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Cuestion 1 (Amplificadores Operacionales, Io del A.Operacional)

Image 2005JunBC1
Solucion:

\begin{displaymath}U_{e}=8\,V\end{displaymath}


\begin{displaymath}U_{1}=15\,V\end{displaymath}


\begin{displaymath}R_{1}=R_{2}=R_{3}=10\,k\Omega\end{displaymath}


\begin{displaymath}R_{4}=30\,k\Omega\end{displaymath}


\begin{displaymath}R_{5}=15\,k\Omega\end{displaymath}


\begin{displaymath}R_{L}=1.5\,k\Omega\end{displaymath}








\begin{displaymath}I_{R_{1}}=\frac{V_{1}}{R_{1}}=\frac{6.5}{10\,10^3}=650\,\mu A\end{displaymath}


\begin{displaymath}I_{R_{2}}=I_{R_{1}}=650\,\mu A\end{displaymath}




\begin{displaymath}U_{s}=I_{R_{2}}\,R_{2}+V_{1}=650\,10^{-6}\cdot 10\,10^3+6.5=13\,V\end{displaymath}


\begin{displaymath}I_{R_{L}}=\frac{U_{s}}{R_{L}}=\frac{13}{1.5\,10^3}=8.67\,mA\end{displaymath}


\begin{displaymath}I_{o}=I_{R_{2}}+I_{R_{L}}=9.32\,mA\end{displaymath}

La corriente es saliente.




Comprobacion con el Micro-Cap.

Image 2005JunAC5b

cajael