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Power Electronics - Mohan (Solution)

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Prévia do material em texto

Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest 
doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 10
14 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
Ti
!-!
1
300 
Solving for Ti using Eg = 1.1 eV, k = 1.4x10
-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014)
 = 43.5 ohm-cm
P-side resistivity rp = 
1
q!mp!Na
 = 
1
(1.6x10-19)(500)(1018)
 = 0.013 ohm-cm
19-3. Material is n-type with Nd = 10
13 cm-3 >> ni = 10
10 cm-3. Hence use approximate
formulas given in Chapter 19.
n = Nd = 10
13 cm-3 ; p = 
n
2
i
Nd
 = 
1020
1013
 = 107 cm-3
19-4. po = 
n
2
i [300]
Nd
 ; 2po = 
n
2
i [300!+!T]
Nd
 
2 n
2
i [300] = n
2
i [300 + T] ; 2x1010 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
T!-!
1
300 
Solving for T yields T = 
q!Eg!300
(q!Eg!-!k!300!ln(2))
 = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(
q!V1
k!T ; 10 I1 = Is exp(
q!V1!+!dV
k!T ) ; dV = 
k!T
q ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width 
on p-side at zero bias.
Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest 
doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 10
14 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
Ti
!-!
1
300 
Solving for Ti using Eg = 1.1 eV, k = 1.4x10
-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014)
 = 43.5 ohm-cm
P-side resistivity rp = 
1
q!mp!Na
 = 
1
(1.6x10-19)(500)(1018)
 = 0.013 ohm-cm
19-3. Material is n-type with Nd = 10
13 cm-3 >> ni = 10
10 cm-3. Hence use approximate
formulas given in Chapter 19.
n = Nd = 10
13 cm-3 ; p = 
n
2
i
Nd
 = 
1020
1013
 = 107 cm-3
19-4. po = 
n
2
i [300]
Nd
 ; 2po = 
n
2
i [300!+!T]
Nd
 
2 n
2
i [300] = n
2
i [300 + T] ; 2x1010 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
T!-!
1
300 
Solving for T yields T = 
q!Eg!300
(q!Eg!-!k!300!ln(2))
 = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(
q!V1
k!T ; 10 I1 = Is exp(
q!V1!+!dV
k!T ) ; dV = 
k!T
q ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width 
on p-side at zero bias.
Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest 
doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 10
14 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
Ti
!-!
1
300 
Solving for Ti using Eg = 1.1 eV, k = 1.4x10
-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014)
 = 43.5 ohm-cm
P-side resistivity rp = 
1
q!mp!Na
 = 
1
(1.6x10-19)(500)(1018)
 = 0.013 ohm-cm
19-3. Material is n-type with Nd = 10
13 cm-3 >> ni = 10
10 cm-3. Hence use approximate
formulas given in Chapter 19.
n = Nd = 10
13 cm-3 ; p = 
n
2
i
Nd
 = 
1020
1013
 = 107 cm-3
19-4. po = 
n
2
i [300]
Nd
 ; 2po = 
n
2
i [300!+!T]
Nd
 
2 n
2
i [300] = n
2
i [300 + T] ; 2x1010 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
T!-!
1
300 
Solving for T yields T = 
q!Eg!300
(q!Eg!-!k!300!ln(2))
 = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(
q!V1
k!T ; 10 I1 = Is exp(
q!V1!+!dV
k!T ) ; dV = 
k!T
q ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width 
on p-side at zero bias.
Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest 
doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 10
14 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
Ti
!-!
1
300 
Solving for Ti using Eg = 1.1 eV, k = 1.4x10
-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014)
 = 43.5 ohm-cm
P-side resistivity rp = 
1
q!mp!Na
 = 
1
(1.6x10-19)(500)(1018)
 = 0.013 ohm-cm
19-3. Material is n-type with Nd = 10
13 cm-3 >> ni = 10
10 cm-3. Hence use approximate
formulas given in Chapter 19.
n = Nd = 10
13 cm-3 ; p = 
n
2
i
Nd
 = 
1020
1013
 = 107 cm-3
19-4. po = 
n
2
i [300]
Nd
 ; 2po = 
n
2
i [300!+!T]
Nd
 
2 n
2
i [300] = n
2
i [300 + T] ; 2x1010 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
T!-!
1
300 
Solving for T yields T = 
q!Eg!300
(q!Eg!-!k!300!ln(2))
 = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(
q!V1
k!T ; 10 I1 = Is exp(
q!V1!+!dV
k!T ) ; dV = 
k!T
q ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width 
on p-side at zero bias.
Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest 
doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 10
14 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
Ti
!-!
1
300 
Solving for Ti using Eg = 1.1 eV, k = 1.4x10
-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014)
 = 43.5 ohm-cm
P-side resistivity rp = 
1
q!mp!Na
 = 
1
(1.6x10-19)(500)(1018)
 = 0.013 ohm-cm
19-3. Material is n-type with Nd = 10
13 cm-3 >> ni = 10
10 cm-3. Hence use approximate
formulas given in Chapter 19.
n = Nd = 10
13 cm-3 ; p = 
n
2
i
Nd
 = 
1020
1013
 = 107 cm-3
19-4. po = 
n
2
i [300]
Nd
 ; 2po = 
n
2
i [300!+!T]
Nd
 
2 n
2
i [300] = n
2
i [300 + T] ; 2x1010 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
T!-!
1
300 
Solving for T yields T = 
q!Eg!300
(q!Eg!-!k!300!ln(2))
 = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(
q!V1
k!T ; 10 I1 = Is exp(
q!V1!+!dV
k!T ) ; dV = 
k!T
q ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width 
on p-side at zero bias.
Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest 
doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 10
14 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
Ti
!-!
1
300 
Solving for Ti using Eg = 1.1 eV, k = 1.4x10
-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014)
 = 43.5 ohm-cm
P-side resistivity rp = 
1
q!mp!Na
 = 
1
(1.6x10-19)(500)(1018)
 = 0.013 ohm-cm
19-3. Materialis n-type with Nd = 10
13 cm-3 >> ni = 10
10 cm-3. Hence use approximate
formulas given in Chapter 19.
n = Nd = 10
13 cm-3 ; p = 
n
2
i
Nd
 = 
1020
1013
 = 107 cm-3
19-4. po = 
n
2
i [300]
Nd
 ; 2po = 
n
2
i [300!+!T]
Nd
 
2 n
2
i [300] = n
2
i [300 + T] ; 2x1010 = 1010 exp Î
Í
È
˚
˙
˘
-!
q!Eg
2k !ÓÌ
Ï
˛
˝
¸1
T!-!
1
300 
Solving for T yields T = 
q!Eg!300
(q!Eg!-!k!300!ln(2))
 = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(
q!V1
k!T ; 10 I1 = Is exp(
q!V1!+!dV
k!T ) ; dV = 
k!T
q ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width 
on p-side at zero bias.
xn(0) + xp(0) = Wo = !
2!e!fc!(Na!+!Nd)
q!Na!Nd
 (1)
fc = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
Na!Nd
n
2
i
 = 0.026 ln Î
Í
È
˚
˙
˘
!
1014!1015
1020
 = 0.54 eV
Conservation of charge: q Na xp = q Nd xn (2)
Solving (1) and (2) simultaneously gives using the numerical values given in the problem
statement gives:
Wo = 2.8 microns ; xn(0) = 2.55 microns ; xp(0) = 0.25 microns
(b) Electric field profile triangular-shaped as shown in Fig. 19-9b. Maximum electric at 
zero bias given by
Emax = 
2!fc
Wo
 = 
(2)!(0.54)
(2.8x10-4)
 = 3,900 V/cm
(c) From part a) fc = 0.54 eV
(d)
C(V)
A = 
e
Wo 1!+!
V
fc
 ; C(V) = space-charge capacitance at reverse voltage V.
C(0)
A = 
(11.7)(8.9x10-14)
2.8x10-4
 = 3.7x10-9 F/cm2 
 C(50)
A 
 = 
(11.7)(8.9x10-14)
2.8x10-4 1!+!
50
0.54
 = 3.8x10-10 F/cm2
(e) I = Is exp(
qV
kT ) ; exp( 
qV
kT ) = exp (
0.7
0.026 ) = 5x10
11
Is = q n
2
i ÎÍ
Í
È
˚
˙
˙
˘
!
Dnt
Na!t
!+!
Dpt
Nd!t
 A
= (1.6x10-19)(1020) ÎÍ
Í
È
˚
˙
˙
˘
!
(38)(10-6)
(1015)(10-6)
!+!
(13)(10-6)
(1014)(10-6)
 (2) 2 Is = 6.7x10
-14 A ;
I = (6.7x10-14 )(5x1011) = 34 mA
19-7. Resistance R = 
r!L
A ; 
L
A = 
0.02
0.01 = 2 cm
-1 
At 25 °C, Nd = 10
14 >> ni so r = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014!)
 
= 41.7 W-cm
R(25 °C) = (41.7)(2) = 83.4 ohms
At 250 °C (523 °K), ni[523] = 10
10 exp Î
Í
È
˚
˙
˘
-!
(1.6x10-19)!(1.1)
(2)(1.4x10-23)
!
Ó
Ì
Ï
˛
˝
¸1
523!-!
1
300 = 
(1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 10
14. Thus we 
should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations 
similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields
no = 
Nd
2 Î
Í
È
˚
˙
˘
1!+! 1!+!
4!n
2
i
N
2
d
 and po = 
n
2
i
no
 . Putting in numerical values yields
no = 
1014
2 ÎÍ
Í
È
˚
˙
˙
˘
1!+! 1!+!
(4)(7.6x1013)2!
(1014)2
 = 1.4x1014 and
po = 
5.8x1027
1014
 = 5.8x1013
Assuming temperature-independent mobilities (not a valid assumption but no other 
information is given in text or the problem statement), resistance is
R(250 °C) = 
r(250!° C)!!L
A ; r(250 °C) ≈ 
1
q!mn!no!+!q!mp!po
 
= 
1
(1.6x10-19)(1500)(1.4x1014!)!+!(1.6x10-19)(500)(5.8x1013!)
 = 26.2 W-cm ;
R(250 °C) ≈ (26.2)(2) = 52.4 ohms
19-8. BVBD = 
e!(Na!+!Nd)!E
2
BD
2!q!Na!Nd
 = 
(11.7)(8.9x1014)(1015!+!1014)(3x105)2
(2)(1.6x10-19)(1015)(1014)
 
= 3,340 volts
19-7. Resistance R = 
r!L
A ; 
L
A = 
0.02
0.01 = 2 cm
-1 
At 25 °C, Nd = 10
14 >> ni so r = 
1
q!mn!Nd
 = 
1
(1.6x10-19)(1500)(1014!)
 
