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I am trying to measure light intensity using a photodiode (S1336-18BQ) connected to an op-amp (LF412CP) in transimpedance amplifier configuration. I couldn't get it to work correctly, so I started searching for problems and have found something very peculiar: there is a current flowing from the negative input of the op-amp. The value of the current is the same order of magnitude as the short-circuit current of the photodiode under the same conditions (~1 µA).

schematic

simulate this circuit – Schematic created using CircuitLab

I have very little experience in analog electronics, but I understand that inputs of an op-amp (especially a JFET one) should have very high impedance. I have tried three identical op-amps to ensure that it is not just a damaged chip. Please forgive me if I am asking something obvious.

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3 Answers 3

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enter image description here

The op-amp you have chosen (LF412) is unsuitable for the power supply regime you show in your schematic. The input common-mode voltage range is a couple of volts inside the power rails hence putting the non-inverting input to ground (also your negative supply rail) is a mistake - you need bipolar supplies for this op-amp.

In addition, the minimum supply recommended for this op-amp is +/- 5 volts (a span of 10 volts) and you are showing it on a single 5 volt rail: -

enter image description here

This op-amp won't work properly on a single 5 volt rail.

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It does not work with single 5V supply.

Datasheet says minimum supply voltage for LF412CP is +/- 3.5V

The circuit is also exceeding the common mode input voltage range. It does not go down to negative supply voltage.

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If you still want a single-supplied circuit, try this clever H&H circuit solution (Page 253, Fig. J):

AE Photodiode converter

I have explained it in my answer to a similar question two years ago. The role of the second resistor R2 is to "lift" the voltage drop across the "floating" photodiode above the ground thus ensuring a kind of a "self-biasing".

I have illustrated the circuit operation by means of voltage bars (in red) and current loops (in green). Looking at the picture now I think the voltage across the photodiode should have an opposite polarity.

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  • \$\begingroup\$ Worth noting they'll still need more than 5 V as the supply, as well, unless they use a different op-amp. \$\endgroup\$
    – Hearth
    Dec 29, 2020 at 16:58
  • \$\begingroup\$ @Hearth, True... but the op-amp is single supplied... \$\endgroup\$ Dec 29, 2020 at 17:04
  • \$\begingroup\$ That doesn't change that it still needs a minimum of 7 V across its power terminals to work. \$\endgroup\$
    – Hearth
    Dec 29, 2020 at 17:05
  • \$\begingroup\$ @Hearth, Of course... \$\endgroup\$ Dec 29, 2020 at 17:52
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    \$\begingroup\$ Thanks for the suggestion, I ended up using a 12 V power supply and LDO set at 6 V for the non-inverting input. This way I can variate the feedback resistor value without worrying that a large value will lead to the inputs being “too close” to 0 V. \$\endgroup\$
    – CreFroD
    Dec 30, 2020 at 11:19

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