Question 12.3: For the Miller’s sweep shown in Fig. 12.12(a), VCC = 25 V, R......

(a)
\ V_{s} = \frac{V_{CC}}{R_{C1}C_{s}} × T_{s}

We have\ V_{s} = V_{CC}

Therefore,
\ T_{s} = R_{C1}C_{s}            \ C_{s} = \frac{T_{s}}{R_{C1}}= \frac{5 × 10^{−3}}{10 × 10^{3}} = 0.5μF

(b)   Retrace time\ T_{r} = R_{C2} × C_{s} = 5 × 10^{3} × 0.5 × 10^{−6} = 2.5 ms
(c)
\ A_{I} = \frac{−h_{fe}}{1+ hoeR_{C2}} = \frac{−80}{1 + \frac{5}{40}}= −71.11

\ R_{i} = h_{ie} + h_{re}A_{I}R_{C2} = 1 + (2.5 × 10^{−4})(−71.11)(5) = (1 − 0.0889) = 0.91 kΩ
\ A = A_{I}\frac{R_{C2}}{R_{i}}= −71.11 × \frac{5}{0.91} = −390.71

\ e_{s(Miller)} = \frac{V_{s}}{V_{CC}} × \frac{1}{|A|} × \left(1 + \frac{R_{C1}}{R_{i}}\right)= \frac{25}{25} × \frac{1}{390.71} × \left(1 + \frac{10}{0.91}\right)= 0.0307 = 3.07 %.