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I am trying to build a transimpedance amplifier (shown below) using the LT1077 Op Amp IC to amplifier a signal from the BPW34B photodiode by OSRAM. My feedback resistor is a value of 2.2k. From what I know from Transimpedance amps, the op-amp output voltage is the product of the constant current source and feedback resistor b.c. the current flowing into the op-amp terminal near zero. Also, the voltage at the two op amp terminals should be the same. However, I am getting 2.4 V at the negative terminal instead of the expected virtual ground. I thought maybe this has to do with the Op amp characteristics and the inability to pull current through the output but I have no way to prove this. What could be causing such a big voltage difference in the two terminals of the op amp

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    \$\begingroup\$ Calculate the outpu voltage : note that it is negative. Re-run this experiment with +/- supplies to the opamp instead of +/0. Or, reverse the photodiode. \$\endgroup\$
    – user16324
    Commented Dec 19, 2020 at 0:35

1 Answer 1

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Using the inverting input on a single (+ve) supply demands that your input goes negative without feedback so that with negative feedback it is a virtual ground to match Vin+=0V.

Thus reverse the diode.

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  • \$\begingroup\$ Ok so since the op-amp most negative rail is 0 volts and the diode is providing a voltage less than that then the negative terminal is forced to the diode voltage? \$\endgroup\$
    – Aaron
    Commented Dec 19, 2020 at 22:01
  • \$\begingroup\$ Not quite. This OPAmp works slightly below 0V and the voltage is reduced by the DC gain feedback to near null voltage. So it works fine in reverse. Not all OA’s work down to Vee or Vss. \$\endgroup\$
    – D.A.S.
    Commented Dec 20, 2020 at 20:20

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