Voltage Output in subtractor photocircuit

In summary, the voltage output of the circuit is the sum of the voltages measured by the two PIN diode detectors.
  • #1
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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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  • #2
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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Related to Voltage Output in subtractor photocircuit

1. What is a subtractor photocircuit?

A subtractor photocircuit is an electronic circuit that uses light-sensitive components, such as photocells, to generate an output voltage proportional to the difference between two input signals.

2. How does voltage output vary in a subtractor photocircuit?

The voltage output in a subtractor photocircuit varies based on the intensity of light hitting the photocells. As the difference between the two input signals changes, the amount of light hitting the photocells also changes, resulting in a corresponding change in the output voltage.

3. What is the purpose of using a subtractor photocircuit?

The purpose of using a subtractor photocircuit is to accurately measure the difference between two input signals. This can be useful in applications such as light intensity control, color detection, and distance measurement.

4. What factors can affect the voltage output in a subtractor photocircuit?

The voltage output in a subtractor photocircuit can be affected by a variety of factors, including the sensitivity of the photocells, the intensity of the light source, and the distance between the photocells and the light source. Changes in these factors can result in variations in the output voltage.

5. How can I calibrate the voltage output in a subtractor photocircuit?

To calibrate the voltage output in a subtractor photocircuit, you can use a known input signal and adjust the sensitivity of the photocells or the distance between the photocells and the light source until the output voltage matches the expected value. It is important to perform regular calibrations to ensure accurate measurements.

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