Power Electronics - Mohan (Solution)

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 = 1014 = 1010 exp ÎÍÈ

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

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 = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.

n = Nd = 1013 cm-3 ; p = n2i

1013 = 107 cm-3

19-4. po = n2i [300]Nd

; 2po = n2i [300!+!T]

2 n2i [300] = n

2i [300 + T] ; 2x1010 = 1010 exp Î

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

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!V1k!T ; 10 I1 = Is exp(

q!V1!+!dVk!T ) ; dV =

k!Tq 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.

ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

Ti = 262 °C or 535 °K.

q!mn!Nd =

1(1.6x10-19)(1500)(1014)

= 43.5 ohm-cm

q!mp!Na =

1(1.6x10-19)(500)(1018)

= 0.013 ohm-cm

n = Nd = 1013 cm-3 ; p = n2i

1013 = 107 cm-3

19-4. po = n2i [300]Nd

; 2po = n2i [300!+!T]

2 n2i [300] = n

2i [300 + T] ; 2x1010 = 1010 exp Î

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

(q!Eg!-!k!300!ln(2)) = 305.2 °K

DT = 305.2 - 300 = 5.2 °K.

q!V1!+!dVk!T ) ; dV =

k!Tq ln(10) = 60 mV

ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

Ti = 262 °C or 535 °K.

q!mn!Nd =

1(1.6x10-19)(1500)(1014)

= 43.5 ohm-cm

q!mp!Na =

1(1.6x10-19)(500)(1018)

= 0.013 ohm-cm

n = Nd = 1013 cm-3 ; p = n2i

1013 = 107 cm-3

19-4. po = n2i [300]Nd

; 2po = n2i [300!+!T]

2 n2i [300] = n

2i [300 + T] ; 2x1010 = 1010 exp Î

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

(q!Eg!-!k!300!ln(2)) = 305.2 °K

DT = 305.2 - 300 = 5.2 °K.

q!V1!+!dVk!T ) ; dV =

k!Tq ln(10) = 60 mV

ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

Ti = 262 °C or 535 °K.

q!mn!Nd =

1(1.6x10-19)(1500)(1014)

= 43.5 ohm-cm

q!mp!Na =

1(1.6x10-19)(500)(1018)

= 0.013 ohm-cm

n = Nd = 1013 cm-3 ; p = n2i

1013 = 107 cm-3

19-4. po = n2i [300]Nd

; 2po = n2i [300!+!T]

2 n2i [300] = n

2i [300 + T] ; 2x1010 = 1010 exp Î

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

(q!Eg!-!k!300!ln(2)) = 305.2 °K

DT = 305.2 - 300 = 5.2 °K.

q!V1!+!dVk!T ) ; dV =

k!Tq ln(10) = 60 mV

ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

Ti = 262 °C or 535 °K.

q!mn!Nd =

1(1.6x10-19)(1500)(1014)

= 43.5 ohm-cm

q!mp!Na =

1(1.6x10-19)(500)(1018)

= 0.013 ohm-cm

n = Nd = 1013 cm-3 ; p = n2i

1013 = 107 cm-3

19-4. po = n2i [300]Nd

; 2po = n2i [300!+!T]

2 n2i [300] = n

2i [300 + T] ; 2x1010 = 1010 exp Î

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

(q!Eg!-!k!300!ln(2)) = 305.2 °K

DT = 305.2 - 300 = 5.2 °K.

q!V1!+!dVk!T ) ; dV =

k!Tq ln(10) = 60 mV

ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

Ti = 262 °C or 535 °K.

q!mn!Nd =

1(1.6x10-19)(1500)(1014)

= 43.5 ohm-cm

q!mp!Na =

1(1.6x10-19)(500)(1018)

= 0.013 ohm-cm

n = Nd = 1013 cm-3 ; p = n2i

1013 = 107 cm-3

19-4. po = n2i [300]Nd

; 2po = n2i [300!+!T]

2 n2i [300] = n

2i [300 + T] ; 2x1010 = 1010 exp Î

˚˙˘

-!q!Eg2k !

ÓÌÏ

˛˝¸1

(q!Eg!-!k!300!ln(2)) = 305.2 °K

DT = 305.2 - 300 = 5.2 °K.

q!V1!+!dVk!T ) ; dV =

k!Tq ln(10) = 60 mV

xn(0) + xp(0) = Wo = !2!e!fc!(Na!+!Nd)

q!Na!Nd (1)

fc = k!Tq ln

ÎÍÍÈ

˚˙˙˘

!Na!Nd

= 0.026 ln ÎÍÈ

˚˙˘

!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 problemstatement 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!fcWo

= (2)!(0.54)(2.8x10-4)

= 3,900 V/cm

(c) From part a) fc = 0.54 eV

(d)C(V)

Wo 1!+!Vfc

; 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(qVkT ) ; exp(

qVkT ) = exp (

0.70.026 ) = 5x1011

Is = q n2i ÎÍ

˚˙˙˘

Na!t !+!Dpt

Nd!t A

= (1.6x10-19)(1020) ÎÍÍÈ

˚˙˙˘

!(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!LA ;

0.020.01 = 2 cm-1

At 25 °C, Nd = 1014 >> 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] = 1010 exp ÎÍÈ

˚˙˘

-!(1.6x10-19)!(1.1)(2)(1.4x10-23)

!ÓÌÏ

˛˝¸1

523!-!1

(1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 1014. 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 = Nd2 Î

˚˙˘

1!+! 1!+!4!n

and po = n2i

no . Putting in numerical values yields

no = 1014

2 ÎÍÍÈ

˚˙˙˘

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

(1.6x10-19)(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!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!LA ;

0.020.01 = 2 cm-1

At 25 °C, Nd = 1014 >> 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] = 1010 exp ÎÍÈ

˚˙˘

-!(1.6x10-19)!(1.1)(2)(1.4x10-23)

!ÓÌÏ

˛˝¸1

523!-!1

(1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 1014. 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 = Nd2 Î

˚˙˘

1!+! 1!+!4!n

and po = n2i

no . Putting in numerical values yields

no = 1014

2 ÎÍÍÈ

˚˙˙˘

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

(1.6x10-19)(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!q!Na!Nd =

(11.7)(8.9x1014)(1015!+!1014)(3x105)2

(2)(1.6x10-19)(1015)(1014)

= 3,340 volts

19-9. E2max = E

2BD ≈

4!f2c!BVBD

W2o!fc

; Eq. (19-13); or W

!fc = 4!BVBD

W2(BVBD) = W

2o!BVBD

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

!fc = 4!BVBD

and taking

the square root yields W (BVBD) ≈ 2!BVBD

19-10. Lp = Dp!t = (13)(10-6) = 36 microns ; Ln = Dn!t = (39)(10-6) = 62 microns

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

I1 = q n2i A

Dp!t1Nd!t1

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

Dp!t2Nd!t2

exp(q!Vk!T )

= 2 = t1t2 ; Thus 4 t2 = t1

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

n2in + q mn n

dsdn = 0 = - q mp

n2 + q mn ; Solving for n yields n = ni mpmn and p = ni

p = 1010 1500500 = 1.7x1010 cm-3; n = 1010

5001500 = 6x109 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. E2max = E

2BD ≈

4!f2c!BVBD

W2o!fc

; Eq. (19-13); or W

!fc = 4!BVBD

W2(BVBD) = W

2o!BVBD

!fc = 4!BVBD

and taking

I1 = q n2i A

Dp!t1Nd!t1

Dp!t2Nd!t2

exp(q!Vk!T )

= 2 = t1t2 ; Thus 4 t2 = t1

n2in + q mn n

dsdn = 0 = - q mp

p = 1010 1500500 = 1.7x1010 cm-3; n = 1010

5001500 = 6x109 cm-3;

19-9. E2max = E

2BD ≈

4!f2c!BVBD

W2o!fc

; Eq. (19-13); or W

!fc = 4!BVBD

W2(BVBD) = W

2o!BVBD

!fc = 4!BVBD

and taking

I1 = q n2i A

Dp!t1Nd!t1

Dp!t2Nd!t2

exp(q!Vk!T )

= 2 = t1t2 ; Thus 4 t2 = t1

n2in + q mn n

dsdn = 0 = - q mp

p = 1010 1500500 = 1.7x1010 cm-3; n = 1010

5001500 = 6x109 cm-3;

19-9. E2max = E

2BD ≈

4!f2c!BVBD

W2o!fc

; Eq. (19-13); or W

!fc = 4!BVBD

W2(BVBD) = W

2o!BVBD

!fc = 4!BVBD

and taking

I1 = q n2i A

Dp!t1Nd!t1

Dp!t2Nd!t2

exp(q!Vk!T )

= 2 = t1t2 ; Thus 4 t2 = t1

n2in + q mn n

dsdn = 0 = - q mp

p = 1010 1500500 = 1.7x1010 cm-3; n = 1010

5001500 = 6x109 cm-3;

20-1. Nd = 1.3x1017BVBD! =

1.3x10172500 = 5x1013 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 previousproblem (20-1) for the same drift region doping density. Hence this must be a punch-through structure and Eq. (20-9) applies.

BVBD = (2x105)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2

(2)(11.7)(8.9x10-14) = 900 V

20-3. Von = Vj + Vdrift ; Vj = k!Tq ln ÎÍ

È˚˙!

; For one-sided step junction Is = q!A!n

2i !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

q2!m3o!n

2b!A2!to

= 3 (5x10-3)4

(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)

= 7.5x10-4

20-1. Nd = 1.3x1017BVBD! =

1.3x10172500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250

microns

BVBD = (2x105)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2

(2)(11.7)(8.9x10-14) = 900 V

È˚˙!

2i !Lp

!Nd!to ;

!(5x1013)(2x10-6) = 1.6x10-9 A

K1 = Wd

q!mo!A!nb =

5x10-3

(1.6x10-19)(900)(2)(1017) = 1.7x10-4

K2 = 3 W

q2!m3o!n

2b!A2!to

= 3 (5x10-3)4

(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)

= 7.5x10-4

20-1. Nd = 1.3x1017BVBD! =

1.3x10172500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250

microns

BVBD = (2x105)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2

(2)(11.7)(8.9x10-14) = 900 V

È˚˙!

