Linear regulator question

[Dannyx]

Good day folks. Pretty simple task at hand here, so go ahead and laugh at me if your want, but only if you also provide a bit of help I need to step down a 12Vdc supply from a power adapter to 9Vdc to power up some wireless routers I found in our box of trash at work despite being functional. Before you say I should just get a 9v adapter or a 7809 regulator, hey where's the fun in that ? People like me like to make things more complicated than they should be just for the sake of learning stuff, so I decided to learn more about linear regulators by building one myself. The final product does not require too high a build quality and precision, since it's just about powering some crap devices which at the end of the day I'm sure would run just fine if I were to power them with the 12v directly, but I don't really wanna do that. So I was thinking of building the simplest linear regulator possible with a transistor, a zener diode and a resistor. I know what a linear regulator is and what it does and even had the schematic in my mind, but the problem is the actual choice of parts I should use. At this point I could've just sourced the exact parts used in one of the many examples I found, but I didn't actually want to BUY anyhting for this project, so I rummaged through my box of transistors and picked the one with the highest current rating I found (1A), the S8050 (datasheet here), since that's what the router requires. I also got myself the C9v1 zener diode (datahseet here). Now it's time for the resistor and this is where I ran into a bit of trouble and got tangled up in numbers and calculations and formulas, all of which I suck at unfortunately, so please bare with me as I'm an absolute disaster at math. Here goes nothing: good ol' Wiki tells me that I should use this formula to calculate the value of the resistor I should place between the base of the transistor and Vcc (12v)....after trying my best, I ended up with a value of 2500 Ohms, which is probably wrong, so let's run through the formulas together like in second grade and see what we get: so
*Vs-Vz is 12v-9v (let's keep it round) = 3
*Going below the fraction, I tried figuring out what Ibase should be for that 8050 I chose. The wiki article tells me Ib is equal to Ir2 divided by the hFe of the transistor. The router (which is the load R2) requires 1A, divide that by the minimum hFe which the datasheet tells me is 40 and it equals 0.025.
*Now, what's the deal with that K ? Is it like a constant ? The wiki article just tells me it should be between 1.2 and 2....ok, I'll just pick 1.2 I guess, so 1.2 times 0.025 (Ib) = 0.03
*Ok, assuming everything so far is correct (which is probably anything but) I then hit the first hurdle: Iz, zener current...what is this ? Is it the maximum current the diode allows ? If so, the datasheet gives me 43mA (which would be 0.04A ?), so we add that to the formula and we get 0.04+0.03 = 0.0012
*3 divided by 0.0012 gives me 2500....2.5 kOhms ? Was I even close ? Again, my main point is not so much to get the damn thing to work, as that could easily be solved , but to learn about stuff which lies dormant in the back of my head.

[Dannyx]

Re: Linear regulator question

Correction: 0.04+0.03 does not equal 0.0012 (I multiplied by accident instead of adding), so 0.04+0.03 = 0.07. 3 divided by 0.07 equals 42.85, so let's say 43 ohms (which I don't think exists as an actual resistor, but we'll round that to 45/40).

[PeteS in CA]

Re: Linear regulator question

43 ohms is a standard 5% resistor value. Keep in mind that a transistor's beta is not the same near its maximum current as it is at low current. That datasheet should have a graph showing how beta varies with collector current. A transistor with an absolute maximum collector current may have a beta so low at the current that it really has a useful maximum current in the 650-850mA range.

A zener needs a certain minimum current for its voltage to be in its specified range. You need to pick the resistor from your unregulated voltage to the cathode of your zener and transistor base so that you can have good base current at high load current, but not burn up your zener at zero load current.

[Dannyx]

Re: Linear regulator question

43 ohms is a standard 5% resistor value. Keep in mind that a transistor's beta is not the same near its maximum current as it is at low current. That datasheet should have a graph showing how beta varies with collector current. A transistor with an absolute maximum collector current may have a beta so low at the current that it really has a useful maximum current in the 650-850mA range.

