Page 47 - Photonics Essentials an introduction with experiments

Photodiodes

Photodiodes 41

This is a differential equation of the type
2
d f(x)
= kf(x) + M
dx 2
where M is a constant driving term. The solution is f(x) = Ae x k +
Be –x k + C, which we will verify presently. The constant k = 1/D e e .
This is just mathematics. The most important part of the solution,
however is the physics of the problem. This is summarized in the
boundary conditions that allow us to solve for A, B, and C.

a. When no light is present, n p at (x p = ) = 0.
b. When light is present, n p at (x p = ) = G L e 0. To see that this
must be so, set the second derivative = 0 in Eq. 3.6.
c. At x p = 0, n p (x = 0) = n p0 (e qV A /kT – 1). (3.7)

First, note that k must have units of 1/L, where L is length. Then,
n p (x) = Ae x/L e + Be –(x/L e ) + C (3.8)
where L e = D e e = diffusion length for electrons
Then apply the boundary condition at x p = , n p (x p = ) = G L e :
Ae + + Be – + C = G L e
If this equation is true, A must be zero. As a result,

(3.9)
C = G L e
However, nothing is learned about B. Next, apply the boundary condi-
tion for n p (x p = 0).

n p (x p = 0) = 0 + Be –(0/L e ) + G L e = n p (e qV A /kT – 1)
B = n p (e qV A /kT – 1) – G L e (3.10)
The solution for n p (x) is written:
n p (x) = e (–x/L e ) [n p (e qV A /kT – 1) – G L e ] + G L e (3.11)

B C
The diffusion current in the photodiode is calculated from the diffu-
sion equation:
d D e
J n = qD e n p (x)| x=0 = –(–1)q [n p (e qV A /kT – 1) – G L e ] (3.12)
dx L e
The extra factor of –1 comes from a change of variable from x p to x n .
The derivative is evaluated at x = 0 because at that point all the cur-
rent is carried by diffusion.

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