Simple question on Transresistance Amplifier circuit

[mkim083]

Something that makes me confused while reading a tutorial on the Transresistance amplifier circuit.

Here is the picture of the circuit. It basically measures the current across the photo device and produces Vout accordingly.

My question is that: if the virtual ground rule applies, that is the voltage on the negative termianal is 0V (or very close to 0V) then there is no voltage difference across the photo device. If there is no voltage difference, how come the current flows across the photo device to be measured first place??

[KlausST]

Hi,

You may simply see the photodiode as current source .... without generating voltage.

Now you may say: no current without voltage.
Then you need to go the more complex, but more realistic way and
See the photodiode as "current source with integrated series resistance".

Then - inside the photodiode - all the generated voltage is compensated as voltage across the resistor .... making the external voltage to be zero.

Klaus

[BradtheRad]

In some cases you may wish to apply some amount of bias to your photodetector, to bring it to a threshold where it responds reliably to a trigger event. The goal is to adjust your circuit to a 'good operating point'.

[FvM]

In some cases you may wish to apply some amount of bias to your photodetector.

The typical case where you apply bias is for increasing the photo diode speed (reduced junction capacitance with reverse bias voltage). Linearity will be typically worse, also you' ll face increased leakage current.

[asdf44]

Yes I think Klauss described it well.

The light provides energy so inside the diode there is a source that does produce voltage. That source is then shorted out by the opamp circuit and the resulting current is measured.

[SunnySkyguy]

Consider A as a high gain error amplifier with high input impedance.

As PD is a high impedance current source it starts to generate an input error voltage from some current but the error amplifier at same time, drives the feedback resistor to absorb this current to null the input error voltage with null input current.

Thus Vout =-IsRf. As long as the output is in the linear range. the input will be 0 V due to 1e6 high gain.