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Question 12.12: Calculate the collector–emitter saturation voltage of a bipo...

Calculate the collector-emitter saturation voltage of a bipolar transistor at T=300 \mathrm{~K}.

Assume that \alpha_{F}=0.99, \alpha_{R}=0.20, I_{C}=1 \mathrm{~mA}, and I_{B}=50 \mu \mathrm{A}.

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Substituting the parameters into Equation (12.77), we have

\begin{array}{l}V_{C E}(\mathrm{sat})=V_{t} \ln \left[\frac{I_{C}\left(1-\alpha_{R}\right)+I_{B}}{\alpha_{F} I_{B}-\left(1-\alpha_{F}\right) I_{C}} \cdot \frac{\alpha_{F}}{\alpha_{R}}\right] \\ \end{array}     (12.77)

\begin{array}{l}V_{C E}(\mathrm{sat})=(0.0259) \ln \left[\frac{(1)(1-0.2)+(0.05)}{(0.99)(0.05)-(1-0.99)(1)}\left(\frac{0.99}{0.20}\right)\right]=0.121 \mathrm{~V}\end{array}

Comment

This V_{C E} (sat) value is typical of collector-emitter saturation voltages. Because of the \log function, V_{C E} (sat) is not a strong function of I_{C} or I_{B}.

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