Chapter 1
1.1 Using the equations leading to Eq. (1.7) we find
2s V
I 1
1
1
1 I (2  2 )


 
 
(2  2) I
2 s
2s
2s s  2
s
V
1.2 Consider the following geometry.
I
r
1
r2
J

V1
0
V2
dV   dV  
0
V3 
1
I
s
t
2
4
3
s
s
I
I
dV
I

; E  J; E  

A 2rt
dx 2rt
I
2t

r1

dr I r2 dr
I 
r 

 V21  V2  V1 
ln  2 

r 2t  r
2t r1 
I 2s 
I  s 
I
I
ln
; V2 
ln
; V32 
ln( 4)  ln( 2)




2t
s
2t
2s
2t
t

t V32
 V32
; s 
ln( 2) I
ln( 2) I
1.3 The resistance is given by R=t/A + 2c/A. Hence =(dR/dt)A and 2c=RinterceptA. The
resistance, plotted as a function of wafer thickness t, is:
1600
y = 12.7 + 15280x R= 1
R (ž)
1200
800
400
0
0
0.02
0.04
0.06
t (cm)
From this plot we find
1
0.08
0.1
= 1.2 Ω·cm, c = 5x10-4 Ω·cm2
1.4 The resistance is generally given by R=L/A; in this case, since A varies with x (vertical
dimension), we write dR=(/A(x))dx where A(x)=(r+x)2. So we have

 1
0dR   0 (r  x)2  R   r  r  t 
R
dx
t
1
For r<<t, we can approximate this equation as
R
 1 1
dR


and
m


 r t 
d(1/ r) 
The approximation r<<t holds for small r or large 1/r. From the R versus 1/r plot provided, we
find the slope m at large 1/r to be m≈2.4, giving =7.5 Ω·cm. Can also solve the first equation
for , i.e., =Rr(r+t)/t, then choose r and R to get the same answer.
= 7.5 Ω·cm
1.5 For the x-direction along L the width can be expressed as
W  Wo (1  x / 2L)  102 (1  x / 2L)
The resistance between points A and B is
R
R
L

 dR 
dx
A
tW
 L
dx

2L ln(2)

2L ln(1 x / 2L)L0 

0
t
Wo(1 x / 2L) tWo
tWo
R = 1.39x105 Ω
1.6 The resistance is generally given by R=L/A; in this case, since A varies with r (radial
dimension), we write dR=(/A(r))dr where A(r)=rt. So we have
R
0
dR 
 r2 dr
 
r 

R

ln  2   219.9 ohms

2t r1 r
2t r1 
1.7 From the given s versus x curve take the slope m=d[log(s)]/dx at each datum point. The
resistivity is given by (x)=0.4343s(x)/m. (x) is plotted below. Using Fig. A1.1 (Appendix
1.1) determine NA knowing . NA(x) versus x is also shown below.
2
(x) (ž-cm)
1018
104
1017
103
1016
102
1015
101
1014
100
1013
10-1
1012
0.00005 0.0001 0.00015 0.0002 0.00025
N A (x) (cm -3 )
105
0
x (cm)
1.8 From the given s versus x curve, get the slope m=d[log(s)]/dx. The resistivity is given by
(x)=0.4343s(x)/m. (x) is plotted below. Using Fig. A1.1 determine NA(x) knowing (x).
104
1019
103
1018
1017
 (ž·cm)
1016
101
1015
100
1014
10-1
1013
1012
0.00002 0.00004 0.00006 0.00008
x (cm)
10-2
0
1.9
s 
1
 (1/ )dx
t
0
1.10


o
0 k


dx
t ln(1 k)
0 (1  kx / t)
t
4.532tF(R12,34  R 23,41 )
2
where F is obtained from the relationship
Rr  1
F
exp[ln(2) / F] 

ar cosh

R r  1 ln(2)
2
3
N A (cm -3 )
102
where Rr=R12,34/R23,41. For this problem, Rr=74/6=12.33 leading to F=0.664 and
 = 6 Ω·cm, s = /t = 120 Ω/square

