An asymmetrical periodic pulse train Vin of 10 V amplitude with on-time TON = 1ms and off-time TOFF = 1 μs is applied to the circuit shown in figure. The diode D1 is ideal.The difference between the maximum voltage and minimum voltage of the output wave form V0 (in integer) is _____V.Correct answer is '10'. Can you explain this answer?

Question

An asymmetrical periodic pulse train Vin of 10 V amplitude with on-time TON = 1ms and off-time TOFF = 1 &mu;s is applied to the circuit shown in figure. The diode D1 is ideal.The difference between the maximum voltage and minimum voltage of the output wave form V0 (in integer) is _____V.Correct answer is '10'. Can you explain this answer?

Answer

Given circuit and asymmetrical periodic input V_in is shown below,

Here,T_ON=1msec, T_off=1μ sec and initially capacitor C is uncharged and diode D is ideal diode.

(i) For 0 < />< />_ON :

Maximum value of V_in is 10 V and diode D is ON and will be short circuit, so capacitor will get charged instantly to 10 V as shown below,

Thus, capacitor voltage V_C=10 V and voltage V< =="" 0="" />

(ii) For T_ON < />< />_OFF :

Given input voltage 0 V_in = it means short circuit and diode is OFF and will be open circuit, so circuit becomes as,

Current, I=-10/500=-20mA

Voltage, V₀=I*500=(-20)*500=-10V

Thus output waveform under steady state is,

Here,V₀(maximum)=0V,V₀(minimum)=-10V

Thus, difference between maximum and minimum output voltages,

V₀ (maximum) −V₀(minimum)= 0 − (−10) = +10 V

Hence, the difference between the maximum voltage and minimum voltage of the output wave form V₀is + 10 V.

Question_Type: 4

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