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Ejercicio 8.11 Hambley (Respuesta en Frecuencia, amplificador emisor comun)

Solapas principales

Enviado por pichu en Vie, 08/17/2012 - 22:17

a)

Solucion:

\begin{displaymath}R_{L}=510\,\Omega\end{displaymath}


\begin{displaymath}R_{C}=510\,\Omega\end{displaymath}


\begin{displaymath}R_{B}=10\,k\Omega\end{displaymath}


\begin{displaymath}R_{S}=50\,\Omega\end{displaymath}


\begin{displaymath}I_{C_{Q}}=10\,mA\end{displaymath}


\begin{displaymath}r_{o}=20\,k\Omega\end{displaymath}


\begin{displaymath}f_{t}=400\,MHz\end{displaymath}


\begin{displaymath}C_{\mu}=5\,pF\end{displaymath}


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


\begin{displaymath}r_{x}=10\,\Omega\end{displaymath}


\begin{displaymath}f_{H}=\frac{1}{2\,\pi\,C_{T}\,R'_{S}}\end{displaymath}


\begin{displaymath}r_{\pi}=\frac{\beta\,V_{T}}{I_{C_{Q}}}=\frac{100\cdot 26\,10^{-3}}{10\,10^{-3}}=260\,\Omega\end{displaymath}


\begin{displaymath}R'_{S}=r_{\pi}\vert\vert(r_{x}+R_{B}\vert\vert R_{S})=260\vert\vert(10+10\,10^3\vert\vert 50)=48.58\,\Omega\end{displaymath}


\begin{displaymath}C_{T}=C_{\pi}+C_{\mu}\,(1+g_{m}\,R'_{L})\end{displaymath}


\begin{displaymath}f_{t}=\frac{\beta}{2\,\pi\,r_{\pi}\,(C_{\pi}+C_{\mu})}\end{displaymath}


\begin{displaymath}C_{\pi}=\frac{\beta}{2\,\pi\,r_{\pi}\,f_{t}}-C_{\mu}=148\,pF\end{displaymath}


\begin{displaymath}g_{m}=\frac{\beta}{r_{\pi}}=\frac{100}{260}=\frac{5}{13}\end{displaymath}


\begin{displaymath}R'_{L}=\frac{1}{\frac{1}{r_{o}}+\frac{1}{R_{C}}+\frac{1}{R_{L...<br /> ...{\frac{1}{20\,10^3}+\frac{1}{510}+\frac{1}{510}}=251.79\,\Omega\end{displaymath}


\begin{displaymath}C_{T}=C_{\pi}+C_{\mu}\,(1+g_{m}\,R'_{L})=148\,10^{-12}+5\,10^{-12}\,(1+\frac{5}{13}\cdot 251.79)=637.21\,pF\end{displaymath}


\begin{displaymath}f_{H}=\frac{1}{2\,\pi\,C_{T}\,R'_{S}}=\frac{1}{2\cdot \pi \cdot 637.21\,10^{-12}\cdot 48.58}=5.14\,MHz\end{displaymath}


b)

Solucion:

\begin{displaymath}R_{L}=510\,\Omega\end{displaymath}


\begin{displaymath}R_{C}=510\,\Omega\end{displaymath}


\begin{displaymath}R_{B}=10\,k\Omega\end{displaymath}


\begin{displaymath}R_{S}=50\,\Omega\end{displaymath}


\begin{displaymath}I_{C_{Q}}=10\,mA\end{displaymath}


\begin{displaymath}r_{o}=20\,k\Omega\end{displaymath}


\begin{displaymath}f_{t}=350\,MHz\end{displaymath}


\begin{displaymath}C_{\mu}=2\,pF\end{displaymath}


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


\begin{displaymath}r_{x}=15\,\Omega\end{displaymath}


\begin{displaymath}f_{H}=\frac{1}{2\,\pi\,C_{T}\,R'_{S}}\end{displaymath}


\begin{displaymath}r_{\pi}=\frac{\beta\,V_{T}}{I_{C_{Q}}}=\frac{100\cdot 26\,10^{-3}}{10\,10^{-3}}=260\,\Omega\end{displaymath}


\begin{displaymath}R'_{S}=r_{\pi}\vert\vert(r_{x}+R_{B}\vert\vert R_{S})=260\vert\vert(15+10\,10^3\vert\vert 50)=51.84\,\Omega\end{displaymath}


