tried epoxy solvent clean up, which i think is xylol, and it does a great job.

Interesting.
I guess Kingbo quality can vary quite a bit... which shouldn't be too surprising.
We (me and the techs at a former shop I worked at long time ago) used to think that Kingbo was some high quality flux. But over the years, I've learned that it's just a common name used by many different Chinese manufacturers, and thus its quality and properties can vary somewhat. Overall, though, it does appear to work well (at least mine does), even for SMD and BGA rework. I have no problems cleaning mine with IPA and small swabs/balls of paper towel. It does take at least 99% IPA, though, to clean it. Anything less and I have to go over each joint and clean it several times.
Acetone works great too, but it can be a bit too aggressive on some PCB coatings.
Don't think I have tried Xylol.
Denatured alcohol also worked, I think.
I don't remember much about MEK (Methyl Ethyl Ketone) though.

[socketa]

Re: Task TK-940TX-DF Blowing fuse

Just as well i checked before desoldering and testing the other components,
and just as well:

Also, it just crossed my mind, but....
This PSU may be one of those units that doesn't like to run with no load connected to it, and maybe that's why the 3.3V rail was low.

Good thinking, Batman! (0% alcohol version)
Even though it has "minimum load"(?) resistors on the three rails?

To test if that is the case, add your 8 Watt, 12V bulb on the 5V rail and put the 4.3 Watt, 12V bulb on the 12V rail. Leave the 3.3V rail without a load, other than its 15-Ohm minimum load resistor. Then connected the PSU with the series bulbs on the input (try 2 or 3 60-Watt bulbs first) and see if the voltages are normal.

OK, i put the 15-Ohm minimum load resistor back on, and carried out the above test, with the following results:

I used two 60W bulbs and a 70W bulb,
and connected the 12V bulbs as you suggested, turned on the PSU, and got good voltages on all three rails.
The PSU didn't shut down.
Only the 12V bulbs glowed.

Then i removed one 60W bulb and turned on the PSU, and one of two thing happened:

1. All bulbs glowed and the PSU made a high pitched squealing noise (what causes that?) with poor voltages (3.3V rail=2.3V, 5.5V rail = 2.4V , 12V rail = 6V) The PSU didn't shut down. The fan may or may not start, but turns at about 1/2 speed if i give it a push with my hand


2. All bulbs glowed. No high pitched squeal.The fan spins at about half-speed, (3.3V rail = 3V, 5.5V rail = 3.1V , 12V rail = 7.5V).

------------
Was the reason for the higher wattage bulb across the lower voltage rail, because this PSU is one that "5V heavy"?

Just out of interest, is the 3.3V derived via a mag-amp in this schematic,
http://danyk.cz/s_atx01u.png,
and if so, where is the mag-amp?

During the course of all this soldering/desoldering, i found one tip that someone may find useful:
Cleaning off flux with isopropyl alcohol doesn't do a very good job with "kingbo" flux; so i tried epoxy solvent clean up, which i think is xylol, and it does a great job.

[momaka]

Re: Task TK-940TX-DF Blowing fuse

5VSB is good after the squeak
And it's good irregardless of whether i switch the PSU off and on straight away, or off and then back on, straight after after the unsqueak.
The unsqueak that happens after switching the PSU off, coincides with the 5VSB voltage reading dropping off the multimeter screen.

Ah OK.
Sounds more like a more-or-less normal 2-transistor 5VSB function.

I back-fed 3.3V from another PSU into the Task 3.3V output rail, initially with a 12V 23W lamp (no glow), then a 12V 8W lamp (no glow), then a 12V 3.4W lamp (glowed when i briefly connected it)
So am guessing that that shows that there is no shorts on that rail.

Yes, looks OK.
In addition to backfeeding a rail, you can also monitor the voltage on that rail to see if it's dropping voltage and how much / to what value. But since you tried 3 different bulbs, with the largest being 23W (about 6.26 Ohms when bulb is fully lit and lower when cold) I'm pretty sure you would have seen/felt something overheat and the bulb glow if there was a bad component. Sometimes the transistor that does the regulation on mag-amp 3.3V rails can partially short and run very hot, with 3.3V regulation being all whack... but again, I don't think that's the case here anymore. Still worth mentioning it, though, in case you take other PSUs for repair.

I removed ZD8, and it looks like it has 18C printed on it,
which looks to be a zener voltage of 18V
V_GSS of K2746 = ± 30V
So the zener that i used is quite a lot lower.
I've now replaced it with a C27 Zener instead, but no change in test result

OK, that's good to know. At least we eliminate that as being a possibility as well.

