19-9. E2max = E

2BD

4!f2c!BVBD

W2o!fc

; Eq. (19-13); or W

2o

!fc = 4!BVBD

E2BD

W2(BVBD) = W

2o!BVBD

!fc ; Eq. (19-11) ; Inserting W

2o

!fc = 4!BVBD

E2BD

and taking

the square root yields W (BVBD) 2!BVBD

E2BD.

19-10. Lp = Dp!t = (13)(10-6) = 36 microns ; Ln = Dn!t = (39)(10

-6) = 62 microns

19-11. Assume a one-sided step junction with Na >> Nd

I1 = q n2i A

Dp!t1Nd!t1

exp(q!Vk!T ) ; I2 = q n

2i A

Dp!t2Nd!t2

exp(q!Vk!T )

I2I1

= 2 = t1t2 ; Thus 4 t2 = t1

19-12. s = q mp p + q mn n ; np = n2i ; Combining yeilds s = q mp

n2in + q mn n

dsdn = 0 = - q mp

n2i

n2 + q mn ; Solving for n yields n = ni

mpmn and p = ni

mnmp

p = 1010 1500500 = 1.7x10

10 cm-3; n = 1010 5001500 = 6x10

9 cm-3;

Thus minimum conductivity realized when silicon is slightly p-type.

Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn .

Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500)

= 2.8x10-6 mhos-cm