Question 19.10: Determine the quiescent output voltage, the small-signal vol......
Determine the quiescent output voltage, the small-signal voltage gain, the input resistance and the output resistance of the following circuit.
Quiescent output voltage
If we assume that the base current is negligible, as no constant current can flow through the input capacitor, the quiescent base voltage is determined simply by the supply voltage V_{CC} and by the potential divider formed by R_{1} and R_{2} . Thus
V_{B} \approx V_{CC} \frac{R_{2}}{R_{1} + R_{2}}Therefore in this case
V_{B} \approx 10 \frac{10 kΩ}{27 kΩ + 10 kΩ} \approx 2.7 VThe quiescent emitter voltage is then given by
V_{E} = V_{B} – V_{BE} = 2.7 – 0.7 = 2.0 VHowever, in this circuit V_{E} represents our output voltage, so the quiescent output voltage is given by
V_{o(quiescent)} = V_{E} = 2.0 VSmall-signal voltage gain
As discussed earlier, the input signal is applied to the base of the transistor through the coupling capacitor C, and therefore
v_{e} \approx v_{b} \approx v_{i}However, in this circuit v_{e} is the output voltage and as the output is equal to the input
voltage gain ≈ 1
Input resistance
From Equation 19.25
r_{i} \approx R_{1}//R_{2} = 27 kΩ//10 kΩ \approx 7.3 kΩOutput resistance
From Equation 19.26
r_{o} \approx r_{e} \approx \frac{1}{40I_{E}}In this case, I_{E} = V_{E}/R_{E} = 2.0 V/100 \Omega = 20 mA . Therefore
r_{o} \approx \frac{1}{40I_{E}} = \frac{1}{40 \times 20 \times 10^{-3}} = 1.25 \Omega