53326496 Power Electronics Mohan 2nd Edition Solution

Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density n i equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus n i (T i ) = N d = 10 14 = 10 10 exp qE g 2k 1 T i 1 300 Solving for T i using E g = 1.1 eV, k = 1.4x10 -23 [1/K] yields T i = 262 C or 535 K. 19-2. N-side resistivity n = 1 q n N d = 1 (1.6x10 -19 )(1500)(10 14 ) = 43.5 ohm-cm P-side resistivity p =

1 q p N a = 1 (1.6x10 -19 )(500)(10 18 ) = 0.013 ohm-cm 19-3. Material is n-type with N d = 10 13 cm -3 >> n i = 10 10 cm -3 . Hence use approximate formulas given in Chapter 19. n = N d = 10 13 cm -3 ; p = n 2 i N d = 10 20 10 13 = 10 7 cm -3 19-4. p o = n 2 i [300] N d ; 2p o =

n 2 i [300+T] N d 2 n 2 i [300] = n 2 i [300 + T] ; 10 = 10 10 exp qE g 2k 1 T 1 300 Solving for T yields T = qE g 300 (qE g -k300ln(2)) = 305.2 K T = 305.2 - 300 = 5.2 K. 19-5. I 1 = I s exp( qV 1 kT ; 10 I 1 = I s exp( qV 1 + V kT ) ; V = kT q

2x10

ln(10) = 60 mV 19-6. (a) x n (0) = depletion layer width on n-side at zero bias; x p (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 n i equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus n i (T i ) = N d = 10 14 = 10 10 exp qE g 2k 1 T i 1 300 Solving for T i using E g = 1.1 eV, k = 1.4x10 -23 [1/K] yields T i = 262 C or 535 K. 19-2. N-side resistivity n = 1 q n N d = 1 (1.6x10 -19 )(1500)(10

14 ) = 43.5 ohm-cm P-side resistivity p = 1 q p N a = 1 (1.6x10 -19 )(500)(10 18 ) = 0.013 ohm-cm 19-3. Material is n-type with N d = 10 13 cm -3 >> n i = 10 10 cm -3 . Hence use approximate formulas given in Chapter 19. n = N d = 10 13 cm -3 ; p = n 2 i N d = 10 20 10 13 = 10 7 cm -3 19-4. p o = n 2 i

[300] N d ; 2p o = n 2 i [300+T] N d 2 n 2 i [300] = n 2 i [300 + T] ; 10 = 10 10 exp qE g 2k 1 T 1 300 Solving for T yields T = qE g 300 (qE g -k300ln(2)) = 305.2 K T = 305.2 - 300 = 5.2 K. 19-5. I 1 = I s exp( qV 1 kT ; 10 I 1 = I s exp( qV

2x10

1 + V kT ) kT q

;

V =

ln(10) = 60 mV 19-6. (a) x n (0) = depletion layer width on n-side at zero bias; x p (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 n i equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus n i (T i ) = N d = 10 14 = 10 10 exp qE g 2k 1 T i 1 300 Solving for T i using E g = 1.1 eV, k = 1.4x10 -23 [1/K] yields T i = 262 C or 535 K. 19-2. N-side resistivity n = 1 q n N

d = 1 (1.6x10 -19 )(1500)(10 14 ) = 43.5 ohm-cm P-side resistivity p = 1 q p N a = 1 (1.6x10 -19 )(500)(10 18 ) = 0.013 ohm-cm 19-3. Material is n-type with N d = 10 13 cm -3 >> n i = 10 10 cm -3 . Hence use approximate formulas given in Chapter 19. n = N d = 10 13 cm -3 ; p = n 2 i N d = 10 20 10 13 = 10 7 cm -3

19-4. p o = n 2 i [300] N d ; 2p o = n 2 i [300+T] N d 2 n 2 i [300] = n 2 i [300 + T] ; 10 = 10 10 exp qE g 2k 1 T 1 300 Solving for T yields T = qE g 300 (qE g -k300ln(2)) = 305.2 K T = 305.2 - 300 = 5.2 K. 19-5. I 1 = I s exp( qV 1 kT

