*I am trying to understand how they found out the equation when Case 1,2 finishe

Question

*I am trying to understand how they found out the equation when Case 1,2 finishes and Case 3 kicks in, when does D1 and D2 turn off? How do I find that out? Here the have it as 0.1 Dvi/dt + vi =0*

I dont want an answer just an explanation please for how to find the time or voltage when both Diode 1 and Diode 2 are off

Thanks

*I am trying to understand how they found out the equation when Case 1,2 finishes and Case 3 kicks in, when does D1 and D2 turn off? How do I find that out? Here the have it as 0.1 Dvi/dt + vi =0*

I dont want an answer just an explanation please for how to find the time or voltage when both Diode 1 and Diode 2 are off

Thanks

2] Compute and sketch Vs(t)-V4). The capacitors are initially discharged, the voltage source is turned on at t = 0, and the diodes, forward voltage drop is negligible (Vf= 0). What is the value of V (10s)- V4(10s)? Vi LDi v. RC: 0.1 s LD2

Explanation / Answer

In case 3, both diode is OFF, id1=0v. And V3 not equal to Vi, therefore by using ordinary differential equations we can find value of V3. id1 is the sum of capacitor current and resistor current. Similarly for diode D2 OFF diode current id2 is sum of resistor current and capacitor current and by using ordinary differential equation, we can find value of V4.

Now as both diode ON is not possible, therefore this condition is not feasible. Now when case 1, 2 finishes equation is determined by applying kvl and by putting value of vi we can find time t.

Note- please find me in conment box if you find any query.

I am not getting what is your exact query.

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