# Trouble with obtaining reliable values of the current from the photodiode.

Discussion in 'General Electronics Discussion' started by DX400, Oct 26, 2017.

1. ### DX400

6
0
Oct 26, 2017
I am trying to build transconductance amplifier circuit to obtain values of the reverse biased photodiode (PD) current in order to estimate some of the LED characteristics.

At first, I have used the scheme named “Recommended Zero Biased Circuit” from page http://home.sandiego.edu/~ekim/photodiode/pdtech.htm (placed at the bottom of the page, the fourth one figure from the bottom ). The photodiode is Vishay BPW21R, OpAmp is LM358n, the value of the Rf control resistor of the negative feedback is 2 kOhm. The output voltage is measured by the multimeter. To obtain some signal, I am using flash light with 3W cool white LED like this https://www.sparkfun.com/products/13105. The flashlight is placed near in front of the photodiode to obtain the same luminous flux at each try.

When flashlight is off (no light) the output voltage (Uo) is about 0.

I have found that, the variation of the Rf (the gain of the amplifier) from 300 Ohm to 500 kOhm leads to not quite proportional change in output voltage (Uo). So, the values of the PD current are slightly different for the different Rf. What is the cause of it?

For example for the Rf of 2 kOhm I have obtain Vout of about 3.6 V and the PD current of 1.8 mA.

Then, I have used the scheme named “Positive bias circuit” from page http://home.sandiego.edu/~ekim/photodiode/pdtech.htm (placed at the bottom of the page, the second one figure from the bottom ). I have used 10 V which supply the OpAmp and to make photodiode reverse biased.

The value of Vout for the Rf of 2 kOhm decreases to 0.7 V for the same illumination.

So the problems:

1. What is the reason of output voltage drop.

2. I have something like overflow. When flashlight is off (no light) the Uo is about 0. Under flash The variation of the Rf, which defines the amplification gain of the circuit, from 1 to 100 kOhm does not change the value of the output voltage Uout of about 0.7.

3. The increase of the Rf up to 0.5-1 MOhm leads to much decrease of the Uo.

4. The variation of the Rf also leads to not proportional change of the output voltage. So the Rf does’t work as gain.

5. I don’t understand how to be certain about obtained results? How to calibrate the results of the PD current? How to make a calibrated light source for such measurements to be sure of correct PD current?

I should be very obliged for help to solve this problem.

2. ### Alec_t

2,970
805
Jul 7, 2015
Welcome to EP!
The links to the sandiego site are broken.

6
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Oct 26, 2017
4. ### DX400

6
0
Oct 26, 2017
I uderstand that there are many posts about the photodiodes and amplifiers on this and another forums, but i have not found there any answers to my questions.

So, I should be very obliged for help.

Sorry.

Last edited: Oct 26, 2017
5. ### Petkan

19
2
Feb 9, 2011

Petkan:
All opto-sensitive devices are also...heat sensitive and also non-linear.
The easiest cure is to use a dual photodiode optocoupler olike IL300.
One is used as signal isolation, the other as feedback. The two are very well matched and placed close on the same die i.e. have the same temperature. An op amp can easily linearize such a circuit by keeping the opto in very narrow range on its curve. For more details - contact me at <removed>

Last edited by a moderator: Jan 3, 2018