Quick check of my power supply schematic?

[varialbender]

I just tried building a power supply and am having a few problems. Instead of getting +/-15V from a 16VAC adaptor, i'm getting +1.84 and -2.19V. I want to make sure what I'm trying to build is right in the first place. Here's the schematic:



Any ideas what could be going wrong?
Thanks a lot

[bancika]

what are those 10K resistors for?

[pjwhite]

A 16 VAC transformer may not actually be producing enough voltage for your circuit.  Depending on what kind of regulators you are using, you may need 18 volts or more at the inputs.  You should also allow for variations in the line voltage, which will translate to variations in your transformer output voltage.  For example, if your transformer gives you an 16V RMS output given a 120V RMS input, you'll only get 14.4 volts out with a 108 volt (10% low) line voltage.
If you are using low-dropout regulators, you usually will need a largish capacitor on the output as well as on the input side of the regulator to stabilize the output.  Check the data sheet.
The half-wave rectifiers will cut the effective DC level.  What are your DC (and AC) voltages at the inputs to the regulators?  Are you testing the power supply with any load on the output other than the 10k resistors shown?

[varialbender]

Paul Perry mentioned that I should put a 10k resistor across the output of the negative regulator because some don't like running without a load. I figured for testing I'd do the same to the positive regulator.
The 10k resistors are my only loads right now. I'm hoping to power a phaser and AD633 ring mod with this power supply.
I'm at school now, but I'll measure the input voltages as soon as I get home.
I'll also try adding some output caps.
The regulators are LM7915CT and something like the positive version (can't remmember the number).

Thanks

[sta63bmx]

Should the bottom diode and cap be turned the other way? *scratching head*

[varialbender]

I'm pretty sure that's right. The geofex article on power supplies has the last image showing them oriented this way. Imagine that there's positive and negative voltage alternating out of that source, so when it's positive, it goes up in the normal direction with the diode, and when it's negative, it goes through the other diode the 'wrong' way. Also, you'll notice that the top line for positive voltage is more positive than ground, so the cap is the right way, and the bottom line is more negative, so also arranged properly (so that the + side goes to the side where the voltage is more positive).

[R.G.]

First, take the regulators out. Now measure the +Vdc and the -Vdc across the capacitors to ground (i.e. the midpoint.) Do you get more than +18 and -18?

If so, the rectifier part works, and I'd bet you have a pinout problem with the regulators.

The 7800 (positive) and 7900 (negative) three terminal regulators look just alike, but they have different pinouts. Not only that, but the tab is ground on the TO-220 version of the 7800, and it's the input voltage on the 7900. So it's easy to make mistakes. I've made them... a lot.

[varialbender]

Hmmm... checked the voltages across the 10k resistors after taking out the regulators, and now I'm getting +2.45V and -5.75V...
On my first attempt at breadboarding this today, I made sure to check the pinouts on the regulators, but figured I'd give it a shot anyway. It seems the regulators are bringing the voltage down and regulating like they would when receiving too low a voltage, but something's wrong somewhere... I'm more and more certain it's going to be something ridiculously stupid. I'd say I'm reading my meter wrong, but then it shouldn't give me different values for positive and negative, so I guess it can't be that. I might have to replace the battery in the meter just to cross that off the list of stupid possible mistakes.

Anything jumping out seeing those odd numbers?
The output of the adaptor is 16VAC 1.1A. The caps are 1000uF up to 50V... that shouldn't have anything to do with it, though, cause that just means that they can handle up to that voltage, right? Diodes, again, 1N4001, shouldn't drop that much voltage, and resistor is pretty standard 10k (from NTE.... expensive bastards). Dayum, this is discouraging.

[R.G.]

Divide and conquer is always a good strategy. That's what I was doing suggesting taking the regulators out.

How about you remove the negative-side diode? The positive side diode/cap should still work, right? Then see what voltage you get out on the positive side cap.  Should be the same as you'd get with the negative side hooked up, and it's simpler to test.

[pjwhite]

Hmmm... checked the voltages across the 10k resistors after taking out the regulators, and now I'm getting +2.45V and -5.75V...

You shouldn't be getting ANY voltage across the resistors with the regulators out, unless there's something you're not telling us.

On my first attempt at breadboarding this today, I made sure to check the pinouts on the regulators, but figured I'd give it a shot anyway. It seems the regulators are bringing the voltage down and regulating like they would when receiving too low a voltage, but something's wrong somewhere... I'm more and more certain it's going to be something ridiculously stupid. I'd say I'm reading my meter wrong, but then it shouldn't give me different values for positive and negative, so I guess it can't be that. I might have to replace the battery in the meter just to cross that off the list of stupid possible mistakes.

Grab a fresh battery (any kind) and see if you get a reasonable reading with your meter.

Anything jumping out seeing those odd numbers?
The output of the adaptor is 16VAC 1.1A.