2x10

; 1 = I s

10 I

exp( qV 1 + V kT ) ; V = kT q ln(10) = 60 mV 19-6. (a) x n (0) = depletion layer width on n-side at zero bias; x p (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 n i equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus n i (T i ) = N d = 10 14 = 10 10 exp qE g 2k 1 T i 1 300 Solving for T i using E g = 1.1 eV, k = 1.4x10 -23 [1/K] yields T i = 262 C or 535 K. 19-2. N-side resistivity

n = 1 q n N d = 1 (1.6x10 -19 )(1500)(10 14 ) = 43.5 ohm-cm P-side resistivity p = 1 q p N a = 1 (1.6x10 -19 )(500)(10 18 ) = 0.013 ohm-cm 19-3. Material is n-type with N d = 10 13 cm -3 >> n i = 10 10 cm -3 . Hence use approximate formulas given in Chapter 19. n = N d = 10 13 cm -3 ; p = n 2 i N d = 10 20

10 13 = 10 7 cm -3 19-4. p o = n 2 i [300] N d ; 2p o = n 2 i [300+T] N d 2 n 2 i [300] = n 2 i [300 + T] ; 10 = 10 10 exp qE g 2k 1 T 1 300 Solving for T yields T = qE g 300 (qE g -k300ln(2)) = 305.2 K T = 305.2 - 300 = 5.2 K. 19-5. I 1

2x10

= I s exp( qV 1 kT ; 1 = I s exp( qV 1 + V kT ) ; V = kT q ln(10) = 60 mV 19-6. (a) x n (0) = depletion layer width on n-side at zero bias; x p (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 n i equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus n i (T i ) = N d = 10 14 = 10 10 exp qE g 2k 1 T i 1 300 Solving for T i using E g = 1.1 eV, k = 1.4x10 10 I

-23 [1/K] yields T i = 262 C or 535 K. 19-2. N-side resistivity n = 1 q n N d = 1 (1.6x10 -19 )(1500)(10 14 ) = 43.5 ohm-cm P-side resistivity p = 1 q p N a = 1 (1.6x10 -19 )(500)(10 18 ) = 0.013 ohm-cm 19-3. Material is n-type with N d = 10 13 cm -3 >> n i = 10 10 cm -3 . Hence use approximate formulas given in Chapter 19. n = N d = 10 13 cm -3 ; p = n 2

i N d = 10 20 10 13 = 10 7 cm -3 19-4. p o = n 2 i [300] N d ; 2p o = n 2 i [300+T] N d 2 n 2 i [300] = n 2 i [300 + T] ; 10 = 10 10 exp qE g 2k 1 T 1 300 Solving for T yields T = qE g 300 (qE

2x10

g -k300ln(2)) = 305.2 K T = 305.2 - 300 = 5.2 K. 19-5. I 1 = I s exp( qV 1 kT ; 10 I 1 = I s exp( qV 1 + V kT ) ; V = kT q ln(10) = 60 mV 19-6. (a) x n (0) = depletion layer width on n-side at zero bias; x p (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 n i equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus n i (T i ) = N d = 10 14 = 10 10 exp qE g 2k 1 T i 1 300

Solving for T i using E g = 1.1 eV, k = 1.4x10 -23 [1/K] yields T i = 262 C or 535 K. 19-2. N-side resistivity n = 1 q n N d = 1 (1.6x10 -19 )(1500)(10 14 ) = 43.5 ohm-cm P-side resistivity p = 1 q p N a = 1 (1.6x10 -19 )(500)(10 18 ) = 0.013 ohm-cm 19-3. Material is n-type with N d = 10 13 cm -3 >> n i = 10 10 cm -3 . Hence use approximate formulas given in Chapter 19. n = N d = 10

13 cm -3 ; n 2 i N d = 10 20 10 13 = 10 7 cm -3 19-4. p o = n 2 i [300] N d ; 2p o = n 2 i [300+T] N d 2 n 2 i [300] = n 2 i [300 + T] ; 10 = 10 10 exp qE g 2k 1 T 1 300 p =

