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

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The voltage v(t) across diode D1 varies depending on the cycle of the waveform. During the positive cycle, diode D1 is short-circuited while capacitor 1 charges, and during the negative cycle, D1 opens as capacitor 2 charges. The discussion highlights that the first half acts as a clamper circuit, shifting the waveform negatively, while the second half functions as a peak rectifier. The exact mechanism of the peak rectifier involves the diode allowing current to pass only during the positive half of the waveform, effectively maintaining the peak voltage. Understanding the behavior of the diodes and capacitors in each quarter cycle is crucial for analyzing the waveform.
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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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