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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?
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Most Upvoted Answer
An asymmetrical periodic pulse train Vin of 10 V amplitude with on-tim...
Given circuit and asymmetrical periodic input Vin is shown below,
Here,TON=1msec, Toff=1μ sec and initially capacitor C is uncharged and diode D is ideal diode.
(i) For 0 < />< />ON :
Maximum value of Vin 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 VC=10 V and voltage V< =="" 0="" />
(ii) For TON < />< />OFF :
Given input voltage 0 Vin = 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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Community Answer
An asymmetrical periodic pulse train Vin of 10 V amplitude with on-tim...
Given circuit and asymmetrical periodic input Vin is shown below,
Here,TON=1msec, Toff=1μ sec and initially capacitor C is uncharged and diode D is ideal diode.
(i) For 0 < />< />ON :
Maximum value of Vin 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 VC=10 V and voltage V< =="" 0="" />
(ii) For TON < />< />OFF :
Given input voltage 0 Vin = 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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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?
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