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I'm attempting to build a linear voltage supply. (I'm not touching the mains yet.)

My first thought was to use an adjustable linear voltage regulator. Unfortunately, I found that linear voltage regulators have minimum load requirements as well as a reference voltage which translates to your minimum voltage output. On most linear voltage regulators I found that reference voltage to be ~1.25V.

I want to be able to go from 0V-30V with current limiting from 1-3A. So, I looked at the equivalent circuit of a linear voltage regulator on one of the datasheets and came up with this...

enter image description here

Since op-amps don't pull much current (in ideal op-amps zero current), and can have a maximum input voltage above 30V I thought that this circuit would be perfect because really my only limit was the maximum collector voltage and current of the BJT connected to the op-amp output, and I can get 0V!

Since the op-amp inputs don't draw current anyway* the voltage divider just has to establish a low current voltage reference. I thought that would be easy, but I was wrong.

Assuming I replace R1 and R2 with a potentiometer, when I'm at a high total resistance everything will be fine. When I drop lower power gets in the way. If I'm at 0 resistance (0V reference) the voltage divider will basically be a short and I'll be fine:

$$I=V/R$$ $$I=30/0Ω$$ $$I=0A$$
$$P=VI$$ $$P=30V(0A)$$ $$P=0W$$

But, when I want a very low voltage power becomes a problem.

(Let's assume total resistance is at worst case resistance of 1Ω)

$$I=V/R$$ $$I=30/1Ω$$ $$I=30A$$
$$P=VI$$ $$P=30V(30A)$$ $$P=900W$$

And even if my constant current regulator was working before hand and only let me draw 3A, I'd still be having power troubles.

My second solution was to leave R2 as a high value resistor and substitute a potentiometer for R1. In this case I wouldn't be having power troubles anymore but something would be funky with the control scheme. Let's say the potentiometer replacing R1 is at 0Ω.

In this case R2 acts as a current shunt, and I get my full rail voltage as the reference voltage. But the potentiometer would have to go from 0Ω - ∞Ω...and I don't think that's feasible.

The third solution I thought of was to use a linear voltage regulator as a voltage reference but...that kind of defeats the purpose.

The fourth solution I thought of was to replace R2 with a potentiometer instead of R1. I wouldn't have power issues, but the power supply would go from 30V to 0V turning right, and again the potentiometer would have to go from 0Ω - ∞Ω.

There must be a way to do this, but how exactly?

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  • \$\begingroup\$ How did you get 0 from 30 divided by 0? \$\endgroup\$
    – user253751
    Jan 18, 2016 at 3:33

2 Answers 2

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You are not thinking about the potentiometer correctly - it is a three pin device with the wiper effectively being the junction of R1 and R2 - you appear to be regarding it as a rheostat i.e. a two pin variable resistor and this of course can be a problem across supplies.

Basically use a pot (maybe 10k ohm) to replace R1 and R2. This gives you a variable voltage between 0V and 30V going into the non-inverting input of the Op-amp.

Also R3 can be a short circuit and R4 is not needed.

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  • \$\begingroup\$ I added the feedback resistors because any of my solutions would have resulted half the intended reference at any resistance value. I can't believe I didn't realize this. \$\endgroup\$
    – Allenph
    Jun 23, 2015 at 9:01
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With linear regulators, a high input voltage (30V) and low output voltage (1V) at a high current (3A) will always give you a high dissipation ( (30V- 1V) * 3 A = 87 Watts ). That is the main disadvantage of this design.

To solve this you have some options but they make your design more complicated.

  • Lower the input voltage (select a different voltage at the transformer (some transformers have several output voltages, like 12V, 18V, 24V etc.) Some commercial Lab supplies use this scheme.
  • Lower the input voltage using a switched converter, I designed and build one like this 20 years ago using an L296 switching regulator, it also has a linear regulator after the switching regulator.
  • Go for a switching regulator.

Getting a regulator to go down to 0V can also be done with a standard (adjustable) voltage regulator IC. It can be done with the help of a negative voltage or you can use an opamp that can work with a 0V input voltage (for example: MCP6002 ) and use the opamp as the error amplifier. The output of the opamp then controls the feedback pin of the regulator IC.

Not all solutions above are so simple that I would advise them to a beginner. My first supply used an L200 linear regulator and a negative supply so I could go down to 0 V. Using an integrated solution also gives protection against overheating which is a nice thing to have. ST has a nice document to get you going with the L200: Designers note. The L200 is ancient, I know ;-) For sure there are more modern ICs that would also do the job.

I you prefer not to use an IC, here is a very nice classic "1970s style" ;-) power supply !

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  • \$\begingroup\$ I was going to go with a SMPS at first, but while they're more efficient they're complicated and aren't as clean. That model BJT I have in the diagram can handle 40W. Why not just use a couple of those? \$\endgroup\$
    – Allenph
    Jun 23, 2015 at 8:55
  • \$\begingroup\$ Sure, you can use a couple of those in parallel, then don't forget to include some small (0.1 - 0.5 ohm) resistors in series with the emitter so the load is equally shared, just like here: powersupply88.com/24v-to-12v-converter.html I agree that switched converters are too complicated if you want to build one from scratch. They are much less complicated when you use an integrated solution but PCB layout can be critical. Output is indeed not so clean due to switching spurs but good filtering can help a lot. \$\endgroup\$
    – Bimpelrekkie
    Jun 23, 2015 at 9:09
  • \$\begingroup\$ Ouch...those are some giant resistors. Is there a way around that? \$\endgroup\$
    – Allenph
    Jun 23, 2015 at 9:11
  • \$\begingroup\$ Giant ? The size of the resistor only depends on the required dissipation. For 3A divided over 3 transistors that's 1 A per transistor, let's assume 0.3 V drop so 0.3 ohms, 0.3 W I'd then select 0.5 W resistors, they are not so large. In the link they're using 10W resistors, no need for that here ! 0.5W will do fine \$\endgroup\$
    – Bimpelrekkie
    Jun 23, 2015 at 9:17
  • \$\begingroup\$ Honestly, I'm not following why we need the resistors...can't all the emitters just be electrically common? \$\endgroup\$
    – Allenph
    Jun 23, 2015 at 9:19

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