= 41.7 W-cm
R(25 °C) = (41.7)(2) = 83.4 ohms
At 250 °C (523 °K), ni[523] = 10
10 exp Î
Í
È
˚
˙
˘
-!
(1.6x10-19)!(1.1)
(2)(1.4x10-23)
!
Ó
Ì
Ï
˛
˝
¸1
523!-!
1
300 = 
(1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 10
14. Thus we 
should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations 
similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields
no = 
Nd
2 Î
Í
È
˚
˙
˘
1!+! 1!+!
4!n
2
i
N
2
d
 and po = 
n
2
i
no
 . Putting in numerical values yields
no = 
1014
2 ÎÍ
Í
È
˚
˙
˙
˘
1!+! 1!+!
(4)(7.6x1013)2!
(1014)2
 = 1.4x1014 and
po = 
5.8x1027
1014
 = 5.8x1013
Assuming temperature-independent mobilities (not a valid assumption but no other 
information is given in text or the problem statement), resistance is
R(250 °C) = 
r(250!° C)!!L
A ; r(250 °C) ≈ 
1
q!mn!no!+!q!mp!po
 
= 
1
(1.6x10-19)(1500)(1.4x1014!)!+!(1.6x10-19)(500)(5.8x1013!)
 = 26.2 W-cm ;
R(250 °C) ≈ (26.2)(2) = 52.4 ohms
19-8. BVBD = 
e!(Na!+!Nd)!E
2
BD
2!q!Na!Nd
 = 
(11.7)(8.9x1014)(1015!+!1014)(3x105)2
(2)(1.6x10-19)(1015)(1014)
 
= 3,340 volts
19-9. E
2
max = E
2
BD ≈ 
4!f
2
c!BVBD
W
2
o!fc
 ; Eq. (19-13); or 
W
2
o
!fc = 
4!BVBD
E
2
BD
 
W2(BVBD) = 
W
2
o!BVBD
!fc ; Eq. (19-11) ; Inserting 
W
2
o
!fc = 
4!BVBD
E
2
BD
 and taking 
the square root yields W (BVBD) ≈ 
2!BVBD
E
2
BD.
 
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 n
2
i A 
Dp!t1
Nd!t1
 exp(
q!V
k!T ) ; I2 = q n
2
i A 
Dp!t2
Nd!t2
 exp(
q!V
k!T )
I2
I1
 = 2 = 
t1
t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n
2
i ; Combining yeilds s = q mp 
n
2
i
n + q mn n
ds
dn = 0 = - q mp 
n
2
i
n2
 + q mn ; Solving for n yields n = ni 
mp
mn and p = ni 
mn
mp 
p = 1010 
1500
500 = 1.7x10
10 cm-3; n = 1010 
500
1500 = 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
19-9. E
2
max = E
2
BD ≈ 
4!f
2
c!BVBD
W
2
o!fc
 ; Eq. (19-13); or 
W
2
o
!fc = 
4!BVBD
E
2
BD
 
W2(BVBD) = 
W
2
o!BVBD
!fc ; Eq. (19-11) ; Inserting 
W
2
o
!fc = 
4!BVBD
E
2
BD
 and taking 
the square root yields W (BVBD) ≈ 
2!BVBD
E
2
BD.
 
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 n
2
i A 
Dp!t1
Nd!t1
 exp(
q!V
k!T ) ; I2 = q n
2
i A 
Dp!t2
Nd!t2
 exp(
q!V
k!T )
I2
I1
 = 2 = 
t1
t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n
2
i ; Combining yeilds s = q mp 
n
2
i
n + q mn n
ds
dn = 0 = - q mp 
n
2
i
n2
 + q mn ; Solving for n yields n = ni 
mp
mn and p = ni 
mn
mp 
p = 1010 
1500
500 = 1.7x10
10 cm-3; n = 1010 
500
1500 = 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
19-9. E
2
max = E
2
BD ≈ 
4!f
2
c!BVBD
W
2
o!fc
 ; Eq. (19-13); or 
W
2
o
!fc = 
4!BVBD
E
2
BD
 
W2(BVBD) = 
W
2
o!BVBD
!fc ; Eq. (19-11) ; Inserting 
W
2
o
!fc = 
4!BVBD
E
2
BD
 and taking 
the square root yields W (BVBD) ≈ 
2!BVBD
E
2
BD.
 
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 n
2
i A 
Dp!t1
Nd!t1
 exp(
q!V
k!T ) ; I2 = q n
2
i A 
Dp!t2
Nd!t2
 exp(
q!V
k!T )
I2
I1
 = 2 = 
t1
t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n
2
i ; Combining yeilds s = q mp 
n
2
i
n + q mn n
ds
dn = 0 = - q mp 
n
2
i
n2
 + q mn ; Solving for n yields n = ni 
mp
mn and p = ni 
mn
mp 
p = 1010 
1500
500 = 1.7x10
10 cm-3; n = 1010 
500
1500 = 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
19-9. E
2
max = E
2
BD ≈ 
4!f
2
c!BVBD
W
2
o!fc
 ; Eq. (19-13); or 
W
2
o
!fc = 
4!BVBD
E
2
BD
 
W2(BVBD) = 
W
2
o!BVBD
!fc ; Eq. (19-11) ; Inserting 
W
2
o
!fc = 
4!BVBD
E
2
BD
 and taking 
the square root yields W (BVBD) ≈ 
2!BVBD
E
2
BD.
 
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 n
2
i A 
Dp!t1
Nd!t1
 exp(
q!V
k!T ) ; I2 = q n
2
i A 
Dp!t2
Nd!t2
 exp(
q!V
k!T )
I2
I1
 = 2 = 
t1
t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n
2
i ; Combining yeilds s = q mp 
n
2
i
n + q mn n
ds
dn = 0 = - q mp 
n
2
i
n2
 + q mn ; Solving for n yields n = ni 
mp
mn and p = ni 
mn
mp 
p = 1010 
1500
500 = 1.7x10
10 cm-3; n = 1010 
500
1500 = 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
Chapter 20 Problem Solutions
20-1. Nd = 
1.3x1017
BVBD!
 = 
1.3x1017
2500 = 5x10
13 cm-3 ; W(2500 V) = (10-5)(2500) = 250
microns
20-2. Drift region length of 50 microns is much less than the 250 microns found in the previous
problem (20-1) for the same drift region doping density. Hence this must be a punch-
through structure and Eq. (20-9) applies.
BVBD = (2x10
5)(5x10-3) - 
(1.6x10-19)(5x1013)(5x10-3)2
(2)(11.7)(8.9x10-14)
 = 900 V
20-3. Von = Vj + Vdrift ; Vj = 
k!T
q ln ÎÍ
È
˚
˙˘!
I
Is
 ; For one-sided step junction Is = 
q!A!n
2
i !Lp
!Nd!to
 ;
Evaluating Is yields 
(1.6x10-19)(2)(1010)2! (13)(2x10-6)
!(5x1013)(2x10-6)
 = 1.6x10-9 A
Vd = K1 I + K2 (I)
2/3 Eq. (20-16) with I = forward bias current through the diode.
K1 = 
Wd
q!mo!A!nb
 = 
5x10-3
(1.6x10-19)(900)(2)(1017)
 = 1.7x10-4
K2 = 
3 W
4
d
q2!m
3
o!n
2
b!A2!to
 = 
3 (5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)
 
= 7.5x10-4
Chapter 20 Problem Solutions
20-1. Nd = 
1.3x1017
BVBD!
 = 
1.3x1017
2500 = 5x10
13 cm-3 ; W(2500 V) = (10-5)(2500) = 250
microns
20-2. Drift region length of 50 microns is much less than the 250 microns found in the previous
problem (20-1) for the same drift region doping density. Hence this must be a punch-
through structure and Eq. (20-9) applies.
BVBD = (2x10
5)(5x10-3) - 
(1.6x10-19)(5x1013)(5x10-3)2
(2)(11.7)(8.9x10-14)
 = 900 V
20-3. Von = Vj + Vdrift ; Vj = 
k!T
q ln ÎÍ
È
˚
˙˘!
I
Is
 ; For one-sided step junction Is = 
q!A!n
2
i !Lp
!Nd!to
 ;
Evaluating Is yields 
(1.6x10-19)(2)(1010)2! (13)(2x10-6)
!(5x1013)(2x10-6)
 = 1.6x10-9 A
Vd = K1 I + K2 (I)
2/3 Eq. (20-16) with I = forward bias current through the diode.
K1 = 
Wd
q!mo!A!nb
 = 
5x10-3
(1.6x10-19)(900)(2)(1017)
 = 1.7x10-4
K2 = 
3 W
4
d
q2!m
3
o!n
2
b!A2!to
 = 
3 (5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)
 