\begin{displaymath}C_{T}=C_{\pi}+C_{\mu}\,(1+g_{m}\,R'_{L})\end{displaymath}


\begin{displaymath}f_{t}=\frac{\beta}{2\,\pi\,r_{\pi}\,(C_{\pi}+C_{\mu})}\end{displaymath}


\begin{displaymath}C_{\pi}=\frac{\beta}{2\,\pi\,r_{\pi}\,f_{t}}-C_{\mu}=\frac{100}{2\cdot \pi\cdot 260\cdot 350\,10^{6}}-2\,10^{-12}=172.89\,pF\end{displaymath}


\begin{displaymath}g_{m}=\frac{\beta}{r_{\pi}}=\frac{100}{260}=\frac{5}{13}\end{displaymath}


\begin{displaymath}R'_{L}=\frac{1}{\frac{1}{r_{o}}+\frac{1}{R_{C}}+\frac{1}{R_{L...<br /> ...{\frac{1}{20\,10^3}+\frac{1}{510}+\frac{1}{510}}=251.79\,\Omega\end{displaymath}


\begin{displaymath}C_{T}=C_{\pi}+C_{\mu}\,(1+g_{m}\,R'_{L})=172.89\,10^{-12}+2\,10^{-12}\,(1+\frac{5}{13}\cdot 251.79)=368.58\,pF\end{displaymath}


\begin{displaymath}f_{H}=\frac{1}{2\,\pi\,C_{T}\,R'_{S}}=\frac{1}{2\cdot \pi \cdot 368.58\,10^{-12}\cdot 51.58}=8.33\,MHz\end{displaymath}


c)

Solucion:

\begin{displaymath}R_{L}=510\,\Omega\end{displaymath}


\begin{displaymath}R_{C}=510\,\Omega\end{displaymath}


\begin{displaymath}R_{B}=10\,k\Omega\end{displaymath}


\begin{displaymath}R_{S}=50\,\Omega\end{displaymath}


\begin{displaymath}I_{C_{Q}}=10\,mA\end{displaymath}


\begin{displaymath}r_{o}=20\,k\Omega\end{displaymath}


\begin{displaymath}f_{t}=500\,MHz\end{displaymath}


\begin{displaymath}C_{\mu}=2\,pF\end{displaymath}


\begin{displaymath}\beta=50\end{displaymath}


\begin{displaymath}r_{x}=5\,\Omega\end{displaymath}


\begin{displaymath}f_{H}=\frac{1}{2\,\pi\,C_{T}\,R'_{S}}\end{displaymath}


\begin{displaymath}r_{\pi}=\frac{\beta\,V_{T}}{I_{C_{Q}}}=\frac{50\cdot 26\,10^{-3}}{10\,10^{-3}}=130\,\Omega\end{displaymath}


\begin{displaymath}R'_{S}=r_{\pi}\vert\vert(r_{x}+R_{B}\vert\vert R_{S})=130\vert\vert(5+10\,10^3\vert\vert 50)=38.52\,\Omega\end{displaymath}


\begin{displaymath}C_{T}=C_{\pi}+C_{\mu}\,(1+g_{m}\,R'_{L})\end{displaymath}


\begin{displaymath}f_{t}=\frac{\beta}{2\,\pi\,r_{\pi}\,(C_{\pi}+C_{\mu})}\end{displaymath}


\begin{displaymath}C_{\pi}=\frac{\beta}{2\,\pi\,r_{\pi}\,f_{t}}-C_{\mu}=\frac{50}{2\cdot \pi\cdot 130\cdot 500\,10^{6}}-2\,10^{-12}=120.42\,pF\end{displaymath}


\begin{displaymath}g_{m}=\frac{\beta}{r_{\pi}}=\frac{50}{130}=\frac{5}{13}\end{displaymath}


\begin{displaymath}R'_{L}=\frac{1}{\frac{1}{r_{o}}+\frac{1}{R_{C}}+\frac{1}{R_{L...<br /> ...{\frac{1}{20\,10^3}+\frac{1}{510}+\frac{1}{510}}=251.79\,\Omega\end{displaymath}


\begin{displaymath}C_{T}=C_{\pi}+C_{\mu}\,(1+g_{m}\,R'_{L})=120.42\,10^{-12}+2\,10^{-12}\,(1+\frac{5}{13}\cdot 251.79)=316.11\,pF\end{displaymath}


\begin{displaymath}f_{H}=\frac{1}{2\,\pi\,C_{T}\,R'_{S}}=\frac{1}{2\cdot \pi \cdot 316.11\,10^{-12}\cdot 38.52}=13.07\,MHz\end{displaymath}



 

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