Question 5.22: For the network of Fig. 5.118 , determine the following para...
For the network of Fig. 5.118 , determine the following parameters using the complete hybrid equivalent model and compare to the results obtained using the approximate model.
a. Z_i and Z_{i}^{\prime}.
b. A_{v} .
c. A_i = I_o/I_i .
d. Z_{o}^{\prime} (within R_C ) and Z_o (including R_C ).
Now that the basic equations for each quantity have been derived, the order in which they are calculated is arbitrary. However, the input impedance is often a useful quantity to know, and therefore will be calculated first. The complete common-emitter hybrid equivalent circuit has been substituted and the network redrawn as shown in Fig. 5.119 . A Thévenin equivalent circuit for the input section of Fig. 5.119 results in the input equivalent of Fig. 5.120 because E_{ Th } \cong V_{s} and R_{ Th } \cong R_{s}=1 k \Omega (a result of R_{B}=470 k \Omega being much greater than R_{s}=1 k \Omega). In this example, R_L = R_C, and I_o is defined as the current through R_C as in previous examples of this chapter. The output impedance Z_o as defined by Eq. (5.170) is for the output transistor terminals only. It does not include the effects of R_C . Z_o is simply the parallel combination of Z_o and R_L . The resulting configuration of Fig. 5.120 is then an exact duplicate of the defining network of Fig. 5.117 , and the equations derived above can be applied.
a. Eq. (5.169):
Z_{i}=\frac{V_{i}}{I_{i}}=h_{i}-\frac{h_{f} h_{r} R_{L}}{1+h_{o} R_{L}}Z_{i}=\frac{V_{i}}{I_{i}}=h_{i e}-\frac{h_{f e} h_{r e} R_{L}}{1+h_{o e} R_{L}}
=1.6 k \Omega-\frac{(110)\left(2 \times 10^{-4}\right)(4.7 k \Omega)}{1+(20 \mu S )(4.7 k \Omega)}
= 1.6 k Ω- 94.52 Ω
= 1.51 k Ω
versus 1.6 kΩ using simplyh_{i e}; and
Z_{i}^{\prime}=470 k \Omega \| Z_{i} \cong Z_{i}=1.51 k \Omegab. Eq. (5.168):
A_{v}=\frac{V_{o}}{V_{i}}=\frac{-h_{fe} R_{L}}{h_{i}+\left(h_{i} h_{o}-h_{f} h_{r}\right) R_{L}}=\frac{-(110)(4.7 k \Omega)}{1.6 k \Omega+\left[(1.6 k \Omega)(20 \mu S )-(110)\left(2 \times 10^{-4}\right)\right] 4.7 k \Omega}
=\frac{-517 \times 10^{3} \Omega}{1.6 k \Omega+(0.032-0.022) 4.7 k \Omega}
=\frac{-517 \times 10^{3} \Omega}{1.6 k \Omega+47 \Omega}
= – 313.9
versus -323.125 using A_{v} \cong-h_{f e} R_{L} / h_{i e}.
c. Eq. (5.167):
A_{i}=\frac{I_{o}}{I_{i}}=\frac{h_{f}}{1+h_{o} R_{L}}A_{i}^{\prime}=\frac{I_{o}}{I_{i}^{\prime}}=\frac{h_{f e}}{1+h_{o e} R_{L}}=\frac{110}{1+(20 \mu S)(4.7 k \Omega)}
=\frac{110}{1+0.094}= 1 0 0 . 5 5
versus 110 using simply h_{f e}. Because 470 k \Omega \gg Z_{i}^{\prime}, I_{i} \cong I_{i}^{\prime} and A_{i} \cong 100.55 also.
d. Eq. (5.170):
Z_{o}=\frac{V_{o}}{I_{o}}=\frac{1}{h_{o}-\left[h_{f} h_{r} /\left(h_{i}+R_{s}\right)\right]} (5.170)
Z_{o}^{\prime}=\frac{V_{o}}{I_{o}}=\frac{1}{h_{o e}-\left[h_{f e} h_{r e} /\left(h_{i e}+R_{s}\right)\right]}
=\frac{1}{20 \mu S -\left[(110)\left(2 \times 10^{-4}\right) /(1.6 k \Omega+1 k \Omega)\right]}
=\frac{1}{20 \mu S-8.46 \mu S}
=\frac{1}{11.54 \mu S }
= 86.66 kΩ
which is greater than the value determined from 1 / h_{o e}, 50 k \Omega; and
Z_{o}=R_{C}\left\|Z_{o}^{\prime}=4.7 k \Omega\right\| 86.66 k \Omega=4.46 k \Omegaversus 4.7 kΩ using only R_C .