= 7.5x10-4
Chapter 20 Problem Solutions
20-1. Nd = 
1.3x1017
BVBD!
 = 
1.3x1017
2500 = 5x10
13 cm-3 ; W(2500 V) = (10-5)(2500) = 250
microns
20-2. Drift region length of 50 microns is much less than the 250 microns found in the previous
problem (20-1) for the same drift region doping density. Hence this must be a punch-
through structure and Eq. (20-9) applies.
BVBD = (2x10
5)(5x10-3) - 
(1.6x10-19)(5x1013)(5x10-3)2
(2)(11.7)(8.9x10-14)
 = 900 V
20-3. Von = Vj + Vdrift ; Vj = 
k!T
q ln ÎÍ
È
˚
˙˘!
I
Is
 ; For one-sided step junction Is = 
q!A!n
2
i !Lp
!Nd!to
 ;
Evaluating Is yields 
(1.6x10-19)(2)(1010)2! (13)(2x10-6)
!(5x1013)(2x10-6)
 = 1.6x10-9 A
Vd = K1 I + K2 (I)
2/3 Eq. (20-16) with I = forward bias current through the diode.
K1 = 
Wd
q!mo!A!nb
 = 
5x10-3
(1.6x10-19)(900)(2)(1017)
 = 1.7x10-4
K2 = 
3 W
4
d
q2!m
3
o!n
2
b!A2!to
 = 
3 (5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)
 
= 7.5x10-4
I Vj Vdrift Von
0 A 0 V 0 V 0 V
1 0.53 0.001 0.53
10 0.59 0.005 0.59
100 0.65 0.033 0.68
1000 0.71 0.25 0.96
3000 0.74 0.67 1.41 • •
•
•
•
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1 10 100 1000 10000
Von in volts
Forward current in amperes
20-4. a) Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r 
L
A ; r = 
1
q!mn!Nd
 
r = 
1
(1.6x10-19)(1500)(5x1013)
 = 85 ohm-cm ; 
L
A = 
5x10-3
2 = 2.5x10
-3
 Rdrift = (85)(2.5x10
-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds
Von(t) = (0.21)(2.5x10
8 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds
Von(4 ms) = (5.3x10
7)(4x10-6) = 212 volts
b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x10
7 t]
Von(t) = {0.21[1 - 2.5x10
7 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds
I Vj Vdrift Von
0 A 0 V 0 V 0 V
1 0.53 0.001 0.53
10 0.59 0.005 0.59
100 0.65 0.033 0.68
1000 0.71 0.25 0.96
3000 0.74 0.67 1.41 • •
•
•
•
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1 10 100 1000 10000
Von in volts
Forward current in amperes
20-4. a) Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r 
L
A ; r = 
1
q!mn!Nd
 
r = 
1
(1.6x10-19)(1500)(5x1013)
 = 85 ohm-cm ; 
L
A = 
5x10-3
2 = 2.5x10
-3
 Rdrift = (85)(2.5x10
-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds
Von(t) = (0.21)(2.5x10
8 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds
Von(4 ms) = (5.3x10
7)(4x10-6) = 212 volts
b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x10
7 t]
Von(t) = {0.21[1 - 2.5x10
7 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds
0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von
 in
volts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + 
IF
diR/dt
 = trr + 
2000
2.5x108
 = trr + 8 ms
trr = 
2!t!IF
diR/dt
 ; t = 4x10-12(BVBD)
2 = 4x10-12(2000)2 = 16 ms
trr = 
(2)(1.6x10-5)(2x103)
2.5x108
 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 
1.3x1017
BVBD
 = 
1.3x1017
150 = 8.7x10
14 cm-3
Wd = 10
-5 BVBD = (10
-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 
1
q!mn!Nd
 
L
A 
A = 
2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2)
 = 0.42 cm2
0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von
 in
volts
With carrier injection
Nocarrier injection
20-5. toff = trr + t3 = trr + 
IF
diR/dt
 = trr + 
2000
2.5x108
 = trr + 8 ms
trr = 
2!t!IF
diR/dt
 ; t = 4x10-12(BVBD)
2 = 4x10-12(2000)2 = 16 ms
trr = 
(2)(1.6x10-5)(2x103)
2.5x108
 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 
1.3x1017
BVBD
 = 
1.3x1017
150 = 8.7x10
14 cm-3
Wd = 10
-5 BVBD = (10
-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 
1
q!mn!Nd
 
L
A 
A = 
2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2)
 = 0.42 cm2
0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von
 in
volts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + 
IF
diR/dt
 = trr + 
2000
2.5x108
 = trr + 8 ms
trr = 
2!t!IF
diR/dt
 ; t = 4x10-12(BVBD)
2 = 4x10-12(2000)2 = 16 ms
trr = 
(2)(1.6x10-5)(2x103)
2.5x108
 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 
1.3x1017
BVBD
 = 
1.3x1017
150 = 8.7x10
14 cm-3
Wd = 10
-5 BVBD = (10
-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 
1
q!mn!Nd
 
L
A 
A = 
2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2)
 = 0.42 cm2
0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von
 in
volts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + 
IF
diR/dt
 = trr + 
2000
2.5x108
 = trr + 8 ms
trr = 
2!t!IF
diR/dt
 ; t = 4x10-12(BVBD)
2 = 4x10-12(2000)2 = 16 ms
trr = 
(2)(1.6x10-5)(2x103)
2.5x108
 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 
1.3x1017
BVBD
 = 
1.3x1017
150 = 8.7x10
14 cm-3
Wd = 10
-5 BVBD = (10
-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 
1
q!mn!Nd
 
L
A 
A = 
2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2)
 = 0.42 cm2
20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] 
Î
Í
Í
È
˚
˙
˙
˘2!e
q!W
2
d
 
= [(2x105)(2x10-3) - 300] 
(2)(11.7)(8.9x10-14)
(1.6x10-19)(2x10-3)2
 = Nd = 3.4x10
14 cm-3
20-9.
R
A npt
 = 
Wd(npt)
q!mn!Nnpt
 ; Nnpt = Nd of non-punch-throuth (npt) diode
R
A pt
 = 
Wd(pt)
q!mn!Npt
 ; Npt = Nd of punch-through (pt) diode
Wd(npt) = 
e!EBD
!q!Nnpt
 ; Derived from Eqs. (19-11), (19-12), and (19-13)
Wd(pt) = 
e!EBD
!q!Npt
 
Î
Í
Í
È
˚
˙
˙
˘
1!+_ ! 1!-!
2!q!Npt!BVBD
e!E
2
BD
 
2!q!Npt!BVBD
e!E
2
BD
 = 
q!Npt
e!EBD
 
Nnpt
Nnpt
 
2!BVBD
EBD
 = 
q!Nnpt
e!EBD
 
2!BVBD
EBD
 
Npt
Nnpt
 
= 
1
Wd(npt)
 Wd(npt) 
Npt
Nnpt
 = x = 
Npt
Nnpt
 ; Wd(pt) = Wd(npt) 
1
x [ ]1!+_ ! 1!-!x 
If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10)
Limit of 
1
x [ ]1!+_ ! 1!-!x as x approaches is infinite for the plus root and 0.5 for the 
minus root. Hence the minus root is the correct choice to use.
R
A pt
 = 
Wd(npt)!
1
x![ ]1!-!! 1!-!x
q!mn!Npt
 = 
Wd(npt)!
1
x![ ]1!-!! 1!-!x
q!mn!Npt
 
Nnpt
!Nnpt
 
= 
Wd(npt)
!q!mn!Nnpt
 
1
x2
 [1 - 1!-!x ] = 
R
A npt
 
1
x2
 [1 - 1!-!x ]
d
dx ÎÍ
È
˚
˙˘!
1
x2
!! 1!-!x! = 0 = 
-2
x3
 [1 - 1!-!x ] + 
1
2!x2! 1!-!x
 
Solving for x yields x = 
8
9 i.e. Npt = 
8
9 Nnpt ; Wd(pt) = 0.75 Wd(npt)
20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] 
Î
Í
Í
È
˚
˙
˙
˘2!e
q!W
2
d
 
= [(2x105)(2x10-3) - 300] 
(2)(11.7)(8.9x10-14)
(1.6x10-19)(2x10-3)2
 = Nd = 3.4x10
14 cm-3
20-9.
R
A npt
 = 
Wd(npt)
q!mn!Nnpt
 ; Nnpt = Nd of non-punch-throuth (npt) diode
R
A pt
 = 
Wd(pt)
q!mn!Npt
 ; Npt = Nd of punch-through (pt) diode
Wd(npt) = 
e!EBD
!q!Nnpt
 ; Derived from Eqs. (19-11), (19-12), and (19-13)
Wd(pt) = 
e!EBD
!q!Npt
 
Î
Í
Í
È
˚
˙
˙
˘
1!+_ ! 1!-!
2!q!Npt!BVBD
e!E
2
BD
 
2!q!Npt!BVBD
e!E
2
BD
 = 
q!Npt
e!EBD
 
Nnpt
Nnpt
 
2!BVBD
EBD
 = 
q!Nnpt
e!EBD
 
2!BVBD
EBD
 
Npt
Nnpt
 
= 
1
Wd(npt)
 Wd(npt) 
Npt
Nnpt
 = x = 
Npt
Nnpt
 ; Wd(pt) = Wd(npt) 
1
x [ ]1!+_ ! 1!-!x 
If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10)
Limit of 
1
x [ ]1!+_ ! 1!-!x as x approaches is infinite for the plus root and 0.5 for the 
minus root. Hence the minus root is the correct choice to use.
R
A pt
 = 
Wd(npt)!
1
x![ ]1!-!! 1!-!x
q!mn!Npt
 = 
Wd(npt)!
1
x![ ]1!-!! 1!-!x
q!mn!Npt
 
Nnpt
!Nnpt
 
= 
Wd(npt)
!q!mn!Nnpt
 
1
x2
 [1 - 1!-!x ] = 
R
A npt
 
1
x2
 [1 - 1!-!x ]
d
dx ÎÍ
È
˚
˙˘!
1
x2
!! 1!-!x! = 0 = 
-2
x3
 [1 - 1!-!x ] + 
1
2!x2! 1!-!x
 