2x10

Solving for T yields T = qE g 300 (qE g -k300ln(2)) = 305.2 K T = 305.2 - 300 = 5.2 K. 19-5. I 1 = I s exp( qV 1 kT ; 10 I 1 = I s exp( qV 1 + V kT ) ; V = kT q ln(10) = 60 mV 19-6. (a) x n (0) = depletion layer width on n-side at zero bias; x p (0) = depletion layer width on p-side at zero bias. x n (0) + x p (0) = W o = 2 c (N a +N d ) qN a N d (1) c = kT

q ln N a N d n 2 i = 0.026 ln 10 14 10 15 10 20 = 0.54 eV Conservation of charge: q N a x p = q N d x n (2) Solving (1) and (2) simultaneously gives using the numerical values given in the problem statement gives: W o = 2.8 microns ; x n (0) = 2.55 microns ; x p (0) = 0.25 microns (b) Electric field profile triangular-shaped as shown in Fig. 19-9b. Maximum ele ctric at zero bias given by E max = 2 c W o = (2)(0.54) (2.8x10 -4 ) = 3,900 V/cm (c) From part a) c = 0.54 eV (d) C(V)

A = W o 1+ V c ; C(V) = space-charge capacitance at reverse voltage V. C(0) A = (11.7)(8.9x10 -14 ) 2.8x10 -4 = 3.7x10 -9 F/cm 2 C(50) A = (11.7)(8.9x10 -14 ) 2.8x10 -4 1+ 50 0.54 = 3.8x10 -10 F/cm 2 (e) I = I s exp( qV kT ) ; exp( qV kT ) = exp ( 0.7 0.026 ) = 5x10 11 I s = q n 2 i

D n N a + D p N d A = (1.6x10 -19 )(10 20 ) (38)(10 -6 ) (10 15 )(10 -6 ) + (13)(10 -6 ) (10 14 )(10 -6 ) (2) 2 I s = 6.7x10 -14 A ; I = (6.7x10 -14 )(5x10 11 ) = 34 mA 19-7. Resistance R = L A ; L A = 0.02

0.01 = 2 cm -1 At 25 C, N d = 10 14 >> n i so = 1 q n N d = 1 (1.6x10 -19 )(1500)(10 14 ) = 41.7 -cm R(25 C) = (41.7)(2) = 83.4 ohms At 250 C (523 K), n i [523] = 10 10 exp (1.6x10 -19 )(1.1) (2)(1.4x10 -23 ) 1 523 1 300 = (10 10 )(7.6x10 3 ) = 7.6x10 13 which is an appreciable fraction of N d = 10 14 . Thus we should solve Eqs. (19-2) and (19-3) exactly for n o

and p o rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for N d >> N a yields n o = N d 2 1+ 1+ 4n 2 i N 2 d and p o = n 2 i n o . Putting in numerical values yields n o = 10 14 2 1+ 1+ (4)(7.6x10 13 ) 2 (10 14 ) 2 = 1.4x10 14 and p o = 5.8x10 27 10 14

= 5.8x10 13 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) = (250 C)L A ; (250 C) 1 q n n o +q p p o = 1 (1.6x10 -19 )(1500)(1.4x10 14 )+(1.6x10 -19 )(500)(5.8x10 13 ) = 26.2 -cm ; R(250 C) (26.2)(2) = 19-8. BV BD = (N a +N d )E 2 BD 2qN a N d = (11.7)(8.9x10 14 )(10 15 +10 14 )(3x10 5 ) 2 (2)(1.6x10 -19 )(10

52.4 ohms

15 )(10 14 ) = 3,340 volts 19-7. Resistance R = L A ; L A = 0.02 0.01 = 2 cm -1 At 25 C, N d = 10 14 >> n i so = 1 q n N d = 1 (1.6x10 -19 )(1500)(10 14 ) = 41.7 -cm R(25 C) = (41.7)(2) = 83.4 ohms At 250 C (523 K), n i [523] = 10 10 exp (1.6x10 -19 )(1.1) (2)(1.4x10 -23 ) 1 523 1 300

= (10 10 )(7.6x10 3 ) = 7.6x10 13 which is an appreciable fraction of N d = 10 14 . Thus we should solve Eqs. (19-2) and (19-3) exactly for n o and p o rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for N d >> N a yields n o = N d 2 1+ 1+ 4n 2 i N 2 d and p o = n 2 i n o . Putting in numerical values yields n o = 10 14 2 1+ 1+ (4)(7.6x10 13 ) 2

