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I'm trying to build a voltage regulator that takes in a reference voltage and an unregulated voltage (to power the components, and this voltage is always greater than the reference voltage) and outputs the reference voltage over any capacitive, inductive or resistive loads.

Here's what I have so far: my circuit schematic

The circuit has to be capable of outputting up to 5A at 6 volts and should have a relatively fast slew rate even at a high capacitive load (nearly 200 V/ms over 4uF load). It also has to have good load regulation (output voltage shouldn't drop over 200 mV at max load in reference to the reference voltage).

To satisfy these requirements, I've added a darlington pair since the 2N3445 is definitely capable of outputting that power, but requires a large base current, which the LM358 can't provide.

The issue I have right now is that when I attach a current source that emulates a 1 Amp load on and off at 1000 Hz, the overshoot and undershoot errors I'm getting are terrible. The output voltage jumps up to nearly 10V above the reference voltage for about 50us before crashing down to near 8V below the reference voltage before finally coming to rest at the reference. Is there anyway I can modify my circuit so this error is fixed?

I tried attaching a larger capacitance to the output (I have a 1 uF right now, this is the design capacitance, and not a load capacitance), but that didn't help out too much. Also attaching a larger capacitance over the smaller transistor's emitter (where the 2uF is hanging right now) destroys the slew rate (I crash down to nearly 8 V/ms when I hang a 22 uF there.)

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"over any capacitive, inductive or resistive loads" - please get real dude. – Andy aka Jun 12 '14 at 16:48
up vote 6 down vote accepted

I only read your first paragraph and looked at the schematic, but there are some obvious problems already:

  1. Get rid of C1. That slows down the plant response, which is always a bad thing for stability. Worse, its effect is asymmetric with rise and fall.

  2. Get rid of C2. This is slowing down the feedback signal, which is a really bad thing to do to any control loop.

  3. Add a little resistance from the base of Q2 to ground. This will help the symmetry of the response a little, and will make the plant more linear at low voltages. It will also overcome leakage current thru Q1 to keep the system fully off when the base of Q1 is driven low. I don't know what currents this is intended for, but 1 kΩ to 10 kΩ should work.

  4. Add a little compensation immediately around the opamp. Put a small cap directly between the opamp output and its negative input. Then put some resistance in series with the output voltage feedback signal into the opamp negative input. Start with 100 pF for the cap and 10 kΩ in series with the feedback to see what happens. Adjust the feedback series resistor to get the step response you want. Lower values will make a faster step response but will cause overshoot. Higher values will decrease and then eliminate overshoot but at the expense of response time.

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Agree with poster above. In addition, put a large capacitor at the output so that it can source/sink current transients – user45513 Jun 12 '14 at 23:15
@user: Depends on what kind of current transients you expect and what voltage transients you can tolerate. A large output cap will help with that, but will also make things less stable, requiring a slower overall control response. Everything is a tradeoff. – Olin Lathrop Jun 14 '14 at 15:50

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