Question 6.SP.3: The transistor of a CE amplifier can be modeled with the tee......
The transistor of a CE amplifier can be modeled with the tee-equivalent circuit of Fig. 6-3 if the base and emitter terminals are interchanged, as shown by Fig. 6-10(a); however, the controlled source is no longer given in terms of a port current—an analytical disadvantage. Show that the circuits of Fig. 6-10(b) and (c), where the controlled variable of the dependent source is the input current i_b, can be obtained by application of Thévenin’s and Norton’s theorems to the circuit of Fig. 6-10(a).
The Thévenin equivalent for the circuit above terminals 1,2 of Fig. 6-10(a) has
v_{th} = \alpha r_{c}ie\quad\quad Z_{th} = r_{c}
By KCL, i_e = i_c + i_b, so that
v_{th} = \alpha r_{c}i_c + \alpha r_{c}i_b (1)
We recognize that if the Thévenin elements are placed in the network, the first term on the right side of (1) must be modeled by using a ‘‘negative resistance.’’ The second term represents a controlled voltage source. Thus, a modified Thévenin equivalent can be introduced, in which the ‘‘negative resistance’’ is combined with Z_{th} to give
v_{th}^{\prime} = \alpha r_{c}i_{b} = r_{m}i_{b}\quad\quad Z_{th}^{\prime} = (1 – \alpha)r_{c} (2)
With the modified Thévenin elements of (2) in position, we obtain Fig. 6-10(b).
The elements of the Norton equivalent circuit can be determined directly from (2) as
Z_{N} = {\frac{1}{Y_{N}}} = Z_{th}^{\prime} = (1 – \alpha)r_{c}\qquad I_{N} = {\frac{v_{th}^{\prime}}{Z_{t h}^{\prime}}} = {\frac{\alpha r_{c}i_{b}}{(1 – \alpha)r_{c}}} = \beta i_{b} (3)
The elements of (3) give the circuit of Fig. 6-10(c).