Mohan Sm Ch19 12
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Transcript of Mohan Sm Ch19 12
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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