Solving for x yields x = 
8
9 i.e. Npt = 
8
9 Nnpt ; Wd(pt) = 0.75 Wd(npt)
R
A npt
 = 0.84 
R
A npt
20-10. Cjo = 
eA
Wd(0)
 ; Area A determined by on-state voltage and current. Depletion-layer 
width Wd(0) set by breakdown voltage. Wd(0) same for both diodes.
Wd(150) = Wd(0) 1!+!
150
0.7 = (10
-5)(150) = 15 mm ;
Wd(0) = 
15
(150)/(0.7) ≈ 1 mm
Schottky diode area ; Vdrift = 2 volts = 
Wd(150)
q!mn!Nd!ASchottky
 IF
Drift region doping density Nd same for both diodes. Nd = 
1.3x1017
150 
= 8.7x1014 cm-3
ASchottky = 
(1.5x10-3)(300)
(2)(1.6x10-19)(1500)(8.7x1014)
 = 1.07 cm2
PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF)
2/3 ; Eq. (20-16)
K1 = 
Wd(150)
q!mo!Apn!nb
 = 
1.5x10-3
(1.6x10-19)(900)(Apn)(10
17)
 = 
10-4
Apn
 
K2 = 
3 W
4
d(150)
q2!m
3
o!n
2
b!A
2
pn!to
 = 
3 (1.5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(A
2
pn)!(2x10-6)
 
K2 = 2.4x10
-4 (Apn)
-0.67 ; 2 volts = 
10-4
Apn
 (300) + 2.4x10-4 (Apn)
-0.67 (300)0.67
Apn = 0.015 + 0.00154 (Apn)
0.333 ; Solve by successive approximation ;
Apn ≈ 0.017 cm
2
Schottky diode Cjo = 
(11.7)(8.9x10-14)(1.07)
10-4
 ≈ 11 nF = 0.011 mF
R
A npt
 = 0.84 
R
A npt
20-10. Cjo = 
eA
Wd(0)
 ; Area A determined by on-state voltage and current. Depletion-layer 
width Wd(0) set by breakdown voltage. Wd(0) same for both diodes.
Wd(150) = Wd(0) 1!+!
150
0.7 = (10
-5)(150) = 15 mm ;
Wd(0) = 
15
(150)/(0.7) ≈ 1 mm
Schottky diode area ; Vdrift = 2 volts = 
Wd(150)
q!mn!Nd!ASchottky
 IF
Drift region doping density Nd same for both diodes. Nd = 
1.3x1017
150 
= 8.7x1014 cm-3
ASchottky = 
(1.5x10-3)(300)
(2)(1.6x10-19)(1500)(8.7x1014)
 = 1.07 cm2
PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF)
2/3 ; Eq. (20-16)
K1 = 
Wd(150)
q!mo!Apn!nb
 = 
1.5x10-3
(1.6x10-19)(900)(Apn)(10
17)
 = 
10-4
Apn
 
K2 = 
3 W
4
d(150)
q2!m
3
o!n
2
b!A
2
pn!to
 = 
3 (1.5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(A
2
pn)!(2x10-6)
 
K2 = 2.4x10
-4 (Apn)
-0.67 ; 2 volts = 
10-4
Apn
 (300) + 2.4x10-4 (Apn)
-0.67 (300)0.67
Apn = 0.015 + 0.00154 (Apn)
0.333 ; Solve by successive approximation ;
Apn ≈ 0.017 cm
2
Schottky diode Cjo = 
(11.7)(8.9x10-14)(1.07)
10-4≈ 11 nF = 0.011 mF
PN junction diode Cjo = 
(11.7)(8.9x10-14)(0.017)
10-4
 ≈ 180 pF
20-11.
BVcyl
BVp
 = 2 r2 (1 + 
1
r ) ln (1 + 
1
r ) - 2r
•
•
•
•
•
• • •
•
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1 1 10
BVcyl / BVp
r = R/(2W n)
20-12.
BVcyl
BVp
 = 
950
1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = 
R
2!Wn
 
R ≈ 12 Wn
PN junction diode Cjo = 
(11.7)(8.9x10-14)(0.017)
10-4
 ≈ 180 pF
20-11.
BVcyl
BVp
 = 2 r2 (1 + 
1
r ) ln (1 + 
1
r ) - 2r
•
•
•
•
•
• • •
•
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1 1 10
BVcyl / BVp
r = R/(2W n)
20-12.
BVcyl
BVp
 = 
950
1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = 
R
2!Wn
 
R ≈ 12 Wn
PN junction diode Cjo = 
(11.7)(8.9x10-14)(0.017)
10-4
 ≈ 180 pF
20-11.
BVcyl
BVp
 = 2 r2 (1 + 
1
r ) ln (1 + 
1
r ) - 2r
•
•
•
•
•
• • •
•
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1 1 10
BVcyl / BVp
r = R/(2W n)
20-12.
BVcyl
BVp
 = 
950
1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = 
R
2!Wn
 
R ≈ 12 Wn
Chapter 21 Problem Solutions
21-1. NPN BJT ; BVCEO = 
BVCBO
b1/4
 ; PNP BJT ; BVCEO = 
BVCBO
b1/6
 
B
*
*
*
* * * * * *
*
* * * * * * *
J
•
•
• • • ••••
•
• • • • •••
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100
BVCEO
BVCBO
PNP
NPN
21-2. The flow of large reverse base currents when emitter current is still flowing in the 
forward direction will lead to emitter current crowding towards the center of the emitter. 
This accenuates the possibility of second breakdown. See Fig. 21-8b.
When emitter-open switching is used, there is no emitter current and thus no emitter 
current crowding and second breakdown is much less likely to occur.
21-3. a) Idealized BJT current and voltage waveforms in step-down converter
Chapter 21 Problem Solutions
21-1. NPN BJT ; BVCEO = 
BVCBO
b1/4
 ; PNP BJT ; BVCEO = 
BVCBO
b1/6
 
B
*
*
*
* * * * * *
*
* * * * * * *
J
•
•
• • • ••••
•
• • • • •••
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100
BVCEO
BVCBO
PNP
NPN
21-2. The flow of large reverse base currents when emitter current is still flowing in the 
forward direction will lead to emitter current crowding towards the center of the emitter. 
This accenuates the possibility of second breakdown. See Fig. 21-8b.
When emitter-open switching is used, there is no emitter current and thus no emitter 
current crowding and second breakdown is much less likely to occur.
21-3. a) Idealized BJT current and voltage waveforms in step-down converter
Chapter 21 Problem Solutions
21-1. NPN BJT ; BVCEO = 
BVCBO
b1/4
 ; PNP BJT ; BVCEO = 
BVCBO
b1/6
 
B
*
*
*
* * * * * *
*
* * * * * * *
J
•
•
• • • ••••
•
• • • • •••
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100
BVCEO
BVCBO
PNP
NPN
21-2. The flow of large reverse base currents when emitter current is still flowing in the 
forward direction will lead to emitter current crowding towards the center of the emitter. 
This accenuates the possibility of second breakdown. See Fig. 21-8b.
When emitter-open switching is used, there is no emitter current and thus no emitter 
current crowding and second breakdown is much less likely to occur.
21-3. a) Idealized BJT current and voltage waveforms in step-down converter
T/2
I oVd
V (t)
CEi (t)C
t d,off t rv t fit d,on t ri t fv
BJT power dissipation Pc = 
1
T ıÛ
0
T
!vCE(t)!iC(t)!dt = {Eon + Esw }fs ; fs = 
1
T 
Eon = VCE,on Io {
T
2 + td,off - td,on} ≈ (2)(40) 
1
2!fs
 = 
40
fs
 Joules
 Esw = Eri + Efv + Erv + Efi 
Eri = 
Vd!Io!tri
2 = 
(100)(40)(2x10-7)
2 = 4x10
-4 Joules
Efv = 
Vd!Io!tfv
2 = 
(100)(40)(1x10-7)
2 = 2x10
-4 Joules
Erv = 
Vd!Io!trv
2 = 
(100)(40)(1x10-7)
2 = 2x10
-4 Joules
Efi = 
Vd!Io!tfi
2 = 
(100)(40)(2x10-7)
2 = 4x10
-4 Joules
Esw = 1.2x10
-3 Joules ; Pc = [
40
fs
 + 1.2x10-3] fs = 40 + 1.2x10
-3 fs
40
80
0
33 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 
150!-!25
1 = 125 watts = 40 + 1.2x10
-3 fs,max
fs,max = 
(125!-!40)
1.2x10-3
 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching 
times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 
0.4
(125!-!25) = 4x10
-3
110 - 50 = Rqja {40 + 1.2x10
-3[1+0.004(110-25)] (2.5x104)}
 Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to 
the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = 
q!Dn!Na!A
!Wb
 = 
(1.6x10-19)(38)(1016)(1)
(3x10-4)
 = 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the 
maximum current that the transistor can carry. These lateral voltage drops lead to emitter 
current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion 
layer protrudes a significant amount into the base. This encroachment may stretch across 
the base and reach the EB depletion layer before the desired blocking voltage is reached. 
At reach-through xp will equal the difference between the base width WB and the base-
side protusion of the base-emitter depletion layer width WEB,depl.The reach-through 
voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
40
80
0
33 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 
150!-!25
1 = 125 watts = 40 + 1.2x10
-3 fs,max
fs,max = 
(125!-!40)
1.2x10-3
 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching 
times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 
0.4
(125!-!25) = 4x10
-3
110 - 50 = Rqja {40 + 1.2x10
-3[1+0.004(110-25)] (2.5x104)}
 Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to 
the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = 
q!Dn!Na!A
!Wb
 = 
(1.6x10-19)(38)(1016)(1)
(3x10-4)
 = 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the 
maximum current that the transistor can carry. These lateral voltage drops lead to emitter 
current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion 
layer protrudes a significant amount into the base. This encroachment may stretch across 
the base and reach the EB depletion layer before the desired blocking voltage is reached. 
At reach-through xp will equal the difference between the base width WB and the base-
side protusion of the base-emitter depletion layer width WEB,depl.The reach-through 
voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
40
80
0
33 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 
150!-!25
1 = 125 watts = 40+ 1.2x10
-3 fs,max
fs,max = 
(125!-!40)
1.2x10-3
 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching 
times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 
0.4
(125!-!25) = 4x10
-3
110 - 50 = Rqja {40 + 1.2x10
-3[1+0.004(110-25)] (2.5x104)}
 Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to 
the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = 
q!Dn!Na!A
!Wb
 = 
(1.6x10-19)(38)(1016)(1)
(3x10-4)
 = 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the 
maximum current that the transistor can carry. These lateral voltage drops lead to emitter 
current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion 
layer protrudes a significant amount into the base. This encroachment may stretch across 
the base and reach the EB depletion layer before the desired blocking voltage is reached. 
At reach-through xp will equal the difference between the base width WB and the base-
side protusion of the base-emitter depletion layer width WEB,depl.The reach-through 
voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
40
80
0
33 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 
150!-!25
1 = 125 watts = 40 + 1.2x10
-3 fs,max
fs,max = 
(125!-!40)
1.2x10-3
 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching 
times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 
0.4
(125!-!25) = 4x10
-3
110 - 50 = Rqja {40 + 1.2x10
-3[1+0.004(110-25)] (2.5x104)}
 Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to 
the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = 
q!Dn!Na!A
!Wb
 = 
(1.6x10-19)(38)(1016)(1)
(3x10-4)
 = 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the 
maximum current that the transistor can carry. These lateral voltage drops lead to emitter 
current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion 
layer protrudes a significant amount into the base. This encroachment may stretch across 
the base and reach the EB depletion layer before the desired blocking voltage is reached. 
At reach-through xp will equal the difference between the base width WB and the base-
side protusion of the base-emitter depletion layer width WEB,depl.The reach-through 
voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
40
80
0
33 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 
150!-!25
1 = 125 watts = 40 + 1.2x10
-3 fs,max
fs,max = 
(125!-!40)
1.2x10-3
 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching 
times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 
0.4
(125!-!25) = 4x10
-3
110 - 50 = Rqja {40 + 1.2x10
-3[1+0.004(110-25)] (2.5x104)}
 Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to 
the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = 
q!Dn!Na!A
!Wb
 = 
(1.6x10-19)(38)(1016)(1)
(3x10-4)
 = 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the 
maximum current that the transistor can carry. These lateral voltage drops lead to emitter 
current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion 
layer protrudes a significant amount into the base. This encroachment may stretch across 
the base and reach the EB depletion layer before the desired blocking voltage is reached. 
At reach-through xp will equal the difference between the base width WB and the base-
side protusion of the base-emitter depletion layer width WEB,depl.The reach-through 
voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
Wo,EB = 
2!e!fc!(NaB!+!NdE)
!q!NaB!NdE
 ; fcE = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!NdE
n
2
i
 
