Operation of diode Rectifier

Consider these equations for diode rectifier;

I(V) = Is(e^(alpha*V) -1) -----------(1)
Let diode voltage be V=Vo+v --------------(2)

Then, Using taylor series for the diode current I(V) [equation (1)] we have;
; where Io=I(Vo)

; where Gd is the dynamic conductance (Gd=1/Rj, where Rj is the juction resistance at Vo)

Then for diode rectifier we have;
----------(3)
then diode current;


Therefore we can say is the rectified current. Using a low pass filter we can simply get rid of the higher
frequency components

[Microwave Engineering 2nd Ed, David M. Pozar, paged 560-563]

What I cannot understand is these equations are derived from the simple assumption that Input Voltage [equation (3)] is across the diode. But in a
rectifier circuit like this [figure below] will not have Vin across the diode. Then how can I use these equations in this situation?

bhathi123 said:

What I cannot understand is these equations are derived from the simple assumption that Input Voltage [equation (3)] is across the diode. But in a
rectifier circuit like this [figure below] will not have Vin across the diode. Then how can I use these equations in this situation?
View attachment 72497

Click to expand...

Why do you say it wont have a Vin ? it shows a Vin.
In this case its inferring a DC voltage as there is a + and - rail, so nothing is being rectified and you will just loose ~ 0.7 V across the diode
A diode is often used that way for reverse polarity protection of a piece of equipment

If instead you had an AC Vin, then you will have a half wave rectifier and again still a ~ 0.7V drop across the diode

Dave

Vin is a AC small signal with very high frequency. There will be a ~0.7 drop across the diode if the diode is idle. In a practical situation the drop will vary. I think that is the basis for the derivation given in the book.

Here voltage across diode will be; V=Vin-Vout [V=Vo+voCos(wt) - Vout]