Diode Voltage Across D1: Solving for v(t) in Positive and Negative Cycles

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SUMMARY

The voltage v(t) across diode D1 is determined by analyzing the behavior of the circuit during positive and negative cycles. In the positive cycle, diode D2 acts as an open circuit, allowing capacitor 1 to charge while D1 is short-circuited. Conversely, in the negative cycle, capacitor 2 charges, D2 becomes short-circuited, and D1 opens, resulting in the voltage across capacitor 2 being applied against D2. The discussion clarifies that the first half of the waveform functions as a clamper circuit, while the second half operates as a peak rectifier circuit.

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Homework Statement



What will be the voltage v(t) across diode D1 ?


Homework Equations



The Figure is shown in the attachment.


The Attempt at a Solution



When there is a positive cycle D2 will be open circuit and capacitor 1 will be charged and diode D1 will be short circuited ,while in negative cycle the capacitor 2 will be charged D2 will be short circuited and D1 will be open so all the voltage across Capacitor 2 will be shown against D2 i guess but i am confused with waveform !
 

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lazy,

Determine what the diode is doing during every quarter cycle of the wave. Don't forget that capacitors accumulate voltage.

Ratch
 
What i am able to understand is that 1 st half is a clamper circuit and second half is peak rectifier circuit but i am confused about the waveform , for clamping waveform will be shifted to negative axis but what will happen for "peak rectifier" ? what is the exact mechanism of peak rectifier ?

I am considering each capacitor and diode to be ideal !
 

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