If receiver is a transimpedance amplifier with an open loop gain of 200 and the average amplifier noise current is 2.7 pA/
Hz^0.5. The input resistance of the amplifier is 27 kΩ and the feedback resistance is 56 kΩ. The junction capacitance of the photodiode is 10 pF. Determine the photocurrent for the diode to maintain the required SNR of 50dB

 

If receiver is a transimpedance amplifier with an open loop gain of 200 and the average amplifier noise current is 2.7 pA/
. The input resistance of the amplifier is 27 kΩ and the feedback resistance is 56 kΩ. The junction capacitance of the photodiode is 10 pF. Determine the photocurrent for the diode to maintain the required SNR of 50dB

Sorry, we don't do your homework for you. But if you show us what you've done and where you have questions we may be able to help.

 

Sorry, we don't do your homework for you. But if you show us what you've done and where you have questions we may be able to help.

Hey I've got the same question as OP. What I've done is I've used two equations:

B = G/(2*pi*Rf*C)

Ip = (2*e*I*B)^0.5

B - Bandwidth
Rf- Feedback resistance
C - capacitance. Although the equation says to use total capacitance. I'm not sure about Junction capacitance. Help?
I - Average amplifier noise current
G - Gain

Let me know what you think.

 

Hi did this formula work out correct for the question?