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Optical Fiber Question

Discussion in 'Electronics Homework Help' started by Diaboltz, Sep 14, 2020.

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  1. Diaboltz


    Sep 14, 2020
    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.

    Ive used the formula B = G/(2*pi*Rf*C) to get the value of B, but I am unsure how to get Ip. I see the formula - Ip = (2*e*I*B)^0.5 , but not sure if this is the correct formula? How do I test if the satisfies my SNR value?

    *This is an assignment question, don't want my work to be done for me, but I can't find the answer to this anywhere. Hope someone can help
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Where do you get this equation from? What does it tell you? How would it be connected to SNR, if at all?

    How do you calculate SNR? Insert the data you have into the equation and solve for Ip.
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