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Power, voltage and current gain

  1. Jan 18, 2014 #1

    I know [itex]G_P (dB) = 10log(G_P)[/itex] and [itex]G_V (dB) = 20log(G_V)[/itex] and [itex]G_I (dB) = 20log(G_I)[/itex].

    I also know that [itex]G_P (dB) = G_V (dB) + G_I (dB)[/itex].

    So, if I have a common base amplifier whose current gain is [itex]G_I (dB) = 0[/itex], then [itex]G_P (dB) = G_V(dB)[/itex].

    [itex]G_V (dB) = 20 dB[/itex]. So [itex]G_V = 10^{20/20} = 10[/itex]
    [itex]G_I (dB) = 0 dB[/itex]. So [itex]G_I = 10^{0/20} = 1[/itex]
    As I've written above, then [itex]G_P (dB)= G_V (dB) = 20 dB[/itex]. So [itex]G_P = 10^{20/10} = 100[/itex].

    But, in linear scale (not in dB), [itex]G_P = G_V\cdot G_I[/itex], so [itex]G_P = 10\cdot 1 = 10[/itex], which is not [itex]100[/itex].

    What's wrong?

  2. jcsd
  3. Jan 18, 2014 #2


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    Staff: Mentor

    Where did your expressions 10^(20/20) come from? That is not part of the definition of dB, AFAIK.
  4. Jan 18, 2014 #3
    Thank you for answering.

    They come from [itex]G_V (dB) = 20log(G_V)[/itex]. If [itex]G_V(dB)= 20 dB[/itex], then [itex]G_V = 10^{G_V(dB)/20} = 10^{20/20} = 10[/itex], isn't it?
  5. Jan 18, 2014 #4
    The voltage gain formula GV(dB)=20log(GV) assumes the input and output impedances are the same. If the impedances are the same, then a 10 times increase in voltage will result in a 10 times increase in current and 10 x 10 = 100. In your example you specify a current gain of 1 or 0dB. The requires that the output impedance be 10 times that of the input. That increase in impedance reduces the power gain to 10.
    Last edited: Jan 18, 2014
  6. Jan 18, 2014 #5
    I don't understand it.

    Isn't the formula [itex]G_P (dB) = G_V (dB) + G_I (dB)[/itex] always valid?
  7. Jan 18, 2014 #6
    No, it's only valid if the input and output impedances are the same. Your example proves it's not valid if the impedances are different.
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