A zener needs a certain minimum current for its voltage to be in its specified range. You need to pick the resistor from your unregulated voltage to the cathode of your zener and transistor base so that you can have good base current at high load current, but not burn up your zener at zero load current.

To my mind, assuming V/R=i, 12/43 would allow a max of 0.2 A to flow through that zener under no-load conditions (theoretically, as it varies drastically in practice I'm sure).

[petehall347]

Re: Linear regulator question

top of my head says 220 - 330 ohms 1/2 watt ..
zener 9.6v ... 10v if not available ...

[redwire]

Re: Linear regulator question

Um, there's going to be some smoke.

Calculate the power dissipation in the regulator (transistor)- with about 3V drop at 1A that is 3W.
For a SS8050 TO-92 transistor, max. PD is 1W. Also, RθJA is 125°C/W, so 3W will heat it up 375°C. The little guy can't take that much.

Linear regulators always waste the extra voltage as heat. You need a TO-220 part instead, with a heatsink.

A 9.1V zener you get about 8.4VDC output. Because you lose VBE, the zener must be a higher voltage by 0.7V I think 10V zener is next value up.
For the zener's resistor, always OK to give extra current there, even 47-68R is OK and a 1W zener.

[budm]

Re: Linear regulator question

"A 9.1V zener you get about 8.4VDC output. Because you lose VBE, the zener must be a higher voltage by 0.7V I think 10V zener is next value up." You got that right, so he can add one or two general diode such as 1n4000 series in forward bias mode in series with the Zener to compensate for the Vbedrops of the transistor.

[Dannyx]

Re: Linear regulator question

Um, there's going to be some smoke.

Calculate the power dissipation in the regulator (transistor)- with about 3V drop at 1A that is 3W.
For a SS8050 TO-92 transistor, max. PD is 1W. Also, RθJA is 125°C/W, so 3W will heat it up 375°C. The little guy can't take that much.

Linear regulators always waste the extra voltage as heat. You need a TO-220 part instead, with a heatsink.

A 9.1V zener you get about 8.4VDC output. Because you lose VBE, the zener must be a higher voltage by 0.7V I think 10V zener is next value up.
For the zener's resistor, always OK to give extra current there, even 47-68R is OK and a 1W zener.

Didn't take the transistor's power rating into account, true...it seems like such a low power application at first, but it does seem I need to step it up after all. True, the router would never actually get close to 1 full amp - it's more like 600mA at 9v, since that's what my bench PSU told me a while back when I tried it out to see if the thing was even working.

[PeteS in CA]

Re: Linear regulator question

To my mind, assuming V/R=i, 12/43 would allow a max of 0.2 A to flow through that zener under no-load conditions (theoretically, as it varies drastically in practice I'm sure).

Your zener is a half watt zener, but .2 A X 9.1 V is 1.82 watts. The maximum zener current, is rated at 43 mA, and you probably need at least 1 mA of zener current. That means you have up to 42 mA of base current. Throwing out round numbers, you would need a beta of 30 to get a collector current of 1.25 A. Not only do you need a TO-220 transistor, but you probably need one rated for something like 5 A to get a beta of 30 at 1 or 1.25 A (TIP41 is rated for 6A).

Now, if you want more gain and can give up an extra .7 V, maybe try a 12V zener (or the 9.1 V zener plus 2 1N4001 rectifiers in series) and a darlington (TIP120 is 5 A).

IIRC, the tempco of a zener is the opposite of a rectifier, so having a zener and a rectifier in series might give better temperature stability.

[momaka]

Re: Linear regulator question

Um, there's going to be some smoke.

Calculate the power dissipation in the regulator (transistor)- with about 3V drop at 1A that is 3W.
For a SS8050 TO-92 transistor, max. PD is 1W. Also, RθJA is 125°C/W, so 3W will heat it up 375°C. The little guy can't take that much.

