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Regarding this question -> Transimpedance amplifier with a non-ground reference

What are the benefits of having an ideal absolute static reference voltage instead of just simple GND on the non-inverting input of a transimpedance amplifier using it with a photodiode?

Does it somehow eliminate the problem with dark current interference or are there any other benefits?

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2 Answers 2

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The voltage on the non-inverting input of the opamp can be used to provide a bias for the photodiode when there is no negative supply available.

The inverting input will be driven to the same voltage as the non-inverting through the feedback resistor.

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  • \$\begingroup\$ So basically it is pretty much the design like in the figure 1b from here: ti.com/lit/an/slyt703/… ,correct? \$\endgroup\$
    – Dakalaom
    Feb 4 at 1:03
  • \$\begingroup\$ @Dakalaom - agreed. \$\endgroup\$ Feb 4 at 1:06
  • \$\begingroup\$ Thanks. What would happen, if you would change the anode with the cathode on figure 1b, so that the anode is connected to IN- and the cathode is grounded, but still V_ref (i.e. 2V) > V_bias (GND) would apply? \$\endgroup\$
    – Dakalaom
    Feb 4 at 1:18
  • \$\begingroup\$ @Dakalaom If you forward bias the diode it will conduct current and possibly burn up. \$\endgroup\$ Feb 4 at 2:21
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Using photodiode in photo-conductive mode (reversely biased) improves dynamic response due to the lower junction capacitance. The bias can be achieved as you described. Additional benefit may be that you are adding an offset to your signal. Small positive offset is often beneficial in the case you are using unipolar ADC to convert the signal. If your op-amp (and other parts of the chain all the way to the ADC) have negative offset you wouldn't have a way to track small signals around zero. With a positive bias this would be possible and you may be able to subtract the measured offset at dark state of the photodiode.

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  • \$\begingroup\$ So inserting for example 2V to the non-inverting input would ideally result in 2V minimum output while no photodiode current is flowing, correct? \$\endgroup\$
    – Dakalaom
    Feb 4 at 16:22
  • \$\begingroup\$ @Dakalaom Correct: 2V bias on the non-inverting input will make the output of the op-amp match this when there is no PD current ==> 2V at the output. And also at the PD cathode - assuming the cathode is connected to the inverting input. If the anode is grounded then the PD is reverse biased by the said 2V and in photo-conductive mode. \$\endgroup\$ Feb 5 at 2:47

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