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Ejercicio 1.15 Hambley (Amplificacion, ganancia modo comun y ganancia diferencial)

Solapas principales

Enviado por pichu en Lun, 07/23/2012 - 13:36

(a) Ganancia diferencial

Solucion:

$\begin{displaymath}v_{i1}=\frac{1}{2}\end{displaymath}$

$\begin{displaymath}v_{i2}=-\frac{1}{2}\end{displaymath}$

$\begin{displaymath}v_{o}=A_{1}\,v_{i1}-A_{2}\,v_{i2}=\frac{A_{1}-(-A_{2})}{2}=\frac{A_{1}+A_{2}}{2}\end{displaymath}$

$\begin{displaymath}v_{icm}=\frac{v_{i1}+v_{i2}}{2}=\frac{\frac{1}{2}-\frac{1}{2}}{2}=0\end{displaymath}$

$\begin{displaymath}v_{id}=v_{i1}-v_{i2}=\frac{1}{2}-(-\frac{1}{2})=1\end{displaymath}$

$\begin{displaymath}v_{o}=A_{d}\,v_{id}+A_{cm}\,v_{icm}=A_{d}\,1+A_{cm}\,0=A_{d}=\frac{A_{1}+A_{2}}{2}\end{displaymath}$

(b) Ganancia en modo comun

Solucion:

$\begin{displaymath}v_{i1}=1\end{displaymath}$

$\begin{displaymath}v_{i2}=1\end{displaymath}$

$\begin{displaymath}v_{o}=A_{1}\,v_{i1}-A_{2}\,v_{i2}=A_{1}-A_{2}\end{displaymath}$

$\begin{displaymath}v_{icm}=\frac{v_{i1}+v_{i2}}{2}=\frac{1+1}{2}=1\end{displaymath}$

$\begin{displaymath}v_{id}=v_{i1}-v_{i2}=1-1=0\end{displaymath}$

$\begin{displaymath}v_{o}=A_{d}\,v_{id}+A_{cm}\,v_{icm}=A_{d}\,0+A_{cm}\,1=A_{cm}=A_{1}-A_{2}\end{displaymath}$

(c) CMRR

Solucion:

$\begin{displaymath}A_{1}=100\end{displaymath}$

$\begin{displaymath}A_{2}=101\end{displaymath}$

$\begin{displaymath}A_{cm}=A_{1}-A_{2}=100-101=-1\end{displaymath}$

$\begin{displaymath}A_{d}=\frac{A_{1}+A_{2}}{2}=\frac{201}{2}=100.5\end{displaymath}$

$\begin{displaymath}CMRR=20\,log\vert\frac{A_{d}}{A_{cm}}\vert=20\,log\vert\frac{100.5}{-1}\vert=40.04\,dB\end{displaymath}$

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