Question

Sketch the small-signal common base BJT model, assuming the output resistance is high enough to be ignored. If $r_{i e}=45 \Omega$ and $\alpha=0.98$, (a) determine the dc emitter current, $I_E$, in the bias circuit. (b) Find the dc base current, $I_B$. (c) Construct a small-signal equivalent circuit for an amplifier using this model and with $v_s$ as the voltage input source, $R_s=50 \Omega, R_{\text {Leq }}=2 \mathrm{k} \Omega, \quad$ and assume $R_i=r_{i e}$. (d) Determine the small-signal voltage gain, including any loss in the input circuit; that is, find $v_o / v_s$.

   Sketch the small-signal common base BJT model, assuming the output resistance is high enough to be ignored. If $r_{i e}=45 \Omega$ and $\alpha=0.98$, (a) determine the dc emitter current, $I_E$, in the bias circuit. (b) Find the dc base current, $I_B$. (c) Construct a small-signal equivalent circuit for an amplifier using this model and with $v_s$ as the voltage input source, $R_s=50 \Omega, R_{\text {Leq }}=2 \mathrm{k} \Omega, \quad$ and assume $R_i=r_{i e}$. (d) Determine the small-signal voltage gain, including any loss in the input circuit; that is, find $v_o / v_s$.

Essentials of Electrical and Computer Engineering

David V. Kerns, Jr.,… 1st Edition

Chapter 12, Problem 36

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Step 1

The emitter current can be expressed as: \[ I_E = \frac{I_B}{1 - \alpha} \] where $ \alpha $ is the common base current gain. Rearranging this gives: \[ I_B = I_E (1 - \alpha) \] However, we need to know either $ I_E $ or $ I_B $ to proceed. Assuming we have Show more…

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Sketch the small-signal common base BJT model, assuming the output resistance is high enough to be ignored. If $r_{i e}=45 \Omega$ and $\alpha=0.98$, (a) determine the dc emitter current, $I_E$, in the bias circuit. (b) Find the dc base current, $I_B$. (c) Construct a small-signal equivalent circuit for an amplifier using this model and with $v_s$ as the voltage input source, $R_s=50 \Omega, R_{\text {Leq }}=2 \mathrm{k} \Omega, \quad$ and assume $R_i=r_{i e}$. (d) Determine the small-signal voltage gain, including any loss in the input circuit; that is, find $v_o / v_s$.

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