1.11
4.532tF(R12,34  R 23,41 )
2
where F is obtained from the relationship
Rr  1
F
exp[ln(2) / F] 

ar cosh

R r  1 ln(2)
2
where Rr=R12,34/R23,41. For this problem, Rr=59/11=5.634 leading to F=0.8055 and
 = 4.47 Ω·cm, s = /t = 127.8 Ω/square

1.12
4.532tF(R12,34  R 23,41 )
2
where F is obtained from the relationship
Rr  1
F
exp[ln(2) / F] 

ar cosh

R r  1 ln(2)
2
where Rr=R12,34/R23,41. For this problem, Rr=90/9=10 leading to F=0.7 and
 = 7.84 Ω·cm, s = /t = 157 Ω/square
1.13
s 
4.532V12 4.532x58
 1.84x10 2
5

 263; t 

 7x10
I12
1
s
263
W
s LI13 263x0.05x104
4

 7.5x10
V23
1.75
s = 263 Ω/square, t = 0.7 µm, W = 7.5 µm
1.14
N D (x) 

R p
 x  R 2 
p
 
exp 0.5
2
R p 



Junction depth xj is where ND=NA, i.e.,



  Rp
x j  R p 2ln 
R p N A 2 
4
The sheet resistance is
s 
1
xj
q qn(x)n dx
 62.6 / square
0
s = 62.6 /square
1.15 (a)
s 
4.532V12 4.532x18
 4x10 3
5

 81.5; t 

 4.9x10
I12
1
s
81.5
W
(b)
V23 
s LI13 81.5x0.1x10 3
3

 5.1x10
V23
1.6
s I13L / 2 s I13L / 2
4.08x10 3


0.8

 W'  1.84x10 3
W
W'
W'
s = 81.5 /square, t = 0.49 µm, W = 51 µm, W' = 18.4 µm
s 
1.16
4.532V12 4.532x58
 1.84x10 2
5

 263; t 

 7x10
I12
1
s
263
W
s LI13 263x0.05x104
4

 7.5x10
V23
1.75
s = 263 /square, t = 0.7 µm, W = 7.5 µm
We have W=sL/R. Let us consider three cases
(a) s is uniform, hence W=sL/R0.
(b) Fig.(a) of the problem:
R1 
R' R"
(2s L / W)(2s1L / W)
2s L / W
s L
W


1.33R 0  W1 

R'  R" (2s L / W)  (2s1L / W) (1  s / s1)
1.33R 0 1.33
(c) Fig. (b) of the problem:
R 2  R' "R""
s L s1L s L 
1  s1 
  1.5R 0  W2  s L  W


2W 2W
W 2 2s 
1.5R0 1.5
W1/W = 0.75; W2/W = 0.67
5
1.17
s 
1
qp 
104
0
19  kx
10 e

dx
6.25x10 3 x105
625 / square
1  e 10
s I12 625x103
s LI16 625x5x102 x103
V34 

 0.18V;V45 

 31.3V
4.532
4.532
W
103
s = 625 /square, V34 = 0.18 V, V45 = 31.3 V
1.18 The thinnest layer is 0.1 µm=10-5 cm, the probe steps 2 µm and we want 20 points or 19
intervals. ∆z=∆xsin(). ∆z=10-5/20=5.26x10-7. ∆x=2x10-4 cm. Hence
5.26x10 7
3
sin() 
   0.15
4  2.63x10
2x10
 = 0.15°
1.19
s 
4.532V34 4.532x7.5
 1.7x10 3
5

