Voltage Output in subtractor photocircuit

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The discussion revolves around determining the output voltage of a subtractor circuit used in Faraday magnetometry. The circuit involves a laser, SF-59 crystal, and two Thor Labs FDS-100 photodetectors wired to produce opposing voltages. The user is confused about how to handle both AC and DC components in the output voltage, particularly regarding their interaction and cancellation. It is suggested that simply reversing the terminals of the detectors won't effectively subtract the photo-currents; instead, they should be wired in series with a current-to-voltage converting amplifier for accurate results. Understanding the relationship between AC and DC components is crucial for achieving the desired output voltage in this setup.
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Hello All, I work in an optical lab and am currently attempting to determine the voltage output of a subtractor circuit I've wired up (as I'd like to work back from my output voltage and FFT magnitudes to read out the value of a magnetic field using faraday magnetometry.)

So my circuit is as follows. I have a laser which send light through a SF-59 crystal with a transverse magnetic field, which rotates the plane of polarization of the light, attenuating it's intensity. The resultant light enters into a beamsplitter cube which I've oriented in such a way so as to send equal amounts of light into a Thor Labs FDS-100 photodetector which is in the back of the cube mount and also to another FDS-100 PD which is on the side of the cube mount.

The photo detectors are wired up in such a way where one PD will have the black wire on it's cathode and the red wire on it's anode while the other is wired opposite and both PD wires come down such that both red wires (1 anode and 1 cathode) are on one side of the circuit while both black wires are on the other side of the circuit. This should make one voltage negative and then add the two of them together, i.e., subtracting one voltage from the next.

I'm just having issues determining what the output voltage should be. When I run a DC current (from say a 6V battery, which feeds the laser) into the circuit then I read out a voltage of 0V on the multimeter, which makes sense as 1/2 of the light goes into one detector and the detectors output opposite signed voltages. The idea is that when we modulate that laser voltage through the magnetic field and introduce an AC component (at frequency 60 hz) we perform the same subtraction in order to 'subtract away' the common noise amongst both photodetectors.

What I don't understand exactly Is how would I treat those input voltages with both an AC and DC component and how they transform. I know with just a DC voltage we get simple cancellation but with the addition of an AC component can we treat it like DC + AC where the DC components subtract then so do the AC components? In that case I'd need to know why equal amts of AC signal aren't being sent to each detector and cancelling each other out. Or would the AC and DC components be inextricably linked in this calculation of the output voltage of this subtractor photocircuit? Thanks in advance everyone.

Relevant equations

Basic Electronics/Circuits equations

V_out = V_1 - V_2?
 
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PIN diode detectors are operated with reverse bias. You cannot simply reverse the terminals to subtract the photo-currents. A PIN diode is normally used with a current to voltage converting amplifier.

To subtract currents you should wire the two detector diodes in series, reverse biassed, say between -5V and +5V. The difference current at the common junction is then converted to a voltage by an op-amp referenced to 0V virtual earth.

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