Question 6.4: In the CC amplifier of Fig. 6-7(b), let hic = 1 kΩ, hrc = 1,......
In the CC amplifier of Fig. 6-7(b), let h_{ic} = 1 kΩ, h_{rc} = 1, h_{fc} = -101, h_{oc} = 12 μS, and R_L = 2 kΩ. Drawing direct analogies with the CE amplifier of Example 6.2, find expressions for the (a) current-gain ratio A_i, (b) voltage-gain ratio A_v, (c) input impedance Z_{in}, and (d) output impedance Z_o. (e) Evaluate this typical CC amplifier.
(a) In parallel with (6.43),
A_{i} = {\frac{i_{L}}{i_{b}}} = -{\frac{h_{f e}}{1 + h_{o e} R_{L}}} = -{\frac{100}{1 + (12 \times 10^{-6})(2 \times 10^{3})}} = -97.7 (6.43)
A_{i} = -{\frac{h_{f c}}{1 + h_{o c} R_{L}}} = -{\frac{-100}{1 + (12 \times 10^{-6})(2 \times 10^{3})}} = 98.6 (6.55)
Note that A_{i} \approx -h_{f c}, and that the input and output currents are in phase because h_{fc} < 0.
(b) In parallel with (6.46),
= -{\frac{(100)(2 \times 10^{3})}{1 \times 10^{3} + (2 \times 10^{3})[(1 \times 10^{3})(12 \times 10^{-6}) – (100)(1 \times 10^{-4})]}} = -199.2 (6.46)
A_{v} = -{\frac{h_{fc}R_{L}}{h_{ic} + R_{L}(h_{ic} h_{o c} – h_{fc}h_{r c})}} = -{\frac{(-101)(2 \times 10^{3})}{1 \times 10^{3} + (2 \times 10^{3})[(1 \times 10^{3})(12 \times 10^{-6}) – (-101)(1)]}} = 0.995 (6.56)
Observe that A_{v} \approx 1/(1 – h_{ic}h_{oc}/h_{fc} )\approx 1. Since the gain is approximately 1 and the output voltage is in phase with the input voltage, this amplifier is commonly called a unity follower.
(c) In parallel with (6.47),
Z_{\mathrm{in}} = {\frac{v_{s}}{i_{b}}} = h_{i e} – {\frac{h_{re}h_{f e}R_{L}}{1 + h_{oe}R_{L}}} = 1 \times 10^{3} – {\frac{(1 \times 10^{-4})(100)(2 \times 10^{3})}{1 + (12 \times 10^{-6})(2 \times 10^{3})}} = 980.5 \Omega (6.47)
Z_{\mathrm{in}} = h_{i c} – {\frac{h_{rc}h_{fc}R_{L}}{1 + h_{oc}R_{L}}} = 1 \times 10^{3} – {\frac{(1)(-101)(2 \times 10^{3})}{1 + (12 \times 10^{-6})(2 \times 10^{3})}} = 8.41 M\Omega (6.57)
Note that Z_{\mathrm{in}} \approx -h_{fc}/h_{oc}.
(d) In parallel with (6.50),
Z_{o} = {\frac{v_{d p}}{i_{d p}}} = {\frac{1}{h_{o e} – h_{fe}h_{r e}/h_{i e}}} = {\frac{1}{12 \times 10^{-6} – (100)(1 \times 10^{-4})/(1 \times 10^{3})}} = 500\, Ω (6.50)
Z_{o} = {\frac{1}{h_{o c} – h_{fc}h_{r c}/h_{ic}}} = {\frac{1}{12 \times 10^{-6} – (-101)(1)/(1 \times 10^{3})}} = 9.9\, Ω
Note that Z_{o} \approx -h_{ic}/h_{fc}.
(e) Based on the typical values of this example, the characteristics of the CB amplifier can be summarized as follows:
1. High current gain
2. Voltage gain of approximately unity
3. Power gain approximately equal to current gain
4. No current or voltage phase shift
5. Large input impedance
6. Small output impedance