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Converting voltage to power in a receiver

  1. Dec 11, 2008 #1


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    Hi everyone. I'll describe my setup first, then state my question.

    I have a receiver chain consisting of the following (where all conversion factors are assumed to be scalar and constant under the conditions in which I use them):
    1) An APD (avalanche photodiode): Input: Optical power P1in [W]; Output: Current I1out [A]; Conversion factor: C1 [A/W].
    2) A transimpedance amplifier: Input: Current I2in = I1out [A]; Output: Voltage U2out [V]; Conversion factor: C2 [V/A = ohm].
    3) A monolithic high-frequency amplifier, 50 ohm differential input. (~50 ohm resistor connected across the amplifier inputs, negative input is AC-coupled to GND, positive input is AC-coupled to the transimpedance amplifier output.) Use Z0 = 50 ohm. Input: P3in [dBm]; Output: P3out [dBm]. Gain factor is C3 [dB].

    My question is: How should I convert the input optical power to an electrical power, at the input of component 3) above? (Or more specifically, how do I convert U2out to P3in?)

    I want to use power expressed as dBm when making the necessary calculations, since I will treat the receiver as a RF system. For example, the amplifier gain is expressed in dB.

    Assume I have the input optical power and all conversion factors, no point in confusing the argument with a bunch of numbers.


    My thoughts on the matter are as follows:

    I am confident that the output from the transimpedance amplifier is U2out = P1in * C1 * C2, and the units seem to be correct.

    But the only way I can think of to convert U2out to P3in is to take P3in = (U2out)^2 / Z0.

    I have a gut feeling that this is incorrect, since (look at the formulas) this means that the original input power is squared. In my case this is a problem since the required dynamic range of the system, when calculated at the input (the APD), is _half_ of the dynamic range when calculated at the 3:rd component.

    Since I do not care about the DC components of U2out, the Z0 resistor seems to be where the voltage drop occurs (but I suspect I should treat the differential input of the 3:rd component as a parallel connection of the external matching resistor and the input impedance, but that would only introduce a factor of 2 in the calculations).

    I can't find the error. I suspect the root cause is in the C1 conversion factor, [A/W].

    Can anyone please point me in the right direction? Or if I am already pointed in the right direction, tell me why? ;-)
  2. jcsd
  3. Dec 11, 2008 #2


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    P3in = (U2out)^2 / Z0 should work as long as you really have a 50 Ohm system.
    Remember that you can always think of the amplifier as a VOLTAGE amplifier if you want to as long as all the components have the same impedance; e.g. a gain of 20 dB simply means that it amplifies the voltage 10 times.
  4. Dec 12, 2008 #3


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    Thanks for the reply, but I thought of that too. But I still have twice the dynamic range (in dB) compared to the dynamic range of the optical input signal.

    I think I might be on to an explanation:
    The APD transforms a power to a current.
    If one considers a grounded load, Zl, connected to a feed line carrying the current I, the power dissipated by the load is Pl=I^2*Zl, or conversely, I=sqrt(Pl/Zl).
    But in my assumptions above, I used a linear transformation from power to current. It seems as it is here my "squaring" occurs.

    I must investigate further if the linear transformation is valid or not (it is not a resistive load so it might be valid), but if I use the resistive load analogy, my formulas make sense.
  5. Dec 12, 2008 #4


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    It appears as if I did not do anything wrong. I just got it confirmed that power is actually squared in the APD, and that I will simply have to deal with the fact that the dynamic range is doubled inside my receiver than outside of it.

    Perhaps my problem with the inital findings was mostly wishful thinking.
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