Current Sensing with Op-Amp, do I need a shunt resistor in series with the load on the output?

Question

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:

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.

EDIT:

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

Answer 1

In my experience, a base resitor would provide some protection for the opamp. In the case of a short circuit of the output to ground (I mean the overall output, a load of 0Ω), the base of the transistor is pinned to 0.7V. With the transistor base near zero volts, the opamp output is effectively short circuited to ground too. It can dissipate a lot of power.

My approach to calculating some sort of current limit would be to figure out what maximum base current should flow under full load (\$I_B = \frac{I_C}{\beta}\$), and also how much voltage across the base resistor you can afford to "lose". Choose a resistor which easily meets both those criteria.

I suspect that 5A is too much for a single pass transistor. At 5A, with β of 50, that's 100mA of base current that your opamp needs to supply. And that amount of base current will certainly give your opamp a hard time, especially when the output is set very low. Maybe you need a darlington pair, and even parallel pass transistors.

Consider something like this:

^{simulate this circuit – Schematic created using CircuitLab}

This drops the current your opamp needs to supply by a factor of 100 or so, so it doesn't need to be a beast any more.

I've used the sense resistors to also act as ballast to overcome transistor mismatch. It's just an idea, and will need work to figure out the exact values. R5 and R6 just "average" the voltages drop across the sense resistors, so you have some signal to represent current through them.

Also, you'll need to add frequency compensation if you expect any capacitive loads, because both your original circuit and this one will oscillate to hell and back without it.

Answer 2

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.