Solving KVL Equations for Clamper Circuits

[ranju]

In the attached circuit , I I am confused about the corresponding KVL equations which will be used to draw the waveform ..this is what I am doing ..please see if I am going right or not..??
when Vi >0 , the diode will be reverse biased so , the diode will be open , so the equation should be -Vi+Vc+Vo =0 , Vo=Vi-Vc & since the capacitor will be fully charged so Vc=Vm..! then Vo =Vi-Vm.>!

And when Vi<0 , it is forward biased and diode is short-circuited.. in that case would'nt the V0 zero..?

 

[rude man]

You don't write KVL equations for this circuit. You have to reason it out.

Start with C discharged and Vin = Vsin(wt) = 0. So as Vin goes up, does C charge? What does the cathode (output) voltage Vd do?
Then, Vin goes back to zero. Did C charge? Discharge? Where is Vd now?
Then, Vin goes negative. Does C charge now? What does Vd do?
Then Vd hits negative max Vin = -V, where is Vd now?
Then, Vin goes in the + direction. Does C charge or discharge? What does Vd do?
What does Vd do after that until t = infinity? Does C charge or discharge at any time?

 

[ranju]

if Vin>0 , C charge with positive potential & diode will be off & open-circuited.. & if negative it charges with positive potential..diode will conduct & will be short-circuited..the polarity of signal will decide whether diode will conduct or not..!
If Vin is goes to zero then C will discharge.>! what is Vd. is it potential across diode? how we will infer Vd..?? Can you please explain..??

 

[rude man]

if Vin>0 , C charge with positive potential & diode will be off & open-circuited..

How can C charge when there can be no current flowing through the diode from cathode to anode?
Hint: nothing happens until Vin goes negative.

If Vin is goes to zero then C will discharge.>!

How can C discharge? Current can only flow in one direction! C can only charge, not discharge.

 

[ranju]

nothing happens until Vin goes negative.

so all you mean that when Vi>0 , since diode is non-conducting & hence the terminals are open & so no current..& hence no charging of capacitor..!
and when Vi<0 ..the capacitor charges with negative potential upto a maximum value (peak) ' -Vm' ..
Is that so..?? what happens then??

 

[rude man]

so all you mean that when Vi>0 , since diode is non-conducting & hence the terminals are open & so no current..& hence no charging of capacitor..!
and when Vi<0 ..the capacitor charges with negative potential upto a maximum value (peak) ' -Vm' ..
Is that so..??

Right. Assuming an ideal diode. If not, the capacitor charges to Vm - VD.

what happens then??

What do you think can happen after the capacitor is fully charged to Vm? Does anything happen when Vin goes positive again? Does it get additional charge when Vin goes negative again?

 

[ranju]

nothing would happen again it'll be open-circuited when Vi becomes positive...! and when it again goes negative ..I guess it'll be charged further i.e., 2Vm..! Is it..??

 

[rude man]

nothing would happen again it'll be open-circuited when Vi becomes positive...!

Correct.

and when it again goes negative ..I guess it'll be charged further i.e., 2Vm..! Is it..??[/QUOTE]

Bad guess. When the input voltage goes back to +Vm, what is the voltage at the diode?

 

[ranju]

As I said that nothing would happen so the voltage will remain the same in positive cycle... but what happens when it goes to negative then?

 

[rude man]

As I said that nothing would happen so the voltage will remain the same in positive cycle... but what happens when it goes to negative then?

What is the cathode voltage when Vin is +Vm?

 

[ranju]

why are we considering cathode voltage.>?? In positive cycle diode is off.. so what is the use of this?

 

[rude man]

why are we considering cathode voltage.>?? In positive cycle diode is off.. so what is the use of this?

I have my reasons as you will see.

The input has gone 0 to +Vm, back to 0, to -Vm, and now went back to + Vm for the second time. What is the cathode voltage?

 

[ranju]

when it goes to +Vm , the diode is open so there will be no charging..! so cathode will be charged to -Vm only..! & that will be cathode voltage (-Vm)

 

[rude man]

when it goes to +Vm , the diode is open so there will be no charging..! so cathode will be charged to -Vm only..! & that will be cathode voltage (-Vm)

When you got to Vin = -Vm the first time, the capacitor voltage was Vcath - Vin = Vm. The Vin side of C was at -Vm and the cathode side was at Vcath = 0. Then, when you got to Vin = +Vm the second time, there was no current flowing any time since you left Vin = -Vm so there was no change in the capacitor voltage. So the Vin end is now at +Vm and the charge Q = C(Vcath -Vin) hasn't changed since Vin = - Vm, so what must be the cathode voltage when Vin = +Vm?

 

[rude man]

when it goes to +Vm , the diode is open so there will be no charging..! so cathode will be charged to -Vm only..! & that will be cathode voltage (-Vm)

How can the cathode voltage be negative? The diode clamps the voltage to Vcath > 0.

 

[ranju]

I am not getting this cathode voltage thing..! what do yo mean by cathode voltage?

 

[rude man]

I am not getting this cathode voltage thing..! what do yo mean by cathode voltage?

The voltage on the cathode.

 

[ranju]

Can't you clear all this in one time only.. I am getting confused now asking queries again and again..?