Fundamentals of Microelectronics Solutions Manual

1115 Pages·028.782 MB·English

by  Behzad Razavi

Solutions Manual to Accompany 
Fundamentals of Microelectronics, 1st Edition 
Book ISBN: 978‐0‐471‐47846‐1 
Razavi 

All materials copyrighted and published by John Wiley & Sons, Inc.  
Not for duplication or distribution
2.1 (a)
k =8.617×10−5 eV/K
0.66eV
n (T =300K)=1.66×1015(300K)3/2exp − cm−3
i 2(8.617×10−5 eV/K)(300K)
(cid:20) (cid:21)
= 2.465×1013 cm−3
0.66eV
n (T =600K)=1.66×1015(600K)3/2exp − cm−3
i 2(8.617×10−5 eV/K)(600K)
(cid:20) (cid:21)
= 4.124×1016 cm−3
ComparedtothevaluesobtainedinExample2.1,wecanseethattheintrinsiccarrierconcentration
in Ge at T = 300K is 2.465×1013 = 2282 times higher than the intrinsic carrier concentration in
1.08×1010
SiatT =300K. Similarly,atT =600K, the intrinsic carrierconcentrationin Ge is 4.124×1016 =
1.54×1015
26.8 times higher than that in Si.
(b) Since phosphorus is a Group V element, it is a donor, meaning N = 5×1016 cm−3. For an
D
n-type material, we have:
n=N = 5×1016 cm−3
D
[n (T =300K)]2
p(T =300K)= i = 1.215×1010 cm−3
n
[n (T =600K)]2
p(T =600K)= i = 3.401×1016 cm−3
n

2.3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only to
the drift component.
I =I
tot drift
=q(nµ +pµ )AE
n p
n=1017 cm−3
p=n2/n=(1.08×1010)2/1017 =1.17×103 cm−3
i
µ =1350cm2/V·s
n
µ =480cm2/V·s
p
1V
E =V/d=
0.1µm
=105 V/cm
A=0.05µm×0.05µm
=2.5×10−11 cm2
Since nµ ≫pµ , we can write
n p
I ≈qnµ AE
tot n
= 54.1µA
(b) All of the parameters are the same except n , which means we must re-calculate p.
i
n (T =400K)=3.657×1012 cm−3
i
p=n2/n=1.337×108 cm−3
i
Since nµ ≫ pµ still holds (note that n is 9 orders of magnitude larger than p), the hole
n p
concentration once again drops out of the equation and we have
I ≈qnµ AE
tot n
= 54.1µA
2.4 (a) From Problem 1, we can calculate n for Ge.
i
n (T =300K)=2.465×1013 cm−3
i
I =q(nµ +pµ )AE
tot n p
n=1017 cm−3
p=n2/n=6.076×109 cm−3
i
µ =3900cm2/V·s
n
µ =1900cm2/V·s
p
1V
E =V/d=
0.1µm
=105 V/cm
A=0.05µm×0.05µm
=2.5×10−11 cm2
Since nµ ≫pµ , we can write
n p
I ≈qnµ AE
tot n
= 156µA
(b) All of the parameters are the same except n , which means we must re-calculate p.
i
n (T =400K)=9.230×1014 cm−3
i
p=n2/n=8.520×1012 cm−3
i
Since nµ ≫ pµ still holds (note that n is 5 orders of magnitude larger than p), the hole
n p
concentration once again drops out of the equation and we have
I ≈qnµ AE
tot n
= 156µA
2.5 Sincethere’snoelectricfield,thecurrentisdueentirelytodiffusion. Ifwedefinethecurrentaspositive
when flowing in the positive x direction, we can write
dn dp
I =I =AJ =Aq D −D
tot diff diff n p
dx dx
(cid:20) (cid:21)
A=1µm×1µm=10−8 cm2
D =34cm2/s
n
D =12cm2/s
p
dn 5×1016 cm−3
=− =−2.5×1020 cm−4
dx 2×10−4 cm
dp 2×1016 cm−3
= =1020 cm−4
dx 2×10−4 cm
I = 10−8 cm2 1.602×10−19 C 34cm2/s −2.5×1020 cm−4 − 12cm2/s 1020 cm−4
tot
=(cid:0)−15.54µA(cid:1)(cid:0) (cid:1)(cid:2)(cid:0) (cid:1)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1)(cid:3)

2.8 AssumethediffusionlengthsL andL areassociatedwiththeelectronsandholes,respectively,inthis
n p
material and that L ,L ≪ 2µm. We can express the electron and hole concentrations as functions
n p
of x as follows:
n(x)=Ne−x/Ln
p(x)=Pe(x−2)/Lp
2
# of electrons= an(x)dx
Z0
2
= aNe−x/Lndx
Z0
2
=−aNL e−x/Ln
n
0
=−aNL (cid:16)e−2/Ln(cid:17)−(cid:12)(cid:12)1
n (cid:12)
2 (cid:16) (cid:17)
# of holes= ap(x)dx
Z0
2
= aPe(x−2)/Lpdx
Z0
2
=aPL e(x−2)/Lp
p
0
=aPL (cid:16)1−e−2/L(cid:17)p(cid:12)(cid:12)
p (cid:12)
(cid:16) (cid:17)
Due to our assumption that L ,L ≪2µm, we can write
n p
e−2/Ln ≈0
e−2/Lp ≈0
# of electrons≈ aNL
n
# of holes≈ aPL
p