Question

A CB amplifier is required to amplify a signal delivered by a coaxial cable having a characteristic resistance of 50Ω.50 \Omega. What bias current ICI_{C} should be utilized to obtain RinR_{\mathrm{in}} that is matched to the cable resistance? To obtain an overall voltage gain of GvG_{v} of 40 V/V, what should the total resistance in the collector (i.e.,RCRL)\left( i.e., R_{C} \| R_{L}\right) be?

Solution

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Due to the resistance of the co-axial cable R=50ΩR=50\Omega. To get an impedance match then

Rin=R\begin{equation} R_{i n}=R \end{equation}

So,

Rin=50Ω\begin{equation} R_{i n}=50 \Omega \end{equation}

To calculate bias current IcI_c, we have:

Rin=re=1gm\begin{equation} \begin{aligned} R_{i n}&=r_{e}\\&=\frac{1}{g_{m}} \end{aligned} \end{equation}

so,

re=VTIEVTIc\begin{equation} r_{e}=\frac{V_{T}}{I_{E}} \approx \frac{V_{T}}{I_{c}} \end{equation}

Since,

VT=25mV=25.103V\begin{equation} V_{T}=25 \mathrm{mV}=25.10^{-3} \mathrm{V} \end{equation}

we get,

Ic=VTre=25103V50Ω=0.5103A=0.5mA\begin{equation} \begin{aligned} I_{c}&=\frac{V_{T}}{r_{e}}\\&=\frac{25 \cdot 10^{-3} V}{50 \Omega}\\&=0.5 \cdot 10^{-3} A\\&=0.5 \mathrm{mA} \end{aligned} \end{equation}

Ic=0.5mA\begin{equation} \boxed{I_{c}=0.5 \mathrm{mA}} \end{equation}

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