 34; t 

 5x10
I 12
1
s
34
sLI 16 34x5x10 2 x103
3
W

 210
V45
0.85
s = 34 /square, t = 0.5 µm, W = 20 µm
1.20 First find an expression for p(x) from the relationship p(x)-n(x)-NA(x)+ND(x)=0.
Solving for p(x) gives:
p(x) 
N(x)  N(x) 1 (2n i / NA )2
2
where N=NA(x)-ND. Then n(x)=ni2/p(x).
With the given mobility expressions, find the sheet resistance from the expression
s 
1
t
q [p(x) p (x)  n(x)n (x)]dx
0
(a) t is the wafer thickness, which is 400 µm or 4x10-2 cm.
(b) t is the p-layer thickness, i.e., the depth at which NA(t)=ND, t=Lln(NA0/ND). For this case
t=11.5 µm.
The results are:
6
s(a) = 326 /square; s(b) = 520.5 /square
1.21 For the hot probe positive with respect to the cold probe: n-type
Ec
EF
Ei
Ev
Equilibrium
Non-equilibrium
The current in an n-type semiconductor is J n   nndEF / dx  q nnPndT / dx 
For Jn=0 (measured with high-impedance voltmeter)
dE F
dT dT
dE
 qPn
;
 0, Pn  0  F  0
dx
dx dx
dx
7
Chapter 2
2.1 First plot 1/C2 versus V.
23
1/C 2 (F -2 )
3 10
23
2 10
23
1 10
0
0
10
20
30
40
V (V)
From this plot find the slope m=d[(1/C2)/dV] and then determine p and W from the expressions
p
2
Ks  o A
2 ; W
qKs  o A m
C
Then plot p versus W.
17
10
-3
p (cm )
16
10
15
10
14
10
0
1
2
3
4
W (µm)
2.2
C inv

C ox
1
Kox
Ks t ox
1
K s o kT ln(NA / ni )
qNA
8
5
(a) In this problem, Cox=Cins. With R=0.32, Kox=Kins=8 and tox=tins=30 nm, solving the “NA”
equation gives:
NA = 1.26x1017 cm-3
(b) for NA=1016 cm-3:
Cinv/Cins = 0.126
(c) Using Eq. (2.19):
NA = 1.41x1017 cm-3
C inv

C ox
2.3
1
Kox
Ks t ox
1
K s o kT ln(NA / ni )
qNA
(a) In this problem, Cox=Cins. With R=0.3, Kox=Kins=6 and tox=tins=100 nm, solving the “NA”
equation gives:
NA = 4.2x1015 cm-3
(b) for NA=1018 cm-3:
Cinv/Cox = 0.85
2.4
G
C G
m
Cm
rs
Cb
(a)
Ya 
(a)
1
rs 

(b)
1
1

jCb G  jC

(b)
jCb (G  jC)
jrsC b (G  jC)  (G  jC)  jCb
(jG  C)[(G / C b  rs C)  j(1 C / C b  rsG)]
(G / C b  rs C)2  (1  C / C b  rs G)2
Yb  G m  jC m
Equating real and imaginary terms, gives
9
Gm 
G(1 rsG)  rs (C)2
C  [C2  (G/ )2 ]/ C b
and
C

m
(1  C / C b  rs G)2  (G/ C b  rs C)2
(1  C / C b  rs G)2  (G/ C b  rs C)2
Limit check: when Cb  ∞, Gm and Cm become Eq.(2.32). Therefore OK.
Cb (F)
Cm (F)
Gm (S)
≥10-5
2x10-8
10-8
9.959x10-11
9.91x10-11
9.861x10-11
4.031x10-5
3.991x10-5
3.952x10-2
@ Cb = 2x10-8 F, Gm has decreased by 1%. Cb = 2x10-8 = KoxoA/tox = 3.9x8.854x10-14A/10-5
giving
A = 0.58 cm2
2.5 (i) From the 1/C2 versus VG plot, we get the slope m=d(1/C2)/dVG=1.21x1021 giving
NA 
2
 4x1016 cm 3
2
qKs  o A m
1/C 2 (F -2 )
2 1021
1 1021
0
0
(ii) Using Eq. (2.18) we have N A 
0.5
1
V G (V)
1.5
2
4 F
C2inv
4 F
R 2C2ox

qKs  oA2 (1  Cinv / Cox ) 2 qKs  oA2 (1  R)2
Solving the first equation iteratively gives NA=3.7x1016 cm-3.
(iii) Using Eq. (2.19) gives NA=4.2x1016 cm-3.
NA(i) = 4x1016 cm-3, NA(ii) = 3.7x1016 cm-3, NA(iii) = 4.2x1016 cm-3
10
2.6 In (a) we have a conventional MOS capacitor. In (b), the back ohmic contact is replaced by a
capacitive contact. The total capacitance is
C
CtopC bottom
C top

