Question 6.6: For the amplifier of Example 6.5, use SPICE methods to deter......
For the amplifier of Example 6.5, use SPICE methods to determine (a) the input impedance Z_{in}, (b) the current gain A_i, and (c) the voltage gain A_v.
Netlist code that describes the amplifier circuit follows:
| Ex6_6.CIR vi 1 0 AC 1V Ri 1 2 100ohm CC1 2 3 1000uF CC2 4 7 1000uF R1 3 0 6kohm R2 3 6 50kohm RC 6 4 1kohm RE 5 0 100ohm RL 7 0 1kohm VCC 6 0 15V Q 4 3 5 QNPNG .MODEL QNPNG NPN(Is=10fA Ikf=150mA Isc=10fA Bf=150 + Br=3 Rb=1ohm Rc=1ohm Va=75V Cjc=10pF Cje=15pF) .AC LIN 1 100Hz 100Hz .PRINT AC Vm(1) Vp(1) Vm(7) Vp(7) .PRINT AC Im(Ri) Ip(Ri) Im(RL) IP(RL) .END |
(a) Execute 〈Ex6_6.CIR〉 and poll the output file to find the values of input voltage and current. Thus,
Z_{\mathrm{in}} = {\frac{\mathrm{V}(1)}{\mathrm{I(\mathrm{vi})}}} = {\frac{1}{2.465 \times 10^{-4}}} = 4.056\,{\mathrm{k}}\Omega
(b) The output file contains the magnitudes and phase angles of the input and output voltages. Hence,
A_{v} = -{\frac{\mathrm{Vm}(7)}{\mathrm{Vm}(1)}} = -{\frac{4.649}{1}} = -4.649
The negative sign accounts for the 180° phase shift [see Vp(7)] of V(7) with respect to V(1).
(c) The output file values of Ip(Ri) and Ip(RL) show the two signals to be 1808 out of phase. The current gain is found as
A_{i} = -{\frac{\mathrm{Im}(\mathrm{RL})}{\mathrm{Im}(\mathrm{Ri})}} = -{\frac{4.649 \times 10^{-3}}{2.465 \times 10^{-4}}} = -18.86