 fcE = 0.26 ln Î
Í
È
˚
˙
˘
!
1034
1020
 = 0.84 V
Wo,EB = 
(2)(11.7)(8.9x10-14)(0.84)(1019!+!1015)
(1.6x10-19)(1019)(1015)
 = 0.33 microns
Estimate of collector-base base-side protusion of WCB,depl.
CB depletion layer thickness W(V) = Wo,CB 1!+!
V
fc = xp + xn ;
xp = protrusion of CB depletion layer into p-type base region. xp = 
W(V)
11 using xp Na
= xn Nd (charge neutrality).
Wo,CB = 
2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC
 ; fcC = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!NdC
n
2
i
 
 fcC = 0.26 ln Î
Í
È
˚
˙
˘
!
1029
1020
 = 0.54 V
Wo,CB = 
(2)(11.7)(8.9x10-14)(0.54)(1014!+!1015)
(1.6x10-19)(1014)(1015)
 = 2.8 microns
{(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+!
V
0.54 : Solving for V yields V = 59 volts. 
Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V.
21-8. BVEBO = 10 V = 
1.3x1017
!NaB
 ; NaB = acceptor doping density in base = 1.3x10
16 cm-3
BVCBO = b
1/4 BVCEO = (5)
1/4(1000) = ≈ 1500 Volts ≈ 
1.3x1017
!NdC
 
NdC = collector drift region donor density = 8.7x10
13 cn -3
Wo,EB = 
2!e!fc!(NaB!+!NdE)
!q!NaB!NdE
 ; fcE = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!NdE
n
2
i
 
 fcE = 0.26 ln Î
Í
È
˚
˙
˘
!
1034
1020
 = 0.84 V
Wo,EB = 
(2)(11.7)(8.9x10-14)(0.84)(1019!+!1015)
(1.6x10-19)(1019)(1015)
 = 0.33 microns
Estimate of collector-base base-side protusion of WCB,depl.
CB depletion layer thickness W(V) = Wo,CB 1!+!
V
fc = xp + xn ;
xp = protrusion of CB depletion layer into p-type base region. xp = 
W(V)
11 using xp Na
= xn Nd (charge neutrality).
Wo,CB = 
2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC
 ; fcC = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!NdC
n
2
i
 
 fcC = 0.26 ln Î
Í
È
˚
˙
˘
!
1029
1020
 = 0.54 V
Wo,CB = 
(2)(11.7)(8.9x10-14)(0.54)(1014!+!1015)
(1.6x10-19)(1014)(1015)
 = 2.8 microns
{(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+!
V
0.54 : Solving for V yields V = 59 volts. 
Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V.
21-8. BVEBO = 10 V = 
1.3x1017
!NaB
 ; NaB = acceptor doping density in base = 1.3x10
16 cm-3
BVCBO = b
1/4 BVCEO = (5)
1/4(1000) = ≈ 1500 Volts ≈ 
1.3x1017
!NdC
 
NdC = collector drift region donor density = 8.7x10
13 cn -3
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the 
base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO)
Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10
-5)(1500) 
= 150 mm
xp(BVCBO) = 
xn(BVCBO)!NdC
NaB
 = xn(BVCBO) 
8.7x1013
1.3x1016
 = xn(BVCBO) 6.7x10
-3
xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm
Hence xp(BVCBO) = (1.5x10
-2 cm)(6.7x10-3) = 1 micron = Wbase
21-9. Beta = 150 = bD bM + bD + bM =20 bM + 20 + bM
bM = 
150!-!20
21 = 6.2
21-10. Must first ascertain the operating states of the two transistors.Two likely choices 
including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT 
both saturated. Initially assume driver BJT saturated and main BJT active.
10 I B,M
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
 
10 IB,M + IB,M = 100 A
IB,M = IC,D = 9.1A 
IC,M = 91 A
VCE,D = 0.2 +(.02)(9.1) = 0.382 V
VCE,M = VCE,D + 0.8 V = 1.18 v
But a saturated main BJT with IC,M = 91 A
going through 0.02 ohms generates
a voltage drop of 1.8 V which is > 1.18 V.
Hence main BJT must be saturated.
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the 
base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO)
Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10
-5)(1500) 
= 150 mm
xp(BVCBO) = 
xn(BVCBO)!NdC
NaB
 = xn(BVCBO) 
8.7x1013
1.3x1016
 = xn(BVCBO) 6.7x10
-3
xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm
Hence xp(BVCBO) = (1.5x10
-2 cm)(6.7x10-3) = 1 micron = Wbase
21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM
bM = 
150!-!20
21 = 6.2
21-10. Must first ascertain the operating states of the two transistors.Two likely choices 
including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT 
both saturated. Initially assume driver BJT saturated and main BJT active.
10 I B,M
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
 
10 IB,M + IB,M = 100 A
IB,M = IC,D = 9.1A 
IC,M = 91 A
VCE,D = 0.2 +(.02)(9.1) = 0.382 V
VCE,M = VCE,D + 0.8 V = 1.18 v
But a saturated main BJT with IC,M = 91 A
going through 0.02 ohms generates
a voltage drop of 1.8 V which is > 1.18 V.
Hence main BJT must be saturated.
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the 
base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO)
Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10
-5)(1500) 
= 150 mm
xp(BVCBO) = 
xn(BVCBO)!NdC
NaB
 = xn(BVCBO) 
8.7x1013
1.3x1016
 = xn(BVCBO) 6.7x10
-3
xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm
Hence xp(BVCBO) = (1.5x10
-2 cm)(6.7x10-3) = 1 micron = Wbase
21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM
bM = 
150!-!20
21 = 6.2
21-10. Must first ascertain the operating states of the two transistors.Two likely choices 
including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT 
both saturated. Initially assume driver BJT saturated and main BJT active.
10 I B,M
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
 
10 IB,M + IB,M = 100 A
IB,M = IC,D = 9.1A 
IC,M = 91 A
VCE,D = 0.2 +(.02)(9.1) = 0.382 V
VCE,M = VCE,D + 0.8 V = 1.18 v
But a saturated main BJT with IC,M = 91 A
going through 0.02 ohms generates
a voltage drop of 1.8 V which is > 1.18 V.
Hence main BJT must be saturated.
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
0.02 W
0.6 V
-
+
I C,M
 
Neglecting IB,D
IC,M + IC,D = 100 A
(.02)IC,M -0.6 = (.02) I C,D +0.2
IC,D = 30 A ; IC,M = 70 A
VCE,M = 0.2 + (70)(.02) = 1.6 V
PDarl = VCE,M [IC,M + IC,D]
PDarl = (1.6 V)(100 A) = 160 W
21-11. CEBO = 
e!AE
WEBO 
 ; CCBO = 
e!AC
WCBO
 
WEBO = zero-bias emitter-base depletion layer thickness
WCBO = zero-bias collector-base depletion layer thickness
WEBO = 
2!e!fcE!(NaB!+!NdE)
!q!NaB!NdE
 ; fcE = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!NdE
n
2
i
 
fcE = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(1016)
1020
 = 0.89 V ;
WEBO = 
(2)(11.7)(8.9x10-14)(0.89)(1019!+!1016)
(1.6x10-19)(1019)(1016)
 = 0.34 microns
CEBO = 
(11.7)(8.9x10-14)(0.3)
3.4x10-5
 = 9.2 nF
WCBO = 
2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC
 ; fcC = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!!NdC
n
2
i
 
fcC = 0.026 ln Î
Í
È
˚
˙
˘
!
(1.3x1016)(8.7x1013)
1020
 = 0.6 V ;
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
0.02 W
0.6 V
-
+
I C,M
 