2i !Lp

!Nd!to ;

!(5x1013)(2x10-6) = 1.6x10-9 A

K1 = Wd

q!mo!A!nb =

5x10-3

(1.6x10-19)(900)(2)(1017) = 1.7x10-4

K2 = 3 W

q2!m3o!n

2b!A2!to

= 3 (5x10-3)4

(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)

= 7.5x10-4

I Vj Vdrift Von0 A 0 V 0 V 0 V1 0.53 0.001 0.5310 0.59 0.005 0.59100 0.65 0.033 0.681000 0.71 0.25 0.963000 0.74 0.67 1.41 • •

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 LA ; r =

1q!mn!Nd

(1.6x10-19)(1500)(5x1013) = 85 ohm-cm ;

5x10-32 = 2.5x10-3

Rdrift = (85)(2.5x10-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds

Von(t) = (0.21)(2.5x108 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds

Von(4 ms) = (5.3x107)(4x10-6) = 212 volts

b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x107 t]

Von(t) = {0.21[1 - 2.5x107 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds

I Vj Vdrift Von0 A 0 V 0 V 0 V1 0.53 0.001 0.5310 0.59 0.005 0.59100 0.65 0.033 0.681000 0.71 0.25 0.963000 0.74 0.67 1.41 • •

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 LA ; r =

1q!mn!Nd

(1.6x10-19)(1500)(5x1013) = 85 ohm-cm ;

5x10-32 = 2.5x10-3

Rdrift = (85)(2.5x10-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds

Von(t) = (0.21)(2.5x108 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds

Von(4 ms) = (5.3x107)(4x10-6) = 212 volts

b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x107 t]

Von(t) = {0.21[1 - 2.5x107 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds

0 0.5 1 1.5 2 2.5 3 3.5 4

Von involts

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!IFdiR/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 = 1.3x1017BVBD

= 1.3x1017

150 = 8.7x1014 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

A = 2x10-3

!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2

0 0.5 1 1.5 2 2.5 3 3.5 4

Von involts

diR/dt = trr + 2000

2.5x108 = trr + 8 ms

trr = (2)(1.6x10-5)(2x103)

2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms

Nd = 1.3x1017BVBD

= 1.3x1017

150 = 8.7x1014 cm-3

Wd = 10-5 BVBD = (10-5) (150) = 15 microns

q!mn!Nd

A = 2x10-3

!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2

0 0.5 1 1.5 2 2.5 3 3.5 4

Von involts

diR/dt = trr + 2000

2.5x108 = trr + 8 ms

trr = (2)(1.6x10-5)(2x103)

2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms

Nd = 1.3x1017BVBD

= 1.3x1017

150 = 8.7x1014 cm-3

Wd = 10-5 BVBD = (10-5) (150) = 15 microns

q!mn!Nd

A = 2x10-3

!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2

0 0.5 1 1.5 2 2.5 3 3.5 4

Von involts

diR/dt = trr + 2000

2.5x108 = trr + 8 ms

trr = (2)(1.6x10-5)(2x103)

2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms

Nd = 1.3x1017BVBD

= 1.3x1017

150 = 8.7x1014 cm-3

Wd = 10-5 BVBD = (10-5) (150) = 15 microns

q!mn!Nd

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] ÎÍÍÈ

˚˙˙˘2!e

= [(2x105)(2x10-3) - 300] (2)(11.7)(8.9x10-14)(1.6x10-19)(2x10-3)2

= Nd = 3.4x1014 cm-3

20-9.RA

Wd(npt)q!mn!Nnpt

; Nnpt = Nd of non-punch-throuth (npt) diode

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

ÎÍÍÈ

˚˙˙˘

1!+_ ! 1!-!2!q!Npt!BVBD

e!E2BD

2!q!Npt!BVBD

e!E2BD

= q!Npte!EBD

NnptNnpt

2!BVBD

q!Nnpte!EBD

2!BVBD

NptNnpt

Wd(npt) Wd(npt) NptNnpt

= x = NptNnpt

; Wd(pt) = Wd(npt) 1x [ ]1!+_ ! 1!-!x

If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10)

Limit of 1x [ ]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.

Wd(npt)!1x![ ]1!-!! 1!-!x

q!mn!Npt =

Wd(npt)!1x![ ]1!-!! 1!-!x

q!mn!Npt

Nnpt!Nnpt

= Wd(npt)

!q!mn!Nnpt

1x2 [1 - 1!-!x ] =

1x2 [1 - 1!-!x ]

ddx ÎÍ

È˚˙!

1x2!! 1!-!x! = 0 =

-2x3 [1 - 1!-!x ] +

12!x2! 1!-!x

Solving for x yields x = 89 i.e. Npt =

89 Nnpt ; Wd(pt) = 0.75 Wd(npt)

20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] ÎÍÍÈ

˚˙˙˘2!e

= [(2x105)(2x10-3) - 300] (2)(11.7)(8.9x10-14)(1.6x10-19)(2x10-3)2

= Nd = 3.4x1014 cm-3

20-9.RA

Wd(npt)q!mn!Nnpt

; Nnpt = Nd of non-punch-throuth (npt) diode

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

ÎÍÍÈ

˚˙˙˘

1!+_ ! 1!-!2!q!Npt!BVBD

e!E2BD

2!q!Npt!BVBD

e!E2BD

= q!Npte!EBD

NnptNnpt

2!BVBD

q!Nnpte!EBD

2!BVBD

NptNnpt

Wd(npt) Wd(npt) NptNnpt

= x = NptNnpt

; Wd(pt) = Wd(npt) 1x [ ]1!+_ ! 1!-!x

If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10)

Limit of 1x [ ]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.

Wd(npt)!1x![ ]1!-!! 1!-!x

q!mn!Npt =

Wd(npt)!1x![ ]1!-!! 1!-!x

q!mn!Npt

Nnpt!Nnpt

= Wd(npt)

!q!mn!Nnpt

1x2 [1 - 1!-!x ] =

1x2 [1 - 1!-!x ]

ddx ÎÍ

È˚˙!

1x2!! 1!-!x! = 0 =

-2x3 [1 - 1!-!x ] +

12!x2! 1!-!x

Solving for x yields x = 89 i.e. Npt =

89 Nnpt ; Wd(pt) = 0.75 Wd(npt)

npt = 0.84

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!+!1500.7 = (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

= 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)(1017) =

10-4Apn

K2 = 3 W

4d(150)

q2!m3o!n

2pn!to

= 3 (1.5x10-3)4

(1.6x10-19)2!(900)3!(1017)2!(A2pn)!(2x10-6)

K2 = 2.4x10-4 (Apn)-0.67 ; 2 volts = 10-4Apn

(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 cm2

Schottky diode Cjo = (11.7)(8.9x10-14)(1.07)

10-4 ≈ 11 nF = 0.011 mF

npt = 0.84

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!+!1500.7 = (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

= 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)(1017) =

10-4Apn

K2 = 3 W

4d(150)

q2!m3o!n

2pn!to

= 3 (1.5x10-3)4

(1.6x10-19)2!(900)3!(1017)2!(A2pn)!(2x10-6)

K2 = 2.4x10-4 (Apn)-0.67 ; 2 volts = 10-4Apn

(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 cm2

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.BVcylBVp

= 2 r2 (1 + 1r ) ln (1 +

1r ) - 2r

••

• • • •

00.10.20.30.40.50.60.70.80.9

0.1 1 10

BVcyl / BVp

r = R/(2W n)

20-12.BVcylBVp

= 9501000 = 0.95 ; From graph in problem 20-11, r ≈ 6 =

R ≈ 12 Wn

10-4 ≈ 180 pF

20-11.BVcylBVp

= 2 r2 (1 + 1r ) ln (1 +

1r ) - 2r

••

• • • •

00.10.20.30.40.50.60.70.80.9

0.1 1 10

BVcyl / BVp

r = R/(2W n)

20-12.BVcylBVp

R ≈ 12 Wn

10-4 ≈ 180 pF

20-11.BVcylBVp

= 2 r2 (1 + 1r ) ln (1 +

1r ) - 2r

••

• • • •

00.10.20.30.40.50.60.70.80.9

0.1 1 10

BVcyl / BVp

r = R/(2W n)

20-12.BVcylBVp

R ≈ 12 Wn

21-1. NPN BJT ; BVCEO = BVCBO

b1/4 ; PNP BJT ; BVCEO = BVCBO

** * * * * *

** * * * * * *

••

• • • •••••

• • • • •••

1 10 100

BVCEOBVCBO

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

** * * * * *

** * * * * * *

••

• • • •••••

• • • • •••

1 10 100

BVCEOBVCBO

** * * * * *

** * * * * * *

••

• • • •••••

• • • • •••

1 10 100

BVCEOBVCBO

V (t)CEi (t)

t d,off t rv t fit d,on t ri t fv

BJT power dissipation Pc = 1T ıÛ

T!vCE(t)!iC(t)!dt = {Eon + Esw }fs ; fs =

Eon = VCE,on Io {T2 + td,off - td,on} ≈ (2)(40)

= 40fs

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 = [40fs

+ 1.2x10-3] fs = 40 + 1.2x10-3 fs

033 kHz

[watts]

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

(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.

033 kHz

[watts]

1 = 125 watts = 40 + 1.2x10-3 fs,max

fs,max = (125!-!40)1.2x10-3 = 71 kHz

(125!-!25) = 4x10-3

IC = q!Dn!Na!A

(1.6x10-19)(38)(1016)(1)(3x10-4)

= 200 A

033 kHz

[watts]

1 = 125 watts = 40 + 1.2x10-3 fs,max

fs,max = (125!-!40)1.2x10-3 = 71 kHz

(125!-!25) = 4x10-3

IC = q!Dn!Na!A

(1.6x10-19)(38)(1016)(1)(3x10-4)

= 200 A

033 kHz

[watts]

1 = 125 watts = 40 + 1.2x10-3 fs,max

fs,max = (125!-!40)1.2x10-3 = 71 kHz

(125!-!25) = 4x10-3

IC = q!Dn!Na!A

(1.6x10-19)(38)(1016)(1)(3x10-4)

= 200 A

033 kHz

[watts]

1 = 125 watts = 40 + 1.2x10-3 fs,max

fs,max = (125!-!40)1.2x10-3 = 71 kHz

(125!-!25) = 4x10-3

IC = q!Dn!Na!A

(1.6x10-19)(38)(1016)(1)(3x10-4)

= 200 A

Wo,EB = 2!e!fc!(NaB!+!NdE)

!q!NaB!NdE ; fcE =

k!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!NdE

fcE = 0.26 ln ÎÍÈ

˚˙˘

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!+!Vfc = 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!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!NdC

fcC = 0.26 ln ÎÍÈ

˚˙˘

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 = 1.3x1016 cm-3

BVCBO = b1/4 BVCEO = (5)1/4(1000) = ≈ 1500 Volts ≈ 1.3x1017

NdC = collector drift region donor density = 8.7x1013 cn -3

Wo,EB = 2!e!fc!(NaB!+!NdE)

!q!NaB!NdE ; fcE =

k!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!NdE

˚˙˘

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!+!Vfc = 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!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!NdC

˚˙˘

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 = 1.3x1016 cm-3

BVCBO = b1/4 BVCEO = (5)1/4(1000) = ≈ 1500 Volts ≈ 1.3x1017

NdC = collector drift region donor density = 8.7x1013 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

0.02 W

10 IB,M + IB,M = 100 AIB,M = IC,D = 9.1A IC,M = 91 AVCE,D = 0.2 +(.02)(9.1) = 0.382 VVCE,M = VCE,D + 0.8 V = 1.18 vBut a saturated main BJT with IC,M = 91 Agoing through 0.02 ohms generatesa voltage drop of 1.8 V which is > 1.18 V.Hence main BJT must be saturated.

NaB = xn(BVCBO)

8.7x1013

1.3x1016 = xn(BVCBO) 6.7x10-3

bM = 150!-!20

21 = 6.2

10 I B,M

0.02 W

NaB = xn(BVCBO)

8.7x1013

1.3x1016 = xn(BVCBO) 6.7x10-3

bM = 150!-!20

21 = 6.2

10 I B,M

0.02 W

0.6 V-

+I C,M

Neglecting IB,DIC,M + IC,D = 100 A(.02)IC,M -0.6 = (.02) I C,D +0.2IC,D = 30 A ; IC,M = 70 AVCE,M = 0.2 + (70)(.02) = 1.6 VPDarl = VCE,M [IC,M + IC,D]

PDarl = (1.6 V)(100 A) = 160 W

21-11. CEBO = e!AE

WEBO ; CCBO = e!AC

WEBO = zero-bias emitter-base depletion layer thicknessWCBO = zero-bias collector-base depletion layer thickness

WEBO = 2!e!fcE!(NaB!+!NdE)

!q!NaB!NdE ; fcE =

k!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!NdE

˚˙˘

!(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!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!!NdC

˚˙˘

!(1.3x1016)(8.7x1013)

1020 = 0.6 V ;

0.02 W

0.6 V-

+I C,M

Neglecting IB,DIC,M + IC,D = 100 A(.02)IC,M -0.6 = (.02) I C,D +0.2IC,D = 30 A ; IC,M = 70 AVCE,M = 0.2 + (70)(.02) = 1.6 VPDarl = VCE,M [IC,M + IC,D]

PDarl = (1.6 V)(100 A) = 160 W

21-11. CEBO = e!AE

WEBO ; CCBO = e!AC

WEBO = zero-bias emitter-base depletion layer thicknessWCBO = zero-bias collector-base depletion layer thickness

WEBO = 2!e!fcE!(NaB!+!NdE)

!q!NaB!NdE ; fcE =

k!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!NdE

˚˙˘

!(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!Tq ln

ÎÍÍÈ

˚˙˙˘

!NaB!!NdC

˚˙˘

!(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 WCCB

CBEVBEVin

VBE(t) = 8 - 16 exp ÎÍÈ

˚˙-t

!t ; t = (10 W)(CBE + CCB)

At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln ÎÍÈ

˚˙16

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!+!VCBfcC

= 1.5x10-8

1!+!1000.6

= 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 WCCB

CBEVBEVin

VBE(t) = 8 - 16 exp ÎÍÈ

˚˙-t

!t ; t = (10 W)(CBE + CCB)

At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln ÎÍÈ

˚˙16

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!+!VCBfcC

= 1.5x10-8

1!+!1000.6

= 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.

