Cuestion 10 (Efecto en corriente de colector de la variacion de las areas en los transistores)

Cuestion 10 (Efecto en corriente de colector de la variacion de las areas en los transistores)

$\begin{displaymath}\beta=100\end{displaymath}$

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

$\begin{displaymath}A_{2}=1\pm 5\,\%\end{displaymath}$

$\begin{displaymath}I_{ES_{2}}=I_{ES_{1}}\,(1\pm 5\,\%)\end{displaymath}$

$\begin{displaymath}I_{E}=I_{ES}\,(exp(\frac{V_{BE}}{V_{t}})-1)\end{displaymath}$

$\begin{displaymath}I_{E_{2}}=I_{E_{1}}\,(1\pm 5\,\%)\end{displaymath}$

$\begin{displaymath}I_{C_{2}}=I_{C_{1}}\,(1\pm 5\,\%)\end{displaymath}$

$\begin{displaymath}I_{B_{2}}=I_{B_{1}}\,(1\pm 5\,\%)\end{displaymath}$

$\begin{displaymath}I_{ref}=I_{C_{1}}+I_{B_{1}}+I_{B_{2}}=\frac{I_{C_{2}}}{(1\pm ... ...\pm 0.05)}+\frac{1}{100\,(1\pm 0.05)}+\frac{1}{100})\,I_{C_{2}}\end{displaymath}$

$\begin{displaymath}I_{C_{2}}=\frac{I_{ref}}{(\frac{1}{(1\pm 0.05)}+\frac{1}{100\,(1\pm 0.05)}+\frac{1}{100})}\end{displaymath}$

$\begin{displaymath}I_{ref}(+0.05)=(\frac{1}{(1+ 0.05)}+\frac{1}{100\,(1+ 0.05)}+\frac{1}{100})\,I_{C_{2}}=\frac{2100}{2041}\,I_{C_{2}}\end{displaymath}$

$\begin{displaymath}I_{ref}(-0.05)=(\frac{1}{(1- 0.05)}+\frac{1}{100\,(1- 0.05)}+\frac{1}{100})\,I_{C_{2}}=\frac{1900}{2039}\,I_{C_{2}}\end{displaymath}$

$\begin{displaymath}\% (+0.05)=\frac{\frac{2100}{2041}-\frac{100}{102}}{\frac{100}{102}}\,100=4.948\,\%\end{displaymath}$

$\begin{displaymath}\% (-0.05)=\frac{\frac{1900}{2039}-\frac{100}{102}}{\frac{100}{102}}\,100=-4.95\approx-5\,\%\end{displaymath}$

La solucion es la d).

Siguiente: Examen 2006 Junio 2 Subir: Problema 2 (Transistor, espejo Anterior: Cuestion 9 (Efecto en Índice General