Green is the waveform of v1(input) and blue is the waveform for voltage across R2.
During the positive cycle (i.e when v1 is greater than 0v) diode will be short circuit and R2 is parallel to the short circuit branch it,s voltage will also be 0
but in this blue waveform the voltage is more than 0v(peak=2.4v)
Can anyone explain why it is 2.4v not 0v during positive cycle???

 

Diodes have a forward voltage drop that is current and temperature dependent. You are driving 6 amps thru D1. The datasheet Fig. 1 shows forward voltage drop of 1.6V at 1 amp.

1N914 datasheet https://www.vishay.com/docs/85622/1n914.pdf

 

Ignore the V2 voltage as it is 0v

 

look at the datasheet... 1N914 is not an ideal diode...



In the real world 6A would kill the diode..

 

If you want an ideal diode add this line to your simulation:

.model idealDiode D(ron=0 roff=1G Vfwd=0)

and label your diode " idealDiode"

 

deleted reply

 

If you want an ideal diode add this line to your simulation:

.model idealDiode D(ron=0 roff=1G Vfwd=0)

and label your diode " idealDiode"

View attachment 322792

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You're so helpful this cleared my mind but ive one more doubt why is that it is that in the practical waveform's peak value remains 2.4v.Does that mean the voltage remains constant across the diode (no more than 2.4v for 5v peak input)?? This doubt might be silly but i really wanna know what tspice did for its calculations.

 

deleted reply

 

Please use a thread title that provides some information on the nature of the thread.
"need help!!!!" is of no use to anyone.

 

Please use a thread title that provides some information on the nature of the thread.
"need help!!!!" is of no use to anyone.

ikr, im new to this i dont even know how to delete replies (crying emoji)

 

@
You're so helpful this cleared my mind but ive one more doubt why is that it is that in the practical waveform's peak value remains 2.4v.Does that mean the voltage remains constant across the diode (no more than 2.4v for 5v peak input)?? This doubt might be silly but i really wanna know what tspice did for its calculations.

When the diode is non-conducting the output voltage is 0.5 x input, due to the voltage divider R1:R2, Vout = R2/(R1+R2) * Vin = 2.5v peak

 

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View attachment 322796

and when the diode is conducting , the output voltage is 2.3v peak(practically) because it remains constant at 2.3 peak because of input characteristics of the diode,right?

 

and when the diode is conducting , the output voltage is 2.3v peak(practically) because it remains constant at 2.3 peak because of input characteristics of the diode,right?

It will be whatever the forward voltage of the diode is at the current through the diode. Though the simulation is inaccurate because the diode is only rated for 300mA max, typically 100mA. If you changed the resistors to, say, 4k that would be a fairy accurate simulation.