Ctop  C bottom 1  C top / C bottom
 Ctop  CMOSC
Hence, configuration (b) will give the same curve as (a), but only if Abottom>>Atop.
2.7 The doping density is obtained from the slope of the VT vs. (2F-VBS)1/2 plot. The slope is m.
m 2C2ox
VT
16 2
NA 
 5.76x10 m where m 
2qKs  o
(2F  VBS )1/ 2
Not knowing NA, assume 2F=0.6 V. Plot VT versus (2F-VBS)1/2. This is shown as (1) on the
figure. take the slope, find a new NA and repeat one or two times. Curve fitting gives:
y=0.04+0.71x, giving a slope of m=0.71 and NA=2.89x1016 cm-3. Using this value, 2F=0.768
and the new curve has the equation y=-0.025+0.72x, giving NA=3x1016 cm-3 and 2F=0.77. This
gives for the third case y=-0.027+0.72x, which is almost the same as for the second case. From
the intercept we find VT(intercept)=VFB+2F=-0.027 V and VFB=-0.8 V.
NA = 3x1016 cm-3, VFB = -0.8 V
4
V T (V)
3
(1) y=0.04+0.71x
2
1
(2) y=-0.025+0.72x
0
0
1
2
(2  F-V
3
1/2
BS)
4
5
(V 1/2 )
2.8 The measured capacitance is related to the true capacitance at low and high frequencies by
C m (lf )  C; Cm (hf) 
C
 rs 
(2frsC)2
C
Cm (2fC)2
This gives C = 600 pF and rs = 560 
11
2.9 In cases (a), (b) and (c) there is no problem with the measurement. If the Schottky diode
space-charge region is confined to the p-layer, only the p-layer is profiled, as desired. In (c) there
may be a problem with series resistance if the p- layer is very lightly doped, but that is neglected.
For (d) there is a problem because the Schottky diode space-charge region capacitance is in series
with the pn junction space-charge region capacitance. The simple analysis is no longer valid
because it was derived for only the Schottky diode capacitance. In principle, one should be able
to take this series capacitance problem into account. In practice it is difficult, since the active
portion of the pn junction capacitance is not well known. It is best to avoid profiling structures of
this type.
2.10 (a) We need to solve Poisson’s equation for the two regions.
• For W ≤ d:
E
 dE  
0
0
qNA1 W
qNA1W 2
qN A1 x
qNA1
;
dx

E(x)


(x

W)
dV

(x

W)dx

V

V Ks o 0
Ks  o 0
Ks  o
2Ks  o
Considering the built-in potential as well, gives W 
2Ks  o (V  Vbi )
qN A1
(1)
• For W > d:
E1
E
0 ≤ x ≤ d:
E2
0
d ≤ x ≤ W:
dE  
S
dE  
qNA1 x
qN A1
dx  E1(x)  ES 
x

Ks  o 0
Ks o
qNA2 x
qN
dx  E2 (x)   A2 (x  W)

W
Ks o
Ks  o
At x = d, the two electric field are equal, giving ES 
q
[(N A1  NA2 )d  NA2 W] (2)
Ks  o
The potential is
0
d
W
V
0
d
  dV   E1(x)dx   E2 (x)dx; V 
q
Ks  o
 [(N
d
0
A1 