ıÙÙÛ

! 1!+!VEBfcE

= CEBO!fcE

(-8) ÎÍÈ

˚˙˘

fcE!!-!! 1!+!8

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 ÎÍÈ

˚˙16

7.3 = (3.4x10-8)(0.78) ≈ 27 nanoseconds

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.

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.

V (t)CE

i (t)C

t d,off t rv t fit d,on t ri t fv

VGS(th)

VGS,Io

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) = - VGGSolution 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 ÎÍÈ

˚˙˘

!2!VGG

!VGG!-!VGS(th)

td,on = (50) (1.15x10-9) ln ÎÍÈ

(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 ÎÍÈ

˚˙˘

!(VGG!-!VGS(th)){gm!+ !

CgdRG(Cgs!+!Cgd)}

gm(VGG!-!VGS(th))!-!Io

tri = (50)(1.15x10-9) ln ÎÍÍÈ

˚˙˙˘

!(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 = Iogm

+ VGS(th) during this interval.

Governing equation is Cgd dvDS

dt = - ÎÍÈ

˚˙˘

!VGG!-!VGS(th)!-!

RG with vDS(0) = Vd.

Solution is given by

vDS(t) = Vd - ÎÍÈ

˚˙˘

VGG!-!VGS(th)!-!Iogm

RGCgd ; At t = tfv, vDS = 0.

Solving for tfv yields

tfv = RG!Cgd!Vd

VGG!-!VGS(th)!-!Iogm

= (50)(1.5x10-10) 300

(15!-!4!-!3) = 300 ns

td,off estimate - use equivalent circuit of Fig. 22-12d with the input voltagge VGGreversed.

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

ÎÍÈ

˚˙˘

!2!VGG

VGG!+!VGS(th)!+!Iogm

= (50)(1.15x10-9) ln ÎÍÍÈ

˚˙˙˘

!(2)(15)

103.3!+!4!+!15

= 18 ns

trv estimate - Use equivalent circuit of Fig. 22-12c with the input voltage VGGreversed. 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 VGGreversed. 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 ÎÍÈ

˚˙˘

!VGG!+!VGS,IoVGG+!VGS(th)

= (50)(1.15x10-9)ln ÎÍÈ

15!+!715!+!4 = 9 ns

b) Estimate the power dissipated in the MOSFET in the same manner as was done for theBJT in problem 21-3.Waveforms for the MOSFET are the same as for the BJT except forappropriate 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 JoulesEon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 VT >> td,on and td,offEon = (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 JoulesEon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 VT >> td,on and td,offEon = (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

V (t)GS

VGS(th)

VGS,Io

t d,ont = t r i fv t = tf i rv

td,offt

V (t)DS

i (t)D

Equivalent circuit during voltage and current rise and fall intervals:

+- Cgs

V (t)G

g (V - V )m GS GS(th)

Governing equation using Miller capacitance approximation:

dvGSdt +

vGSt =

!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

ÎÍÈ

˚˙˘

!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 ÎÍÈ

˚˙˘

!VGS(th)!+!

Vd!gm!RD

VGS(th) . Invert equation for tri to find Cgd. Result is

ÓÔÌÔÏ

Ô˝Ô

ÎÍÈ

˚˙˘

!VGG!-!VGS(th)

VGG!-!VGS(th)!-!!!Vd

!gm!RD

!!-!!RG!Cgs ÓÌÏ

˛˝¸1

RG(1!+!gmRD)

Cgd = ÓÔÌÔÏ

Ô˝Ô3x10-8

!ln!ÎÍÈ

15!-!415!-!4!-!1

!!-!!5x10-9 ÓÌÏ

˛˝¸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 ÎÍÈ

15!-!415!-!4!-!2 = 7 ms ; trv = 3.5x10-5 ln ÎÍ

È˚˙!

4!+!24 = 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 = 1T ; Eon = [ID]2 rDS,on(Tj)

dvGSdt +

vGSt =

!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

ÎÍÈ

˚˙˘

!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 ÎÍÈ

˚˙˘

!VGS(th)!+!

Vd!gm!RD

VGS(th) . Invert equation for tri to find Cgd. Result is

ÓÔÌÔÏ

Ô˝Ô

ÎÍÈ

˚˙˘

!VGG!-!VGS(th)

VGG!-!VGS(th)!-!!!Vd

!gm!RD

!!-!!RG!Cgs ÓÌÏ

˛˝¸1

RG(1!+!gmRD)

Cgd = ÓÔÌÔÏ

Ô˝Ô3x10-8

!ln!ÎÍÈ

15!-!415!-!4!-!1

!!-!!5x10-9 ÓÌÏ

˛˝¸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 ÎÍÈ

15!-!415!-!4!-!2 = 7 ms ; trv = 3.5x10-5 ln ÎÍ

È˚˙!

4!+!24 = 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 = 1T ; Eon = [ID]2 rDS,on(Tj)

ID = VdRD

= 300150 = 2 A ; rDS,on(Tj) = 2 ÎÍ

È˚˙1!+!

Tj!-!25150 = 2 ÎÍ

È˚˙0.833!+!

Eon = (4)(2) ÎÍÈ

˚˙0.833!+!

= {3.32 + 0.027 Tj} 1fs

Esw = 1T ıÙ

Vd!ID(1!-!ttri

)(ttri

)dt + 1T ıÙ

Vd!ID(1!-!ttfi

)(ttfi

)dt = Vd!ID

6 [tri + tfi]

Esw = (300)(2)

6 [7x10-6 + 14x10-6] = 2.1x10-3 joules

<PMOSFET> = {3.32 + 0.027 Tj} 1fs

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 10 20 30 40 50 60 70 80 90 100

PMOSFET

Temperature [ °K]

22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff

ID = VdRD

= 300150 = 2 A ; rDS,on(Tj) = 2 ÎÍ

È˚˙1!+!

Tj!-!25150 = 2 ÎÍ

È˚˙0.833!+!

Eon = (4)(2) ÎÍÈ

˚˙0.833!+!

= {3.32 + 0.027 Tj} 1fs

Esw = 1T ıÙ

Vd!ID(1!-!ttri

)(ttri

)dt + 1T ıÙ

Vd!ID(1!-!ttfi

)(ttfi

)dt = Vd!ID

6 [tri + tfi]

Esw = (300)(2)

6 [7x10-6 + 14x10-6] = 2.1x10-3 joules

<PMOSFET> = {3.32 + 0.027 Tj} 1fs

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 10 20 30 40 50 60 70 80 90 100

PMOSFET

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) ÎÍÈ

˚˙1!+!0.8!!

Tj!-!25100 ; 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!ri =

Io2!reff

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) ÎÍÈ

˚˙1!+!0.8!!

Tj!-!25100 ; 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!ri =

Io2!reff

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

VDS,on VCE,on

I or I ; r < 1o

(1 - r)I o

22-7. BVDSS ≈ 1.3x1017

Ndrift = 750 volts ; Ndrift = 1.7x1014 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!Tq ln

ÎÍÍÈ

˚˙˙˘

!Na!Nd

v = vDS CE

VDS,on VCE,on

I or I ; r < 1o

(1 - r)I o

22-7. BVDSS ≈ 1.3x1017

Ndrift = 750 volts ; Ndrift = 1.7x1014 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!Tq ln

ÎÍÍÈ

˚˙˙˘

!Na!Nd

fc = 0.026 ln ÎÍÈ

˚˙˘

!(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

dt > 0.7

Rbody!Cgd will turn on the BJT.

22-9. VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts

22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th))

Cox = e

(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 = 1005800 = 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

˚˙˘

!(1019)(5x1016)

(1020) = 0.94

Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)

(1.6x10-19)(5x1016) ≈ 0.16 microns

dt ≈ Cgd dvDS

dt = 0.7 V

dt > 0.7

Cox = e

(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2

N = 2!iD!L

N = !(2)(100)(10-4)

(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells

Ron = Wdrift

˚˙˘

!(1019)(5x1016)

(1020) = 0.94

Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)

(1.6x10-19)(5x1016) ≈ 0.16 microns

dt ≈ Cgd dvDS

dt = 0.7 V

dt > 0.7

Cox = e

(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2

N = 2!iD!L

N = !(2)(100)(10-4)

(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells

Ron = Wdrift

˚˙˘

!(1019)(5x1016)

(1020) = 0.94

Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)

(1.6x10-19)(5x1016) ≈ 0.16 microns

dt ≈ Cgd dvDS

dt = 0.7 V

dt > 0.7

Cox = e

(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2

N = 2!iD!L

N = !(2)(100)(10-4)

(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells

Ron = Wdrift

˚˙˘

!(1019)(5x1016)

(1020) = 0.94

Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)

(1.6x10-19)(5x1016) ≈ 0.16 microns

dt ≈ Cgd dvDS

dt = 0.7 V

dt > 0.7

Cox = e

(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2

N = 2!iD!L

N = !(2)(100)(10-4)

(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells

Ron = Wdrift

Nd = 1.3x1017BVDSS

= 1.3x1017

800 ≈ 1.6x1014 cm-3

A = 8x10-3

(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2

10!A0.5!cm2 = 20

Acm2 << the allowable maximum of 200

Acm2 , 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 didt = 100 + (10-7) ÎÍ

È˚˙!

1005x10-8 = 300 V > BVDSS = 150 V

Check for excessive power dissipation.

Pallowed = Tj,max!-!Ta

Rq,j-a =

150!-!501 = 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 = 1.3x1017BVDSS

= 1.3x1017

800 ≈ 1.6x1014 cm-3

A = 8x10-3

(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2

10!A0.5!cm2 = 20

È˚˙!

1005x10-8 = 300 V > BVDSS = 150 V

Rq,j-a =

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)]

Nd = 1.3x1017BVDSS

= 1.3x1017

800 ≈ 1.6x1014 cm-3

A = 8x10-3

(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2

10!A0.5!cm2 = 20

È˚˙!

1005x10-8 = 300 V > BVDSS = 150 V

Rq,j-a =

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)]

Nd = 1.3x1017BVDSS

= 1.3x1017

800 ≈ 1.6x1014 cm-3

A = 8x10-3

(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2

10!A0.5!cm2 = 20

È˚˙!

1005x10-8 = 300 V > BVDSS = 150 V

Rq,j-a =

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)]

Current rise/fall times proportional to [Cgs + Cgd] VGGIG

Voltage rise/fall times proportional to Cgd VdIG

Cgs roughly the same size as Cgd and Vd >> VGG

Hence voltage switching times much greater than current switching times.

