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Q. With the help of circuit diagram explain a Miller Sweep generator
Figure given below shows the circuit of a Miller integrator or a sweep circuit. Transistor Q1 acts as a switch and transistor Q2 is a common - emitter amplifier. i.e. a high gain amplifier. Consider the case when Q1 is ON and Q2 is OFF. At this condition, the voltage across the capacitor C and the output voltage V0 is equal to Vcc.
When a negative pulse is applied to the base of Q1, the emitter - base junction of Q1 is reverse biased and hence Q1 is turned OFF. Thus, the collector voltage (Vc1) of Q1 increases which increases the bias to Q2 and as a result Q2 is turned ON. Since Q2 conducts, Vout begins to decrease. Because the capacitor is coupled to the base of transistor Q2, the rate of decrease of output voltage is controlled by rate of discharge of capacitor. The time constant of the discharge is given by td = RB2C.
As the value of time constant is very large, the discharge current practically remains constant. Hence, the run down of the collector voltage is linear. When the input pulse is removed, Q1 turns ON and Q2 turns OFF. The capacitor charges quickly to +Vcc through Rc with the time constant t =R C.
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Q. The block diagram of Figure represents a multi loop control system. (a) Determine the transfer function C(s)/R(s). (b) For G 2 G 3 H 2 = 1, evaluate C(s)/R(s).
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