Neglecting IB,D
IC,M + IC,D = 100 A
(.02)IC,M -0.6 = (.02) I C,D +0.2
IC,D = 30 A ; IC,M = 70 A
VCE,M = 0.2 + (70)(.02) = 1.6 V
PDarl = VCE,M [IC,M + IC,D]
PDarl = (1.6 V)(100 A) = 160 W
21-11. CEBO = 
e!AE
WEBO 
 ; CCBO = 
e!AC
WCBO
 
WEBO = zero-bias emitter-base depletion layer thickness
WCBO = zero-bias collector-base depletion layer thickness
WEBO = 
2!e!fcE!(NaB!+!NdE)
!q!NaB!NdE
 ; fcE = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!NdE
n
2
i
 
fcE = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(1016)
1020
 = 0.89 V ;
WEBO = 
(2)(11.7)(8.9x10-14)(0.89)(1019!+!1016)
(1.6x10-19)(1019)(1016)
 = 0.34 microns
CEBO = 
(11.7)(8.9x10-14)(0.3)
3.4x10-5
 = 9.2 nF
WCBO = 
2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC
 ; fcC = 
k!T
q ln 
Î
Í
Í
È
˚
˙
˙
˘
!
NaB!!NdC
n
2
i
 
fcC = 0.026 ln Î
Í
È
˚
˙
˘
!
(1.3x1016)(8.7x1013)
1020
 = 0.6 V ;
WCBO = 
(2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013)
(1.6x10-19)(1.3x1016)(8.7x1013)
 = 2.1 microns
CCBO = 
(11.7)(8.9x10-14)(3)
2.1x10-4
 = 14.8 nF
21-12. Equivalent circuit for turn-on delay time, td,on, calculation.
+
-
10 W
CCB
CBE
VBEVin
+
-
-8 V
8 V
Vin
t
VBE(t) = 8 - 16 exp Î
ÍÈ
˚
˙˘-t
!t ; t = (10 W)(CBE + CCB)
At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln Î
ÍÈ
˚
˙˘16
7.3 
The space-charge capacitances are nonlinear functions of the voltages across them. Need 
to find an average value for each of the two capacitors. During the td,on interval, the 
voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. 
Hence CCB will be given by
CCB = 
CCBO
! 1!+!
VCB
fcC
 = 
1.5x10-8
1!+!
100
0.6
 = 1.2 nF
The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find 
the average value of CBE.
WCBO = 
(2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013)
(1.6x10-19)(1.3x1016)(8.7x1013)
 = 2.1 microns
CCBO = 
(11.7)(8.9x10-14)(3)
2.1x10-4
 = 14.8 nF
21-12. Equivalent circuit for turn-on delay time, td,on, calculation.
+
-
10 W
CCB
CBE
VBEVin
+
-
-8 V
8 V
Vin
t
VBE(t) = 8 - 16 exp Î
ÍÈ
˚
˙˘-t
!t ; t = (10 W)(CBE + CCB)
At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln Î
ÍÈ
˚
˙˘16
7.3 
The space-charge capacitances are nonlinear functions of the voltages across them. Need 
to find an average value for each of the two capacitors. During the td,on interval, the 
voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. 
Hence CCB will be given by
CCB = 
CCBO
! 1!+!
VCB
fcC
 = 
1.5x10-8
1!+!
100
0.6
 = 1.2 nF
The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find 
the average value of CBE.
CBE = 
ıÙ
ÙÛ
8
0
!
CEBO
! 1!+!
VEB
fcE
!dVEB
ıÛ
8
0
dVEB
 = 
CEBO!fcE
(-8) Î
Í
È
˚
˙
˘
1!+!
0
fcE!!-!! 1!+!
8
fcE 
CBE = 
(9.2x10-9)(0.89)
8 [ 1!+!
8
0.89 - 1] = 2.2 nF
td,on = (10) [2.2x10
-9 + 1.2x10-9] ln ÎÍ
È
˚
˙˘16
7.3 = (3.4x10
-8)(0.78) ≈ 27 nanoseconds
Chapter 22 Problem Solutions
22-1. The capacitance of the gate-source terminals can be modeled as two capacitors connected 
electrically in series. Cox is the capacitance of the oxide layer and is a constant 
independent of vGS. Cdepl(vGS) isthe capacitance of the depletion layer which 
increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the 
depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gate-
source capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), 
the depletion layer thickness becomes constant because the formation of the inversion 
layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS
(any additional increase in vGS is dropped across the oxide layer). Thus both 
components of Cgs are constant for vGS > VGS(th).
22-2. a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we 
need the numerical values of the various waveform parameters. The voltage and current 
amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) 
= 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the 
various switching times.
Chapter 22 Problem Solutions
22-1. The capacitance of the gate-source terminals can be modeled as two capacitors connected 
electrically in series. Cox is the capacitance of the oxide layer and is a constant 
independent of vGS. Cdepl(vGS) is the capacitance of the depletion layer which 
increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the 
depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gate-
source capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), 
the depletion layer thickness becomes constant because the formation of the inversion 
layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS
(any additional increase in vGS is dropped across the oxide layer). Thus both 
components of Cgs are constant for vGS > VGS(th).
22-2. a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we 
need the numerical values of the various waveform parameters. The voltage and current 
amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) 
= 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the 
various switching times.
T/2
I oVd
V (t)
CE
i (t)
C
t d,off t rv t fit d,on t ri t fv
-VGG
0
VGS(th)
VGS,Io
VGG
V (t)GS
td,on estimate - Use equivalent circuit of Fig. 22-12a.
Governing equation is 
dvGS
dt + 
vGS
RG(Cgs!+!Cgd)
 = 
VGG
RG(Cgs!+!Cgd)
 ;
Boundary condition vGS(0) = - VGG
Solution is vGS(t) = VGG - 2 VGG e
-t/t ; t = RG(Cgs + Cgd) ;
At t = td,on , vGS = VGS(th). Solving for td,on yields
td,on = RG(Cgs + Cgd) ln Î
Í
È
˚
˙
˘
!
2!VGG
!VGG!-!VGS(th)
 
td,on = (50) (1.15x10
-9) ln ÎÍ
È
˚
˙˘!
(2)(15)
(15!-!4) = 58 ns
tri estimate - Use equivalent circuit of Fig. 22-12b.
vGS(t) still given by governing equation given above in td,on estimate. Changing time 
origin to when vGS = VGS(th) yields;
vGS(t) = VGG + [VGG - VGS(th)] e
-t/t . The drain current is given by
iD(t) = Cgd 
d(Vd!-!vGS)
dt + gm[vGS(t) - VGs(th)] ; gm = 
10
7!-!4 = 3.3 mhos
At t = tri, iD = Io. Substituting vGS(t) into iD(t) and solving for tri yields
tri = RG(Cgs + Cgd) ln Î
Í
È
˚
˙
˘
!
(VGG!-!VGS(th)){gm!+ !
Cgd
RG(Cgs!+!Cgd)
}
gm(VGG!-!VGS(th))!-!Io
 
tri = (50)(1.15x10
-9) ln Î
Í
Í
È
˚
˙
˙
˘
!
(15!-!4)(3.3!+!
1.5x10-10
(50)(1.15x10-9)
)
(3.3)(15!-!4)!-!10 = 21 ns
tfv estimate - Use equivalent circuit of Fig. 22-12c.
vGS approximately constant at VGS,Io = 
Io
gm
 + VGS(th) during this interval.
Governing equation is Cgd 
dvDS
dt = - Î
Í
È
˚
˙
˘
!
VGG!-!VGS(th)!-!
Io
gm
RG
 with vDS(0) = Vd.
Solution is given by
vDS(t) = Vd - Î
Í
È
˚
˙
˘
VGG!-!VGS(th)!-!
Io
gm
 
t
RGCgd
 ; At t = tfv, vDS = 0.
Solving for tfv yields
tfv = 
RG!Cgd!Vd
VGG!-!VGS(th)!-!
Io
gm
 = (50)(1.5x10-10) 
300
(15!-!4!-!3) = 300 ns
td,off estimate - use equivalent circuit of Fig. 22-12d with the input voltagge VGG
reversed.
vGS(t) = - VGG + 2 VGG e
-t/t ; At t = td,off, vGS = VGS,Io. Solving for td,off
td,off = RG(Cgs + Cgd) ln 
Î
Í
È
˚
˙
˘
!
2!VGG
VGG!+!VGS(th)!+!
Io
gm
 