23-1. vs(t) = 2 Vs sin(wt) ; iL(t) = vs(t)RL

; a < wt < p

<PSCR> = 12p ıÛ

[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt)

<PSCR> = Vs

2!pRL {1 + cos(a)} +

1π Î

˚˙˘

2Ron [π - a +

sin(2a)2 ]

23-2. 120 °F = 49 °C ; Tj,max = 125 °C ; <PSCR>|max = Tj,max!-!Ta,max

<PSCR>|max = 125!-!49

0.1 = 760 Watts

Check <PSCR> at a = 0

<PSCR> = 2202π!(1) [1 + cos(0)] +

1π ÎÍ

È˚˙!

2(2x10-3)[π - 0 +

sin(0)2 ] = 107 watts

<PSCR> = 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible.

Average load power <PL> = 12π ıÛ

π{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt)

<PL> = 12π ıÛ

π{ 2!(220)}2sin2(wt)!d(wt) =

(220)26.28 [3.14 - a +

sin(2a)2 ]

For a = 0 <PL> = 24.2 kW

23-3. PSCR(t) = instantaneous power dissipated in the SCR during turn-on.

PSCR(t) = vAK(t) iA(t) = VAK {1 - ttf

} dIdt t during tf

P(t) = power density = PSCR(t)

A(t) = watts per cm2 ; A(t) = conducting area of SCR

; a < wt < p

<PSCR> = 12p ıÛ

<PSCR> = Vs

2!pRL {1 + cos(a)} +

1π Î

˚˙˘

2Ron [π - a +

sin(2a)2 ]

<PSCR>|max = 125!-!49

0.1 = 760 Watts

<PSCR> = 2202π!(1) [1 + cos(0)] +

1π ÎÍ

È˚˙!

2(2x10-3)[π - 0 +

sin(0)2 ] = 107 watts

<PL> = 12π ıÛ

π{ 2!(220)}2sin2(wt)!d(wt) =

(220)26.28 [3.14 - a +

sin(2a)2 ]

For a = 0 <PL> = 24.2 kW

} dIdt t during tf

; a < wt < p

<PSCR> = 12p ıÛ

<PSCR> = Vs

2!pRL {1 + cos(a)} +

1π Î

˚˙˘

2Ron [π - a +

sin(2a)2 ]

<PSCR>|max = 125!-!49

0.1 = 760 Watts

<PSCR> = 2202π!(1) [1 + cos(0)] +

1π ÎÍ

È˚˙!

2(2x10-3)[π - 0 +

sin(0)2 ] = 107 watts

<PL> = 12π ıÛ

π{ 2!(220)}2sin2(wt)!d(wt) =

(220)26.28 [3.14 - a +

sin(2a)2 ]

For a = 0 <PL> = 24.2 kW

} dIdt t during tf

A(t) = π [ro + us t]2 - π ro2 = π [2 ro us t + (us t)2]

dTj = 1Cv

tfP(t)!dt =

ıÙÛ

VAK!{1!-!!ttf

!}!dIdt !!t!

π![2!ro!us!t!+!(us!t)2]!dt

dTj = 1Cv

2πrous dIdt

ıÙÙÛ

!ÎÍÍÈ

˚˙˙˘

!1!-!!

1!+!ust2ro

!dt ; Let a = a' = 1, b = -1tf

, and b' = us2ro

Integral becomes ıÙÛ

!ÎÍÈ

a!+!bta'!+!b't !dt ; Using integral tables

ıÙÛ

!ÎÍÈ

a!+!bta'!+!b't !dt =

btfb' +

[ab'!-!ab][b']2

ln[a' + b' tf] ;

b = - 1

2x10-5 = - 5x104 sec-1 ; b' = 104

(2)(0.5) = 104 sec-1

Evaluating the integral yields

ıÙÛ

!ÎÍÈ

a!+!bta'!+!b't !dt =

-1104 + Î

˚˙˘

10-4!+!!5x104

108 ln[1 + (104)(2x10-5)] = 9.4x10-6 sec

With Cv = 1.75 Joule/(°C-cm3), the expression for dTj becomes

dTj = (103)(!9.4x10-6)

(2π)(1.75)(104)(0.5) dIdt = 125 - 25 = 100 °C ; Solving for

dIdt yields

dIdt =

1001.7x10-7 = 590 A/ms

23-4. Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times.

Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability.

23-5. ton = (4!-!0.5)!cm

3.5104 = 350 microseconds

23-6. Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct.

In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.

23-7. a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt.

Nd = 1.3x1017

2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns

b) t = qWd

kT(mn!+!mp) = (1.6x10-19)(2x10-2)2

(1.4x10-23)(300)(900) = 17 microseconds

Used (mn + mp) = 900 cm2/V-sec which is value appropriate to large excess carrier

densities (approaching 1017 cm-3)

c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term.

Rdrift ≈ Wd

q!(mn!+!mp)!nb!A ; 2

2000 = 10-3 = 0.02

(1.6x10-19)(900)(1017)!A ;

Solving for A gives A = 0.02

(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2

23-5. ton = (4!-!0.5)!cm

Nd = 1.3x1017

2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns

b) t = qWd

kT(mn!+!mp) = (1.6x10-19)(2x10-2)2

(1.4x10-23)(300)(900) = 17 microseconds

Rdrift ≈ Wd

2000 = 10-3 = 0.02

(1.6x10-19)(900)(1017)!A ;

(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2

23-5. ton = (4!-!0.5)!cm

Nd = 1.3x1017

2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns

b) t = qWd

kT(mn!+!mp) = (1.6x10-19)(2x10-2)2

(1.4x10-23)(300)(900) = 17 microseconds

Rdrift ≈ Wd

2000 = 10-3 = 0.02

(1.6x10-19)(900)(1017)!A ;

(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2

23-5. ton = (4!-!0.5)!cm

Nd = 1.3x1017

2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns

b) t = qWd

kT(mn!+!mp) = (1.6x10-19)(2x10-2)2

(1.4x10-23)(300)(900) = 17 microseconds

Rdrift ≈ Wd

2000 = 10-3 = 0.02

(1.6x10-19)(900)(1017)!A ;

(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2

Resulting current density is 20001.4 = 1430 A/cm2 which is excessively large. Probably

should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.

23-8. Cathode area = 3000200 = 15 cm2 = 0.65 Asi ; Asi =

150.65 = 23 cm2

23 cm2 = π Rsi2 ; Rsi =

23π = 2.7 cm

23-9.dvAK

dt |max ≈ IBO

Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance

Cj2(0) ≈ e!A

Wdepl(0) ; Wdepl(0) ≈ 2!e!fj2q!Nd

; fj2 = kTq ln[

NaNdni

fj2 = 0.026 ln[(1014)(1017)

(1020) ] = 0.66 V ;

Wdepl(0) ≈ (2)(11.7)(8.9x10-14)(0.66)

(1.6x10-19)(1014) = 2.9 microns

Cj2(0) = (11.7)(8.9x10-14)(10)

2.9x10-4 = 36 nF

dvAKdt |max =

0.053.6x10-8 = 1.4x106 V/sec or 1.4 V per microsecond

150.65 = 23 cm2

23 cm2 = π Rsi2 ; Rsi =

23π = 2.7 cm

23-9.dvAK

dt |max ≈ IBO

Cj2(0) ≈ e!A

; fj2 = kTq ln[

NaNdni

fj2 = 0.026 ln[(1014)(1017)

(1020) ] = 0.66 V ;

Wdepl(0) ≈ (2)(11.7)(8.9x10-14)(0.66)

(1.6x10-19)(1014) = 2.9 microns

Cj2(0) = (11.7)(8.9x10-14)(10)

2.9x10-4 = 36 nF

dvAKdt |max =

150.65 = 23 cm2

23 cm2 = π Rsi2 ; Rsi =

23π = 2.7 cm

23-9.dvAK

dt |max ≈ IBO

Cj2(0) ≈ e!A

; fj2 = kTq ln[

NaNdni

fj2 = 0.026 ln[(1014)(1017)

(1020) ] = 0.66 V ;

Wdepl(0) ≈ (2)(11.7)(8.9x10-14)(0.66)

(1.6x10-19)(1014) = 2.9 microns

Cj2(0) = (11.7)(8.9x10-14)(10)

2.9x10-4 = 36 nF

dvAKdt |max =

24-1. Cross-sectional view of GTO gate-cathode area with reverse gate current flowing.

GateCathode

Center Line

t++ -- vGKvGK

The negative gate current, -iG, flowing through the P2 layher beneath the N2 cathode layer develops a lateral voltage drop vGK as indicated. Maximum negative vGK = BVJ3

|vGK| = |IG,max| RGK < BVJ3 in order to avoid breakdown or IG,max < BVJ3RGK

RGK = R

2!N = rp2!W4!N!L!t ; N = number of cathode islands in parallel.

IG,max = BVJ3RGK

= IA,maxboff : Solving for IA,max yields

IA,max = 4!N!L!t!boff!BVJ3

24-2. Assume that the current is communtated from the GTO to the turn-off snubber and associated stray inductance linearly as a function of time. That is the inductor current

24-1. Cross-sectional view of GTO gate-cathode area with reverse gate current flowing.

GateCathode

Center Line

t++ -- vGKvGK

The negative gate current, -iG, flowing through the P2 layher beneath the N2 cathode layer develops a lateral voltage drop vGK as indicated. Maximum negative vGK = BVJ3

|vGK| = |IG,max| RGK < BVJ3 in order to avoid breakdown or IG,max < BVJ3RGK

RGK = R

2!N = rp2!W4!N!L!t ; N = number of cathode islands in parallel.

IG,max = BVJ3RGK

= IA,maxboff : Solving for IA,max yields

IA,max = 4!N!L!t!boff!BVJ3

24-2. Assume that the current is communtated from the GTO to the turn-off snubber and associated stray inductance linearly as a function of time. That is the inductor current

iLs = Io ttfi

; Assume that just prior to the end of the current fall time interval, the

voltage across the snubber capacitor has built up to approximately Vd.

vAK,max = 1.5 Vd = Ls diLs

dt + vcap = Ls

+ Vd ; Solving for Ls yields

Ls = Vd!tfi2!Io

24-3. Equivalent circuit during tgq shown below.

PNPN J3 forward

biased during tgq

i (t)G

LGdiGdt = - VGG- ; iG(t) =

-!VGG-LG

t ; At t = tgq want iG = - Io

boff ; Solve for LG

LG = boff!VGG-!tgq

(5)(15)(5x10-6)(500) = 0.75 microhenries

Equivalent circuit during tw2 interval.

iLs = Io ttfi

; Assume that just prior to the end of the current fall time interval, the

voltage across the snubber capacitor has built up to approximately Vd.

vAK,max = 1.5 Vd = Ls diLs

dt + vcap = Ls

+ Vd ; Solving for Ls yields

Ls = Vd!tfi2!Io

24-3. Equivalent circuit during tgq shown below.

PNPN J3 forward

biased during tgq

i (t)G

LGdiGdt = - VGG- ; iG(t) =

-!VGG-LG

t ; At t = tgq want iG = - Io

boff ; Solve for LG

LG = boff!VGG-!tgq

(5)(15)(5x10-6)(500) = 0.75 microhenries

Equivalent circuit during tw2 interval.

i (t)G

LGdiGdt = BVJ3 - VGG- ; iG(0) = -

Ioboff ; iG(t) = -

Ioboff +

!BVJ3!-!VGG-!LG

At t = tw2 , iG = 0 ; solving for tw2 yields

tw2 = Io!LG

boff![BVJ3!-!VGG-] =

(500)(7.5x10-7)(5)(25!-!15) = 7.5 microseconds

25-1.Ron(MOS)

A proprotional to 1

mmajority ; mn = 3 mp ; Hence

Ron(p-channel)A = 3

Ron(n-channel)A

Ron(IGBT)A proportional to

1dn(mn!+!mp) ; dn = excess carrier density

dn = dp so p-channel IGBTs have the same Ron as n-channel IGBTs

25-2. Turn-off waveforms of short versus long lifetime IGBTs

long lifetime

short lifetimeI (long)BJT

I (short)BJTt

Long lifetime IGBT -

a. BJT portion of the device has a larger beta and thus the BJT section carries the largest fraction of the IGBT current. Thus IBJT(long) > IBJT(short).

b. Longer lifetime leads to longer BJT turn-off times.

