Variable Bipolar Power Supply

[Hero999]

Good idea, I'll add those 100nF capacitors, because I'm aware that the large electrolytics have a high inductance and theie resonant frequency is often way under 1MHz. I didn't include them becase the datasheet says they aren't required if it's under a certain distance from the filter.

I'm just using a plain electrolytic for the ADJ bypass capacitor. I didn't want to use a tantulum because I've heard they can be unreliable. I might add a 100nF capacitor in parallel if adding one on the filter doesn't help.

 

[Sig239]

That's it? Did you get the supply stabilized or did it get shoved to the back burner? Good luck, I hope you worked it out.

 

[Hero999]

It got shoved to the back burner I'm afraid, my oscilloscope popped its clogs and I've been meaning to get around to having a go at repairing it.

 

[Sig239]

I'll be rooting for you..lol Keep us posted

 

[cadstarsucks]

Here's the plans for a power supply I'm intending to build. By my claculations it should provide me with +-15.7V@1.5A. The mains transformer I'm using is a 15V-0-15V torroid rated to 2.67A which is slightly over-rated which is perfect as the regulators might not current limit untill 2.2A.

Anyway, I just thought I'd ask you lot if you can think of any improvements before I etch the PCB and build this.

This is probably late and might have been mentioned already, but oh well

Briefly if you are on 115V:
the mains normal operating point is 100-130V
the transformer is probably 15%
the mains peak voltage is RMSx1.414 or 141-184V
The safe transformer current is rating/1.8 or 1.48A
The regulators need 3V to operate - so you need 18 volts at the valley of the filtered waveform, or preferably an 18V transformer
The diodes need to be rated for 2A or better
and the caps should have a 4.5A RMS rating...

That is what is supposed to be done. For a complete discussion see the end of this catalog and cut corners as you like:
**broken link removed**

I prefer offline switchers myself but most people do not have the resources that I do.

D.

 

[Hero999]

Briefly if you are on 115V:

230V.

the mains normal operating point is 100-130V

220V to 250V.

the transformer is probably 15%

I don't know what the regulation is but it's certainly better than that, it's an 80VA toroidal transformer and these give pretty good regulation, about twice as good as an E-core transformer so let's say 7.5%. Anyway isn't the voltage of a transformer specified at the full load?

the mains peak voltage is RMSx1.414 or 141-184V

About 311V to 354V but what difference it makes is beyond me, I normally use the output voltage to calculate ripple.

The safe transformer current is rating/1.8 or 1.48A

It's rated for 2.67A, but I disagree, I want to be able to draw at least 1.5A so the transformer needs to be rated for at least [latex]1.5 \times sqrt{2} = 2.21A[/latex].

The regulators need 3V to operate - so you need 18 volts at the valley of the filtered waveform, or preferably an 18V transformer

You're mistaken, providing it's not too hot or not too cold the LM217 has a dropout voltage of 2.25V@1.5A so minimum DC voltage needs to be 17.25V,therefore a 15V transformer will do, let's calculate the ripple assuming a 4700µF capacitor.

[latex]V_{RIPPLE} = \frac{I_{LOAD}}{2FC}= \frac{1.5A}{2 \times 50 \times 4700 \times 10^{-6}} = 3.2V[/latex]

The peak voltage is [latex]15 \times \sqrt{2}= 21.2V[/latex]

The minimum voltage in-between the ripple is [latex]21.2 - 3.2 - 0.7 = 17.3V[/latex]

The diodes need to be rated for 2A or better

They only conduct for half the cycle so they only need to be rated for 1.11A, either way I've used a 3A 400V bridge from an old PC power supply so it doesn't matter.

and the caps should have a 4.5A RMS rating...

Why?

Sure, the peak current might be 4.5A but the RMS current is probably nearer to I/√2.

That is what is supposed to be done. For a complete discussion see the end of this catalog and cut corners as you like:
**broken link removed**

I dissagree with some of the formulae and figures.

For a start the formula for the capacitor is incorrect if you use a linear regulator, it says "2000µF/amp for 3V p-p ripple", this assumes a resistive load rather than a constant current load. A 2000µF capacitor will give 4.167V of ripple @60Hz when given a 1A constant current load. The formula used in my calculation above if pretty accurate for constant current loads, try simulating it in a SPICE simulator.

The claim that rectifiers can't handle surges and need to be rated to the full output current is rubbish, a 1N4001 rectifier can withstand a non repetitive peak surge of 30A and the WOO5 can take 50A for a 8.3ms half sine wave (this is just a ballpark figure I didn't use 1N4001s or a WOO5 in my design). I might see the point if this was a huge power supply but not in my little power supply where the internal impedance of the transformer will limit the surge current to a safe level.

