Question 10.4: It is required to find the midband gain and the upper 3-dB f...
It is required to find the midband gain and the upper 3-dB frequency of the common-emitter amplifier of Fig. 10.9(a) for the following case: IE = 1 mA, RB = RB1 || RB2 = 100 kΩ, RC = 8 kΩ, Rsig = 5 kΩ, RL = 5 kΩ, β0 = 100, VA = 100 V, Cμ = 1 pF, fT = 800 MHz, and rx = 50 Ω. Also, determine the frequency of the transmission zero.
The transistor is biased at IC \simeq 1 mA. Thus the values of its hybrid-π model parameters are
g_{m} = \frac{I_{C}}{V_{T}} = \frac{1 mA}{25 mV} = 40 mA/V
r_{π} = \frac{β_{0}}{g_{m}} = \frac{100}{40 mA/V} = 2.5 kΩ
r_{o} = \frac{V_{A}}{I_{C}} = \frac{100 V}{1 mA} = 100 kΩ
C_{π} + C_{μ} = \frac{g_{m}}{ω_{T}} = \frac{40 × 10^{−3}}{2π × 800 × 10^{6}} = 8 pF
Cμ = 1 pF
Cπ = 7 pF
rx = 50 Ω
The midband voltage gain is
A_{M} = − \frac{R_{B}}{R_{B} + R_{sig}} \frac{r_{π}}{r_{π} + r_{x} + (R_{B} || R_{sig})} g_{m}R_{L}^{′}
where
R_{L}^{′} = r_{o} || R_{C} || R_{L}
= (100 || 8 || 5) kΩ = 3 kΩ
Thus,
g_{m}R_{L}^{′} = 40 × 3 = 120 V/V
and
A_{M} = −\frac{100}{100 + 5} × \frac{2.5}{2.5 + 0.05 + (100 || 5)} × 120
= −39 V/V
and
20 log |AM| = 32 dB
To determine fH we first find Cin,
C_{in} = C_{π} + C_{μ} (1 + g_{m}R_{L}^{′})
= 7 + 1 (1 + 120) = 128 pF
and the effective source resistance R_{sig}^{′}
R_{sig}^{′} = r_{π} || [ r_{x} + (R_{B} || R_{sig})]
= 2.5 || [0.05 + (100 || 5)]
= 1.65 kΩ
Thus,
f_{H} = \frac{1}{2πC_{in} R_{sig}^{′}} = \frac{1}{2π × 128 × 10^{−12} × 1.65 × 10^{3}} = 754 kHz
Finally, as in the case of the CS amplifier, it can be shown that the CE amplifier has a transmission zero with frequency
f_{Z} = \frac{g_{m}}{2πC_{μ}} = \frac{40 × 10^{−3}}{2π × 1 × 10^{−12}} = 6.37 GHz
which is much higher than fH.