Short lifetime IGBT -

a. BJT beta smaller. MOSFET section of the device carries most of the current.

b. Shorter lifetime means less stored charge in the BJT section and thus faster turn-off.

25-1.Ron(MOS)

A proprotional to 1

mmajority ; mn = 3 mp ; Hence

Ron(p-channel)A = 3

Ron(n-channel)A

Ron(IGBT)A proportional to

1dn(mn!+!mp) ; dn = excess carrier density

dn = dp so p-channel IGBTs have the same Ron as n-channel IGBTs

25-2. Turn-off waveforms of short versus long lifetime IGBTs

long lifetime

short lifetimeI (long)BJT

I (short)BJTt

Long lifetime IGBT -

a. BJT portion of the device has a larger beta and thus the BJT section carries the largest fraction of the IGBT current. Thus IBJT(long) > IBJT(short).

b. Longer lifetime leads to longer BJT turn-off times.

Short lifetime IGBT -

a. BJT beta smaller. MOSFET section of the device carries most of the current.

b. Shorter lifetime means less stored charge in the BJT section and thus faster turn-off.

P P+N -

Body region of MOSFET section of IGBT

Collector junction of pnp BJT section of IGBT Base of pnp BJT

Emitter of pnp BJT

V > VDS1DS2

Depletion layer Effective base width

Drain of IGBT

Significant encroachment intoa the base of the PNP BJT section by the depletion layer of the blocking junction. The effective base width is thus lowered and the beta increases as vDS increases. This is base width modulation and it results in a lower output resistance ro (steeper slope in the active region of the iD-vDS characteristics).

P P+N -

Emitter of pnp BJT

Depletion layer

Drain of IGBT

Effective base width independent of VDS

Depletion encroaches into the N- layer but the advance is halted at moderate vDS values

by the N+ buffer layer. The PNP base width becomes constant and so the effective resistance ro remains large.

25-4. One dimensional model of n-channel IGBT

P P+N- N+Source

1017 1014

25 m m

Reverse blocking junction is the P+ - N+ junction because of body-source short.

BVRB ≈ 1.3x1017

1019 < 1 volt. No reverse blocking capability.

Forward breakdown - limited by P - N- junction.

P P+N -

Emitter of pnp BJT

Depletion layer

Drain of IGBT

Effective base width independent of VDS

Depletion encroaches into the N- layer but the advance is halted at moderate vDS values

by the N+ buffer layer. The PNP base width becomes constant and so the effective resistance ro remains large.

25-4. One dimensional model of n-channel IGBT

P P+N- N+Source

1017 1014

25 m m

Reverse blocking junction is the P+ - N+ junction because of body-source short.

BVRB ≈ 1.3x1017

1019 < 1 volt. No reverse blocking capability.

Forward breakdown - limited by P - N- junction.

BVFB = 1.3x1017

1014 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns

130 microns > 25 micron drift region length. Hence forward blocking limited bypunch-through.

BVFB = (2x105)(2.5x10-3) - (1.6x10-19)(1014)(2.5x10-3)2

(2)(11.7)(8.9x10-14) = 453 volts

25-5. IGBT current - Ion,IGBT ; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT

Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT.

Ion,IGBT ≈ 3!-!0.8

Ron,IGBT! ; Ron,IGBT ≈

Wdq!(mn!+!mp)!nb!A

Wd = (10-5)(750) = 75 mm ; Ron,IGBT = 7.5x10-3

(1.6x10-19)(900)(1016)(2) = 2.6x10-3 W

Ion,IGBT ≈ 2.2

2.6x10-3 ≈ 850 amps

MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS

Ion,MOS = Von(MOS)Ron,MOS

; Ron,MOS = Wd

q!mn!!Nd!A ; Wd = 75 mm

Nd = 1.3x1017

750 = 1.7x1014 cm-3

Ron,MOS = 7.5x10-3

(1.6x10-19)(1.5x103)(1.7x1014)(2) = 0.09 ohms

Ion,MOS = 3

0.09 = 33 amps

25-6. Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT

Ion,PTIon,NPT

≈ Ron,NPTRon,PT

since Vj,NPT ≈ Vj,PT

BVFB = 1.3x1017

BVFB = (2x105)(2.5x10-3) - (1.6x10-19)(1014)(2.5x10-3)2

(2)(11.7)(8.9x10-14) = 453 volts

Ion,IGBT ≈ 3!-!0.8

Wdq!(mn!+!mp)!nb!A

Wd = (10-5)(750) = 75 mm ; Ron,IGBT = 7.5x10-3

(1.6x10-19)(900)(1016)(2) = 2.6x10-3 W

Ion,IGBT ≈ 2.2

2.6x10-3 ≈ 850 amps

; Ron,MOS = Wd

Nd = 1.3x1017

750 = 1.7x1014 cm-3

Ron,MOS = 7.5x10-3

(1.6x10-19)(1.5x103)(1.7x1014)(2) = 0.09 ohms

Ion,MOS = 3

0.09 = 33 amps

Ion,PTIon,NPT

≈ Ron,NPTRon,PT

BVFB = 1.3x1017

BVFB = (2x105)(2.5x10-3) - (1.6x10-19)(1014)(2.5x10-3)2

(2)(11.7)(8.9x10-14) = 453 volts

Ion,IGBT ≈ 3!-!0.8

Wdq!(mn!+!mp)!nb!A

Wd = (10-5)(750) = 75 mm ; Ron,IGBT = 7.5x10-3

(1.6x10-19)(900)(1016)(2) = 2.6x10-3 W

Ion,IGBT ≈ 2.2

2.6x10-3 ≈ 850 amps

; Ron,MOS = Wd

Nd = 1.3x1017

750 = 1.7x1014 cm-3

Ron,MOS = 7.5x10-3

(1.6x10-19)(1.5x103)(1.7x1014)(2) = 0.09 ohms

Ion,MOS = 3

0.09 = 33 amps

Ion,PTIon,NPT

≈ Ron,NPTRon,PT

Ron,NPTRon,PT

= Wd,NPT!Wd,PT

≈ 2 assuming doping level in PT drift region is much less than

the doping level in the NPT drift region.

Hence Ion,PT

Ion,NPT ≈ 2

25-7. Cv dT = dQ ; dQ = PV dt ; P = power dissipated in IGBT during overcurrent transient.

V = volume in IGBT where power is dissipated.Duration of transient = dt.

P = Iov2 Ron ; Iov =

!V!Cv!dT

dt!Ron ; V ≈ A Wdrift

Ron = Wdrift

q!(mn!+!mp)!nb!A ; Iov = q!(mn!+!mp)!nb!A2!Cv!dT

Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)

(10-5) ≈ 2.5x103 amps

Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived.

25-8. The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance.

25-9. Check dvDS

dt at turn-off ; dvDS

dt = Vd!trv

; trv < 0.75 microseconds

dt > Vd!toff

!3x10-7 = 2330 V/ms > 800 V/ms limit.

Device is overstressed by an overly large dvDS

Ron,NPTRon,PT

= Wd,NPT!Wd,PT

Hence Ion,PT

Ion,NPT ≈ 2

!V!Cv!dT

Ron = Wdrift

Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)

(10-5) ≈ 2.5x103 amps

25-9. Check dvDS

dt = Vd!trv

dt > Vd!toff

!3x10-7 = 2330 V/ms > 800 V/ms limit.

Ron,NPTRon,PT

= Wd,NPT!Wd,PT

Hence Ion,PT

Ion,NPT ≈ 2

!V!Cv!dT

Ron = Wdrift

Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)

(10-5) ≈ 2.5x103 amps

25-9. Check dvDS

dt = Vd!trv

dt > Vd!toff

!3x10-7 = 2330 V/ms > 800 V/ms limit.

Ron,NPTRon,PT

= Wd,NPT!Wd,PT

Hence Ion,PT

Ion,NPT ≈ 2

!V!Cv!dT

Ron = Wdrift

Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)

(10-5) ≈ 2.5x103 amps

25-9. Check dvDS

dt = Vd!trv

dt > Vd!toff

!3x10-7 = 2330 V/ms > 800 V/ms limit.

Check switching losses Psw = Vd Io !tri!+!trv!+!!tfi!+!tfv

Psw = (700)(100) (3x10-7!+!7.5x10-7)

2 (2.5x104) = 875 W

Allowable power loss = Tj,max!-!Ta

Rqja =

150!-!250.5 = 250 watts

Switching losses exceed allowable losses. Module is overstressed by too muchpower dissipation.

26-1. Equivalent circuit for JFET in active region.

vGSvDS

Equivelent circuit for JFET Linearized I-V characteristicsin the blocking state.

CGSvGS

0 -V GS2GS1-Vi D

VDS1 VDS20

26-2. Drive circuit configuration

Vdrive

MOSFET off

VDS = VKG = - VGK = VDD!R2R1!+!R2

Negative enough to insure thatthe FCT is off.

MOSFET on

VDS = VKG = - VGK = 0

and FCT is on.

26-1. Equivalent circuit for JFET in active region.

vGSvDS

Equivelent circuit for JFET Linearized I-V characteristicsin the blocking state.

CGSvGS

0 -V GS2GS1-Vi D

VDS1 VDS20

26-2. Drive circuit configuration

Vdrive

MOSFET off

VDS = VKG = - VGK = VDD!R2R1!+!R2

Negative enough to insure thatthe FCT is off.

MOSFET on

VDS = VKG = - VGK = 0

and FCT is on.

MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss

26-3. Vdrift = Wd

(mn!+!!mp)!t ; BVBD = EBD!Wd

2 ; Vdrift,GaAs = Vdrift,Si

2(Si)(mn!+!!mp)|Si!tSi

2(GaAs)(mn!+!!mp)|GaAs!tGaAs

tGaAtSi =

(mn!+!!mp)|Si(mn!+!!mp)|GaAs

ÎÍÈ

˚˙˘EBD(Si)

EBD(GaAs) 2

(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec

EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm

tGaAtSi =

29 ÎÍ

È˚˙3

= 0.125 ; GaAs has the shorter lifetime.

26-4. EBD = 107 V/cm ; BVBD = EBD tox

tox = 103

107 = 10-4 cm = 1 micron

26-5. IA,max = (105)(1.5x10-2) = 1500 amperes

26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence

IA,max = (3)(105)(1.5x10-2) = 4500 amperes

26-7. Assume an n-type drift region since mn > mp.