(10 14 ) 2 = 1.4x10 14 and p o = 5.8x10 27 10 14 = 5.8x10 13 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) = (250 C)L A ; (250 C) 1 q n n o +q p p o = 1 (1.6x10 -19 )(1500)(1.4x10 14 )+(1.6x10 -19 )(500)(5.8x10 13 ) = 26.2 -cm ; R(250 C) (26.2)(2) = 19-8. BV BD = (N a +N d )E 2 BD 2qN a N d

52.4 ohms

= (11.7)(8.9x10 14 )(10 15 +10 14 )(3x10 5 ) 2 (2)(1.6x10 -19 )(10 15 )(10 14 ) = 3,340 volts 19-9. E 2 max = E 2 BD 4f 2 c BV BD W 2 o f c ; Eq. (19-13); W 2 o f c = 4BV BD E 2 BD W 2 (BV BD ) = W 2 o or

BV BD f c ; W 2 o f c = 4BV BD E 2 BD and taking the square root yields W (BV BD ) 2BV BD E 2 BD . 19-10. L p = D p t = (13)(10 -6 ) = 36 microns ; L n = D n t = (39)(10 -6 ) = 62 microns 19-11. Assume a one-sided step junction with N a >> N d I 1 = q n 2 i A D p t 1 N d E q. (19- 11) ; Inserting

t 1 exp( qV kT ) ; I 2 = q n 2 i A D p t 2 N d t 2 exp( qV kT ) I 2 I 1 = 2 = t 1 t 2 ; Thus 4 t 2 = t 1 19-12. s = q m p p + q m n n ; np = n 2 i ; Combining yeilds s p n 2 i n + q m n n ds dn = 0 = - q m p n 2

= q m

i n 2 + q m n ; Solving for n yields n = n i m p m n and p = n i m n m p p = 10 10 1500 500 = 1.7x10 10 cm -3 ; n = 10 10 500 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 s min = 2 q n i m p m n . Putting in numerical values s min = (2)(1.6x10 -19 )(10 10 ) (500)(1500)

= 2.8x10 -6 mhos-cm 19-9. E 2 max = E 2 BD 4f 2 c BV BD W 2 o f c ; Eq. (19-13); W 2 o f c = 4BV BD E 2 BD W 2 (BV BD ) = W 2 o BV BD f c ; W 2 o f c = 4BV BD E 2 E q. (19- 11) ; Inserting or

BD and taking the square root yields W (BV BD ) 2BV BD E 2 BD . 19-10. L p = D p t = (13)(10 -6 ) = 36 microns ; L n = D n t = (39)(10 -6 ) = 62 microns 19-11. Assume a one-sided step junction with N a >> N d I 1 = q n 2 i A D p t 1 N d t 1 exp( qV kT ) ; I 2 = q n 2 i A D p t 2 N d

t 2 exp( qV kT ) I 2 I 1 = 2 = t 1 t 2 ; 2 = t 1 19-12. s = q m p p + q m n n ; np = n 2 i ; Combining yeilds s p n 2 i n + q m n n ds dn = 0 = - q m p n 2 i n 2 + q m n ; Solving for n yields n = n i m p m n and p = n i m n Thus 4 t

= q m

m p p = 10 10 1500 500 = 1.7x10 10 cm -3 ; n = 10 10 500 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 s min = 2 q n i m p m n . Putting in numerical values s min = (2)(1.6x10 -19 )(10 10 ) (500)(1500) = 2.8x10 -6 mhos-cm 19-9. E 2 max = E 2 BD 4f 2 c BV BD W 2

o f c ; Eq. (19-13); W 2 o f c = 4BV BD E 2 BD W 2 (BV BD ) = W 2 o BV BD f c ; W 2 o f c = 4BV BD E 2 BD and taking the square root yields W (BV BD ) 2BV BD E 2 BD . 19-10. L p = D E q. (19- 11) ; Inserting or

p t = (13)(10 -6 ) = 36 microns ; L n = D n t = (39)(10 -6 ) = 62 microns 19-11. Assume a one-sided step junction with N a >> N d I 1 = q n 2 i A D p t 1 N d t 1 exp( qV kT ) ; I 2 = q n 2 i A D p t 2 N d t 2 exp( qV kT ) I 2 I 1 = 2 = t 1 t 2 ; Thus 4 t 2

