[Amplifiers] Calculating source-to-load voltage gain

Thread Starter

nDever

Joined Jan 13, 2011
153
Hi guys,

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

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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.
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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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\(\frac{v_{s}}{v_{L}}=\frac{1.1v_{in}}{0.91v_{out}}=\frac{1.1}{0.91}*\frac{1}{100}=\frac{1.1}{91}=0.0121\)
 
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Thread Starter

nDever

Joined Jan 13, 2011
153
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.
 
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