Dear all

The attachment is the design of a photodiode preamplifier.

The signal generated by the photodiode I=I(dc)+I(ac), which means DC current and AC current respectively.

However, only the ac part is needed. So a capacitor is used to block the DC current from going through the opamp.

Can I design in this way? If so, the path for DC cannot be provided actually.

If not, how can I get rid of the dc current? Thanks in advance.


Richie

 

Something seems inherently wrong with a DC current generator in series with a capacitor. I think taking out the DC component should happen after the first stage of amplification.
Exactly which photodiode you are using is relevant.
I have looked up some datasheets and found that some photodiodes have an amplifier in their package. Some datasheets have circuit examples.
Take a peek at these.

 

You can add a resistor from the photodiode output to ground for the DC current. Since the op amp is connected as a transimpedance current amplifier it has essentially zero input impedance and all the AC current will go through the capacitor and appear as a voltage at the op amp output.

 

It depends on what you are trying to measure. If you are measuring static light levels, this circuit would not work since the photodiode is AC coupled.

If you are measuring photo pulses, you still need to bias the photodiode through a resistor.