No doubt

The first thing one needs to do when re-using components is to make a rough calculation of whether or not the components will meet the power requirements. And by meet, I mean that each component should have at least 25-50% of safety margin on it's ratings if you want the circuit to be reliable and last. Running components near 100% of their rating is always asking for trouble.

[Dannyx]

Re: Linear regulator question

I found this. It uses a NTE184 transistor, which is indeed much larger that what I originally found in my parts box. Still, not wanting to actually buy parts for this little experiment, I'll try sourcing a TO-220 or similar transistor. Or perhaps a simple bloody resistor in series with the 12v supply is all I need

[Dannyx]

Re: Linear regulator question

Back on this project (hope people are still around ): I tried various combinations of resistors and played around with this thing all day, but I could NOT get it to work for the life of me...I AM able to get 9v out of 12v when nothing is attached to the circuit, so at least I know it works correctly, but as soon as I try to hook up the router to this contraption....jack sh!t. The output drops to 3v, which of course is not enough. I can't seem to get the resistor right, which is actually annoying the hell out of me, because despite turning into a mathematician overnight and going over the numbers over and over again, it does not work in practice. Goes to show electronics is not always about the numbers. Found an even better article explaining how the thing is supposed to work here, though I couldn't replicate it in practice, despite being fairly sure my calculations were spot-on. Part of the problem I believe is that the power requirement is too great for such a thing, because I need at least 400-500mA to run the router, while all the examples are for low power applications (stuff like 100mA which is pathetic), so I'd need components that are so large and inefficient I'd be better off getting a 317 or 7809 instead....

[budm]

Re: Linear regulator question

Why don't you show us the exact circuit that you draw up and built? How about using light load first and go up to see how it reacts?
"I can't seem to get the resistor right, which is actually annoying the hell out of me, because despite turning into a mathematician overnight and going over the numbers over and over again, it does not work in practice. Goes to show electronics is not always about the numbers." 1A through 1 Ohm will develop 1V drops, nothing wrong with the math in circuit or in practice, if you only see 0.5V and you measure 1A of current flowing through that 1 Ohm (suppose to be) then you will have to investigate and find out why, the answer will be obvious as to what the problem is.
When something does not make logical sense that DATA and the design has to be checked.
"so I'd need components that are so large and inefficient I'd be better off getting a 317 or 7809 instead."
What do you think is inside the 7809, 317? The Pass Element in your discreet circuit or in the IC is acting like Variable resistor to maintain the constant Voltage on the load when the Load resistance changes.

https://cdn.badcaps-static.com/pdfs/...4077910403.pdf
You still have Pass Element inside the IC, the IC version is just more precise and added protection version of the simple discreet circuit.
You still have to take care of Vdrops on the device just like with your discreet circuit, nothing change there. If the input is 12v and the output is 9V so you will still have 3v drops on the 7809/317 and at 1A it will still dissipating 3W.
Input Voltage to the regulator stays at 12V or it drops down with load?
BTW, did you actually verify the current draw of the Router when running on 9V? Make and Model of the Router?

[Dannyx]

Re: Linear regulator question

Update: this thing's a cock-up already, but I almost got it working, more through trial and error than through math...I just suck at it. I'm able to follow up to a certain point but then I just lose it :|The circuit is this thing, just so you don't have to scroll through my first post again.
The transistor: again, not wanting to buy anything for the cause, I arbitrarily went with a MJE13007 I found in my parts bin and just followed the schematic. The diode: 1n5347 (I don't have it in front of me right now, but I'm fairly sure this is the one).
Now for that bloody resistor: a 100 ohm one gets it pretty close at around 8v, the power LED does come on on the device, but it doesn't actually boot, since it's still too low. As for Budm's question: yes, the 0.3A current draw at 9v was measured, not guessed. The label says 1A, though I'm pretty sure it never reaches that value - it's just for headroom for the AC adapter which I don't have.
I'm not sure what the next step is. I was thinking of dropping R even further to see what happens. As far as I can make out, the lower R is, the higher the base current is, allowing more current to flow through C>E. I understand I can't drop it TOO low either, because it'll burn a lot of current (makes sense, since I=v/r, so the lower R is, the higher I will be). This is what I was talking about when I mentioned "big bulky parts" - the simplest solution to this whole deal would be a resistor in series with the 12v input and feed that straight into the router, but it'd have to be something like a 6w resistor which is totally impractical and inefficient (correct me if I'm wrong, since I probably am, but e.g 12v/22ohm for instance=0.5A which would be enough to run the router, but at the same time 0.5A*12v = 6W). Cheers.

