Question 6.2: In the CE amplifier of Fig. 6-5(b), let hie = 1 kΩ, hre = 10......
In the CE amplifier of Fig. 6-5(b), let h_{i e} = 1\,\mathrm{k\Omega}, h_{r e} = 10^{-4},h_{f e} = 100, h_{o e} = 12\,\mu\mathrm{S}, and R_{L}=2\,\mathrm{k\Omega}. (These are typical CE amplifier values.) 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 CE amplifier.
(a) By current division at node C,
i_{L} = {\frac{1/h_{o e}}{1/h_{o e} + R_{L}}}(-h_{f e}i_{b}) (6.42)
and 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)
Note that A_{i} \approx -h_{i e}, where the minus sign indicates a 180° phase shift between input and output currents.
(b) By KVL around B, E mesh,
v_{s}=v_{b e}=h_{i e}i_{b}+h_{r e}v_{c e} (6.44)
Ohm’s law applied to the output network requires that
v_{c e} = -h_{fe}i_{b}\left(\frac{1}{h_{o e}}\|R_{L}\right) = \frac{-h_{fe}R_{L}i_{b}}{1 + h_{o e}R_{L}} (6.45)
Solving (6.45) for i_b, substituting the result into (6.44), and rearranging yield
= -{\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)
Observe that A_{v} \approx -h_{fe}R_{L}/h_{ie}, where the minus sign indicates a 180° phase shift between input and output voltages.
(c) Substituting (6.45) into (6.44) and rearranging yield
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)
Note that for typical CE amplifier values, Z_{\mathrm{in}} \approx h_{i e}.
(d) We deactivate (short) v_s and replace R_L with a driving-point source so that v_{dp} = v_{ce}. Then, for the input mesh,
Ohm’s law requires that
i_{b} = -{\frac{h_{r e}}{h_{i e}}}\,v_{d p} (6.48)
However, at node C (with, now, i_{c} = i_{d p}), KCL yields
i_{c} = i_{d p} = h_{fe}i_{b} + h_{o c}v_{d p} (6.49)
Using (6.48) in (6.49) and rearranging then yield
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)
The output impedance is increased by feedback due to the presence of the controlled source h_{re}v_{ce}.
(e) Based on the typical values of this example, the characteristics of the CE amplifier can be summarized as follows:
1. Large current gain
2. Large voltage gain
3. Large power gain (A_iA_v)
4. Current and voltage phase shifts of 180°
5. Moderate input impedance
6. Moderate output impedance