My goal is to amplify AC signal (~ nA) generated by a device resembling FET, where light changes the gate voltage, consequently changing the drain-source current. Here I drew it as a photodiode.

Drain-source bias voltage needs to be applied, which generates relatively large DC component (10 -100 uA) to the output.

The pre-amplifier (the box in the figure, don't ask what's inside) has very low noise characteristics and input impedance of 50 ohm. I am tempted to use it due to its good performance. However, the problem is this: the DC current exceeds the maximum input current (1 uA) for the amplifier. Thus, the circuit cannot be made as a simple transimpedance amplifier, where the detector would be directly connected to the amplifier input (virtual ground).

Blocking the DC component with capacitor would work, but in my circuit, the bias voltage of the detector is now divided between the resistor and the detector. Most likely, almost all the voltage would end up being over the resistor.

Is there any method of removing / reducing the DC component instead of using a different amplifier? Please note that the amplifier should be as near to the detector as possible to keep the noise level down.

 

Do you need DC response or can you periodically zero the output?

 

Hi Dick,
I'm afraid I don't understand what 'periodically zeroing the output' means.
The circuit only needs to respond to AC.

 

My question about perodically zeroing the output was about, for example, setting the output to zero every ten seconds.

I think your resistor-capacitor idea will work as long as the input impedance of the transimpedance amplifier is significantly lower than the resistor, that way you won't loose much signal.

Another approach that probably adds too much noise is to low pass the output of the transimpedance amplifier then apply it to an amplifier dedicated to zeroing the offset. The in offset correcting amplifier the inverting input is grounded and the low passed output of the transimpedance amplifier is applied to the non-inverting input, which drives the summing nose through a large resistor or a voltage controlled current source.

The low pass filter has to have a very low corner frequency such that the combined open loop gain of the zeroing amplifier and the low pass filter is such that you only get 135 degrees of phase shift in the feedback (45° phase margin) when the loop gain crosses 0 db so as to prevent oscillation of the zeroing loop. This is also necessary so that the zeroing loop does not significantly attenuate the AC signal from the photodetector. As I said, I suspect that the zeroing amplifier might contribute too much noise since your incoming signal is so small. I used the approach once but the incoming signal was on the order of a microamp.

Go with the capacitor if you can.

 

I see, unfortunately the output cannot be zeroed.

If the detector in question can be modeled as a resistor, I was worried how the voltage would be divided over it and the resistor R1. The larger the R1 is, the less bias is over the detector. However, if R1 is given a very small value, the signal is lost to ground, as you said. Since you felt the circuit could work, I'm encouraged to try it out. Thank you for your thoughts and solutions.

 

You might be able to adapt the technique shown in the circuit at the bottom of page 1 of this op amp data sheet. There, a 2N3904 transistor is used as a controlled current sink to counteract the photodiode current caused by ambient light; the transistor is driven by an integrator (the LT1097) which monitors the output of the amplifier (LT1793) and acts to keep it centered on zero volts. The time constant of the integrator (determined by R2 and C2) dictates how fast this negative feedback loop operates, which in turn determines the lower frequency limit of your AC signal. The upper frequency limit is determined by the amplifier's frequency response, of course.

Just a thought...

 

(some text removed) However, if R1 is given a very small value, the signal is lost to ground, as you said.

To be clear: The input resistance of your transimpedance amp is the value of the feedback resistor divided by the amplifier's open loop gain +1 (from memoery), so with sufficient open loop gain the great bulk of the signal current will be seen by the amplifier.

Let us know how it works!