Cuestion 6 (Transistor, corrientes)

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Cuestion 6

Suponemos el transistor en activa

$\begin{displaymath}I_{1}=I_{B}+I_{C}=(\beta+1)\, I_{B}=I_{E}\end{displaymath}$

$\begin{displaymath}U_{CC}=R_{C}\,I_{1}+R_{B}\,I_{B}+U_{BE}+U_{E}=R_{C}\,(\beta+1)\, I_{B}+R_{B}\,I_{B}+U_{BE}+U_{E}=\end{displaymath}$

$\begin{displaymath}=(R_{C}\,(\beta+1)+R_{B})\, I_{B}+U_{BE}+U_{E}\end{displaymath}$

$\begin{displaymath}I_{B}=\frac{U_{CC}-U_{BE}-U_{E}}{R_{C}\,(\beta+1)+R_{B}}\end{displaymath}$

Suponiendo diodo conduciendo y transistor en activa.

$\begin{displaymath}I_{B}=\frac{12-0.7-0.6}{101\cdot 3\,10^{3}+60\,10^{3}}=29.48\,\mu A\end{displaymath}$