1
\$\begingroup\$

I saw this circuit the other day on here: What is the basic circuit to wire up a position-sensitive detector, or PSD?

enter image description here

From what I understand from the answer given, the voltages X1 and X2 can be used to figure out a position of a light beam. My question though is: what is the range of the output voltages, meaning the polarity with respect to ground?

I think the photodiode current flows from cathode to anode. If so, the current is going into virtual ground (non-inverting terminal), but I think the rule is that no current is going into the op-amp inputs (except very little to bias it), so most of it goes into the output the op-amp, meaning the op-amp output is sinking current. I think that the output voltage will be from 0 V (no light) to -V (max light). This means the op-amp should have a split supply.

Is my analysis correct or am I missing something?

\$\endgroup\$

2 Answers 2

2
\$\begingroup\$

My question though is what is the range of the output voltages? Meaning the polarity with respect to ground?

The output polarity will be a negative voltage.

The range cannot be determined without knowing the light source and the detectors and how much each receive in terms of light. Guesswork really.

I think the photodiode current flows from cathode to anode.

Yes, photo-current flows from cathode to anode.

I think that the output voltage will be from 0V (no light) to -V (max light).

Correct.

This means the opamp should have a split supply.

Or you make a mid-rail point and call it 0 volts.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Thanks for info! \$\endgroup\$
    – joe
    Nov 15, 2022 at 14:46
0
\$\begingroup\$

For position-sensitive-photodetector, photocurrents are often compared - one subtracted from the other. You can do this with one opamp (powered by split-supply). Output voltage in the single opamp circuit (below) can still compare light balance from a very bright light source, whereas the the OP's two op-amp circuit saturate their output voltage at -Vsupply. You might consider this an advantage or a disadvantage, depending on your application.
In the circuit below, since both diodes are zero-biased (or nearly so, within a few millivolts), their photo-currents sum at the op-amp inverting input. If both photodiodes see equal light, opamp output voltage will be zero volts (Vref). Output voltage can swing both above and below Vref, depending on which photodiode sees more light.

schematic

simulate this circuit – Schematic created using CircuitLab
Disadvantage? For example, if output is used in a feedback/control system, gain (defined as output voltage divided by photocurrent) might be required. This simple single-opamp method gives no indication of this photo-gain whereas the two-opamp method does provide photo-gain.

\$\endgroup\$
1
  • \$\begingroup\$ Interesting idea. Though it seems like in practice the PSDs are common cathode or common anode. But if I find one that is not like this, will keep this in mind! \$\endgroup\$
    – joe
    Nov 15, 2022 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.