Clipper Circuits: Vi vs. Vr Understanding

[ranju]

In the given circuit. It was given that when Vi>Vr diode will be reverse biased or off..and when Vi<Vr diode will be forward biased... but I am not getting how we infer this..??It may sound a very basic question but its not clear...I am not able to decide it

 

[Jony130]

For Vin = 0V the diode is ON or OFF?

 

[ranju]

I have no idea about this.. actually I am not getting the use of the battery Vr..!

 

[Jony130]

Battery (Vr) is nothing more than a voltage source. You are not familiar with a voltage source?

As for the diode. Diode allows current to flow only in one direction (called the diode's forward direction), while blocking current in the opposite direction (the reverse direction).

Forward-biased diode (diode is ON)

Reverse-biased diode (diode is OFF)

 

[zoki85]

I_R isn't 0. It is just small.

 

[Jony130]

I_R isn't 0. It is just small.

Well yes, but I don't want to confuse the OP. And you can read from the first picture the reverse current is typically in nanoamps range

 

[ranju]

yes I know Vr is voltage source ,but here it is affecting the bias of diode whether forward or reverse..but how its being decided.

 

[Jony130]

If we set Vin = 0V, then you can replace Vin with a short wire.
So the circuit will looks like this:

And now can you tell me in which state the diode is, ON or OFF? And Vo is ?

 

[ranju]

It seems to be in reverse bias..!

 

[Jony130]

Yes, diode is reverse biased. But what about Vo voltage? Any clues?

 

[ranju]

its in rever biase ..so it'll be open , so output V0 will be Vr..>!

 

[ranju]

no..actually considering the polarity of battery , V0 will be -Vr.

 

[Jony130]

Yes, very good Vo = -vr. So if we assume that Vr = 2V then Vo = -2V

OK, so now let as try to find Vo if Vi is larger than 0V. For example Vi = 5V

The voltage at cathode is 5V and the voltage at anode is -2V. So voltage across the diode is equal to Vd = -2V - 5V = -7V.
And this means that diode is OFF(Reverse-biased) and Vo = -2V.

Now let as change the polarity of the Vi voltage source from +5V into -5V.

This time diode is forward biased because Vd is positive. Vd = Vo - Vi = -2V - (-5V) = 3V. And this is why the diode will conduct current.
And Vo = -4.3V.
So we can conclude that the diode will be ON only if input voltage (Vi) is 0.7V voltage lower than Vr voltage.
Diode is ON if Vin < -2.7V

 

[ranju]

but if we take Vi=1V then diode will be reverse biase even if Vi=-1V.. since Vd= -2-(-1)=-1V..so it will not conduct..!

 

[ranju]

So we can conclude that the diode will be ON only if input voltage (Vi) is 0.7V voltage lower than Vr voltage.
Diode is ON if Vin < -2.7V

I did'nt get this point..how can you generalise it.>??

 

[Jony130]

but if we take Vi=1V then diode will be reverse biase even if Vi=-1V.. since Vd= -2-(-1)=-1V..so it will not conduct..!

Yes exactly. Diode will conduct only if Vin is lower than -2.7V

I did'nt get this point..how can you generalise it.>??

What you don't understand here?
When diode conduct, the voltage drop across the diode is around 0.5V...0.7V.
This means that Vin voltage must be 0.7V lower than Vr voltage to force diode to conduct.
For Vr = 2V input voltage need to be lower than -2.7V

 

[ranju]

ohkk ..so for solving the clipper circuits.. we should proceed in this way only..?? I mean assuming values of unknowns ( Vr & Vin)..then find the bias of diode..??

 

[Jony130]

No it is not necessary, but this "method" help us visualize the situation much better.