W
NA2 )d  NA2 W  NA1x]dx   N A2 (x  W)dx
V
q
[(N A1  NA2 )d 2  N A2 W2 ]
2Ks  o
W
2Ks  o (V  Vbi ) (N A2  NA1 )d 2

qNA2
NA2
d
leading to
From Eq. (1): for W = d = 3 m, Vbi = 0.4 V, hence V = 6.55 V. Eq. (1) is for V ≤ 6.55.V,
(3) is for V > 6.55 V. Using C=KsoA/W, plot C and 1/C2 vs. V.
12
(3)
Eq.
1 10 -11
2 10 24
C-2 (F -2)
C (F)
N A2 =1014 cm -3
5 10 -12
N A2 =1016 cm -3
1 10 24
1015 cm -3
1015 cm -3
1014 cm -3
0
0
10
20
1016 cm -3
30
40
0
50
0
10
20
V (V)
30
40
V (V)
(b) Breakdown: At breakdown Es = Eaval = 3x105 V/cm. From Eq. (2) find WBD at Es = Eaval; then
from Eq. (3) find VBD at W = WBD.
NA2 = 1014 cm-3: WBD = 1.67x10-2 cm, VBD = 2165 V.
NA2 = 1015 cm-3: WBD = 1.94x10-3 cm, VBD = 291 V.
NA2 = 1016 cm-3: WBD = 4.64x10-4 cm, VBD = 103 V.
2.11 This problem is similar to Problem 2.10.
• For W ≤ d:
E
x kx
qN
K  kx  kW
qN
0dE   KsA1o 0 e dx  E(x)   k1 e  e ; K1  KsA1o

0
VdV  

K1 W  kx
K 

(e
 e kW)dx  V   1 e kWW 

k 0
k 
Considering the built-in potential as well, gives V  Vbi  
1  1 

k  k 
K1   kW
1
1
e
W  
 


k 
k
k 
• For W > d:
E1
E
0 ≤ x ≤ d:
d ≤ x ≤ W:
E2
0
S
x kx
dx
0
dE  K1  e
 E1(x)  ES 
K1
 kx
(1  e )
k
x
dE  K2  dx  E2(x)  K2 (x  W); K2 
W
qN A2
Ks  o
At x = d, the two electric field are equal, giving ES  K2 (W  d)  (K1 / k)(1  e kd )
The potential is


0
d
W
d
W
K
 kx 
  dV   E1(x)dx   E2 (x)dx; V   ES  1 1  e
dx   K 2(W  x)dx
V
0
d
0 
d

k
13
50
V  Vbi  ESd 


K1 1
K
K
kd
 K
2
 kd
2
2
(1 e )  d  2 (W  d)  21 1 e (1 kd)  2 (W  d )
 2
k k
k
2
5 10-11
1 1024
4 10-11
8 1023
3 10-11
6 1023
2 10-11
4 1023
1 10-11
2 1023
0
0
5
10
15
20
25
30
1/C2 (F -2)
C (F)
Choose W, then calculate C using C=KsoA/W, find V and plot.
0
V (V)
2.12 (a)  
(b) C 
Ks  oA C W
1.4pW
0.14W
;  1.4W(m)%
;

;  

W
C
W
W
10 5
(c) W2 
2.13
145
1.4p
K A
1.4p Ks o A 0.0145
%
; 
; C s o 

W(m)
C / C
W
C W
W(cm)
2Ks  o V W Ks o V
1.4pqNAW2
2
2
9 2
2
;

;



 1.44x10 W (cm ) ;  14.4W (m )%
2
qNA
W
qN AW
K s oV
Cp 
C
G(1  rs G)  rs (2fC)2
;
G