I assume that's what's printed on it. 

The caps are 1000uF up to 50V... that shouldn't have anything to do with it, though, cause that just means that they can handle up to that voltage, right?

Right.

Diodes, again, 1N4001, shouldn't drop that much voltage, and resistor is pretty standard 10k (from NTE.... expensive bastards). Dayum, this is discouraging.

The diodes should drop about 0.7 volts each.

I'll ask again -- what is the voltage with respect to ground at the input pin of each regulator?

[varialbender]

I removed the negative half of the circuitry, just leaving one power supply wire going to ground, the other to the diode in the positive direction, then 1000uF to ground, with a 10k in parallel with the cap. The voltage across the 10k is now 24.5V! That seems a bit more reasonable, eh? A nice high number. I'll have to rebuild it step by step and see if I run into the same problems.

So why shouldn't I get voltage across the resistors when removing the regulators? (Of course, I didn't *just* take it out, I connected the top of the cap and resistor together, though I thought that was implied).
I'll put in a fresh battery as my next move, but my last reading seemed reasonable, right? So I don't think it's the battery.
Yes, what's printed out on the adaptor is 16VAC 1.1A, that's not my readings.
I'll give you that reading when I put it back together. Sorry, I'm a bit tied up, family is coming over and I'm not able to spend hours on the PS like I'd love to right now.

Thanks a lot

[Transmogrifox]

1N4001's are rated at about 17A peak for 1 cycle (or short period of time).

Let's assume your transformer is unlimited in its ability to supply current at the rated voltage, thus the AC at the transformer is this:

v(t)=16*cos(2*pi*60)

The current voltage relationship of a capacitor is this:

i(t)=C*dV(t)/dt

For you non-math buffs, this means that the faster the voltage is changing, the greater the current will be.  This is makes intuitive sense because we expect that there is no current flowing through a capacitor when DC is applied (except for small leakage due to nonideal characteristics of a real capacitor).  When you change the voltage quickly, then the capacitor conducts more current as it is charged and discharged by the changing voltage.  For really high frequencies it looks like a short circuit.  See a problem here?

How much current flows through a 1000uF capacitor at 16 Volts?  Let's continue

dV(t)/dt = d[16*cos(2*pi*60)]/dt = 2*pi*60*16*cos(2*pi*60)= 6032*cos(2*pi*60)

so, i(t)=C*6032*cos(2*pi*60)=1000e-6*6032*cos(2*pi*60)=6*cos(2*pi*60)

You have about 6 amps RMS in those diodes on the startup.  The transient on the charge-up will determine how long you're exceeding the max current.  Each cycle is peaking at 6*sqrt(2)=8.5A  (so that's within spec).

So the only other issue is that the 1N4001 is rated for 1A continuously.

As long as your load is much less than 1 amp, then you're diodes are ok.

Nevermind this post.  I guess it was just good for me to go through the math again for the sake of review--and there it is for those of you who are designing power supplies.

You better measure the transformer's AC output to make sure it's all good.  It doesn't make sense that the voltages are as you reported if it's wired exactly to your schematic and all the parts are good.  I can't remember if you mentioned the cap rating or not, but do make sure your capacitors are rated at 25V or higher.

[varialbender]

Good to see the math sometimes. It's nice to get a blend of different teaching on this forum.
My caps are rated at 50V, so that's not an issue (thank god... I've had enough of caps blowing up in other people's designs, such as a computer monitor and a cable box recently...)
I guess, if there's nothing else wrong with my schematic than adding output caps, there must be something wrong with how I had it wired. I went over it many times, knowing that it's a common problem to think something's wired properly then finding something backwards, but I could have sworn it was all good. I even looked at beginner guides to caps to make sure that my electros were in properly, and, of course, checked the data sheets for the regulators.
I'll try to rebuild.

Thanks again.

[pjwhite]

So why shouldn't I get voltage across the resistors when removing the regulators? (Of course, I didn't *just* take it out, I connected the top of the cap and resistor together, though I thought that was implied).

Sorry, I didn't take that as implied.  Just removing the regulators as you said would have left the resistors with no connection except the grounded side.

I'll put in a fresh battery as my next move, but my last reading seemed reasonable, right?

Yes, the 24 volt no-load reading seems about right.  I just meant that you could measure a known good battery as a check of your voltmeter before you blindly replace the meter battery.

Are you sure that the resistors in your circuit are 10K and not 10 ohms?  If the regulators are wired correctly, that's all I can think of that might be your problem.  Do the resistors or regulators get warm when you turn on the power?

[R.G.]

Theoretical math is good.

However, as long as we're modelling, we should model the more accurate case.

A sub-20VA transformer is not unlimited in its ability to supply current. It has real-world primary and secondary resistances, as well as leakage inductance, and those figure right into the transient solution. So does the ESR of the filter cap and the bulk resistance of the diode. The real world load seen by the transformer is

(16/120)^2 * Rprimary + Rsecondary +Resr+jwLleakage - 1/jwC

discounting the ESL of the cap and the various high frequency parasitic capacitances and the varying diode resitances.

