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I am interested in charging a supercapacitor from a solar panel. I saw a circuit on this page by David Johnson. He doesn't let you copy the circuit, but the gist of it is this:

schematic

simulate this circuit – Schematic created using CircuitLab

I understand most of this. The op-amp is acting as a comparator for V+ through a voltage divider, so that when V+ is 2.654V it will reach the comparator threshold of 1.2V (the LM385 is a 1.2V reference).

(2.654 * 61.9) / (75 + 61.9) = 1.2

When the input voltage exceeds 2.654V the op-amp will turn on the MOSFET which will conduct R5 to V-, thus shunting the input voltage to stop it exceeding 2.654V (by much). (The super-capacitor in this case is a 2.7V one).

I have two questions:

  1. What is C1 for? I presume stability of the op-amp, but am not sure.

  2. What is R2 for? That doesn't seem to be doing much. It isn't acting as a voltage divider, and the op-amp input is already high-impedance. I am guessing it is something to do with circuit stability, but am not sure. What is to stop you connecting the voltage reference directly to the non-inverting input of the op-amp? Why choose 10k and not some other value?

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It's explained in Negative feedback capacitor in Op-Amp comparator circuit. R2 and C1 decrease the high-frequency gain of the comparator, so it won't rapidly flip between V+ and V- when the input is near the threshold.

Edit: I've looked at the circuit some more, and I think a comparator is the wrong way to think about it. In steady state, it won't be fully on or off, but in the middle. The transistor will be partially turned on to hold the supercapacitor voltage around 2.654V. It's only if the voltage is significantly too high or low that it will act as a comparator.

In more detail, it's an op amp integrator acting as a voltage regulator. It integrates the error signal (difference between the capacitor voltage and the reference voltage). If the capacitor voltage is too high, the integrator will slowly turn on the transistor, pulling the voltage back down. It should reach a level where the error is zero and stabilize. If the voltage is too low, the integrator output drops, gradually turning the transistor off.

For your original question, R2*C1 controls how fast the integrator will change; without R2 it wouldn't integrate and would act as a simple comparator.

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    \$\begingroup\$ Hi Ken! Nice to see you here. I'm familiar with your blog, and enjoy reading some of your in-depth analyses. It might be useful for demonstration to imagine that R2=0. Keeping the filter the same requires R2*C1=constant as you point out, so C1 blows up to infinity... but since it can't be infinity, C1 would not have much effect in the circuit! \$\endgroup\$ – Daniel Dec 19 '16 at 5:22
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    \$\begingroup\$ Hi Ken! I loved your description of the TL431 voltage reference. It was interesting to see that fuses were literally links that were "blown" in order to configure the chip. However on the link you gave about the comparator there was no resistor in series with the voltage reference (to the op-amp). \$\endgroup\$ – Nick Gammon Dec 19 '16 at 6:06
  • \$\begingroup\$ Thanks, Nick. You're right about the series resistor not explicitly appearing in the reference. The discussion converts R3 and R4 into an equivalent parallel resistor Rp, which I think has approximately the same role as R2. I have to wave my hands a bit at this point. But note that if you consider the Zener as a fixed voltage reference, if you don't have R2 the negative input will also be a fixed voltage. \$\endgroup\$ – Ken Shirriff Dec 19 '16 at 6:21
  • \$\begingroup\$ Is that bad? Or are you saying that the capacitor will effectively have no effect without R2? I could understand that. Would an alternative be to replace C1 with a (fairly large) resistor to give some hysteresis? \$\endgroup\$ – Nick Gammon Dec 19 '16 at 6:30
  • \$\begingroup\$ Yes, without R2 the capacitor would have no effect. (Well, it might have some weird interaction with the Zener diode.) An output change will quickly charge the capacitor, which will slowly discharge through R2. The time constant depends on R2*C1. Using a resistor for hysteresis is the normal solution suggested everywhere else I looked, so I don't know why the capacitor solution was chosen. \$\endgroup\$ – Ken Shirriff Dec 19 '16 at 6:52

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