1
\$\begingroup\$

I made a Trans Impedance Amplifier circuit using a OPA322 op amp with a photodiode (Hamamatsu-S2386). I use a MCP3421 to convert the analogy signal to I2C digital signal after, and use a MCU to read this signal.

I put negative pin of the photodiode to the ground (like the image below). I think it would give me a negative V_out value. but, what I really got is positive voltage. Then, I switch the direction of the photodiode (positive pin to ground). I got negative voltage output. It confuses me.

The circuit works fine with negative pin of photodiode to ground. It increased the voltage output while I increase the light intensity. I checked the pins of photodiode with multi-meter. The pins are just as the datasheet showed. Can anyone take a look where do I miss? Op Amp Circuit

enter image description here

\$\endgroup\$
7
  • \$\begingroup\$ You must be mistaken because if the lowest supply voltage on the op-amp is 0 volts, you can't get a negative output voltage. \$\endgroup\$
    – Andy aka
    Aug 6, 2023 at 13:21
  • \$\begingroup\$ It is I2C ... @winny ... \$\endgroup\$
    – Antonio51
    Aug 6, 2023 at 13:28
  • \$\begingroup\$ @Andyaka Yea, that is also confused me. I am thinking maybe MCP3421 connected op amp. and MCP3421 support the minimum voltage (VSS-0.3v) , which makes it to detect some tiny negative voltage. \$\endgroup\$ Aug 6, 2023 at 13:33
  • \$\begingroup\$ Read this ... ti.com/lit/ug/tidu535/tidu535.pdf?ts=1691265231334 figure 4 \$\endgroup\$
    – Antonio51
    Aug 6, 2023 at 13:35
  • 1
    \$\begingroup\$ @Antonio51(VSS-0.3v) is the max voltage you can give without damaging the device. It won't actually read that voltage, but the part should survive. Since your circuit cannot produce negative voltage, I guess you are measuring wrong. How did you check the voltage? \$\endgroup\$ Aug 6, 2023 at 15:18

2 Answers 2

1
\$\begingroup\$

You need to do a couple of things here. You want to apply a positive voltage to the inputs of a single supply op-amp in this kind of configuration. This reverse biases the photodiode, which you’ll want to do do improve its response. Reverse bias increases the depletion region of the diode, which is where photons are absorbed. This positive voltage also gets the op-amp off the negative rail. This will improve the op-amp response and allows the output to be closer to actual 0.0V with no photons hitting the diode.

You want the diode reversed of how you drew it. It should be reverse biased with 1V on the inputs.

Based on the diode curves, I’d put 1V on pin 3 and see how it looks.

Check out Fig 31 in the OPA datasheet.

\$\endgroup\$
2
  • \$\begingroup\$ I’d be interested to know how you got 1V of reverse bias. I scanned the datasheet and nothing jumped out at me - though I don’t work with photodiodes. \$\endgroup\$
    – Bryan
    Aug 6, 2023 at 22:00
  • 1
    \$\begingroup\$ I apologize for the late response here. I'm just kind of guessing at a starting point. We're characterizing some photodiodes in the lab right now actually. We vary the reverse bias as one parameter and then characterize the response. The optimum reverse bias depends on the particular application. \$\endgroup\$
    – 65Roadster
    Sep 8, 2023 at 19:27
0
\$\begingroup\$

Pin 1 of the diode should connect to ground.

It will try to produce a negative voltage at the input of the opamp. The feedback around the opamp provided by R4 will counteract this and keep the input at ground level within a few millivolts.

The current being forced through the feedback resistor will cause the output voltage to be proportional the photocurrent.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.