= (50)(1.15x10-9) ln 
Î
Í
Í
È
˚
˙
˙
˘
!
(2)(15)
10
3.3!+!4!+!15
 = 18 ns
trv estimate - Use equivalent circuit of Fig. 22-12c with the input voltage VGG
reversed. vGS approximately constant at VGS,Io as in previous of tfv. Governing 
equation is
Cgd 
d{vDS!-!VGS,Io}
dt = 
VGG!+!VGS,Io
RG
 with vDS(0) = 0. Solution given by
vDS(t) = 
VGG!+!VGS,Io
RG!Cgd
 t . At t = trv , vDS = Vd . Solving for trv yields
trv = 
Vd!RG!Cgd
VGG!+!VGS,Io
 = 
(300)(50)(1.5x10-10)
(15!+!7) = 100 ns
tfi estimate - use equivalent circuit of Fig. 22-12b with the input voltage VGG
reversed. Governing equation the same as in previous calculation of tri. At t = 0, vGS(0) 
= VGS,Io. Solution in this caae is given by
vGS(t) = - VGG + [VGS,Io + VGG] e
-t/t ; At t = tfi, vGS = VGS(th). Solving for tfi
tfi = RG(Cgs + Cgd) ln Î
Í
È
˚
˙
˘
!
VGG!+!VGS,Io
VGG+!VGS(th)
 = (50)(1.15x10-9)ln ÎÍ
È
˚
˙˘!
15!+!7
15!+!4 = 9 ns
b) Estimate the power dissipated in the MOSFET in the same manner as was done for the
BJT in problem 21-3.Waveforms for the MOSFET are the same as for the BJT except for
appropriate re-labeling of the currents and voltages.
Eri = (0.5)(300)(10)(2.1x10
-8) = 3x10-5 Joules
Efv = (0.5)(300)(10)(3x10
-7) = 4.5x10-4 Joules
Eon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 V
T >> td,on and td,off
Eon = (10)(300)(0.5)(5x10
-5) = 1.25x10-3 Joules
Erv = (0.5)(300)(10)(10
-7) = 1.5x10-4 Joules
Efi = (0.5)(300)(10)(9x10
-9) = 1.5x10-5 Joules
Pc = (1.95x10
-3)(2x104) = 39 watts
22-3. Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with 
the 150 ohm load. Test circuit waveforms are shown below.
Eri = (0.5)(300)(10)(2.1x10
-8) = 3x10-5 Joules
Efv = (0.5)(300)(10)(3x10
-7) = 4.5x10-4 Joules
Eon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 V
T >> td,on and td,off
Eon = (10)(300)(0.5)(5x10
-5) = 1.25x10-3 Joules
Erv = (0.5)(300)(10)(10
-7) = 1.5x10-4 Joules
Efi = (0.5)(300)(10)(9x10
-9) = 1.5x10-5 Joules
Pc = (1.95x10
-3)(2x104) = 39 watts
22-3. Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with 
the 150 ohm load. Test circuit waveforms are shown below.
V (t)G
VGG
V (t)GS
VGS(th)
VGS,Io
t
t d,on
t = t r i fv t = tf i rv
td,off
t
VGG
Vd
V (t)DS
I o
i (t)
D
t
Equivalent circuit during voltage and current rise and fall intervals:
+
- Cgs
Cgd
V (t)G
R D
V
d
g (V - V )
m GS GS(th)
RG
Governing equation using Miller capacitance approximation:
dvGS
dt + 
vGS
t = 
!VG(t)
t ; t = RG [Cgs + Cgd{1 + gmRD}] ;
During tri = tfv interval, VG(t) = VGG. Solution is
vGS(t) = VGG + {VGS(th) - VGG} e
-t/t ; At t = tri, VGS = VGS(th) + 
Vd
gmRD
 ;
Solving for tri = tfv yields
tri = tfv = t ln 
Î
Í
È
˚
˙
˘
!
VGG!-!VGS(th)
VGG!-!VGS(th)!-!!
Vd
!gm!RD
 
During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e
-t/t .
At t = trv, vGS(t) = VGS(th). Solving for trv yields
trv = t ln Î
Í
È
˚
˙
˘
!
VGS(th)!+!
Vd
!gm!RD
VGS(th)
 . Invert equation for tri to find Cgd. Result is
 Cgd = 
ÓÔ
Ì
ÔÏ
Ô˛
˝
Ô¸
tri
!ln!
Î
Í
È
˚
˙
˘
!
VGG!-!VGS(th)
VGG!-!VGS(th)!-!!
!Vd
!gm!RD
!!-!!RG!Cgs ÓÌ
Ï
˛
˝
¸1
RG(1!+!gmRD)
 
 Cgd= 
ÓÔ
Ì
ÔÏ
Ô˛
˝
Ô¸3x10-8
!ln!ÎÍ
È
˚
˙˘!
15!-!4
15!-!4!-!1
!!-!!5x10-9 
Ó
Ì
Ï
˛
˝
¸1
5(1!+!25) = 2.3x10
-9 F = 2.3 nF
Solving for switching times in circuit with RD = 150 ohms.
t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms
tri = 3.5x10
-5 ln ÎÍ
È
˚
˙˘!
15!-!4
15!-!4!-!2 = 7 ms ; trv = 3.5x10
-5 ln ÎÍ
È
˚
˙˘!
4!+!2
4 = 14 ms
22-4. Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in 
MOSFET given by
<PMOSFET> = [Eon + Esw] fs ; fs = 
1
T ; Eon = [ID]
2 rDS,on(Tj) 
T
2 ;
dvGS
dt + 
vGS
t = 
!VG(t)
t ; t = RG [Cgs + Cgd{1 + gmRD}] ;
During tri = tfv interval, VG(t) = VGG. Solution is
vGS(t) = VGG + {VGS(th) - VGG} e
-t/t ; At t = tri, VGS = VGS(th) + 
Vd
gmRD
 ;
Solving for tri = tfv yields
tri = tfv = t ln 
Î
Í
È
˚
˙
˘
!
VGG!-!VGS(th)
VGG!-!VGS(th)!-!!
Vd
!gm!RD
 
During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e
-t/t .
At t = trv, vGS(t) = VGS(th). Solving for trv yields
trv = t ln Î
Í
È
˚
˙
˘
!
VGS(th)!+!
Vd
!gm!RD
VGS(th)
 . Invert equation for tri to find Cgd. Result is
 Cgd = 
ÓÔ
Ì
ÔÏ
Ô˛
˝
Ô¸
tri
!ln!
Î
Í
È
˚
˙
˘
!
VGG!-!VGS(th)
VGG!-!VGS(th)!-!!
!Vd
!gm!RD
!!-!!RG!Cgs ÓÌ
Ï
˛
˝
¸1
RG(1!+!gmRD)
 
 Cgd = 
ÓÔ
Ì
ÔÏ
Ô˛
˝
Ô¸3x10-8
!ln!ÎÍ
È
˚
˙˘!
15!-!4
15!-!4!-!1
!!-!!5x10-9 
Ó
Ì
Ï
˛
˝
¸1
5(1!+!25) = 2.3x10
-9 F = 2.3 nF
Solving for switching times in circuit with RD = 150 ohms.
t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms
tri = 3.5x10
-5 ln ÎÍ
È
˚
˙˘!
15!-!4
15!-!4!-!2 = 7 ms ; trv = 3.5x10
-5 ln ÎÍ
È
˚
˙˘!
4!+!2
4 = 14 ms
22-4. Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in 
MOSFET given by
<PMOSFET> = [Eon + Esw] fs ; fs = 
1
T ; Eon = [ID]
2 rDS,on(Tj) 
T
2 ;
ID = 
Vd
RD
 = 
300
150 = 2 A ; rDS,on(Tj) = 2 Î
ÍÈ
˚
˙˘1!+!
Tj!-!25
150 = 2 Î
ÍÈ
˚
˙˘0.833!+!
Tj
150 
Eon = (4)(2) Î
ÍÈ
˚
˙˘0.833!+!
Tj
150 
1
2fs
 = {3.32 + 0.027 Tj} 
1
fs
 
Esw = 
1
T ı
ÙÛ
0
tri
Vd!ID(1!-!
t
tri
)(
t
tri
)dt + 
1
T ı
ÙÛ
0
tfi
Vd!ID(1!-!
t
tfi
)(
t
tfi
)dt = 
Vd!ID
6 [tri + tfi]
Esw = 
(300)(2)
6 [7x10
-6 + 14x10-6] = 2.1x10-3 joules
<PMOSFET> = {3.32 + 0.027 Tj} 
1
fs
 fs + 2.1x10
-3 fs
<PMOSFET> = {3.32 + 0.027 Tj} +[ 2.1x10
-3][104] = 24.3 + 0.027 Tj
B B
B B
B
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80 90 100
PMOSFET
Watts
Temperature [ °K]
22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff
ID = 
Vd
RD
 = 
300
150 = 2 A ; rDS,on(Tj) = 2 Î
ÍÈ
˚
˙˘1!+!
Tj!-!25
150 = 2 Î
ÍÈ
˚
˙˘0.833!+!
Tj
150 
Eon = (4)(2) Î
ÍÈ
˚
˙˘0.833!+!
Tj
150 
1
2fs
 = {3.32 + 0.027 Tj} 
1
fs
 
