Can someone please explain why ideal voltage/current sources are short circuit for voltage source and open circuit for open circuit in small signal analysis? Any mathematical proof to this?
I am wondering why the law of mass action in semiconductors is this relation: n*p=ni^2. Any proof of this why is it not n*p=ni^3 or something...Also why doesn't dopants affect this relationship since doesn't dopants increase the overall number of carriers?
Q=CV for a capacitor.
The Q is for the amount of charge on one plate right(for parallel capacitor case)? So is the total charge 2Q on the capacitor?
Does it always have +Q on one plate and -Q on the other? Will it ever happen when it's +Q on one plate and not -Q on the other or vice...
What you have said makes a lot of sense. However do you know why Va is the only parameter that appears to the "outside" world? Why doesn't the Phi(built-in) kind of cancel the Va since Phi is the drop across the deletion region under low injection conditions.
Let me know what you think...
I believe it's the top band that's the conduction band and bottom band is valance band.
I see. If Va is the voltage drop across the diode(like when you solve a circuit problem you assume 0.7V drop) then why do they define the Phi variable as the drop across the semiconductor? I think that's...
Thank you for replying and writing such a detailed explanation. I just have a couple of questions that I still don't get.
I'm not sure what this means. The diagram indicates for the electrons to wander to the p side it's a higher potential. And for the holes to wander to the n side it's a...
Hmm my semiconductor book defines as
Phi = Phi(built-in)-Va(voltage applied)
And Va as the voltage applied. If it is the way u described shouldn't it be Phi=Va-Phi(built-in) and if Phi > 0 then it's forward biased?
The book also defines when Va is about Phi(built-in) then it's considered high...
For a PN junction diode, the total potential across the semiconductor equals the built-in potential minus the applied voltage.
Phi = Phi(built-in)-Va(voltage applied)
My question is what happens when the Va exceeds the Phi(built-in)? Will the Phi = negative? But how come they always...