Question 12.4: The transistor bootstrap circuit in Fig. 12.16(a) has the fo......

Referring to the circuit in Fig. 12.16(a):

(a)   Sweep speed =\frac{ I_{1}}{C_{1}}= \frac{V_{CC}}{R_{1}C_{1}} = \frac{15}{10 × 10^{3} × 0.005 × 10^{−6}} =  3 × 10^{5}V/s

i.e.,
\ 15 = (3 × 10^{5})T_{s}
Therefore, sweep time,
\ T_{s} = \frac{15}{3 × 10^{5}} = 50 μs

(b) At the end of the input pulse,\ Q_{1} once again goes into saturation.

\ i_{B1} = \frac{V_{CC}}{R_{B}}= \frac{15}{30 × 10^{3}} = 0.5 mA        \ i_{C1} = h_{FE}i_{B1} = 50 × 0.5 = 25 mA

The retrace time\ T_{r} is,

\ T_{r} = \frac{\frac{V_{s}}{V_{CC}}C_{1}}{\frac{h_{FE}}{R_{B}}− \frac{1}{R_{1}}}= \frac{\frac{15}{15} × 5 × 10^{−9}}{\frac{50}{30 × 10^{3}} − \frac{1}{10 × 10^{3}}}= 3.18 μs

\ T = T_{g} + T_{r} = (60 + 3.18) = 63.18 μs

Recovery time\ T_{1} = \frac{V_{CC}}{V_{EE}} × \frac{R_{E}}{R_{1}} × T = \frac{15}{10} × \frac{5}{10} × 63.18 × 10^{−6} = 47.385 μs

(c) To find the slope error:
The current gain of the emitter follower is given by:

Therefore,
\ A_{I} = \frac{1 + h_{fe}}{1 + h_{oe}R_{E}}= \frac{1 + 50}{1 + \frac{1}{40} × 5}= \frac{51}{1.5} = 45.33

Input impedance of the emitter follower is given by:
\ R_{i} = h_{ie} + A_{I} R_{E}
Therefore,

A is the voltage gain of the emitter follower.
\ R_{i} = h_{ie} + A_{I} R_{E} = 1kΩ + 45.33 × 5kΩ = 227.67 kΩ

\ 1 − A = \frac{h_{ie}}{R_{i}}= \frac{1}{227.67} = 0.00439

The slope error,

\ e_{s} = \left(1 − A + \frac{R_{1}}{R_{i}} \right) \frac{V_{s}}{V_{CC}}= \left[0.00439 + \frac{10}{227.67} \right] \frac{15}{15 }= 0.0483 = 4.83%

(d) Using the above calculations, the waveforms can be sketched as shown in Fig. 12.19.