26-3. Vdrift = Wd

tGaAtSi =

ÎÍÈ

˚˙˘EBD(Si)

EBD(GaAs) 2

tGaAtSi =

29 ÎÍ

È˚˙3

tox = 103

107 = 10-4 cm = 1 micron

26-5. IA,max = (105)(1.5x10-2) = 1500 amperes

IA,max = (3)(105)(1.5x10-2) = 4500 amperes

26-3. Vdrift = Wd

tGaAtSi =

ÎÍÈ

˚˙˘EBD(Si)

EBD(GaAs) 2

tGaAtSi =

29 ÎÍ

È˚˙3

tox = 103

107 = 10-4 cm = 1 micron

26-5. IA,max = (105)(1.5x10-2) = 1500 amperes

IA,max = (3)(105)(1.5x10-2) = 4500 amperes

26-3. Vdrift = Wd

tGaAtSi =

ÎÍÈ

˚˙˘EBD(Si)

EBD(GaAs) 2

tGaAtSi =

29 ÎÍ

È˚˙3

tox = 103

107 = 10-4 cm = 1 micron

26-5. IA,max = (105)(1.5x10-2) = 1500 amperes

IA,max = (3)(105)(1.5x10-2) = 4500 amperes

26-3. Vdrift = Wd

tGaAtSi =

ÎÍÈ

˚˙˘EBD(Si)

EBD(GaAs) 2

tGaAtSi =

29 ÎÍ

È˚˙3

tox = 103

107 = 10-4 cm = 1 micron

26-5. IA,max = (105)(1.5x10-2) = 1500 amperes

IA,max = (3)(105)(1.5x10-2) = 4500 amperes

26-3. Vdrift = Wd

tGaAtSi =

ÎÍÈ

˚˙˘EBD(Si)

EBD(GaAs) 2

tGaAtSi =

29 ÎÍ

È˚˙3

tox = 103

107 = 10-4 cm = 1 micron

26-5. IA,max = (105)(1.5x10-2) = 1500 amperes

IA,max = (3)(105)(1.5x10-2) = 4500 amperes

Rdrift = Wd

q!mn!Nd!A ; Rdrift,sp = Rdrift A = Wd

q!mn!Nd

Using Eq. (20-1) Nd = e!EBD

2!q!BVBD ; Using Eq. (20-3) Wd =

2!!BVBDEBD

Substituting into the expression for Rdrift,sp yields

Rdrift,sp = 2!!BVBD

2!q!BVBDe!EBD

2 = !!!4!!(BVBD)2

e!mn!(EBD)3

26-8. Silicon : Rdrift,sp = (4)(500)2

(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2

GaAs: Rdrift,sp = (4)(500)2

(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2

6H-SiC: Rdrift,sp = (4)(500)2

(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2

Diamond: Rdrift,sp = (4)(500)2

(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2

26-9. Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at anygiven temperature. This statement presumes that the phase change listed for diamond inthe table of material properties exceeds the sublimation temperature of SiC (1800 °C).

26-10. Eq. (20-1): Nd = e!EBD

2!q!BVBD

For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2

(2)(1.6x10-19)(BVBD) =

5.7x1017BVBD

For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2

(2)(1.6x10-19)(BVBD) =

1.1x1019BVBD

Rdrift = Wd

q!mn!Nd

2!!BVBDEBD

Rdrift,sp = 2!!BVBD

2!q!BVBDe!EBD

2 = !!!4!!(BVBD)2

e!mn!(EBD)3

(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2

(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2

(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2

(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2

26-10. Eq. (20-1): Nd = e!EBD

2!q!BVBD

For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2

(2)(1.6x10-19)(BVBD) =

5.7x1017BVBD

For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2

(2)(1.6x10-19)(BVBD) =

1.1x1019BVBD

Rdrift = Wd

q!mn!Nd

2!!BVBDEBD

Rdrift,sp = 2!!BVBD

2!q!BVBDe!EBD

2 = !!!4!!(BVBD)2

e!mn!(EBD)3

(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2

(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2

(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2

(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2

26-10. Eq. (20-1): Nd = e!EBD

2!q!BVBD

For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2

(2)(1.6x10-19)(BVBD) =

5.7x1017BVBD

For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2

(2)(1.6x10-19)(BVBD) =

1.1x1019BVBD

Rdrift = Wd

q!mn!Nd

2!!BVBDEBD

Rdrift,sp = 2!!BVBD

2!q!BVBDe!EBD

2 = !!!4!!(BVBD)2

e!mn!(EBD)3

(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2

(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2

(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2

(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2

26-10. Eq. (20-1): Nd = e!EBD

2!q!BVBD

For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2

(2)(1.6x10-19)(BVBD) =

5.7x1017BVBD

For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2

(2)(1.6x10-19)(BVBD) =

1.1x1019BVBD

For diamond: Nd = (5.5)(8.9x10-14)(107)2

(2)(1.6x10-19)(BVBD) =

1.5x1020BVBD

Eq. (20-3): Wd = 2!!BVBD

For GaAs: Wd = 2!!BVBD

4x105 = 5x10-6 BVBD [cm]

For 6H-SiC: Wd = 2!!BVBD

2x106 = 10-6 BVBD [cm]

For diamond: Wd = 2!!BVBD

107 = 2x10-7 BVBD [cm]

26-11. Use equations from problem 26-10.

Material Nd Wd

GaAs 2.9x1015 cm-3 10-2 cm

6H-SiC 5.5x1016 cm-3 2x10-3 cm

Diamond 7.5x1017 cm-3 4x10-5 cm

26-12. Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k)-1

k = thermal conductivity and C = constant

Using silicon diode data: C = (Tj!!-!!Tcase)!k

Pdiode =

(150!-!50)(!1.5)200 = 0.75 cm-1

Rqjc(GaAs) = 0.750.5 = 1.5 °C/W : Rqjc(SiC) =

0.755 = 0.15°C/W

Rqjc(diamond) = 0.7520 = 0.038°C/W

Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C

For diamond: Nd = (5.5)(8.9x10-14)(107)2

(2)(1.6x10-19)(BVBD) =

1.5x1020BVBD

Eq. (20-3): Wd = 2!!BVBD

4x105 = 5x10-6 BVBD [cm]

2x106 = 10-6 BVBD [cm]

107 = 2x10-7 BVBD [cm]

Material Nd Wd

GaAs 2.9x1015 cm-3 10-2 cm

6H-SiC 5.5x1016 cm-3 2x10-3 cm

Pdiode =

(150!-!50)(!1.5)200 = 0.75 cm-1

0.755 = 0.15°C/W

Rqjc(diamond) = 0.7520 = 0.038°C/W

Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C

For diamond: Nd = (5.5)(8.9x10-14)(107)2

(2)(1.6x10-19)(BVBD) =

1.5x1020BVBD

Eq. (20-3): Wd = 2!!BVBD

4x105 = 5x10-6 BVBD [cm]

2x106 = 10-6 BVBD [cm]

107 = 2x10-7 BVBD [cm]

Material Nd Wd

GaAs 2.9x1015 cm-3 10-2 cm

6H-SiC 5.5x1016 cm-3 2x10-3 cm

Pdiode =

(150!-!50)(!1.5)200 = 0.75 cm-1

0.755 = 0.15°C/W

Rqjc(diamond) = 0.7520 = 0.038°C/W

Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C

Tj(diamond) = (0.038)(200) + 50 = 57.4 °C

26-13. Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3.

W2 = Wo 1!+!

Vpfc ; fc =

kTq ln

ÎÍÍÈ

˚˙˙˘Na!Nd

ni2 ; Wo =

2!e!fcq!Nd

Solving for Vp yields Vp = fc ÎÍÈ

2!Wo 2 - fc

˚˙˘(1019)(2x1014)

1010 = 0.8 V

Wo = (2)(11.7)(8.9x10-14)(0.8)

(1.6x10-19)(2x1014) = 2.3 microns

Vp = (0.8) ÎÍÈ

˚˙10

(2)(2.3) 2 - 0.8 = 3.8 - 0.8 = 3 V

26-14. Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.

Tj(diamond) = (0.038)(200) + 50 = 57.4 °C

W2 = Wo 1!+!

Vpfc ; fc =

kTq ln

ÎÍÍÈ

˚˙˙˘Na!Nd

ni2 ; Wo =

2!e!fcq!Nd

2!Wo 2 - fc

˚˙˘(1019)(2x1014)

1010 = 0.8 V

Wo = (2)(11.7)(8.9x10-14)(0.8)

(1.6x10-19)(2x1014) = 2.3 microns

Vp = (0.8) ÎÍÈ

˚˙10

(2)(2.3) 2 - 0.8 = 3.8 - 0.8 = 3 V

Tj(diamond) = (0.038)(200) + 50 = 57.4 °C

W2 = Wo 1!+!

Vpfc ; fc =

kTq ln

ÎÍÍÈ

˚˙˙˘Na!Nd

ni2 ; Wo =

2!e!fcq!Nd

2!Wo 2 - fc

˚˙˘(1019)(2x1014)

1010 = 0.8 V

Wo = (2)(11.7)(8.9x10-14)(0.8)

(1.6x10-19)(2x1014) = 2.3 microns

Vp = (0.8) ÎÍÈ

˚˙10

(2)(2.3) 2 - 0.8 = 3.8 - 0.8 = 3 V

source

P+P+ W - 2Wo

W + Wg

W + W /2o g

l gsWg

l - W - W /2gd o g

Rs = lgs

q!mn!Nd!d!W = 10-3

(1.6x10-19)(1500)(2x1014)(0.07)(10-3) = 298 ohms

Rc = lc

q!mn!Nd!d!(W!-!2Wo)

= 10-3

(1.6x10-19)(1500)(2x1014)(0.07)(10-3!-!4.6x10-4) = 552 ohms

Rt estimate. Treat the region of thickness Wo + Wg/2 as though it has an average width

given by (W!-!2!Wo)!+!(W!+!Wg)

2 = W + Wg/2 - Wo. Rt now approximately given by

Rt = Wo!+!Wg/2

q!mn!Nd!d!(W!+!Wg/2!-!Wo)

Rt = (10-3!+!5x10-4)

(1.6x10-19)(1500)(2x1014)(0.07)(10-3!+!5x10-4!-!2.3x10-4) = 351 ohms

Rd = (lgd!-!Wo!-Wg/2)

q!mn!Nd!d!(W!+!Wg) =

Rd = (35x10-4!-!2.3x10-4!-!5x10-4)

(1.6x10-19)(1500)(2x1014)(0.07)(10-3!+10-3) = 412 ohms

Total resistance of a single cell is Rcell = Rs + Rc + Rt + Rd

Rcell = 298 + 552 + 351 + 412 = 1613 ohms

There are N = 28 cells in parallel so the the net on-state resistance is

Ron = Rcell

N = 161328 = 58 ohms

26-15. As the drain-source voltage increases, the reverse-bias on the gate-drain pn junction increases. The depletion region of the two adjacent P+ regions merge and then grow towards the drain. The drift region of length lgd and doping Nd must contain this depletion region and will determine the breakdown voltage. The short length of the drift region suggests that punch-through will limit the breakdown voltage. Check for this possibility first.

Non-punch-through estimate:

BV = 1.3x1017

2x1014 = 650 V ; Wd > (10-5)(6.5x102) = 65 microns > 35 microns

Hence this is a punch-through structure. Use Eq. (21-21) and EBD = 2x105 V/cm

BV = (2x105)(3.5x10-3) - (1.6x10-19)(2x1014)(3.5x10-3)2

(2)(11.7)(8.9x10-14) = 700 - 189

BV = 511 V

Rcell = 298 + 552 + 351 + 412 = 1613 ohms

There are N = 28 cells in parallel so the the net on-state resistance is

Ron = Rcell

N = 161328 = 58 ohms

26-15. As the drain-source voltage increases, the reverse-bias on the gate-drain pn junction increases. The depletion region of the two adjacent P+ regions merge and then grow towards the drain. The drift region of length lgd and doping Nd must contain this depletion region and will determine the breakdown voltage. The short length of the drift region suggests that punch-through will limit the breakdown voltage. Check for this possibility first.

Non-punch-through estimate:

BV = 1.3x1017

2x1014 = 650 V ; Wd > (10-5)(6.5x102) = 65 microns > 35 microns

Hence this is a punch-through structure. Use Eq. (21-21) and EBD = 2x105 V/cm

BV = (2x105)(3.5x10-3) - (1.6x10-19)(2x1014)(3.5x10-3)2

(2)(11.7)(8.9x10-14) = 700 - 189

BV = 511 V

27-1. a. During turn-off of the GTO, Io communtates linearly to Cs.

Cs dvCdt = Io

; dvCdt =

dvAKdt =

Io!tCs!tfi

< 5x107 V/s

Maximum dvAK

dt occurs at tfi. Solving for Cs yields

Cs > Io ÎÍÈ

dvAKdt

5005x107 = 10 microfarads

Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then

ICs,max = 1000 - 500 -100 = 400 A

Rs = !500!400 ≈ 1.3 ohms

Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.

b. Power dissipated in snubber PRs ≈ fsw!Cs!Vd

PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW

27-2. L sdiAdt max = Vd ; Ls ≈

5003x108 ≈ 1.7 microhenries.