[budm]

Re: Linear regulator question

Since you are using 1N5347 (10V Zener) so that means the Voltage on that Zener has to be maintained at 10V and since it is reference Voltage to determine what the Output Voltage will be (10 - Vbe = 9.4V).
So what Voltage do you actually have on the Zener diode without load and with load?
Do you have R2 in place?
R2 is the load.
The hFE of MJE13007 can be as low as 8!
https://cdn.badcaps-static.com/pdfs/...6fa262ef7d.pdf

[budm]

Re: Linear regulator question

"the simplest solution to this whole deal would be a resistor in series with the 12v input and feed that straight into the router, but it'd have to be something like a 6w resistor which is totally impractical and inefficient (correct me if I'm wrong, since I probably am, but e.g 12v/22ohm for instance=0.5A which would be enough to run the router, but at the same time 0.5A*12v = 6W). Cheers."
Think about it, so if you use 22 Ohm in series with the load, so if you have 0.5A of current flowing, then:
What Voltage will you have on the resistor?
What Voltage will you have on the load?
I think you need to understand basic Electronic theory first.

[Dannyx]

Re: Linear regulator question

I think you need to understand basic Electronic theory first.

True that, that's why I'm not trying it out on MY devices Now, measuring at the cathode of the diode+the base of the transistor (red circle is what you mean), I get the correct value: around 9.4-9.5. The emitter (output let's call it) reads 8v with the 100ohm resistor when under load, and around 10.45 open circuit (if memory serves).

[redwire]

Re: Linear regulator question

Dannyx, at least you are having fun.
A couple rules for learning electronics, so you don't get lost or frustrated:
1. Start with parts that are working
2. Don't connect the parts wrong.

The circuit node you circled red, R1 supplies current (12-9.4)/100R = 26mA to it. Of the 26mA, some goes to the zener, some goes into the transistor base.

A transistor should read about 0.7V emitter-base. Since you are not measuring this: base 9.4V and emitter (output) 8V = 1.4V VBE then your transistor is pooched or wrong. Check it with a multimeter on diode test to verify

[Dannyx]

Re: Linear regulator question

Dannyx, at least you are having fun.
A couple rules for learning electronics, so you don't get lost or frustrated:
1. Start with parts that are working
2. Don't connect the parts wrong.

The circuit node you circled red, R1 supplies current (12-9.4)/100R = 26mA to it. Of the 26mA, some goes to the zener, some goes into the transistor base.

A transistor should read about 0.7V emitter-base. Since you are not measuring this: base 9.4V and emitter (output) 8V = 1.4V VBE then your transistor is pooched or wrong. Check it with a multimeter on diode test to verify

Salvaged that tranny from some power supply and although it doesn't read any immediate shorts, it could still be bad, so your 1st point might not apply indeed...point 2 is not too important in this case
Now, question: you say a transistor should read around 0.7V between the emitter and base ? Why ? Does the datasheet tell you something I'm not seeing ? Quite possible, since some people just look at all those numbers and graphs and are like "elementary". I know SOME theory, but some things are a bit over my head like I said.

[petehall347]

Re: Linear regulator question

have a play with this ..
http://www.falstad.com/circuit/e-npn.html