p
(1  rs G)2  (2frs C) 2
(1  rs G)2  (2frs C) 2
At high frequencies: C p 
C
1
 rs  400 
2 ; Gp 
(2frs C)
rs
At low frequencies: C p 
C
G
2 ; Gp 
(1 rs G)
(1  rs G)
From conductance curve, G < 10-4 S at low f, hence, rsG << 1 and with Cp ≈ C/(2frsC)2 at high f.
This gives C = 125 pF. From conductance curve, knowing rs and C, G = 10-5 S.
rs = 400 , C = 125 pF, G = 10-5 S
2.14 The doping density is obtained from the slope of the VT vs. (2F-VBS)1/2 plot. The slope is
14
NA 
m 2C2ox
VT
16 2
 5.76x10 m where m 
2qKs  o
(2F  VBS )1/ 2
Not knowing NA, assume 2F=0.6 V. Plot VT versus (2F-VBS)1/2. This is shown as (1) on the
figure. The slope m=0.92 and NA = m2Cox2/2qKso = 9x1016m2 = 7.6x1016 cm-3. Then use
NA=7.6x1016 cm-3 to find new F. This gives 2F=0.82 V, leading to plot (2) on the figure. The
slope m=0.95 and NA = 7.85x1016 cm-3.
Since the slope is constant, the doping density is uniform into the wafer.
At (2F-VBS)1/2 = 0, we find VT =1.55 V = VFB+2F, giving VFB = 0.73 V.
NA = 7.85x1016 cm-3, VFB = 0.73 V
5
V T (V)
4
y=1.68+0.92x (1)
3
(2) y=1.55+0.95x
2
1
0
0
0.5
1
1.5
2
(2  F -V B S )
15
1/2
2.5
3
Chapter 3
3.1 From the log(I) versus V, the slope is m = 1/(2.3nkT/q) = 33.3; for n=1  T =350 K.
From the IdV/dI versus I plot, the slope is 3 Ω.
0.3
IdV/dI (A·ž)
0.25
0.2
0.15
0.1
0.05
0
y = 0.02919 + 3.0181x R= 0.99777
0
0.02
0.04
0.06
0.08
0.1
I (A)
T = 350 K, rs = 3 
3.2
-1
10
40
y = 2.715e-09 * e^(31.042x) R= 1
-2
y = 5.6 + 26000x R= 1
10
30
-3
-4
10
RT (ž)
I (A)
10
-5
10
20
-6
10
10
-7
10
T=300K
-8
10
0
0.1
0.2
0.3
0.4
0.5
V (V)
0
0
0.0002 0.0004 0.0006 0.0008 0.001
d (cm)
(a) I=Io[exp(qV/nkT)-1]≈Ioexp(qV/nkT) gives ln(I)=ln(Io)+qV/nkT or d[ln(I)]/dV=q/nkT. From
the curve fit we have ln(I)=ln(2.72x10-9)+31.04V. This gives q/nkT=31.04 or n=1.25.
The intercept is Io=AA*T2exp(-qB/kT)=2.72x10-9 using A=7.85x10-6 cm2. This gives B=0.8
V.
n = 1.25, B = 0.8 V
(b) NA(xj)=8x1019exp(-xj/5x10-6)=ND at x=xj. For =0.1 -cm, ND=7x1016 cm-3 (see p. 47 in
book). This gives xj= 3.52x10-5 cm. For the sheet resistance, evaluate the expression
1
s 
 260.6  / square
xj
19  x/ 5x10 6
q p 0 8x10 e
dx
16
xj = 3.52x10-5 cm, s = 260.6 /square
(c) The TLM test structure data are shown above. They give: slope=s/Z=26000 giving s=260
Ω/square; intercept (at d=0)=2Rc=5.6 giving Rc=2.8 ; intercept (at RT=0)=-2LT=-2.15x10-4
giving LT=1.08x10-4 cm or c=sLT2=3x10-6 -cm2.
s = 260 /square, Rc = 2.8 , c = 3x10-6 -cm2
(d) For the one-element contact string, the transfer length is LT=1.08x10-4 cm << L = 25 µm.
Hence, the effective contact area is Ac=LTZ==1.08x10-4x0.01=1.08x10-6 cm2. The resistance
between points A and B is R=2Rc+Rs=2c/Ac+s500/100=2x3x10-6/1.08x10-6+260x500/100=
5.6+1300 =1305.6 .
RAB = 1305.6 
(e) The resistance through the wafer is Rbulk=2Rc+t/A, where now Rc=c/A and A=r2. For r=0.5
cm and c=(kT/q)exp(qB/kT)/A*T2=7.63x104 -cm2. This gives Rbulk=1.9x105+0.0095=
1.9x105 .
Rbulk = 1.9x105 
3.3 From the I-V curve we have:
v deviation from ideal at I=0.04 A is 0.18 V, leading to rs=0.18/0.04=4.5 Ω
v slope=15.2=q/2.3nkT leading to n=1.1.
10-1
-20
-2
10
ln(I o /T 2)
10-3
I (A)
-4
10
10-5
10-6
10-7
I 02
-25
y = -2.2863 + -8123.1x R= 1
y = -2.823 + -7003.7x R= 1
T=300 K
10-8
0
0.1
0.2
0.3
0.4
V (V)
0.5
I 01
-30
2 10-3
0.6
2.5 10-3
3 10-3
1/T (K -1)
For low V, we can neglect the “Irs” in the exponent; furthermore, the Io values are obtained by
extrapolation to V=0. Hence we have
I  I o  AA*T2e q B / kT or ln(I / T2)  ln(AA* )  q B / kT
and a plot of ln(I/T2) vs. 1/T has an intercept of ln(AA*) and a slope of -qB/k.
17
3.5 10-3
v For device 1 we have from the intercept ln(AA*)=-2.2863 or AA*=0.1 leading to A*=100
A/cm2·K2, and from the slope=-8123.1 we find B=-slopek/q giving B=0.7 V with q=1.
v Vbi=B-Vo, where qVo=Eg/2-(EF-Ei). EF-Ei=(kT/q)ln(ND/ni) giving Vo=0.204 V. For device 1
with V=0, Vbi=0.495 V and B1=0.7 V (same as from the I-V data) using the capacitance
expression
CA
Ks o qN D
2(Vbi  V)
v For C2 we have
C2 
A  Ks o qN D
Ks o qND  C1 A
Ks o qND