Mind you, i wasn't super hopeful about that completely fixing it; because why would the 5V and 12V rails briefly spark up as good, but the 3.3V rail be way under spec? You'd think that they would all be affected if either of the main switching transistors were having consequential issues.

Well, the only reason I thought about checking the 3.3V rail is because the mag-amp transistor basically switches On and Off to create "counter currents" through the mag-amp coil, which then is what reduces the voltage from the 5V AC tap to 3.3V.

So that just leaves that mag-amp circuit that you mentioned.
I traced out what i could, removed the minimum load resistor which measured 15 ohms, so that i could check the other resistors (see photo), but when i put one lead of my meter on 3.3V rail and the other one on the earth rail to test the 3 resistors (the first one doesn't count, because the +S wire is connected to the 3.3V rail in the motherboard connector), their total resistance should be (330 + 110 + 22)ohms, but the meter reading keeps on going up to over 1K.

Ok, in that case pull out those resistors as necessary and test if they read proper resistance out of circuit too.

Also, it just crossed my mind, but....
This PSU may be one of those units that doesn't like to run with no load connected to it, and maybe that's why the 3.3V rail was low.

To test if that is the case, add your 8 Watt, 12V bulb on the 5V rail and put the 4.3 Watt, 12V bulb on the 12V rail. Leave the 3.3V rail without a load, other than its 15-Ohm minimum load resistor. Then connected the PSU with the series bulbs on the input (try 2 or 3 60-Watt bulbs first) and see if the voltages are normal.

Before i remove the transistor and the shunt regulator to test out of circuit, would you say that the zener diode that i replaced with the C27 diode was an acceptable replacement (was the closest value zener that i could scavenge of other parts PSUs), and wouldn't be the cause of PSU shutting down?

From what I understood, there was no change in behavior when going from the 18C to C27 Zener, correct?

Actually, there are two factors when considering the Gate protection Zener diodes: one is, as mentioned in the previous post, the maximum voltage the Gate of the MOSFET is rated for. So the Zener needs to have a Zener voltage less than the gate. The other factor is the drive voltage. In the case of ZD1, this would be the Vcc voltage going to the PWM controller (UC3843 IC). Note that the UC384x family of PWM controllers is rated for 30V absolute max for Vcc. So with that in mind, we know the voltage that the PWM controller outputs could be up to 30V, depending on Vcc, but not more than that. So Z1 needs to be selected to have a Zener voltage that is higher than whatever Vcc actually is in the circuit (but not more than 30V, because UC384x PWM IC is not expected to output more than that anyways.)

Thus, the C27 Zener you used should be a good fit. Not only should it protect the MOSFET Gate from going over its maximum V_GSS voltage (30V), but should also protect the PWM IC in cases where the MOSFET can go bad and conduct Drain-to-Gate (opening the possibility of a high voltage leaking back from Drain into Gate.)

[socketa]

Re: Task TK-940TX-DF Blowing fuse

when you try to power on the main PS and you hear the PSU squeak, what happens to the 5VSB?

5VSB is good after the squeak
And it's good irregardless of whether i switch the PSU off and on straight away, or off and then back on, straight after after the unsqueak.
The unsqueak that happens after switching the PSU off, coincides with the 5VSB voltage reading dropping off the multimeter screen.

I back-fed 3.3V from another PSU into the Task 3.3V output rail, initially with a 12V 23W lamp (no glow), then a 12V 8W lamp (no glow), then a 12V 3.4W lamp (glowed when i briefly connected it)
So am guessing that that shows that there is no shorts on that rail.

I removed ZD8, and it looks like it has 18C printed on it,
which looks to be a zener voltage of 18V
V_GSS of K2746 = ± 30V
So the zener that i used is quite a lot lower.
I've now replaced it with a C27 Zener instead, but no change in test result

Mind you, i wasn't super hopeful about that completely fixing it; because why would the 5V and 12V rails briefly spark up as good, but the 3.3V rail be way under spec? You'd think that they would all be affected if either of the main switching transistors were having consequential issues.

So that just leaves that mag-amp circuit that you mentioned.
I traced out what i could, removed the minimum load resistor which measured 15 ohms, so that i could check the other resistors (see photo), but when i put one lead of my meter on 3.3V rail and the other one on the earth rail to test the 3 resistors (the first one doesn't count, because the +S wire is connected to the 3.3V rail in the motherboard connector), their total resistance should be (330 + 110 + 22)ohms, but the meter reading keeps on going up to over 1K.

Those two diodes, that i drew in test good.
No shorts on the shunt regulator, or the transistor

Before i remove the transistor and the shunt regulator to test out of circuit, would you say that the zener diode that i replaced with the C27 diode was an acceptable replacement (was the closest value zener that i could scavenge of other parts PSUs), and wouldn't be the cause of PSU shutting down?