= t 1 19-12. s = q m p p + q m n n ; np = n 2 i ; Combining yeilds s p n 2 i n + q m n n ds dn = 0 = - q m p n 2 i n 2 + q m n

= q m

; Solving for n yields n = n i m p m n and p = n i m n m p p = 10 10 1500 500 = 1.7x10 10 cm -3 ; n = 10 10 500 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 s min = 2 q n i m p m n . Putting in numerical values s min = (2)(1.6x10 -19 )(10 10 ) (500)(1500) = 2.8x10 -6 mhos-cm 19-9. E 2 max = E 2 BD 4f 2 c BV BD W 2 o f c ; Eq. (19-13); W 2 o f c = 4BV BD E 2 BD or

W 2 (BV BD ) = W 2 o BV BD f c ; W 2 o f c = 4BV BD E 2 BD and taking the square root yields W (BV BD ) 2BV BD E 2 BD . 19-10. L p = D p t = (13)(10 -6 ) = 36 microns ; L n = D n t = (39)(10 -6 ) = 62 microns 19-11. Assume a one-sided step junction with N a >> N d I 1 = q n E q. (19- 11) ; Inserting

2 i A D p t 1 N d t 1 exp( qV kT ) ; I 2 = q n 2 i A D p t 2 N d t 2 exp( qV kT ) I 2 I 1 = 2 = t 1 t 2 ; Thus 4 t 2 = t 1 19-12. s = q m p p + q m n n ; np = n 2 i ; Combining yeilds s p n 2 i n + q m

= q m

n n ds dn = 0 = - q m p n 2 i n 2 + q m n ; Solving for n yields n = n i m p m n and p = n i m n m p p = 10 10 1500 500 = 1.7x10 10 cm -3 ; n = 10 10 500 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 s min = 2 q n i m p m n .

Putting in numerical values s min = (2)(1.6x10 -19 )(10 10 ) (500)(1500) = 2.8x10 -6 mhos-cm Chapter 20 Problem Solutions 20-1. N d = 1.3x10 17 BV BD = 1.3x10 17 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 pu nchthrough structure and Eq. (20-9) applies. BV BD = (2x10 5 )(5x10 -3 ) (1.6x10 -19 )(5x10 13 )(5x10 -3 ) 2 (2)(11.7)(8.9x10 -14 ) = 900 V 20-3. V on = V

j + V drift ; V j = kT q ln I I s ; For one-sided step junction I s = qAn 2 i L p N d t o ; Evaluating I s yields (1.6x10 -19 )(2)(10 10 ) 2 (13)(2x10 -6 ) (5x10 13 )(2x10 -6 ) = 1.6x10 - 9 A V d = K 1 I + K 2 (I) 2/3 Eq. (20-16) with I = forward bias current through the diode. K 1 = W

d qm o An b = 5x10 -3 (1.6x10 -19 )(900)(2)(10 17 ) = 1.7x10 -4 K 2 = 3 W 4 d q 2 m 3 o n 2 b A 2 t o = 3 (5x10 -3 ) 4 (1.6x10 -19 ) 2 (900) 3 (10 17 ) 2 (2) 2 (2x10 -6 ) = 7.5x10 -4 Chapter 20 Problem Solutions 20-1. N

d = 1.3x10 17 BV BD = 1.3x10 17 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 pu nchthrough structure and Eq. (20-9) applies. BV BD = (2x10 5 )(5x10 -3 ) (1.6x10 -19 )(5x10 13 )(5x10 -3 ) 2 (2)(11.7)(8.9x10 -14 ) = 900 V 20-3. V on = V j + V drift ; V j = kT q ln I I s