2  2(B1  Vo )
2(B2  Vo )  2 2 2(B2  Vo )
Substituting numerical values gives Vbi=0.395 V and B2=0.6 V.
n = 1.1, rs = 4.5 Ω, A* = 100 A/cm2·K2, B1 = 0.7 V, B2 = 0.6 V
q
/ kT
3.4 Calculate the current for each device, then add the currents; using I  I o  AA*T2e B, eff
and solving for B,eff, using I(intercept)=1.59x10-7 A, gives B,eff= 0.64 V.
C
qK s o ND
Calculate the capacitance for each device, then add capacitances; using
and

A
2(Vbi |V|)
plotting (A/C)2 versus V gives the intercept on the V axis as -Vi=Vbi=B,eff+V0=0.94 V. With
V0=(kT/q)ln(NC/ND)=0.262 V, we find B,eff=0.68 V.
102
8 10 16
101
(A/C) 2 (cm 4 /F 2 )
y = 1.5949e-07 * e^(38.726x) R= 1
I (A)
100
10-1
10-2
10-3
10-4
10-5
10-6
0
0.1
0.2
0.3
0.4
0.5
y = 1.1373e+16 + 1.2062e+16x R= 0.99999
6 10 16
4 10 16
2 10 16
0
-1
V (V)
0
1
2
3
V (V)
B,eff(I–V) = 0.64 V; B,eff(C–V) = 0.68 V
As expected, I–V measurements tend to favor those regions with low B, even if the fractional
area is small, while C-V measurements favor those regions with the largest area.
18
4
5
3.5 (a) A plot of RT versus d is:
20
y = 3.46 + 3000x R= 1
15
RT (ž)
10
5
0
-0.001
0
0.001 0.002 0.003 0.004 0.005
d (cm)
The slope m=d(RT)/d(d)=3000=s/Z  s=30 /square; the intercept on the RT axis is
2Rc=3.46  Rc=1.73 . The intercept on the d axis is -d=2LTc(sc/s)=2(csc)1/2/s= 1.15x103 cm
 (csc)1/2 = 1.73x10-2 ·cm. The end resistance Re=3.4x10-3 . Using the equation
Re 
csc
Z sinh(L / L Tc )
gives L/LTc=6.925  LTc=1.73x10-4 cm. From (csc)1/2 = 1.73x10-2 ·cm and
LTc=(c/sc)1/2 =1.73x10-4 cm  c=3x10-6 ·cm2, sc=100 /square
s = 30 /square, sc = 100 /square, Rc = 1.73 ,
c = 3x10-6 ·cm2, LTc = 1.73x10-4 cm
(b) When the contact metal sheet resistance sm is no longer negligible, then we must use Eq. (3.34)
R cf 
c
L Tcm Z(1 )2

