Question 12.10: Determine the values of R1, RE, RB, C1 and C3 of a bootstrap......

To calculate\ R_{1}:
Applying KVL to collector loop of\ Q_{1}, we have the collector voltage of\ Q_{1} when it is ON:
\ V_{C1} = V_{CC} − V_{D1} − I_{1}R_{1}

= 12.73 kΩ

To calculate\ R_{B}:
Applying KVL to the base loop of\ Q_{1} we have \ V_{CC} − I_{B1}R_{B} − V_{BE(sat)} = 0

Choose,\ I_{B1} = 1.5I_{B1(min)} = 1.5 × \frac{I_{C1}}{h_{FE(min)}}= 1.5 × \frac{1.5 × 10^{−3}}{30}0.075 mA

Therefore,
\ R_{B} = \frac{20 − 0.7}{0.075 × 10^{−3}} = 257.3 kΩ

To calculate\ C_{1} :
Sweep Time\ T_{s} = R_{1}C_{1}
Therefore,

Given
\ T_{s} = 12 ms
\ C_{1} = \frac{12 × 10^{−3}}{12.73 × 10^{3}} = 0.943μF

The value of\ C_{3} should be larger than\ C_{1}, therefore, choose\ C_{3} as 10 times of\ C_{1}.
Therefore,
\ C_{3} = 9.43μF

To Calculate\ R_{E}:
The output voltage\ ν_{o} = V_{CE(sat)} − V_{BE(active)}
Therefore,
\ ν_{o} = 0.3 − 0.6 = −0.3

Applying KVL, to the input side of\ Q_{2} we have:
\ ν_{o} + V_{EE} − R_{E}i_{E2} = 0
Therefore,

\ R_{E} = \frac{ν_{o} + V_{EE}}{I_{E2}}= \frac{−0.3 + 12}{1.75 × 10^{−3}} = 6.68 kΩ