; For one-sided step junction I s = qAn 2 i L p N d t o ; Evaluating I s yields (1.6x10 -19 )(2)(10 10 ) 2 (13)(2x10 -6 ) (5x10 13 )(2x10 -6 ) = 1.6x10 - 9 A V d = K 1 I + K 2 (I) 2/3 Eq. (20-16) with I = forward bias current through the diode. K 1 = W d qm o An b = 5x10 -3 (1.6x10 -19 )(900)(2)(10 17 ) = 1.7x10

-4 K 2 = 3 W 4 d q 2 m 3 o n 2 b A 2 t o = 3 (5x10 -3 ) 4 (1.6x10 -19 ) 2 (900) 3 (10 17 ) 2 (2) 2 (2x10 -6 ) = 7.5x10 -4 Chapter 20 Problem Solutions 20-1. N d = 1.3x10 17 BV BD = 1.3x10 17 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 pu nchthrough structure and Eq. (20-9) applies. BV BD = (2x10 5 )(5x10 -3 ) (1.6x10 -19 )(5x10 13 )(5x10 -3 ) 2 (2)(11.7)(8.9x10 -14 ) = 900 V 20-3. V on = V j + V drift ; V j = kT q ln I I s ; For one-sided step junction I s = qAn 2 i L p N d o ; Evaluating I

s yields (1.6x10 -19 )(2)(10 10 ) 2 (13)(2x10 -6 ) (5x10 13 )(2x10 -6 ) = 1.6x10 9 V d = K 1 I + K 2 (I) 2/3 Eq. (20-16) with I = forward bias current through the diode. K 1 = W d q o An b 5x10 -3 (1.6x10 -19 )(900)(2)(10 17 ) = 1.7x10 -4 K 2 = 3 W 4 d q 2 3 o n

2 b A 2 o = 3 (5x10 -3 ) 4 (1.6x10 -19 ) 2 (900) 3 (10 17 ) 2 (2) 2 (2x10 -6 ) = 7.5x10 -4 I V j V drift V on 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 V on in volts

Forward current in amperes 20-4. a) V on (t) = R drift I(t) >> V j 1 V ; R drift = L A ; = 1 q n N d = 1 (1.6x10 -19 )(1500)(5x10 13 ) = 85 ohm-cm ; L A = 5x10 -3 2 = 2.5x10 -3 R drift = (85)(2.5x10 -3 ) = 0.21 ohms ; I(t) = 2.5x10 8 t ; 0 < t < 4 microseconds V on (t) = (0.21)(2.5x10 8 t ) = 5.3x10 7 t Volts ; 0 < t < 4 microseconds V on (4 s) = (5.3x10 7 )(4x10 -6 ) = 212 volts b) V on (t) = R drift

(t) I(t) ; R drift (t) = 0.21[1 - 2.5x10 7 t] V on (t) = {0.21[1 - 2.5x10 7 t]} { 2.5x10 8 t } = 53[t - 0.25 t 2 ] ; t in microseconds I V j V drift V on 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 V on in volts Forward current in amperes 20-4. a) V on (t) = R drift I(t) >> V j 1 V ; R drift = r L A ; r = 1 qm n

N d r = 1 (1.6x10 -19 )(1500)(5x10 13 ) = 85 ohm-cm ; L A = 5x10 -3 2 = 2.5x10 -3 R drift = (85)(2.5x10 -3 ) = 0.21 ohms ; I(t) = 2.5x10 8 t ; 0 < t < 4 microseconds V on (t) = (0.21)(2.5x10 8 t ) = 5.3x10 7 t Volts ; 0 < t < 4 microseconds V on (4 m s) = (5.3x10 7 )(4x10 -6 ) = 212 volts b) V on (t) = R drift (t) I(t) ; R drift (t) = 0.21[1 - 2.5x10 7 t] V on (t) = {0.21[1 - 2.5x10 7 t]} { 2.5x10 8 t } = 53[t - 0.25 t 2 ] ; 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. t off = t rr + t 3 = t rr + I F di R /dt = t rr + 2000 2.5x10 8 = t rr + 8 m s t rr = 2t I F di R /dt ; t = 4x10 -12 (BV BD ) 2 = 4x10 -12 (2000) 2 = 16 m s t rr = (2)(1.6x10 -5 )(2x10 3 )

