# Working of op-amp based current source

Can anyone please explain me the working of the circuit attached? I believe the op-amp U1 is connected to provide a stable DC voltage at the non inverting terminal of U3. Is my approach correct?

The circuit is able to source a stable DC current of 50uA. However the first challenge is to reduce the output current to levels of about 6nA and 6uA.I was able to get around 6uA while tuning some components randomly. However I am not aware about the effects.

What tuning should I target in order to get the same?

Second challenge is to obtain a AC current out of the circuit. The frequency can be around 30 Hz.

Can I use an op-amp based oscillator instead of VG1 to get the variable frequency current source?

If yes, can someone please guide me for the same?

The input voltage VG1 is 2V DC in this case to obtain 50uA. I know the current required is very less considering the noise that will be induced in the system. However is it even possible using op-amps or do I need to start looking for some other solution? The U1 is a current sense amplifier. It measures the voltage drop on R2 and it gives the actual current measurement to the U2. The U2 is a proportional controller with extremely high loop gain, since there is no feedback.

At the input you have a voltage divider 1:2 and this is a voltage setpoint, which is compared to R2 voltage (subtracted). The U2 shall output such voltage, so that V_R2 and V_in+ should match.

Now it's not very clear that you input 2V and you get 50uA. Let see: 2V is divided by 2, so the voltage on non-inverting input is 1V. The V_R2 would be 1V if the current is I=V/R=1/10k= 100uA.

EDIT:

My bad: The U1 measures the voltage drop on 10k resistor and has a gain of G=2, so correct, you get 50uA at V_in=2V.

To be able to get lower current you can amplify the signal from U1 by replacing R6 and R9 to a higher value.

EDIT2:

In order to get low current such as 1ua and little less, you would need to change the U1 with a JFET instrumetation amplifier, preferably with laser trimmed resistances. With actual schamtics, you have a current path from sensing resistor 100k to inverting input of U1. This is quite a lot of current passing with regard to your goals.

• OR, more simply, change the value of R2. Aug 6, 2020 at 13:03
• AND - U3 is configured as an integrator. As such, it has significant negative feedback. Aug 6, 2020 at 13:04