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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).

cajael