Question 10.13: Consider a CC–CE amplifier such as that in Fig. 10.39(b) wit...
Consider a CC–CE amplifier such as that in Fig. 10.39(b) with the following specifications: I1 = I2 = 1 mA and identical transistors with β = 100, fT = 400 MHz, and Cμ = 2 pF. Let the amplifier be fed with a source Vsig having a resistance Rsig = 4 kΩ, and assume a load resistance of 4 kΩ. Find the voltage gain AM, and estimate the 3-dB frequency, fH. Compare the results with those obtained with a CE amplifier operating under the same conditions. For simplicity, neglect ro and rx .
At an emitter bias current of 1 mA, Q1 and Q2 have
gm = 40 mA/V
re = 25 Ω
r_{π} = \frac{β}{g_{m}} = \frac{100}{40} = 2.5 kΩ
C_{π} + C_{μ} = \frac{g_{m}}{ω_{T}} = \frac{g_{m}}{2π f_{T}}
= \frac{40 × 10^{−3}}{2π × 400 × 10^{6}} = 15.9 pF
Cμ = 2 pF
Cπ = 13.9 pF
The voltage gain AM can be determined from the circuit shown in Fig. 10.40(a) as follows:
Rin2 = rπ2 = 2.5 kΩ
Rin = (β1 + 1) (re1 + Rin2)
= 101 (0.025 + 2.5) = 255 kΩ
\frac{V_{b1}}{V_{sig}} = \frac{R_{in}}{R_{in} + R_{sig}} = \frac{255}{255 + 4} = 0.98 V/V
\frac{V_{b2}}{V_{b1}} = \frac{R_{in2}}{R_{in2} + r_{e1}} = \frac{2.5}{2.5 + 0.025} = 0.99 V/V
\frac{V_{o}}{V_{b2}} = −g_{m2}R_{L} = −40 × 4 = −160 V/V
Thus,
A_{M} = \frac{V_{o}}{V_{sig}} = −160 × 0.99 × 0.98 = −155 V/V
To determine fH we use the method of open-circuit time constants. Figure 10.40(b) shows the circuit with Vsig set to zero and the four capacitances indicated. Capacitance Cμ1 sees a resistance Rμ1,
Rμ1 = Rsig || Rin
= 4 || 255 = 3.94 kΩ
To find the resistance Rπ1 seen by capacitance Cπ1 we refer to the equivalent circuit in Fig. 10.40(c). Analysis of this circuit results in
R_{π1} ≡ \frac{V_{x}}{I_{x}} = \frac{R_{sig} + R_{in2}}{1 + \frac{R_{sig}}{r_{π1}} + \frac{R_{in2}}{r_{e1}}}
= \frac{4000 + 2500}{1 + \frac{4000}{2500} + \frac{2500}{25}} = 63.4 Ω
Capacitance Cπ2 sees a resistance Rπ2,
Rπ2 = Rin2 || Rout1
= r_{π2} || \left[r_{e1} + \frac{R_{sig}}{β1 + 1}\right]
= 2500 || \left[25 + \frac{4000}{101}\right] = 63 Ω
Capacitance Cμ2 sees a resistance Rμ2. To determine Rμ2 we refer to the analysis of the frequency response of the CE amplifier in Section 10.4.4 to obtain
Rμ2 = (1 + gm2RL) (Rin2 || Rout1) + RL
= (1 + 40 × 4) \left[2500 || \left(25 + \frac{4000}{101}\right) \right] + 4000
= 14,143 Ω \ddot{\simeq } 14.1 kΩ
We now can determine τH from
τH = Cμ1Rμ1 + Cπ1Rπ1 + Cμ2Rμ2 + Cπ2Rπ2
= 2 × 3.94 + 13.9 × 0.0634 + 2 × 14.1 + 13.9 × 0.063
= 7.88 + 0.88 + 28.2 + 0.88 = 37.8 ns
We observe that Cπ1 and Cπ2 play minor roles in determining the high-frequency response. As expected, Cμ2 through the Miller effect plays the most significant role. Capacitor Cμ1, which interacts directly with (Rsig || Rin), also plays an important role. The 3-dB frequency fH can be found as follows:
f_{H} = \frac{1}{2π τ_{H}} = \frac{1}{2π × 37.8 × 10^{−9}} = 4.2 MHz
For comparison, we evaluate AM and fH of a CE amplifier operating under the same conditions. Refer to Fig. 10.40(d). The voltage gain AM is given by
A_{M}= \frac{R_{in}}{R_{in} + R_{sig}} (−g_{m}R_{L})
= \frac{r_{π}}{r_{π} + R_{sig}} (−g_{m}R_{L})
= \frac{2.5}{2.5 + 4} (−40 × 4)
= −61.5 V/V
Rπ = rπ || Rsig = 2.5 || 4 = 1.54 kΩ
Rμ = (1 + gmRL) (Rsig || rπ) + RL
= (1 + 40 × 4) (4 || 2.5) + 4
= 251.7 kΩ
Thus,
τH = CπRπ + CμRμ
= 13.9 × 1.54 + 2 × 251.7
= 21.4 + 503.4 = 524.8 ns
Observe the dominant role played by Cμ. The 3-dB frequency fH is
f_{H} = \frac{1}{2π τ_{H}} = \frac{1}{2π × 524.8 × 10^{−9}} = 303 kHz
Thus, including the buffering transistor Q1 increases the gain, |AM|, from 61.5 V/V to 155 V/V—a factor of 2.5—and increases the bandwidth from 303 kHz to 4.2 MHz—a factor of 13.9! The gain–bandwidth product is increased from 18.63 MHz to 651 MHz—a factor of 35!
