Question 12.9: The bootstrap sweep circuit is shown in Fig.12.22. A square ......
The bootstrap sweep circuit is shown in Fig.12.22. A square wave whose amplitude varies between 0 and −4 V and duration 0.5 ms is applied as a trigger. a) Calculate all the quiescent state currents and voltages. b) Determine the sweep amplitude, sweep time and sweep frequency. Assume\ h_{FE(min)} = 30,\ V_{CE(sat)} = 0.3 V,\ V_{BE(sat)} = 0.7 V,\ V_{BE(active)} = V_{D} = 0.6 V
(a) Current through\ R_{1} is:
\ I_{1} = \frac{V_{CC} − V_{D} − V_{CE}(sat)}{R_{1}} = \frac{18 − 0.6 − 0.3}{15 × 10^{3} } = 1.14 mA
Base Current of\ Q_{1} is:
\ i_{B1} = \frac{V_{CC} − V_{BE}(sat)}{R_{B}} = \frac{18 − 0.7}{150 kΩ} = 116 μA
Emitter current of\ Q_{2} = i_{E2} = \frac{ν_{o} + V_{EE}}{R_{E}}
\ ν_{o} = V_{C1} − V_{BE2}(active)
\ ν_{o} = V_{CE}(sat) − V_{BE2}(active) = 0.3 − 0.6 = −0.3 V
Therefore,
\ i_{E2} = \frac{−0.3 + 10}{10 × 10^{3}} = 0.97 mA
Therefore,
\ i_{C2} = \frac{i_{E2}}{1 + \frac{1}{ h_{FE}}} = \frac{0.97 × 10^{−3}}{ 1 + \frac{1}{30}}= 0.938 mA
\ i_{B2} = i_{E2} − i_{C2} = 0.97 × 10^{−3} − 0.938 × 10^{−3} = 0.032 mA
\ i_{C} = I_{1} − i_{B2} = 1.14 × 10^{−3} − 0.032 × 10^{−3} = 1.108 mA
\ V_{CE2} = V_{CC} − ν_{o} = 18 + 0.3 = 18.3 V
(b) Sweep time\ T_{s} = R_{1}C_{1} = 15 × 10^{3} × 0.01 × 10^{−6} = 0.15 ms
Therefore,
\ T_{g} = 0.5 ms
Since
\ T_{s} < T_{g} , \ V_{s} = V_{CC} = 18 V
Return time,\ T_{r} = \frac{\frac{ C_{1}V_{S}}{V_{CC}}}{\frac{h_{FE}}{R_{B}} − \frac{1}{ R_{1}}} = \frac{\frac{0.01 × 10^{−6} × 18}{ 18}}{\frac{30}{150 × 10^{3}} − \frac{1}{15 × 10^{3}}} = 0.075 ms
\ T = T_{g} + T_{r} = 0.5 + 0.075 = 0.575 ms
\ f = \frac{1}{T} = \frac{1000}{0.575} = 1.74 kHz