One could do it time domain, subbing in V=Ldi/dt and I=Cdv/dt as per your example and solving.

The voltage across the 10k is now 24.5V

OK, that's better. Working backwards from that, you have a transformer secondary voltage of (24.5+0.6)/1.414 = 17.75V. That makes sense as the open circuit voltage of a small power transformer that will sag to 16.0V at full load of 1.1A. So we can calculate the transformer impedance as (17.75-16)/1.1 = 1.59 ohms, which is the sum of the secondary and referred primary resistances.

Now go hook up your negative side. If you hook it up properly, you'll see -24.5V.

[varialbender]

I tried adding the negative section, and it doesn't seem to be working. While testing different voltages, I accidentally shorted something with one of the 'probes', and got a nice big pop and spark, scared the hell out of me. Anyway, it's all connected, and the positive side is working properly, but the negative side is reading 0 across the 10k (they're definitely 10k btw, but it would still read about the same voltage if it were 10M eh?). Then, I decided to check something else: on the baord, I've got one horizontal line for V+, two for ground (one on top and one on bottom, and I've connected both) and one for V-, and decided to check the ground at that wire connecting both ground lines and the positive side of the resistor where the 25V is, but it showed me 0, which makes no sense. I usually get 25V from one leg of the resistor to the other, and the one leg at ground should be the same as touching ground anywhere, but apparently on my circuit it's not... wtf? Now I have to check all the holes on my breadboard, I guess... that's strange and pisses me off.... it's a new breadboard from 3M and it may be the source of some problems? Argh!
Is there any secret to 3M breadboards?

[R.G.]

OK, so divide and conquer again. Remove the + side diode and get only the negative side to work.

Are the diode and capcitor polarities correct?

Breadboards are OK, but they won't do high currents very well. Try plugging the parts into other contacts on the breadboard that may have not been fried.


Also check your diode and cap with the ohmmeter setting of your DMM to be sure they aren't shorted.

[Transmogrifox]

Theoretical math is good.

However, as long as we're modelling, we should model the more accurate case.

A sub-20VA transformer is not unlimited in its ability to supply current. It has real-world primary and secondary resistances, as well as leakage inductance, and those figure right into the transient solution. So does the ESR of the filter cap and the bulk resistance of the diode. The real world load seen by the transformer is

(16/120)^2 * Rprimary + Rsecondary +Resr+jwLleakage - 1/jwC

discounting the ESL of the cap and the various high frequency parasitic capacitances and the varying diode resitances.

One could do it time domain, subbing in V=Ldi/dt and I=Cdv/dt as per your example and solving.

The voltage across the 10k is now 24.5V

OK, that's better. Working backwards from that, you have a transformer secondary voltage of (24.5+0.6)/1.414 = 17.75V. That makes sense as the open circuit voltage of a small power transformer that will sag to 16.0V at full load of 1.1A. So we can calculate the transformer impedance as (17.75-16)/1.1 = 1.59 ohms, which is the sum of the secondary and referred primary resistances.

Now go hook up your negative side. If you hook it up properly, you'll see -24.5V.

Thanks RG.  This is true.

My train of thinking (correct me if I'm wrong) is that the addition of ESR, Diode equivalent resistances, secondary & primary coil resistances all serve to reduce the amount of instantaneous current through the diode.  If I can show (as above) that in the case where we ignore all of these parameters and treat the circuit as an ideal voltage source feeding an ideal diode into an ideal capacitor--and the currents come out within specifications for the diode, then that's some easy math to get peace of mind that the diodes aren't being fried by excessive instantaneous current.

In the end, RG's method of divide-and-conquer is about the only way to solve the problem.  The circuit is pretty clearly a good design destined to work.  It's just a simple matter of getting the real thing to look like it does on paper.

For as simple as the circuit is, and the fact that you have it on a breadboard, you should take it all apart (completely off the board), and rebuild it.  Make sure you don't inadvertently short the (-) supply to ground through a trace you didn't think was interconnected on the breadboard.

Sparks and magic smoke are what keep me going sometimes.  It can be a lot of fun to be playing with power supply designs live on a breadboard and put something into the hot circuit backward or wrong pinout or so and watch a plume of smoke rise from the table.

[varialbender]

Yeah, yesterday, I was trying it again, and forgot that I previously turned the cap around, and when I plugged it in, I could smell something was wrong... I unplugged it right away, and wait for a while, then touched the cap, and it was nice and hot. Didn't seem to be damaged though.

[Sir H C]

You also want smaller (10uF - 100uF) caps on the output of the regulators.  The regulators are not that "fast" so they do not compensate for really quick glitches on the supply, the caps catch those.