photodiode circuit

dvd280

Joined Nov 5, 2020
4
I am trying to set up a circuit which should generate a signal according to the voltage generated by two photodiodes by comparing them. I know how to do this in reverse bias, and how to convert a photodiode current into voltage by using an op amp.

When I measure the voltage across a photodiode while hitting it with light (without connecting anything to it) - it does have up to 400-500mV between its leads without an op amp, so I am wondering if it is possible to just plug two photodiodes in forward bias into a comparator, and have it generate a HIGH/LOW signal according to the comparison.

I'm am not sure if this is feasible, and how exactly to set it up without running significant currents through the photodiodes, I am thinking something similar to the attached image. Any ideas about how to implement something like this with minimal circuitry would be appreciated

Attachments

• 13.9 KB Views: 12

Bordodynov

Joined May 20, 2015
3,167
See

SamR

Joined Mar 19, 2019
5,019
without running significant currents through the photodiodes
? That is the neat thing about Op Amps. You don't need a lot of current...

dvd280

Joined Nov 5, 2020
4
Suppose I have two photodiodes, and I wish to compare the light hitting each of them. I found that when I wire them like this:

I actually get a current running across the Amp-meter shown in the circuit, the polarity/direction of the current seems to switch according to which of them has more light on it, and the reading itself seems to have a magnitude which is proportional to the difference in the amount of light that hits each photodiode.

Also, it seems that connecting a voltmeter instead of an Amp meter ( exactly the same way) also yields a voltage reading which polarizes according to which photodiode has more light hitting it.

Is there any way I could take advantage of this in order to generate a voltage signal according to which photodiode has more light hitting it ( i.e according to the direction of current or polarity of the voltage), all without using complicated circuitry ( i.e as few components as possible)?

neonstrobe

Joined May 15, 2009
190
Generally photodiodes turn light into current. In PV mode, small voltage differences can arise between diodes due to process variations and as the forward voltage tends to be limited, (to around 400mV) I suspect you would be better off to use the diodes in reverse bias mode and drop the voltage across two resistors. You can then measure the voltage between them. That should be reasonably linear with light level.