help me with this circuit
the shown diagram i have used as current to voltage converter in the case of photodiode. pin 1 corresponds to cathode of photodiode and pin 2 corresponds to anode which is grounded.
i am using INA128... please help me about the working how the current is converted to voltage in this case...
The INA128 is a differential input instrumentation amplifier. It is designed to amplify a small voltage difference applied to pins 2 and 3. It is not designed to act as a "current amplifier" for a current injected into one of the inputs.
The photo diode should be connected cathode to +12V, anode to a resistor the other end of which is grounded. Connect pin 2 of the INA to the anode of the photodiode.
This way, the reverse leakage of the photodiode is modulated by incident light, and this current creates a small voltage across the resistor. This small voltage is amplified by the INA.
The second way to use a photo-diode is when it has no bias voltage. Then it generates a tiny current the same as a solar cell. Since it has no bias voltage then it has no "dark current" and it capacitance is higher therefore its response is slower than the reversed-biased way.
Uncle $crooge
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