The only corner I have cut is undersizing the filter capacitors slightly so I might see some ripple on the output if the mains voltage is a little too low, the regulator is too hot/cold, the capacitors are in the lowest tolerance band, or the voltage is whacked up to 15.7V (15V is the design maximum but due to component tolerances it can go 0.7V higher). I know I should've used 6800µF capacitors but I didn't have any handy, for a start as the transformer is slightly over-sized the output voltage will be higher than expected at full load, 1A is good enough for most applications and I'm not doing this professionally, besides there's nothing stopping me piggy backing a couple of 2200µF capacitors on there.

I prefer offline switchers myself but most people do not have the resources that I do.

I prefer switchers too, but they can be too noisy and troublesome to build.

 

[Roff]

The minimum voltage in-between the ripple is 21.2 - 3.2 - 0.7 = 17.3V

In your DC voltage and ripple calculations, you need to keep in mind that a diode bridge drops about 1.5V.

A 2000µF capacitor will give 4.167V of ripple @60Hz when given a 1A constant current load. The formula used in my calculation above if pretty accurate for constant current loads, try simulating it in a SPICE simulator.

I ran the simulation, and got 3.06V p-p ripple. Your equation fails to account for the fact that the cap is only discharging for about 6.4ms. For the remaining ~2ms, it's charging.

 

[Hero999]

In your DC voltage and ripple calculations, you need to keep in mind that a diode bridge drops about 1.5V.

That's true when measured from +15V to -15V but when measured from +-15V to 0V it's half that value, 0.75V.

I ran the simulation, and got 3.06V p-p ripple. Your equation fails to account for the fact that the cap is only discharging for about 6.4ms. For the remaining ~2ms, it's charging.

You sure you used a constant current load?

I'll have another go I suppose but if what you're saying is correct then I haven't undersized my filter capacitors so I haven't cut any after all.

 

[Roff]

That's true when measured from +15V to -15V but when measured from +-15V to 0V it's half that value, 0.75V.

I'm saying that your peak DC voltage will be about 19.7V. There are always 2 diodes in series with the transformer between GND and the rectified output.

You sure you used a constant current load?

I'll have another go I suppose but if what you're saying is correct then I haven't undersized my filter capacitors so I haven't cut any after all.

Yep, I used a constant current load. Furthermore, a with resistive load and 3V ripple, the current only changes about 15% for a 20V supply, so the load could be approximated by a constant current without too much error.

 

[Hero999]

I'm saying that your peak DC voltage will be about 19.7V. There are always 2 diodes in series with the transformer between GND and the rectified output.

I still disagree.

Let's look at it like a single rail 30V supply first (load connected from +V to -V).

[latex]V_{OUT+/-} = 30 \sqrt{2}-1.5=40.93V[/latex]

Now let's look at it like it's a +/-15V supply (load connected from +V to 0V)

If [latex]V_{OUT+/-}=40.93V[/latex] Then [latex]V_{OUT+} = \frac{V_{OUT+/-}}{2} = 20.46V = 15 \sqrt{2}-0.75=20.46V[/latex]

Yep, I used a constant current load.

Which is how a linear regulator should behave, providing its load is resistive the current will be the same regardless of the input voltage because the load current it always the same.

Furthermore, a with resistive load and 3V ripple, the current only changes about 15% for a 20V supply, so the load could be approximated by a constant current without too much error.

I agree, I've just simulated it and you're right.

 

[Roff]

Sorry. I missed the part where you said you have a center-tapped (15-0-15) transformer. You're absolutely right.

 

[cadstarsucks]

Sorry, I am forced to short change this a bit...

It's rated for 2.67A, but I disagree, I want to be able to draw at least 1.5A so the transformer needs to be rated for at least [latex]1.5 \times sqrt{2} = 2.21A[/latex].


You're mistaken, providing it's not too hot or not too cold the LM217 has a dropout voltage of 2.25V@1.5A so minimum DC voltage needs to be 17.25V,therefore a 15V transformer will do, let's calculate the ripple assuming a 4700µF capacitor.

[latex]V_{RIPPLE} = \frac{I_{LOAD}}{2FC}= \frac{1.5A}{2 \times 50 \times 4700 \times 10^{-6}} = 3.2V[/latex]

The peak voltage is [latex]15 \times \sqrt{2}= 21.2V[/latex]

The minimum voltage in-between the ripple is [latex]21.2 - 3.2 - 0.7 = 17.3V[/latex]

As long as your valley voltage stays above the drop out sure...

They only conduct for half the cycle so they only need to be rated for 1.11A, either way I've used a 3A 400V bridge from an old PC power supply so it doesn't matter.

Why?

Sure, the peak current might be 4.5A but the RMS current is probably nearer to I/√2.

I dissagree with some of the formulae and figures.

