Hi I am building a circuit to measure current draw from a load, I opted for a voltage follower with a transistor and a shunt in between, The transistor is needed to allow higher currents to flow through the load, here is my schematic:

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The Op-Amp is configured with Negative Feedback, So as long as the Input Voltage plus the shunt voltage plus the transistor Vce is lower than the Collector Voltage, the transistor would be in saturation mode (I'm actually not sure).

If the BJT is in saturation mode, the voltage across the load would be exactly the same as the Input Voltage, also the current across the shunt is the same as the current across the load, allowing to sense the current with a differential amplifier across the shunt.

What I am not really sure is if I need to place a resistor between the Op-Amp Output and the base of the BJT, I think that the Op-Amp would compensate by itself, but idk.

So the actual question is if a resistor needed between the base of the BJT and the Op-Amp output.


This is the second stage for a DC Variable Voltage source.

This stage has three purposes:

-The first, to keep the voltage across the load equal to the input voltage.

-The second is to be able to measure the load current with a differential amplifier across the load

-The third to increase me maximum current that can be supplied to the load. Since the first stage is 1.5 Amps max, I would like to make this at least 3 Amp max

  • \$\begingroup\$ If your aim is only to be able to sense current through the 1k load resistor using the 100m shunt, it's not clear what the purpose of the opamp and transistor are. Are they already part of an existing circuit? \$\endgroup\$ – brhans Jan 22 '20 at 22:44
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    \$\begingroup\$ @brhans I think the circuit shown is intended to be a variable voltage supply where an omitted differential amp circuit is to measure the current through the load via the Shunt voltage. \$\endgroup\$ – DKNguyen Jan 22 '20 at 22:49
  • \$\begingroup\$ You don't need to, but the resistor does make it so the opamp doesn't have to be so squirrely to maintain equilibrium (takes only a small change in output voltage for a large change output current). Things will get more squirrely as your load resistance is reduced (or if it is something other than a resistance) since it helps out in reducing this sensitivity. \$\endgroup\$ – DKNguyen Jan 22 '20 at 22:50
  • \$\begingroup\$ This is the second stage for a DC Variable Voltage source, that I would like to make 5A max, so something like a TIP41 can be used. This stage has two purposes, the first, to keep the voltage across the load equal to the input voltage, and the second is to be able to measure the load current with a differential amplifier across the load. \$\endgroup\$ – Cheche Romo Jan 22 '20 at 22:58
  • \$\begingroup\$ I responded to the body of the question, which does not match the title. When you say "shunt" in the title, people are going to assume it is the current sensing shunt resistor and not a base resistor. You can edit to clarify. \$\endgroup\$ – DKNguyen Jan 22 '20 at 22:58

What it will do is basically waste power, since the opamp will do its best to modify it's output to make the negative terminal to the postive one. So you'll still get 10V on the load, there will still be a voltage drop on the 100mΩ resistor and a slight amount of power wasted.

So you don't need it. One thing you might need is a resistor between the opamp and the transistor because you might exceed the output current on the opamp. Or if the opamp output can source a lot of current, you might exceed Ib on the transistor, so make sure neither of those situations doesn't occur. Another problem with these circuits is they can become unstable at high frequencies, especially if there is a cable adding inductance on the load, so watch out for that.

A useful thing to do with this circuit and a variable load is to use the shunt resistor for current measurement.

  • \$\begingroup\$ The OP actually is asking if a resistor is needed between the op-amp output and BJT base. In the body of the question anyways, which doesn't match the title. \$\endgroup\$ – DKNguyen Jan 22 '20 at 22:54
  • \$\begingroup\$ Thanks for your answer, I know there will be some power loss, but i think is a fair tradeoff for being able to sense the current through the load at a precise voltage, also with a 0.1 ohm resistor the power loss wouldn't be much, at 3A dc , 0.9 watts would be wasted. How can I approach doing the calculations for the base resistor?, I think this resistor would limit the maximum current through the BJT, right? \$\endgroup\$ – Cheche Romo Jan 22 '20 at 23:07
  • \$\begingroup\$ I usually use a spice package and a 100Ω resistor to start, its really only when the transistor is saturated when things become a problem, because the opamp want's to add more and more current. Just make sure the opamp output is smaller than Ib \$\endgroup\$ – Voltage Spike Jan 22 '20 at 23:43

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