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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??

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  • \$\begingroup\$ The diode will work because this current is diode reverse current look here en.wikipedia.org/wiki/Photodiode#/media/… or here fig 5 vishay.com/docs/81521/bpw34.pdf \$\endgroup\$ – G36 Oct 9 at 5:28
  • \$\begingroup\$ As you can see the diode in reverse direction behaves just like a "light controlled" current source. Notice that the voltage does not have any effect on the diode current (Reverse Light Current). \$\endgroup\$ – G36 Oct 9 at 5:32
  • \$\begingroup\$ ahh I see. Thanks! \$\endgroup\$ – Moon Oct 9 at 5:36
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The current is is equal to the current moving through the resistor. This is why we called a virtual ground, the voltage of the negative terminal is the same as the positive terminal if in negative feedback, but no current flows into the terminal in the case of an ideal op amp such as this one.

Because of this the currents are equal and the current flowing through the resistor can only flow through the diode.

There is still a voltage difference in the diode, however the op amp is continually driving the diodes voltage to 0 volts.

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  • \$\begingroup\$ but my question is that how can the current flows when there is no voltage difference between the diode. It just destroys the physics. If the current flows because there is no other path, then at least there should be voltage difference (0.7v or 0.3v or somehing). But because of the virtual ground negative terminal is 0V since positive termianl is 0V and there is no voltage difference between the diode. \$\endgroup\$ – Moon Oct 9 at 5:19
  • \$\begingroup\$ It's because the op amp is driving the dialed to 0 volts while it does this it produces a current to drive it to zero. This is an effect of a negative feedback \$\endgroup\$ – Voltage Spike Oct 9 at 5:22
  • \$\begingroup\$ how would it though? It will just destroy the diode, forcing current while having no voltage difference. \$\endgroup\$ – Moon Oct 9 at 5:28
  • \$\begingroup\$ It's a photo diode, it's designed to operate in reverse bias. \$\endgroup\$ – Voltage Spike Oct 9 at 5:32
  • \$\begingroup\$ incoming photons produce electrons and holes; that current would make the opamp's Vin-- become non-zero; the opamp manipulates its output voltage, to drive the Vin-- toward zero, about the 10 microvolt level. Not quite zero. \$\endgroup\$ – analogsystemsrf Oct 9 at 10:48
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If the op amp is powered from V+ and ground, of course it won't work at all. No matter what the state of the diode, the output will remain at ground in order to best approach the virtual ground state, and the diode will be unbiased.

If the op amp is powered from V+ and V-, the output will remain in the vicinity of ground, keeping the diode current at zero to maintain the virtual ground.

In order for this to work as intended, the op amp's negative power pin and the anode of the photodiode should be tied to a negative supply, or the positive input should be held at a voltage higher than ground.

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