Voltage across GTO at turn-off = Vd + Io Rs : Assume Io

Rs = 0.2 Vd

Rs = (0.2)(500)

(500) = 0.2 ohms.

27-3. vCs(t) = Vd - Vd cos(wot) + Vd Cbase

Cs sin(wot) = Vd + K sin(wot - f)

vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)

Cs dvCdt = Io

; dvCdt =

dvAKdt =

Io!tCs!tfi

< 5x107 V/s

Maximum dvAK

Cs > Io ÎÍÈ

dvAKdt

ICs,max = 1000 - 500 -100 = 400 A

Rs = !500!400 ≈ 1.3 ohms

PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW

Rs = 0.2 Vd

Rs = (0.2)(500)

(500) = 0.2 ohms.

Cs dvCdt = Io

; dvCdt =

dvAKdt =

Io!tCs!tfi

< 5x107 V/s

Maximum dvAK

Cs > Io ÎÍÈ

dvAKdt

ICs,max = 1000 - 500 -100 = 400 A

Rs = !500!400 ≈ 1.3 ohms

PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW

Rs = 0.2 Vd

Rs = (0.2)(500)

(500) = 0.2 ohms.

K sin(wot) cos(f) - K cos(wot) sin(f) = Vd Cbase

Cs sin(wot) - Vd cos(wot)

K cos(f) = Vd Cbase

Cs and K sin(f) = Vd ;

[ ]K!cos(f) 2 + [ ]K!sin(f) 2 = K2 = Vd2

CbaseCs

vCs,max = Vd + K = Vd + Vd 1!+!Cbase

27-4. a. Equivalent circuit after diode reverse recovery.

L = 10 mH

200 V+

iL(0+) = Irr ; During reverse recovery L diRdt = 200 V

diRdt =

Irrtrr

= 20010-5 = 2x107 A/sec ; Irr = (2x107)(3x10-7) = 6 A

b. vCs,max = 500 V = 200 + 200 1!+!Cbase

1!+!Cbase

Cs = 1.5 ;

CbaseCs

= 1.5 ≈ 1.25

Cbase = (10-5) ÎÍÈ

˚˙˘62

(200)2 = 9 nF

Cs = 9!nF1.25 ≈ 7 nF

K sin(wot) cos(f) - K cos(wot) sin(f) = Vd Cbase

Cs sin(wot) - Vd cos(wot)

K cos(f) = Vd Cbase

Cs and K sin(f) = Vd ;

[ ]K!cos(f) 2 + [ ]K!sin(f) 2 = K2 = Vd2

CbaseCs

vCs,max = Vd + K = Vd + Vd 1!+!Cbase

27-4. a. Equivalent circuit after diode reverse recovery.

L = 10 mH

200 V+

iL(0+) = Irr ; During reverse recovery L diRdt = 200 V

diRdt =

Irrtrr

= 20010-5 = 2x107 A/sec ; Irr = (2x107)(3x10-7) = 6 A

b. vCs,max = 500 V = 200 + 200 1!+!Cbase

1!+!Cbase

Cs = 1.5 ;

CbaseCs

= 1.5 ≈ 1.25

Cbase = (10-5) ÎÍÈ

˚˙˘62

(200)2 = 9 nF

Cs = 9!nF1.25 ≈ 7 nF

27-5. Use the circuit shown in problem 27-4.

Cs = Cbase = (10-5) ÎÍÈ

˚˙˘62

(200)2 = 9 nF

Rs = 1.3 Rbase = (1.3) ÎÍÈ

˚˙200

6 = 43 ohms

vCs,max = (1.5)(200) = 300 V

27-6. P = WR fsw = ÎÍÈ

˚˙˘Ls!Irr

2!+!Cs!Vin2

WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules

P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts

27-7. a. BJT waveforms (trv assumed to be zero for Cs = 0)

tt f i

C = 0s

C > 0s

Power dissipation for Cs = 0 is Pc = Vd!Io!tfi

˚˙˘62

(200)2 = 9 nF

Rs = 1.3 Rbase = (1.3) ÎÍÈ

˚˙200

6 = 43 ohms

vCs,max = (1.5)(200) = 300 V

˚˙˘Ls!Irr

2!+!Cs!Vin2

WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules

P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts

tt f i

C = 0s

C > 0s

˚˙˘62

(200)2 = 9 nF

Rs = 1.3 Rbase = (1.3) ÎÍÈ

˚˙200

6 = 43 ohms

vCs,max = (1.5)(200) = 300 V

˚˙˘Ls!Irr

2!+!Cs!Vin2

WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules

P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts

tt f i

C = 0s

C > 0s

Pc = (200)(25)(4x10-7)

2 (2x104) = 20 W

Power dissipation for Cs = Cs1

Pc = Wc fsw ; Wc = ıÙÛ

Io(1!-!ttfi

2!Cs1!t2tfi

!!d!t = Vd!Io!tfi

Pc = (200)(25)(4x10-7)(2x104)

12 = 3.3 watts

Factor of six reduction in the turn-off losses.

b. BJT losses increase at turn-on only becaue of energy stored in Cs being dissipatedin the BJT, but also because the time to complete turn-on is extended as shown in

Fig. 27-14a. This extended duration of traversal of the active region also increases the turn-on losses.

During the turn-on interval, the collector-emitter voltage is given by (assuming that the external circuit dominates the transient)

Cs1 dvCE

dt = - diCdt t - Irr ; vCE(t) = Vd -

t22!Cs1

- Irr t

Seting the expression for vCE(t) equal to zero and solving for the timeDT = t2 - (tri + trr) (see Fig. 27-14a) required for vCE to reach zero yields

DT = - Irr

diC/dt + ÎÍÈ

˚˙˘

diC/dt2!+!

2!Vd!Cs1diC/dt !

Note that DT = 0 if Cs1 = 0 which is consistent with the assumption that the external circuit and not the BJT that dominates the turn-on transient. Extra energy disspated in the BJT at turn-on due to Cs1 is thus

DTvCE(t)!iC(t)!dt = Vd Irr DT + Î

˚˙˘Vd

2 !diCdt !!!-!

[Io!+!Irr]!Irr2!Cs1

- (Io!+!3!Irr)

2!Cs1 diCdt DT3 - ÎÍ

È˚˙diC

dt 2 DT4

The increase in the BJT loss is ÎÍÍÈ

˚˙˙˘

DTvCE(t)!iC(t)!dt fs where fs is the switching

frequency. Numerical evaluation of DT gives DT = 0.29 ms (Irr = 10 A and Cs1 = 25 nF).

Evaluation of the loss gives 1.3x10-3 fs = 26.5 watts

27-8. a. Ds shorts out the snubber resistance during the BJT turn-off. Hence Cs1 is directly across the BJT as in problem 27-7a. Thus the loss reduction is the same as in problem 27-7a.

b. Equivalent circuit during BJT turn-off after free-wheeling diode reverse recovery is shown below.

sCI + tr rdi Cdt

-v (0 ) = V+C d

vCE = CsRs dvCdt + vC ;; Cs

dvCdt = - Irr -

diCdt t

Combining equations and solving for vCE(t) yields

vCE(t) = Vd - Irr Rs - ÎÍÈ

˚˙˘

!IrrCs

!+!Rs!diCdt t -

t22!Cs

At t = D T, vCE = 0 and turn-on is completed.

D T = - Irr!+!Rs!Cs!!

!diCdt !

+ ÎÍÍÈ

˚˙˙˘

Irr!+!Rs!Cs!!diCdt

!diCdt !

2!CsdiCdt

![Vd!-!Irr!Rs]

D T goes to zero when Rs = VdIrr

. hence there is no increase in power dissipation

in the BJT due to the presence of Cs.

The increase in the BJT loss is ÎÍÍÈ

˚˙˙˘

DTvCE(t)!iC(t)!dt fs where fs is the switching

frequency. Numerical evaluation of DT gives DT = 0.29 ms (Irr = 10 A and Cs1 = 25 nF).

Evaluation of the loss gives 1.3x10-3 fs = 26.5 watts

27-8. a. Ds shorts out the snubber resistance during the BJT turn-off. Hence Cs1 is directly across the BJT as in problem 27-7a. Thus the loss reduction is the same as in problem 27-7a.

b. Equivalent circuit during BJT turn-off after free-wheeling diode reverse recovery is shown below.

sCI + tr rdi Cdt

-v (0 ) = V+C d

vCE = CsRs dvCdt + vC ;; Cs

dvCdt = - Irr -

diCdt t

Combining equations and solving for vCE(t) yields

vCE(t) = Vd - Irr Rs - ÎÍÈ

˚˙˘

!IrrCs

!+!Rs!diCdt t -

t22!Cs

At t = D T, vCE = 0 and turn-on is completed.

D T = - Irr!+!Rs!Cs!!

!diCdt !

+ ÎÍÍÈ

˚˙˙˘

Irr!+!Rs!Cs!!diCdt

!diCdt !

2!CsdiCdt

![Vd!-!Irr!Rs]

D T goes to zero when Rs = VdIrr

. hence there is no increase in power dissipation

in the BJT due to the presence of Cs.

27-9. a. Proposed snubber circuit configuration shown below.

I o2 Vs

Equivalent circuit swith SCRs 3 & 4 on and 1&2 going off or vice-versa. Continuous flow of load current formces SCRs 1 & 2 to remain on past the time of natural commutation (when vs(t) goes through zero and becomes negative).

3 or 4

1 or 2

With 3 & 4 on, 1 & 2 are off, and effectively in parallel with the Rs-Cs snubber. same is true when 1 & 2 are on and 3 & 4 are off. Thus the Rs-Cs snubber functions as a turn-off snubber.

b. w Ls = 0.05!Vs

Ia1 ; worst case situation (maximum reverse voltage across SCR

which is turning off) occurs when SCR which is turning on is triggered with a

delay angle of 90°. During reverse recovery of SCR1, Ls didt = 2 Vs and

didt =

Irrtrr

. Solving for Irr yields

Irr = 2!w!trr!Ia1

!0.05 ; Cbase = Ls ÎÍÍÈ

˚˙˙˘

! 2!Vs 2

= !w!!Ia1!trr

!0.05!Vs

Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values

Cs = 0.9933 Ia1 mF

Rbase = 2!VsIrr

= !0.05!Vsw!trr!Ia1

= (0.05)!(230)

(377)!(10-5)!Ia1 =

3050Ia1

Rs,opt = 1.3 Rbase = 4000Ia1

c. Peak line voltage = 2 (230) = 322 V

Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbaseand Rs = 1.3 Rbase. For Ia1 = 100 A

Rs,opt = 1.3 Rbase = 4000100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF

27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required.

The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.

27-11. a. Check dvDS

dt at turn-off; dvDS

dt = Vdtrv

dt > 700

7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed.

Not enough information available to check on power dissipation or overcurrents.

didt =

Irrtrr

Irr = 2!w!trr!Ia1

˚˙˙˘

! 2!Vs 2

= !w!!Ia1!trr

!0.05!Vs

Cs = 0.9933 Ia1 mF

Rbase = 2!VsIrr

= !0.05!Vsw!trr!Ia1

= (0.05)!(230)

(377)!(10-5)!Ia1 =

3050Ia1

dt = Vdtrv

dt > 700

didt =

Irrtrr

Irr = 2!w!trr!Ia1

˚˙˙˘

! 2!Vs 2

= !w!!Ia1!trr

!0.05!Vs

Cs = 0.9933 Ia1 mF

Rbase = 2!VsIrr

= !0.05!Vsw!trr!Ia1

= (0.05)!(230)

(377)!(10-5)!Ia1 =

3050Ia1

dt = Vdtrv

dt > 700

b. dvDS

dt = IoCs

= 400 V/ms ; Cs = 100

4x108 = 0.25 mF

Choose Rs to limit total current ID to less than 150 A

150 A = 100 + 700Rs

; Rs = 700

!150!-!100 = 14 ohms

Check snubber recovery time = 2.3 Rs Cs = (2.3)(14)(2.5x10-7) = 8 ms

Off time of the IGBT is 10 microseconds which is greater than the snubber recovery time. Hence choice of Rs is fine.

