[Amplifiers] Calculating source-to-load voltage gain

Discussion in 'Homework Help' started by nDever, Feb 26, 2015.

  1. nDever

    Thread Starter Active Member

    Jan 13, 2011
    154
    4
    Hi guys,

    I'm doing some elementary amplifier exercises, and decided to do this one.

    In.PNG
    Out.PNG

    An amplifier has a specified voltage gain of 100, input resistance 100k\Omega, and output resistance 10\Omega. Calculate the source-to-load voltage gain for a source resistance of 10k\Omega and a load resistance 100\Omega.

    Correct answer: 82.6

    My answer comes out to be the reciprocal of the correct answer; clearly, I'm going wrong somewhere.
    ____________________________________________

    Work:

    A_v=100
    R_{in}=100k\Omega
    R_{out}=10\Omega
    R_{s}=10k\Omega
    R_{L}=100\Omega
    ____________________________________________

    v_{in}=\frac{R_{in}v_{s}}{R_{in}+R_{s}}=\frac{100k{\Omega}v_{s}}{110k\Omega}=0.91v_{s},

    so

    v_{s}=1.1v_{in}
    ____________________________________________

    v_{L}=\frac{R_{L}v_{out}}{R_{L}+R_{out}}=\frac{100{\Omega}v_{out}}{110\Omega}=0.91v_{out}
    ____________________________________________

    \frac{v_{s}}{v_{L}}=\frac{1.1v_{in}}{0.91v_{out}}=\frac{1.1}{0.91}*\frac{1}{100}=\frac{1.1}{91}=0.0121
     
    Last edited: Feb 26, 2015
  2. Veracohr

    Well-Known Member

    Jan 3, 2011
    552
    76
    Voltage gain is expressed as the ratio of the load/output voltage to the source/input voltage: Vl/Vs.

    You have Vs/Vl.
     
  3. nDever

    Thread Starter Active Member

    Jan 13, 2011
    154
    4
    I thought that too, but when a ratio is expressed as A-to-B, isn't it A : B or A/B? The exercise wanted the source-to-load voltage gain.

    EDIT: Nevermind. I understand; "source-to-load" is specifying the elements it's interested in, not mathematical notation.
     
    Last edited: Feb 27, 2015
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