Question 12.5: Design a transistor bootstrap sweep generator to provide an ......

Refer to the bootstrap circuit shown in Fig. 12.16(a).
\ R_{L} = R_{E} = 1 kΩ

When\ V_{o} = 0,
\ I_{E2} = \frac{0 − (−V_{EE})}{R_{E}}  = \frac{18  V}{1  k Ω} = 18 mA

When\ V_{o} = V_{s}(= 10V),
\ I_{E2} = \frac{10 − (−V_{EE})}{R_{E}} = \frac{28  V}{1  k Ω} = 28 mA

At\ V_{o} = 0,
\ I_{B2} = \frac{I_{E2}}{h_{FE} } = \frac{18  mA}{100} = 0.18 mA

At\ V_{o} = V_{s},
\ I_{B2} = \frac{I_{E2}}{h_{FE}} = \frac{28}{100} = 0.28 mA      \ ΔI_{B2} = 0.28 − 0.18 = 0.10 mA

\ I_{1} is much larger than\ I_{B2}
Let

\ I_{1} = 100 × ΔI_{B2} = 100 × 0.10 mA = 10 mA            \ C_{1} = \frac{I_{1}T_{s}}{V_{s}} = \frac{10 × 10^{−3} × 1 × 10^{−3}}{10V} = 1 μF

\ V_{R1} = V_{CC} − V_{D1} − V_{CE}(sat) = 18 − 0.7 − 0.2 = 17.1 V      \ R_{1} = \frac{V_{R1}}{I_{1}} = \frac{17.1}{10 × 10^{−3}} = 1.71 kΩ

For 1 per cent non-linearity due to discharge of\ C_{3}:
\ ΔV_{C3} = 1 per cent of the initial\ V_{CC} level
Initial
\ V_{C3} = V_{CC} = 18 V        \ ΔV_{C3} = \frac{18 × 1}{100} = 0.18 V

And\ C_{3} discharge current\ I_{1} is equal to 10 mA. Therefore,
\ C_{3} = \frac{I_{1}T_{(spacing)}}{ΔV_{C3}} = \frac{10 × 10^{−3} × 1 × 10^{−3}}{0.18 V }= 55 μF

The discharge time of\ C_{1} is 0.1ms which is 1/10 th of the charging time. For\ Q_{1} to discharge\ C_{1} in 1/10 th of the charging time,
\ I_{C1} = 10 × (C_{1} charging current) = 10 × 10 mA = 100 mA

\ I_{B1} = \frac{I_{C1}}{h_{FE}}= \frac{100  mA}{100} = 1 mA        \ R_{B} = \frac{V_{CC} − V_{BE1}}{I_{B1}} = \frac{18 − 0.7}{1  mA} = \frac{17.3  V}{1  mA} = 17.3 kΩ

Choose\ R_{B} = 20 kΩ,\ Q_{1} is to be biased OFF at the end of the input pulse.
\ |ΔV_{B}| = ν_{o} − (pulse amplitude) = 0.7 − 5 V = −4.3 V

The charging current of\ C_{2} is equal to the current through\ R_{B} when\ Q_{1} is OFF.

\ I = \frac{V_{CC} − V_{i}}{R_{B}} = \frac{18 − (−5)}{20  kΩ} = \frac{23  V}{20  kΩ} = 1.15 mA

\ C_{2} = \frac{I × t}{ΔV_{B}} = \frac{1.15  mA × 1  ms}{4.3  V }= 0.27μF