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Photodiode Amplifier

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lurkingdevil

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I have a photodiode that looks like this :
**broken link removed**

I have set up a circuit like this :
**broken link removed**

that I picked up from here : https://www.robotroom.com/ReversedLED.html

I used a 220k resistor in pace of the 30M. But I can't get anything on the op-amp output.
My meter reads 3-6mV. Even if I remove the photodiode the reading doesn't change.
Honestly I don't know what resistor I should use, I just used the maximum value I had at hand.

I'm using LM358 opamp .

Help please..
 
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I've built plenty of these things. The output is -I*Rf. Because of the sign reversal, you have to use + and - supplies. The fact that removal of the diode really suggests to me that your using a single supply. What is the power supply your using for the OP Amp?
 
Obviously there's a gain reduction of a factor of 320M/220k = 1454 between the two amp configurations. If you are using a dual supply for the op amp, then the incorrect resistor is likely your problem.

If you don't know what resistor to use why not use the value given on the schematic?

You op amp also has too high an input bias current for a 320M feedback resistor. It will cause the output to saturate. You need an amp with a very low input bias current such as the TI JFET TL08X series.
 
True. 320 M is a very high feedback resistor and it's also where the choice of capacitor matters too BUT the output would saturate at the negative rail. Thus, I'm assuming the negative rail is at 0 Volts. Opening the input should saturate to the negative rail. If you used a high-power light source such as a flashlight, it should cause some change in output even with a 320K FB resistor.
 
Yes I am using single power supply. The two rails are GND and 5V.
The diode is put in reverse so that it generates a negative voltage(or negative current?) with respect to GND at the V- terminal of the op amp so I should get a positive output(inverted?), or thats what I think it should be. If you open the link I posted, that is what the author has done, except he used an LED, I have a IR photodiode and it worked for him.
Here: https://www.robotroom.com/ReversedLED2.html

EDIT:
My understanding of this circuit is that the photodiode produces a current that is then pulled through the feedback resistor and pushed into GND. V- is held at GND because V+ is at GND. The op amp output equals the potential drop across the resistor = -(-I*Rf)
Click here for circuit.

Now I understand what you mean by saturation. If the potential drop is more than the rail voltage, the op amp cannot produce that volage so it saturates to negative/positive rail.

This is how I think it works but I could be wrong.
 
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The original circuit uses a Cmos opamp that has NO bias current. You are using a lousy old LM358 dual opamp that has a fairly high bias current.
Your feedback resistor is more than 1000 times too small so the voltage gain is more than 1000 times too small. But an LM358 opamp cannot use a very high value for its feedback.
Therefore your circuit does not work.
 
I have used CA3140 op-amps for such tasks, I never tried a LM358 (input bias of 40 to 200 nA instead of the CA3140's 10 to 40 pA) due to the reasons the Guru stated :)
 
True. 320 M is a very high feedback resistor and it's also where the choice of capacitor matters too BUT the output would saturate at the negative rail. Thus, I'm assuming the negative rail is at 0 Volts. Opening the input should saturate to the negative rail. If you used a high-power light source such as a flashlight, it should cause some change in output even with a 320K FB resistor.
With an open input and a feedback resistor the op amp will go to 0V, not saturation.
 
Ok so the op-amp is at fault here. I didn't know the bias current mattered so much.
Thanks a lot for your help.

I put together the simple circuit in the attatchment. When I put a IR led in place of the diode, I get 4.4 to 4.6 V at out.
If I put a photodiode there, I get an output all the way from 0.1V to 4V! I didn't quite expect such a drastic change. This is usable voltage.

Could I just use the LM358 as a follower and use the output from there? The photodiode will be used to constantly detect small variation of distance at about 10cm. I plan on feeding the output from op-amp straight into a microcontroller. Anything I should be weary of?
 

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Why is there an open input?
KISS was talking about the op amp output going to the negative rail when the input is open (or no signal), but that's not what happens.
 
Your purely resistive circuit puts the diode in photovoltaic mode and not a photocurrent mode. Some detectors have a better response using this mode.
 
Your purely resistive circuit puts the diode in photovoltaic mode and not a photocurrent mode. Some detectors have a better response using this mode.
Au contraire. That configuration, with the diode connected directly to the op amp summing junction and a resistive feedback, is a standard current mode input transimpedance amp. This circuit keeps the voltage across the detector at 0V so it is operating in the photocurrent mode.
 
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Why is there an open input?
When the photodiode is in the dark then it does not conduct, it is open. The (+) input of the opamp is grounded and it will try to make the (-) input also at ground through the feedback resistor by driving its output to ground.
 
With a 100 ohm resistor to +5V and a reversed diode to ground at the (-) input of an opamp and the (+) input grounded then the output of the opamp should be almost 0V, not +4.6V.

A reversed diode with no voltage across it generates a small voltage and current.
A reversed diode biased by a high value resistor leaks a small current. But since your resistor is only 100 ohms then the diode does nothing.
 
Sorry.
Many of us come here to talk GEEK with our peers about a complicated circuit, not to teach the basics of photo-diodes.
Also, your resistor values are very wrong and your opamp is completely wrong.
 
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