Question about the term "Vf*Ron*Goff" in the series voltage in diode block from Simscape Electrical.

2 views (last 30 days)
Hi, everyone,
The diode block has the following explanation in its Help file:
So, according this explanation, the series voltage source is:
V = Vf * (1 − Ron*Goff)
V = Vf − Vf*Ron*Goff
V = Vf − (Vf*Goff)*Ron
V = Vf − (Ioff)*Ron
I don't remember seeing this second voltage component (Vf*Ron*Goff) anywhere else.
And, altough it is said that this second term is to ensure zero current when zero voltage is applied across the diode, I could not understand and draw an equivalent electrical circuit that shows this afirmative in a clear and easy way.
Can anyone help me understand this (Vf*Ron*Goff) voltage component?
Thank you.

Accepted Answer

David John
David John on 16 Dec 2021
In the off-state, the diode equation is:
v = i/Goff
In the on-state, the equation is:
v = i*Ron + Vsource
where Vsource is like a series voltage source. The transition between the two states occurs at v=Vf. In order for the current to be continuous at Vf, we must have Vf*Goff=(Vf-Vsource)/Ron. This implies that Vsource=Vf*(1-Ron*Goff).
  2 Comments
Marcelo
Marcelo on 16 Dec 2021
Edited: Marcelo on 16 Dec 2021
Hi, David,
Thank you for your answer.
I understood your explanation about using the term "Vf*Ron*Goff" in order to not have a discontinuity in diode current.
But, about what it is said in Help file ("The Ron*Goff term ensures that the diode current is exactly zero when the voltage across it is zero"), I still do not understand.
When the voltage across the diode is zero, the diode is backward biased, as voltage is less than Vf. Then:
v = i/Goff
i = v*Goff
i = 0 when v = 0
So, the term "Ron*Goff" does not make any contribution when the voltage across the diode is zero.
Your explanation was really good and answered my question. But the justificative in Help file seems not right for me. It would be better if it was:
The Ron*Goff term ensures that the diode current has no discontinuity when the voltage across it is equal Vf.
David John
David John on 4 Jan 2022
Thanks for this. I'll feed this back to the library team to see about updating the documentation to make it more clear.

Sign in to comment.

More Answers (0)

Products


Release

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!