Why Does the Output Waveform Change in Combination Clipper Circuits?

[ranju]

In clipper circuits the diode clips the input signal upto a particular voltage , say if a battery of V volts is connected in series with the diode then it'll clip it at voltage V..
in the attached fig.. there are 2 cases.. in both cases , the diode clips the input sinusoidal waveform at V volts..but why the shape of the output waveform is changed in case of combination clipper..?
in one case its sinusoidal & in other it has changed to a triangular waveform...!

 

[256bits]

With clipping, the less the clipped voltage is, the more the output waveform approaches square wave.
To obtain a triangular wave, one has to then integrate the output waveform.

 

[ranju]

but here we are'nt given any specific value of Voltage...!

 

[Averagesupernova]

The reason the 2nd pic looks different is because half of the components that are responsible for the actual clipping are missing.

 

[sophiecentaur]

In clipper circuits the diode clips the input signal upto a particular voltage , say if a battery of V volts is connected in series with the diode then it'll clip it at voltage V..
in the attached fig.. there are 2 cases.. in both cases , the diode clips the input sinusoidal waveform at V volts..but why the shape of the output waveform is changed in case of combination clipper..?
in one case its sinusoidal & in other it has changed to a triangular waveform...!

but here we are'nt given any specific value of Voltage...!

It's not triangular; it's just a badly drawn negative going half sinusoid section.
They don't need to give a specific battery voltage. It just needs the un-clipped waveform to have an excursion that takes it above and below the battery+diode volts each side. It's just qualitative.

 

[jim hardy]

but here we are'nt given any specific value of Voltage...!

Sure we are - V1. V2, and V.

Observe the clippers clip at that voltage.

Have you written a KVL equation around the right hand half of either circuit ?

 

[ranju]

Sure we are - V1. V2, and V.

Observe the clippers clip at that voltage.

Have you written a KVL equation around the right hand half of either circuit ?

but 256 bits..is saying that "the less the clipped voltage is, the more the output waveform approaches square"..but we are'nt given magnitude of the voltages..
what is the purpose opf writing those kvl equ...??

 

[jim hardy]

what is the purpose of writing those kvl equ...??

to make oneself realize that the circuit is just a linear voltage divider until a diode conducts.

Which explains "why the shape of the output waveform is changed in case of combination clipper..?"

 

[ranju]

I am not getting this thing..!how does it explain the change in the waveform??

 

[jim hardy]

In each of your attachments, ask yourself "What is the maximum positive and maximum negative voltage that can appear across R_L?
KVL will answer that.

 

[sophiecentaur]

I am not getting this thing..!how does it explain the change in the waveform??

What happens as soon as the volts across those diodes causes them to conduct? Where does any extra current go?

 

[256bits]

but 256 bits..is saying that "the less the clipped voltage is, the more the output waveform approaches square"..but we are'nt given magnitude of the voltages..
what is the purpose opf writing those kvl equ...??

With no clipper circuit, the output wave would be of the same shape as the input, ie a sine wave. The clipper components cut the top off the sine wave to give the wave from as shown.
You mentioned it looks triangular, but not so. The output resembles more a square wave as more and more of the top of a sine wave is cut off.

In the first picture, two clipper circuits are used. One which includes D1 and Va, cut off the sine wave on the positive cycle. The other clipper circuit, of D2 and V2, cut off the sine wave on the negative part of the cycle.

In the second picture, the clipper of D and V, cut off the positive cycle. The negative cycle retains the sine wave shape.

If you analyze a clipper circuit, such as in the second picture, as the input voltage increase from 0, the output voltage will follow along until the voltage reaches a value where the diode begins to conduct.

You can use KVL to determine when this will happen.

 

[jim hardy]

Hint:

When you write KVL, the voltage from cathode to anode of an ideal diode can be either zero or any negative number.

You can write KVL as an inequality.