 L  

2
L 
 

(1  )2 coth


L Tcm  sinh(L / L Tcm ) L Tcm 


where LTcm=(c/(sm+sc))1/2. Solving this equation, gives
sm = 50 /square
3.6 From the I-V curve: the slope at T=300 K is m=d[log(I)/dV]=14.6; n=1/(2.3mkT/q)=
1/(2.3x14.6x0.0258)=1.15
19
The ln(I/T2) versus 1000/T curve at V=0.3 V is:
-12
V=0.3 V
ln(I/T 2 )
-14
-16
-18
-20
-22
y=-2.14-5.68x
-24
-26
1
2
3
1000/T (K
-1)
4
The slope m = d[ln(I/T2)]d[1000/T] =-5.68; the intercept is -2.14; B=V1/n-1000km/q =
0.3/1.15+0.0862x5.68 = 0.75 V. Intercept = ln(AA*)  A* = exp(-2.14)/0.00785 = 15 A/cm2·K2
A* = 15 A/cm2·K2, n = 1.15, B = 0.75 V
3.7 From the log(I) versus V, the T=300K slope is m = 1/(2.3nkT/q) = 16.8  n =1.
The T=? slope is m=24  T=210 K.
From the deviation of the experimental data from the extrapolated straight line, we find for I = 0.3 A
rs= 0.24V/0.3A = 0.8 . Alternately, from the IdV/dI versus I plot, the slope is 0.8 .
T = 210 K, rs = 0.8 
3.8 Can use several methods. Two of these are:
(a) Plot ln(I) versus V. The high-injection device will exhibit a straight line with slope
m=1/2kT/q; the series resistance dominated device will show a curved line, as illustrated in
Problem 3.7.
(b) Plot IdV/dI versus I. For the high-injection device IdV/dI=2kT/q; for the series resistance
dominated device IdV/dI=Irs+kT/q.
3.9
ND
n
(a)
p
x
20
(b)
The specific contact resistivity depends on the doping density at the surface. Since ND(a)<ND(b)
at the surface, hence c(a)<c(b).
The sheet resistance depends on the area under the ND - x curve. Since the areas are the same, the
sheet resistances are the same.
3.10 Since LT<L, Aeff=LTZ for A-B measurements. For vertical current flow, Aeff=LZ for A-C
measurements. Hence o RcA(A-C)<RcA(A-B) is the correct answer.
3.11 What parameters are determined with this test structure?
Slope gives s, RT intercept gives Rc, and d intercept gives LT which gives c.
One day something goes wrong during processing and a thin oxide film remains on the n-layer
before the metal contacts are deposited.
The oxide film adds a resistive layer, hence Rc, c, and LT increase. Hence the line shifts up.
d
RT
With oxide film
n
p
d
3.12 When both contacts are Schottky contacts, one is always forward biased while the other is
reverse biased. The current is determined by the leakage current of the reverse-biased diode.
V
+I
I
A
Two Schottky
diodes
n-type
+V
B
3.13
* 2 qB / kT
ln(I)  ln(AA T e
)
qV
d[ln(I)]
1
; slope 

nkT
dV
nkT / q
Since only the slope is affected, it must be caused by a change in diode ideality factor n.
21