# [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.

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$\Omega$.

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

$A_v=100$
$R_{in}=100k\Omega$
$R_{out}=10\Omega$
$R_{s}=10k\Omega$
$R_{L}=100\Omega$
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$v_{in}=\frac{R_{in}v_{s}}{R_{in}+R_{s}}=\frac{100k{\Omega}v_{s}}{110k\Omega}=0.91v_{s}$,

so

$v_{s}=1.1v_{in}$
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$v_{L}=\frac{R_{L}v_{out}}{R_{L}+R_{out}}=\frac{100{\Omega}v_{out}}{110\Omega}=0.91v_{out}$
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$\frac{v_{s}}{v_{L}}=\frac{1.1v_{in}}{0.91v_{out}}=\frac{1.1}{0.91}*\frac{1}{100}=\frac{1.1}{91}=0.0121$

Last edited: Feb 26, 2015
2. ### Veracohr Well-Known Member

Jan 3, 2011
559
77
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