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I want to build a photodiode light sensing circuit so that the output voltage is inversely proportional to the light intensity going to the photodiode. I.e less light equals more voltage.

I'm thinking of a circuit like this:

schematic

simulate this circuit – Schematic created using CircuitLab

As the light intensifies the current in the circuit get higher and there is more voltage drop in the resistor. That's what I'm thinking.

I have a few questions: Would this work? What should the resistor value be? Do I need an amplifier?

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  • \$\begingroup\$ At first I thought you had the diode backwards, then I noticed the upside-down power supply. That's annoying. Draw the schematic right. Also, just showing "output" is meaningless without a reference node. Either show the reference explicitly or show one node to be ground. \$\endgroup\$ Apr 6, 2018 at 19:43
  • \$\begingroup\$ Is it now correct? \$\endgroup\$
    – Kestis
    Apr 6, 2018 at 19:46
  • \$\begingroup\$ Yes, that's better. Downvote removed. \$\endgroup\$ Apr 6, 2018 at 19:48
  • \$\begingroup\$ This circuit gives \$V_{out}=5-R_1 I_p\$, where \$I_p\$ is the photocurrent. But that isn't the same as being inversely proportional to the photocurrent. \$\endgroup\$
    – The Photon
    Apr 6, 2018 at 19:59
  • \$\begingroup\$ Then the resistor should be something like 100k. \$\endgroup\$
    – Kestis
    Apr 6, 2018 at 20:12

2 Answers 2

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Would this work?

As Dorian has answered and The Photon has commented, this will give you a voltage which decreases with increasing light level.

As The Photon commented, this is technically not the same thing as "inversely proportional". Inversely proportional means that, if the light level doubles the signal out is halved.

What should the resistor value be?

Dude - There is no way to know. Unless you specify the optical power which falls on the active area of the photodiode (and the area of the active area)(and the sensitivity of the PD at the wavelength you are using) you cannot know what the PD current will be, and therefor cannot know what the voltage across the resistor will be.

Do I need an amplifier?

Again, and for the reasons listed above, there is simply no way to tell. Additionally, you have not specified what signal level you need from the circuit. For instance, if you need a 10 volt swing (light to dark), there is no way on God's green earth you will get that from a 5 volt supply. Do you need 10 volts? Then you absolutely need an amp, regardless of anything else. If you need one volt, then it all depends on the other factors.

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Unless you use a digital input I don't see any reason to have the output inverted. Your DAC input is not inverted.

The voltage on the diode will be 5V - kRP where P is the light power and k a coefficient including photodiode sensitivity. This is highly dependent on power supply, any variation of power supply voltage is passed entirely at the output.

In the datasheet the current is given up to 100uA so, considering at least 1V for diode voltage R should be around (5v-1v)/100uA = 40kohm.

It works but with high errors, I would recommend using positive output by switching the resistor and diode and reading the resistor voltage, still dependent on temperature but with less errors due the power supply voltage.

In this case output voltage will be Vo = kRP.

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