For a start the formula for the capacitor is incorrect if you use a linear regulator, it says "2000µF/amp for 3V p-p ripple", this assumes a resistive load rather than a constant current load. A 2000µF capacitor will give 4.167V of ripple @60Hz when given a 1A constant current load. The formula used in my calculation above if pretty accurate for constant current loads, try simulating it in a SPICE simulator.

The claim that rectifiers can't handle surges and need to be rated to the full output current is rubbish, a 1N4001 rectifier can withstand a non repetitive peak surge of 30A and the WOO5 can take 50A for a 8.3ms half sine wave (this is just a ballpark figure I didn't use 1N4001s or a WOO5 in my design). I might see the point if this was a huge power supply but not in my little power supply where the internal impedance of the transformer will limit the surge current to a safe level.

I agree on the diode rating, but that as you pointed out about some of the transformer mfgs equations, is based on a resistive load. Given that your transformer only represents 1hm: impedance you could have some high instantaneous currents. But the 3A bridge will probably not be a problem. The thing is a cap is not a resistive load and there is only one ohm between it and an ideal AC source.

The only corner I have cut is undersizing the filter capacitors slightly so I might see some ripple on the output if the mains voltage is a little too low, the regulator is too hot/cold, the capacitors are in the lowest tolerance band, or the voltage is whacked up to 15.7V (15V is the design maximum but due to component tolerances it can go 0.7V higher). I know I should've used 6800µF capacitors but I didn't have any handy, for a start as the transformer is slightly over-sized the output voltage will be higher than expected at full load, 1A is good enough for most applications and I'm not doing this professionally, besides there's nothing stopping me piggy backing a couple of 2200µF capacitors on there.

I prefer switchers too, but they can be too noisy and troublesome to build.

Then have at it. Personally, I think I would use the transformer and a couple simple buck regulators and if that wasn't clean enough got to the LM317s, but I would not like throwing out 40W at the low end. Make sure they are properly heatsunk.

D.

 

[Peter_wadley]

This looks like a great project!

Before I begin I will need to learn more about Zener Diodes and Op-amps though.

Hero, Unless I missed them or you still dont have your camera, can you please post some snaps of your PCB.

Thanks,
Peter

 

[Hero999]

FWIW, transient load recovery can be improved by using something better than that aweful 741 opamp and you can do away with the bias current error correction resistor (5k1) as well by using a FET input type.

A fast op-amp would not help things as far as transient response is concerned, the LM337 takes care of that.

This was discussed at the end of the first page of replies to this topic.
https://www.electro-tech-online.com/threads/variable-bipolar-power-supply.24291/

 

[mrx23]

Happy new year!

What do you think about my Tracking Bipolar Lab Power Supply?
This can be adjusted between +- 1.25-12V 1.5A, it has an OPA regulator so no need to worry about exceeding +-15V input voltage.
And it's centered around GND, so it can supply +-12V at output if needed. I've simulated it's 0,01V accurate with high stability.
The circuit includes all protection diodes and bypass caps.

 

[Hero999]

I have a few concerns.

1k2 is too high for R1 in the LM317 and LM337, I think you've got the decimal point in the wrong place, use 120R.

With a 12V transformer you might have trouble getting 12V out at 1.5A.

Like my circuit, yours doesn't have any compensation so it might be unstable and oscillate under certain conditions.

 

[mrx23]

The 1k2 resistor and 10K pot are taken from the LM311 datasheet
(page 9 2nd pic, attached):
"1.2V-20V Regulator with minimum Program current"
The original parts are: 1k2 resistor, 20K pot.
So I guess it's not a problem.
Or could you recommend a smaller resistor and pot for 1.25-12V?

You are right the 2x12V transformer is might be too small.
I didn't include any feedback compensation because I don't know how to calc them and this PSU its only for hobby purpose.

Can I load this PSU through + - rails only?
If I can then how much current can it provide (1x1.5A or 2x1.5A)?

Do you have any further simple ideas how to improve the schematic?

Thank you!

 

[Hero999]

The circuit in the datasheet will only work if there's a 4mA load connected to it, you've obviously missed the part where it says "*4mA minimum load current".

I got away witout using a compensation capacitor but you might not be so lucky. If you have any trouble, then don't bother calculating it, just try connecting 100nF between the output of the op-amp to the inverting input.

 

[audioguru]

The LM117 is shown. It is a premium IC and costs more. Its minimum load current is 5mA but the LM317 that you used has a minimum load current of 10ma.
Your 10k resistor on the output isn't a 10mA load and if you used a 100 ohm output load resistor for 10mA when the output is 1V then the current will be 120mA when the output is 12V.

The problem is that the output voltage will rise when the load current is low.
so use a 120 ohm resistor with the LM317.

 

[mrx23]

I recalculated to 240 Ohm R and 2k2 potentiometer so the output voltage will be 1.25V - 12.7V.
Now I don't have to worry about the minimum load current?

Thank you!