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I was wondering how to effectively read a photodiode datasheet. Leds have max current ratings and voltage drops clearly listed but the photodiode one confuses me.

Im trying to calculate a resistance value to add in series with it but am unsure.

I see light current, dark current, and some have forward voltage and some have just reverse on the datasheets, so I am really confused.

I just want my photodiode to be most sensitive to changes in light.

Using this PD

http://www.marktechopto.com/pdf/products/datasheet/MTD5052N.pdf

I wanted to drive it with 5V supply.

UPDATE:

I just tested it with the photodiode in series with a resistor to give its light current rating. I measure the voltage across the photodiode and it changes with more light, but I thought the voltage is supposed to stay constant im confused. Is there a problem with my current setup?

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  • \$\begingroup\$ Trivia. You can actually use an ordinary LED, photodiode is just more sensitive. And for best sensitivity, look for a phototransistor. \$\endgroup\$ – Barleyman Apr 20 '16 at 11:12
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Photodiodes need to be treated in a special way !

1) use the diode in zero-bias or reverse mode (not in forward mode)

2) when dark, no current flows through the photodiode

3) a circuit is needed to "catch" the photocurrent and amplify it.

Here's an example of such a circuit:

enter image description here

This circuit is called a transimpedance amplifier. Through feedback the opamp keeps the voltage across the diode zero so we're in zero-bias mode. Resistor RF converts the very small photocurrent to a voltage which you can measure at the opamp's output.

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  • \$\begingroup\$ It bears to be mentioned that the currents can be tiny, in nanoampere class. So your opamp needs to be cmos/jfet type with pico/femtoamp bias current. Also the feedback resistor is very large, 1 megaohm is a good starting point and this can scale up to 100 megaohms or more. For best sensitivity you would of course add more amplifier stages as there's just so much amplification you can ask from an opamp and things start to get funky when you've got 100M component in your feedback. \$\endgroup\$ – Barleyman Apr 20 '16 at 11:11
  • \$\begingroup\$ thanks! so if I used the photodiode in the description. How do I determine values of Rf? based off its specs. Also Could I drive it with a 5V supply if it says reverse voltage 5v? \$\endgroup\$ – lightro Apr 20 '16 at 23:53
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What FakeMoustache describes is known as unbiased or photovoltaic mode. In this mode, the photodiode generates its own current, similar to a solar cell.

Photodiodes can also be used in biased or photoconductive mode illustrated by the circuit below:

schematic

simulate this circuit – Schematic created using CircuitLab

In this schematic, the value of R1 in series with the photodiode should be chosen so that the voltage drop on dark current alone R1*Id is below the threshold voltage of the transistor (0.6 V or so). That way the transistor stays off in the dark and starts to conducts when the photodiode is exposed to light.

There are also op-amp based photoconductive circuits if you're interested.

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    \$\begingroup\$ The difference being that photovoltaic is more sensitive while photoconductive is faster. Latter suffers from dark current and in general this circuit would require a great deal of light to work properly as the transistor won't work properly with the tiny currents from the LED if there's any kind of "load". A darlington pair might work and opamp definitely would. \$\endgroup\$ – Barleyman Apr 20 '16 at 11:07
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    \$\begingroup\$ It's not unusual for photodiodes to provide tens of uA of light current. Assuming x100 amplification, the load would get several milliamperes, enough to light a LED for example. \$\endgroup\$ – Dmitry Grigoryev Apr 20 '16 at 11:22
  • \$\begingroup\$ See comment about "great deal of light". Of course you get nice current if you have a high power LED right next to your sensor. I've been working on and off on an optical probe that senses a white pixel line on a black background image on a LCD display. And to give some accuracy the aperture is 1mm dia. That doesn't give you a whole a lot of current in the (phototransistor) sensor. \$\endgroup\$ – Barleyman Apr 20 '16 at 12:12
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Photo diodes (PD) (approximately) make one electron for each photon absorbed. There are various circuits that can make the response faster, but the only way to get more sensitivity is to catch more photons... a bigger area or a lens.
For DC light level my favorite (easy) circuit is to use a DMM with a uA current input, just measure the current the PD produces.
The TIA above by fakemoustache is nice. I've used this too, V1 ~5-15V, R1 depends on the light level and area of PD. 10k to 100M ohm

schematic

simulate this circuit – Schematic created using CircuitLab

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