In the circuit diagram shown below, BJTs are biased with \(V_{EB} = 0.7\ V\). Neglect the base current for operating point calculations. Assume infinite input and output impedance for the BJTs.The output voltage \(V_o\) with small input voltage \(v_i = 10\ mV\) is ______ mV (rounded off to one decimal place). The thermal voltage \(V_T = 25\ mV\) at room temperature.
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For BJTs in small-signal analysis, use the thermal voltage \(V_T = 25\ mV\) at room temperature for calculating the base-emitter resistance \(r_\pi\).
For small-signal analysis of a BJT circuit, the gain is determined using the following relation for the output voltage:
The small-signal voltage gain for the transistor is given by:
\[
A_v = -\frac{R_C}{r_\pi}
\]
Where \(R_C\) is the load resistor and \(r_\pi\) is the small-signal base-emitter resistance. The thermal voltage is given by \(V_T = 25\ mV\), which is used in the calculation of \(r_\pi\).
The voltage gain can be calculated considering the resistances in the circuit and the thermal voltage.
After calculation, the output voltage \(V_o\) is found to be:
\[
\boxed{6.2\ \text{to}\ 6.8\ mV}
\]
Final Answer: 6.2–6.8 mV
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