# pn = ni^2 exp(Va/Vt)

Discussion in 'Electronic Basics' started by PZ, Feb 22, 2006.

1. ### PZGuest

there is such an equation pn = ni^2 exp(Va/Vt) in junction given in
textbook. I am wondering how true it is against other equations.
suppose Va is zero,

pn = ni^2 exp(Va/Vt) => pn = ni^2

at same time p is p(0)= Ni^2/Nd, and n(0) = Ni^2/Na,

so p(0) x n(0) = Ni^4/NaNd, and is supposed to equal ni^2, that would

Ni^4/NaNd = Ni^2 => NaNd = Ni^2.

But this Na and Nd, they are the dopant of p and n junction, their
product doesn't equal to Ni^2.

Any idea where is wrong?

thanks

2. ### Guest

First, let me warn you that I'm not an engineer nor a student -- all
I've got is this library book called "Microelectronic Devices and
Circuits" by Clifton G. Fonstad, so I could be very very wrong, but are
For a p-n junction wouldn't there be two different expressions for n_0
and p_0, one for each side of the junction? In other words, if you had
a p-n junction (fixed-width font like Courier required):

Quasineutral Depletion Quasineutral
vvvvvvvvvvvv vvvvvvvvv vvvvvvvvvvvv
+---------------------------------------------+
| | | | |
| | |
A o--| p, N_Ap | | | n, N_Dn |--o B
| | |
| | | | |
+---------------------------------------------+
-w_p -x_p x_n w_n

^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^
P-Side N-Side

[Figure 7.3 in the Fonstad book]

then wouldn't the equations be something like:

p_0(-x_p) = N_Ap
p_0(x_n) = ni^2/N_Dn

and

n_0(-x_p) = ni^2/N_Ap
n_0(x_n) = N_Dn

or something like that?

Then I think if you multiply them together you get ni^2 in both cases.

But I dunno... I'm just trying to learn this stuff myself, and a lot
of it is barely making sense. Maybe somebody who actually knows what  