I have seen this circuit at my workplace and I'm trying to understand the working and the purpose of each component in the below schematic.

I also happen to note that this was a constant current circuit where R5 - 3ohms happens to be the load resistance, whose value can be from 3-50ohms.


simulate this circuit – Schematic created using CircuitLab

To understand about the constant current Op-amp circuit working, I read how it works. But in those circuits, the Op-amp would have a negative feedback configuration and the output of the Op-amp would drive the base of an NPN transistor which would drive a constant current connected to the emitter of the NPN transistor. The Op-amp will try to maintain the voltage between the inverting and non-inverting terminal and try to regulate the load current through the load resistor.

But in my above circuit shown above, can someone help me how, the constant current through the load resistance is achieved? I am not sure why there is a shunt regulator placed in the negative feedback path?

If possible, please also help me understand why there is a capacitor of 100nF between the inverting and non-inverting terminals of the Op-amp? I have not seen this type of a design. Would like to know the role that each component plays so that I can understand the design better.

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    \$\begingroup\$ I can see how it works but it's a real beast to explain and will lead to a hundred questions from you! Good luck to anyone attempting it and remind me to award +1 if done successfully and succinctly. \$\endgroup\$
    – Andy aka
    Oct 18, 2020 at 12:45
  • \$\begingroup\$ Yes, I understand that all the answers to my asked questions in this forum resulted in multiple questions from me in the comments sections. Just wanted some easy and simple explanation for a beginner \$\endgroup\$
    – user220456
    Oct 18, 2020 at 12:51
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    \$\begingroup\$ "I want a simple explanation for this circuit that contains a lot of complexity": sorry, will not happen. Understand the larger picture, then ask specifically about the things you do not understand. voting to close this as too broad, I'm afraid. \$\endgroup\$ Oct 18, 2020 at 13:05
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    \$\begingroup\$ Where did you find this circuit, where does it come from? I know that, since you ask, you didn't design it. So you got it from somewhere. Without explaining what it does consider the possibility that I make a website with lots of circuits that look like they can work but actually they don't (nonsense circuits). \$\endgroup\$ Oct 18, 2020 at 13:09
  • \$\begingroup\$ I want to just understand the working of the circuit . That's what I am trying to understand here. I was not able to understand from the other standard Op-amp circuits available on the internet, so trying to get an answer over here. \$\endgroup\$
    – user220456
    Oct 18, 2020 at 13:10

1 Answer 1


I'll not address the AC behavior of this circuit- I suspect the AC characteristics might not have been 'designed' as such anyway. So ignore the capacitors.

You can equate the voltages at the inputs of the op-amp.

The voltage at the non-inverting input is simply Vload (we can ignore offset voltage and input bias current- note that the designer has reasonably balanced the impedances seen by the two inputs, though 20K would be better for R4).

The voltage at the inverting input is more interesting- assuming the shunt regulator is regulating-

It's Vinv = \$V_{LOAD}+I_{LOAD} \cdot 2.34\Omega - 1.24V\cdot\frac{33K\Omega}{33K\Omega+47K\Omega}\$

So \$I_{LOAD} = k(1.24V/2.34\Omega)\$ where k = 33/(33+47) = 0.4125 or about 216mA.

The shunt regulator, as you can see from the above, is the reference for the constant current. By dividing it down by R3/(R1+R3) less voltage is dropped across the sense resistor R7.

R2 is there to provide adequate bias current for the shunt regulator, a job at which I can't see it quite doing at the lower bound for the load resistance.

216mA * (5.34 ohms) = 1.15V. The reference needs about 55 to 100uA to operate properly, so that's another 55-100mV, so it's unlikely the regulation will work properly below about 3.7 ohms. Upper limit for load resistance calculation is left as an exercise.

  • \$\begingroup\$ Could you please tell me the purpose of the C1 capacitor, R7 resistor and how did you form the equation? Could you also help me what is Vload ? \$\endgroup\$
    – user220456
    Oct 18, 2020 at 16:45
  • \$\begingroup\$ Also please help me to understanding the working of the circuit \$\endgroup\$
    – user220456
    Oct 18, 2020 at 16:46
  • \$\begingroup\$ My main objective is to understand how the shunt regulator helps in regulating \$\endgroup\$
    – user220456
    Oct 18, 2020 at 16:46
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    \$\begingroup\$ Do you understand the equation? The shunt regulator determines the load current directly and R7 is the sense resistor to measure the current. C1 reduces EMI sensitivity and tends to make the circuit less stable and noisier. \$\endgroup\$ Oct 18, 2020 at 17:04
  • \$\begingroup\$ Could you please explain the equation ? I understand that the equation is arrived using KVL. Could you just point out the loop of the equation. And I am not able to understand the shunt regulator regulator. Request you to help with that. And regarding your answer to C1, I thought the C2 capacitor does that. I am not sure on how connecting a capacitor in parallel to both the terminals will help that. Request you to clarify my confusions \$\endgroup\$
    – user220456
    Oct 18, 2020 at 17:08

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