Esw = 
1
T ı
ÙÛ
0
tri
Vd!ID(1!-!
t
tri
)(
t
tri
)dt + 
1
T ı
ÙÛ
0
tfi
Vd!ID(1!-!
t
tfi
)(
t
tfi
)dt = 
Vd!ID
6 [tri + tfi]
Esw = 
(300)(2)
6 [7x10
-6 + 14x10-6] = 2.1x10-3 joules
<PMOSFET> = {3.32 + 0.027 Tj} 
1
fs
 fs + 2.1x10
-3 fs
<PMOSFET> = {3.32 + 0.027 Tj} +[ 2.1x10
-3][104] = 24.3 + 0.027 Tj
B B
B B
B
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80 90 100
PMOSFET
Watts
Temperature [ °K]
22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff
reff = 
r1!r2!r3
r1!r2!+!r2!r3!+!r3!r1
 ; r1 etc. = on-state resistance of MOSFET #1 etc.
r1(Tj) = r1(25 °C) Î
ÍÈ
˚
˙˘1!+!0.8!!
Tj!-!25
100 ; r1(105 °C) = (1.64) r1(25 °C) etc.
r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W
reff(105 °C) = 
(2.95)(3.28)(3.61)
[(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms
For the ith MOSFET, Pi = 
Von
2
2!ri
 = 
Io
2!reff
2
2!ri
 ; Assume a 50% duty cycle and ignore 
switching losses.
P1 = 
(5)2(1.09)2
(2)(2.95) = 5 W ; P2 = 
(5)2(1.09)2
(2)(3.28) = 4.5 W ; P3 = 
(5)2(1.09)2
(2)(3.61) = 4.1 W
22-6. Hybrid switch would combine the low on-state losses of the BJT and the faster switching 
of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on 
before the BJT and turned off after the BJT. The waveforms shown below indicate the 
relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the 
on-state.
reff = 
r1!r2!r3
r1!r2!+!r2!r3!+!r3!r1
 ; r1 etc. = on-state resistance of MOSFET #1 etc.
r1(Tj) = r1(25 °C) Î
ÍÈ
˚
˙˘1!+!0.8!!
Tj!-!25
100 ; r1(105 °C) = (1.64) r1(25 °C) etc.
r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W
reff(105 °C) = 
(2.95)(3.28)(3.61)
[(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms
For the ith MOSFET, Pi = 
Von
2
2!ri
 = 
Io
2!reff
2
2!ri
 ; Assume a 50% duty cycle and ignore 
switching losses.
P1 = 
(5)2(1.09)2
(2)(2.95) = 5 W ; P2 = 
(5)2(1.09)2
(2)(3.28) = 4.5 W ; P3 = 
(5)2(1.09)2
(2)(3.61) = 4.1 W
22-6. Hybrid switch would combine the low on-state losses of the BJT and the faster switching 
of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on 
before the BJT and turned off after the BJT. The waveforms shown below indicate the 
relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the 
on-state.
v = vDS CE
i D
i C
vGS
vBE
VDS,on VCE,on
I o
r I ; r < 1o
(1 - r)I o
22-7. BVDSS ≈ 
1.3x1017
Ndrift
 = 750 volts ; Ndrift = 1.7x10
14 cm-3
Wdrift ≈ (10
-5)(750) = 75 microns ;
Wd,body = protrusion of drain depletion layer into body region
≈ 
Wdrift!Ndrift
Nbody
 = 
(75)(1.7x1014)
5x1016
 ≈ 0.3 microns
Even though body-source junction is shorted, there is a depletion layer associated with it 
which is contained entirely on the body side of the junction. This must be included in the 
estimate of the required length of the body region.
Ws,body ≈ 
2!e!fc
q!Na,body
 ; fc = 
k!T
q ln ÎÍ
Í
È
˚
˙
˙
˘
!
Na!Nd
ni
2 ;
v = vDS CE
i D
i C
vGS
vBE
VDS,on VCE,on
I o
r I ; r < 1o
(1 - r)I o
22-7. BVDSS ≈ 
1.3x1017
Ndrift
 = 750 volts ; Ndrift = 1.7x10
14 cm-3
Wdrift ≈ (10
-5)(750) = 75 microns ;
Wd,body = protrusion of drain depletion layer into body region
≈ 
Wdrift!Ndrift
Nbody
 = 
(75)(1.7x1014)
5x1016
 ≈ 0.3 microns
Even though body-source junction is shorted, there is a depletion layer associated with it 
which is contained entirely on the body side of the junction. This must be included in the 
estimate of the required length of the body region.
Ws,body ≈ 
2!e!fc
q!Na,body
 ; fc = 
k!T
q ln ÎÍ
Í
È
˚
˙
˙
˘
!
Na!Nd
ni
2 ;
fc = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(5x1016)
(1020)
 = 0.94
Ws,body ≈ 
(2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016)
 ≈ 0.16 microns
 In order to avoid reach-through , Wbody > Wd,body + Ws,body 
= 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd 
dvGD
dt ≈ Cgd 
dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd 
dvDS
dt = 0.7 V
 
dvDS
dt > 
0.7
Rbody!Cgd
 will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10
6) (5x10-6) = 16.7 volts
22-10. a) iD = 
mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L 
Cox = 
e
tox
 = 
(11.7)(8.9x10-14)
10-5
 = 1.04x10-7 F/cm2
N = 
2!iD!Lmn!Cox!Wcell!(vGS!-!VGS(th))
 
N = 
!(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4)
 ≈ 5,800 cells
b) Icell = 
100
5800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = 
Wdrift
q!mn!Nd!A
 : Wdrift = 10
-5 BVDSS = (10
-5)(800) = 80 microns
fc = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(5x1016)
(1020)
 = 0.94
Ws,body ≈ 
(2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016)
 ≈ 0.16 microns
 In order to avoid reach-through , Wbody > Wd,body + Ws,body 
= 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd 
dvGD
dt ≈ Cgd 
dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd 
dvDS
dt = 0.7 V
 
dvDS
dt > 
0.7
Rbody!Cgd
 will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10
6) (5x10-6) = 16.7 volts
22-10. a) iD = 
mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L 
Cox = 
e
tox
 = 
(11.7)(8.9x10-14)
10-5
 = 1.04x10-7 F/cm2
N = 
2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
 
N = 
!(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4)
 ≈ 5,800 cells
b) Icell = 
100
5800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = 
Wdrift
q!mn!Nd!A
 : Wdrift = 10
-5 BVDSS = (10
-5)(800) = 80 microns
fc = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(5x1016)
(1020)
 = 0.94
Ws,body ≈ 
(2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016)
 ≈ 0.16 microns
 In order to avoid reach-through , Wbody > Wd,body + Ws,body 
= 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd 
dvGD
dt ≈ Cgd 
dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd 
dvDS
dt = 0.7 V
 
dvDS
dt > 
0.7
Rbody!Cgd
 will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10
6) (5x10-6) = 16.7 volts
22-10. a) iD = 
mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L 
Cox = 
e
tox
 = 
(11.7)(8.9x10-14)
10-5
 = 1.04x10-7 F/cm2
N = 
2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
 
N = 
!(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4)
 ≈ 5,800 cells
b) Icell = 
100
5800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = 
Wdrift
q!mn!Nd!A
 : Wdrift = 10
-5 BVDSS = (10
-5)(800) = 80 microns
fc = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(5x1016)
(1020)
 = 0.94
Ws,body ≈ 
(2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016)
 ≈ 0.16 microns
 In order to avoid reach-through , Wbody > Wd,body + Ws,body 
= 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd 
dvGD
dt ≈ Cgd 
dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd 
dvDS
dt = 0.7 V
 
dvDS
dt > 
0.7
Rbody!Cgd
 will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10
6) (5x10-6) = 16.7 volts
22-10. a) iD = 
mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L 
Cox = 
e
tox
 = 
(11.7)(8.9x10-14)
10-5
 = 1.04x10-7 F/cm2
N = 
2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
 
N = 
!(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4)
 ≈ 5,800 cells
b) Icell = 
100
5800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = 
Wdrift
q!mn!Nd!A
 : Wdrift = 10
-5 BVDSS = (10
-5)(800) = 80 microns
fc = 0.026 ln Î
Í
È
˚
˙
˘
!
(1019)(5x1016)
(1020)
 = 0.94
Ws,body ≈ 
(2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016)
 ≈ 0.16 microns
 In order to avoid reach-through , Wbody > Wd,body + Ws,body 
= 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd 
dvGD
dt ≈ Cgd 
dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd 
dvDS
dt = 0.7 V
 
dvDS
dt > 
0.7
Rbody!Cgd
 will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10
6) (5x10-6) = 16.7 volts
22-10. a) iD = 
mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L 
Cox = 
e
tox
 = 
(11.7)(8.9x10-14)
10-5
 = 1.04x10-7 F/cm2
N = 
2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
 
N = 
!(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4)
 ≈ 5,800 cells
b) Icell = 
100
5800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = 
Wdrift
q!mn!Nd!A
 : Wdrift = 10
-5 BVDSS = (10
-5)(800) = 80 microns
Nd = 
1.3x1017
BVDSS
 = 
1.3x1017
800 ≈ 1.6x10
14 cm-3
A = 
8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4)
 ≈ 0.5 cm2
10!A
0.5!cm2
 = 20 
A
cm2
 << the allowable maximum of 200 
A
cm2
 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10
-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray 
inductance and excessive power dissipation. Check for overvoltage first.
VDS(turn-off) = Vd + L 
di
dt = 100 + (10
-7) ÎÍ
È
˚
˙˘!
100
5x10-8
 = 300 V > BVDSS = 150 V
Check for excessive power dissipation.
Pallowed = 
Tj,max!-!Ta
Rq,j-a
 = 
150!-!50
1 = 100 watts ; Pdissipated = [Eon + Esw] fs
Eonfs = 
Io
2!rDS(on)
2 = 
(100)2(0.01)
2 = 50 watts
Esw = 
Vd!Io
2 [tri + tfi + trv +tfv] = 
(100)(100)
2 [(2)(5x10
-8) + (2)(2x10-7)]
Esw = 2.5x10
-3 joules ; Eswfs = (2.5x10
-3)(3x104) = 75 watts
Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts
MOSFET overstressed by both overvoltages and excessive power dissipation.
22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is 
approximately constant during the four time intervals. However during the current rise 
and fall times the voltages VGS and VGD change by only a few tens of volts. However 
during the voltage rise and fall times VGD changes by approximately Vd which is much 
larger than a few tens of volts. Thus we have:
Nd = 
1.3x1017
BVDSS
 = 
1.3x1017
800 ≈ 1.6x10
14 cm-3
A = 
8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4)
 ≈ 0.5 cm2
10!A
0.5!cm2
 = 20 
A
cm2
 << the allowable maximum of 200 
A
cm2
 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10
-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray 
inductance and excessive power dissipation. Check for overvoltage first.
VDS(turn-off) = Vd + L 
di
dt = 100 + (10
-7) ÎÍ
È
˚
˙˘!
100
5x10-8
 = 300 V > BVDSS = 150 V
Check for excessive power dissipation.
Pallowed = 
Tj,max!-!Ta
Rq,j-a
 = 
150!-!50
1 = 100 watts ; Pdissipated = [Eon + Esw] fs
Eonfs = 
Io
2!rDS(on)
2 = 
(100)2(0.01)
2 = 50 watts
Esw = 
Vd!Io
2 [tri + tfi + trv +tfv] = 
(100)(100)
2 [(2)(5x10
-8) + (2)(2x10-7)]
Esw = 2.5x10
-3 joules ; Eswfs = (2.5x10
-3)(3x104) = 75 watts
Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts
MOSFET overstressed by both overvoltages and excessive power dissipation.
22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is 
approximately constant during the four time intervals. However during the current rise 
and fall times the voltages VGS and VGD change by only a few tens of volts. However 
during the voltage rise and fall times VGD changes by approximately Vd which is much 
larger than a few tens of volts. Thus we have:
Nd = 
1.3x1017
BVDSS
 = 
1.3x1017
800 ≈ 1.6x10
14 cm-3
A = 
8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4)
 ≈ 0.5 cm2
10!A
0.5!cm2
 = 20 
A
cm2
 << the allowable maximum of 200 
A
cm2
 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10
-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities,

Outros materiais