Chapater 28 Problem Solutions

28-1. Schematic of drive circuit shown below.

vDS(t) waveform same as in problem 22-2.

During MOSFET turn-on dvDS

dt = Vdtfv

< 500 V/ms

From problem 22-2, Vdtfv

= ÎÍÈ

˚˙˘

VGG+!-!VGSth!-!!Iogm

RG!Cgd

During MOSFET turn-off, dvDS

dt = Vdtrv

< 500 V/ms

From problem 23-2, Vdtrv

= ÎÍÈ

˚˙˘

VGG-!+!VGSth!+!!Iogm

RG!Cgd

gm = Io

VGS!-!VGSth =

607!-!4 = 20 A/V

Estimate of RG for MOSFET turn-on:

5x108 V/sec > VGG+!-!4!-!

(RG)(4x10-10) ; VGG+,min = VGSth +

= 4 + 10020 = 9 V

Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses.

RG > 15!-!9

(4x10-10)(5x108) = 30 ohms

Estimate of RG at MOSFET turn-off:

5x108 V/sec > VGG-!+!4!+!

(RG)(4x10-10)

Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimizeturn-off times.

RG > 15!+!9

(4x10-10)(5x108) = 115 ohms

Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG-= 15 V and RG > 115 ohms.

28-2. a. Circuit diagram shown below.

RG2Q s

I o= 200 A

Vd = 1000 V

When FCT is off we need VKG = (1.25) 100040 = 31.25 V = (1000)

RG2RG1!+!!RG2

Now RG1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2

RG2 = 31.25 kW and RG1 = 969 kW

Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses.

RG > 15!-!9

(4x10-10)(5x108) = 30 ohms

Estimate of RG at MOSFET turn-off:

5x108 V/sec > VGG-!+!4!+!

(RG)(4x10-10)

Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimizeturn-off times.

RG > 15!+!9

(4x10-10)(5x108) = 115 ohms

Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG-= 15 V and RG > 115 ohms.

28-2. a. Circuit diagram shown below.

RG2Q s

I o= 200 A

Vd = 1000 V

When FCT is off we need VKG = (1.25) 100040 = 31.25 V = (1000)

RG2RG1!+!!RG2

Now RG1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2

RG2 = 31.25 kW and RG1 = 969 kW

b. MOSFET characteristics

- large on-state current capability- low Ron- low BVDSS (BVDSS of 50-100 V should work)

29-1. Assume Ts = 120 °C and Ta = 20 °C.

Rq,rad ≈ 0.12A ; A in m2 ; Eq. (29-18)

Rq,conv ≈ 1

(1.34)(A) ÎÍÈ

˚˙˘dvert

DT 1/4

; Eq. (29-20)

Rq,conv ≈ 0.24A [dvert]

1/4 for DT = 100 °C.

A cube having a side of length dvert has a surface area A = 6 [dvert]2

Rq,conv ≈ 0.4

[dvert]1.75 ; Rq,rad ≈

0.02[dvert]

Rq,sa = net surface-to-ambient thermal resistance = Rq,conv!Rq,rad!

Rq,conv!!+!Rq,rad

Rq,sa = 0.04

![dvert]1.75!+!2![dvert]

Heat Sink # 1 2 3 5 6

Volume [m]3 7.6x10-5 10-4 1.8x10-4 2x10-4 3x10-4

dv = (vol.)1/3 [m] 0.042 0.046 0.057 0.058 0.067

A = 6 [dv]2 [m]2 0.011 0.013 0.019 0.002 0.027

dv1.75 0.004 0.046 0.0066 0.0069 0.0088

dv2 0.0018 0.0021 0.0032 0.0034 0.0045

Rq,sa [°C/W] 5.3 4.5 3.1 2.9 2.3

Rq,sa (measured) 3.2 2.3 2.2 2.1 1.7

Heat Sink # 7 8 9 10 11 12

Volume [m]3 4.4x10-4 6.810-4 6.1x10-4 6.3x10-4 7x10-4 1.4x10-3

dv = (vol.)1/3 [m] 0.076 0.088 0.085 0.086 0.088 0.11

A = 6 [dv]2 [m]2 0.034 0.046 0.043 0.044 0.047 0.072

dv1.75 0.011 0.014 0.013 0.014 0.014 0.021

dv2 0.0058 0.0078 0.0071 0.0073 0.0078 0.012

Rq,sa [°C/W] 1.8 1.4 1.5 1.4 1.3 0.9

Rq,sa (measured) 1.3 1.3 1.25 1.2 0.8 0.65

Heat sink #9 is relatively large and cubical in shape with only a few cooling fins.Heat sink #9 is small and flat with much more surface area compared to its volume.Large surface-to-volume ratios give smaller values of Rq,sa.

29-2. Rq,conv ≈ 0.24A [dvert]

1/4 for DT = 100 °C ; From problem 29-1

Rq,conv ≈ 24 0 [dvert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2

dvert Rq,conv

1 cm 76 °C/W

5 cm 113 °C/W

12 cm 141 °C/W

20 cm 160 °C/W

Volume [m]3 4.4x10-4 6.810-4 6.1x10-4 6.3x10-4 7x10-4 1.4x10-3

dv = (vol.)1/3 [m] 0.076 0.088 0.085 0.086 0.088 0.11

A = 6 [dv]2 [m]2 0.034 0.046 0.043 0.044 0.047 0.072

dv1.75 0.011 0.014 0.013 0.014 0.014 0.021

dv2 0.0058 0.0078 0.0071 0.0073 0.0078 0.012

Rq,sa [°C/W] 1.8 1.4 1.5 1.4 1.3 0.9

Rq,sa (measured) 1.3 1.3 1.25 1.2 0.8 0.65

Heat sink #9 is relatively large and cubical in shape with only a few cooling fins.Heat sink #9 is small and flat with much more surface area compared to its volume.Large surface-to-volume ratios give smaller values of Rq,sa.

29-2. Rq,conv ≈ 0.24A [dvert]

1/4 for DT = 100 °C ; From problem 29-1

Rq,conv ≈ 24 0 [dvert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2

dvert Rq,conv

1 cm 76 °C/W

5 cm 113 °C/W

12 cm 141 °C/W

20 cm 160 °C/W

0 2 4 6 8 10 12 14 16 18 20

dvert [cm]

R ,conv

29-3. Rq,conv ≈ 1

(1.34)(A) ÎÍÈ

˚˙˘dvert

DT 1/4

; Eq. (29-20)

Rq,conv ≈ 353 [DT]-.25 A = 10 cm2 and dvert = 5 cm

DT Rq,conv

60 °C 127 °C/W

80 °C 118 °C/W

100 °C 112 °C/W

120 °C 107 °C/W

0 2 4 6 8 10 12 14 16 18 20

dvert [cm]

R ,conv

29-3. Rq,conv ≈ 1

(1.34)(A) ÎÍÈ

˚˙˘dvert

DT 1/4

; Eq. (29-20)

Rq,conv ≈ 353 [DT]-.25 A = 10 cm2 and dvert = 5 cm

DT Rq,conv

60 °C 127 °C/W

80 °C 118 °C/W

100 °C 112 °C/W

120 °C 107 °C/W

60 70 80 90 100 110 120

R ,convq

DT [°C]

29-4. Rq,rad = !DT

5.1!A!ËÁÊ

¯˜!ËÁ

Ê¯˜Ts

1004!-!ËÁ

Ê¯˜Ta

; Eq. (29-17)

Rq,rad = 196 120!-!Ta(° C)

!ÎÍÈ

˚˙239!-!ÎÍ

È˚˙Ta(° K)

1004! ; A = 10 cm2 and Ts = 120 °C

Ta Rq,rad

0 °C 128 °C/W

10 °C 123 °C/W

20 °C 119 °C/W

40 °C 110 °C/W

60 70 80 90 100 110 120

R ,convq

DT [°C]

29-4. Rq,rad = !DT

5.1!A!ËÁÊ

¯˜!ËÁ

Ê¯˜Ts

1004!-!ËÁ

Ê¯˜Ta

; Eq. (29-17)

Rq,rad = 196 120!-!Ta(° C)

!ÎÍÈ

˚˙239!-!ÎÍ

È˚˙Ta(° K)

1004! ; A = 10 cm2 and Ts = 120 °C

Ta Rq,rad

0 °C 128 °C/W

10 °C 123 °C/W

20 °C 119 °C/W

40 °C 110 °C/W

B B BB

0 5 10 15 20 25 30 35 40

Ta [°C ]

R ,rad

[ °C/W]

29-5. Rq,rad = !DT

5.1!A!ËÁÊ

¯˜!ËÁ

Ê¯˜Ts

1004!-!ËÁ

Ê¯˜Ta

; Eq. (29-17)

Rq,rad = 196 Ts(° C)!-!40

!ÎÍÈ

ÎÍÈ

˚˙Ts(° K)

-!96 ; A = 10 cm2 and Ta= 40 °C

Ts Rq,rad

80 °C 114 °C/W

100 °C 120 °C/W

120 °C 110 °C/W

140 °C 101 °C/W

B B BB

0 5 10 15 20 25 30 35 40

Ta [°C ]

R ,rad

[ °C/W]

29-5. Rq,rad = !DT

5.1!A!ËÁÊ

¯˜!ËÁ

Ê¯˜Ts

1004!-!ËÁ

Ê¯˜Ta

; Eq. (29-17)

Rq,rad = 196 Ts(° C)!-!40

!ÎÍÈ

ÎÍÈ

˚˙Ts(° K)

-!96 ; A = 10 cm2 and Ta= 40 °C

Ts Rq,rad

80 °C 114 °C/W

100 °C 120 °C/W

120 °C 110 °C/W

140 °C 101 °C/W

80 90 100 110 120 130 140

Ts [ °C ]

Rq,rad

[ °C/W]

29-6. PMOSFET,max = 150!° C!!-!50!° C!

!1!° C/W = 100 W ; 100 W = 50 + 10-3 fs

Solving for fs yields fs = 50 kHz

29-7. PMOSFET = 50 + 10-3 • 2.5x104 = 75 W

Rq,ja = Rq,jc + Rq,ca = 150!° C!!-!35!° C!

!75!W = 1.53 °C/W

Rq,ca = 1.53 - 1.00 = 0.53 °C/W

80 90 100 110 120 130 140

Ts [ °C ]

Rq,rad

[ °C/W]

29-6. PMOSFET,max = 150!° C!!-!50!° C!

!1!° C/W = 100 W ; 100 W = 50 + 10-3 fs

29-7. PMOSFET = 50 + 10-3 • 2.5x104 = 75 W

Rq,ja = Rq,jc + Rq,ca = 150!° C!!-!35!° C!

!75!W = 1.53 °C/W

Rq,ca = 1.53 - 1.00 = 0.53 °C/W

80 90 100 110 120 130 140

Ts [ °C ]

Rq,rad

[ °C/W]

29-6. PMOSFET,max = 150!° C!!-!50!° C!

!1!° C/W = 100 W ; 100 W = 50 + 10-3 fs

29-7. PMOSFET = 50 + 10-3 • 2.5x104 = 75 W

Rq,ja = Rq,jc + Rq,ca = 150!° C!!-!35!° C!

!75!W = 1.53 °C/W

Rq,ca = 1.53 - 1.00 = 0.53 °C/W

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