2.5x10 8 = 16 m s ; t off = 8 m s + 16 m s = 24 m s 20-6. Assume a non-punch-through structure for the Schottky diode. N d = 1.3x10 17 BV BD = 1.3x10 17 150 = 8.7x10 14 cm -3 W d = 10 -5 BV BD = (10 -5 ) (150) = 15 microns 20-7. V drift = (100 A) (R drift ) = 2 V ; R drift = 0.02 ohms ; 0.02 = 1 qm n N d L A A = 2x10 -3 (1.6x10 -19 )(1500)(10 15 )(2x10 -2 ) = 0.42 cm 2 0 50






2.5x10 8 = 16 m s ; t off = 8 m s + 16 m s = 24 m s 20-6. Assume a non-punch-through structure for the Schottky diode. N d = 1.3x10 17 BV BD = 1.3x10 17 150 = 8.7x10 14 cm -3 W d = 10 -5 BV BD = (10 -5 ) (150) = 15 microns 20-7. V drift = (100 A) (R drift ) = 2 V ; R drift = 0.02 ohms ; 0.02 = 1 qm n N d L A A = 2x10 -3 (1.6x10 -19 )(1500)(10 15 )(2x10 -2 ) = 0.42 cm 2 20-8. Use Eq. (20-9) to solve for N d

; N d = [E BD W d - BV BD ] 2e qW 2 d = [(2x10 5 )(2x10 -3 ) - 300] (2)(11.7)(8.9x10 -14 ) (1.6x10 -19 )(2x10 -3 ) 2 = N d = 3.4x10 14 cm -3 20-9. R A npt = W d (npt) qm n N npt ; N npt = N d of non-punch-throuth (npt) diode R A pt = W d

(pt) qm n N pt ; N pt = N d of punch-through (pt) diode W d (npt) = e E BD qN npt ; Derived from Eqs. (19-11), (19-12), and (19-13) W d (pt) = e E BD qN pt

1+ _ 12qN pt BV BD e E 2 BD2qN pt BV BD e E 2 BD = qN pt e E BD N npt N npt 2BV BD E BD = qN

npt e E BD 2BV BD E BD N pt N npt = 1 W d (npt) W d (npt) N pt N npt = x = N pt N npt ; W d (pt) = W d (npt) 1 x [ ] 1+ _ 1-x If N pt t d,on and t d,off E on = (10)(300)(0.5)(5x10 -5 ) = 1.25x10 -3 Joules E rv = (0.5)(300)(10)(10 -7 ) = 1.5x10 -4 Joules E fi = (0.5)(300)(10)(9x10 -9 ) = 1.5x10 -5 Joules P c = (1.95x10 -3 )(2x10 4

) = 39 watts 22-3. Use test conditons to estimate C gd . Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below. E ri = (0.5)(300)(10)(2.1x10 -8 ) = 3x10 -5 Joules E fv = (0.5)(300)(10)(3x10 -7 ) = 4.5x10 -4 Joules E on = I o V DS,on [0.5 T - t d,on + t d,off ] ; V DS,on = I o r DS,on = (10)(0.5) = 5 V T >> t d,on and t d,off E on = (10)(300)(0.5)(5x10 -5 ) = 1.25x10 -3 Joules E rv = (0.5)(300)(10)(10 -7 ) = 1.5x10 -4 Joules E fi = (0.5)(300)(10)(9x10 -9 ) = 1.5x10 -5

Joules P c = (1.95x10 -3 )(2x10 4 ) = 39 watts 22-3. Use test conditons to estimate C gd . Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below. V (t) G V GG V (t) GS V GS(th) V GS,Io t t d,on t = t r i f v t = t f i r v t d,off t V GG V d V (t) DS I o i (t) D t Equivalent circuit during voltage and current rise and fall intervals: + - C gs C gd V (t) G R D V d g (V - V ) m GS GS(th) R G

Governing equation using Miller capacitance approximation: dv GS dt + v GS = V G (t) ; G [C gs + C gd {1 + g m R D }] ; During t ri = t fv interval, V G (t) = V GG . Solution is v GS (t) = V GG + {V GS(th) - V GG } e -t/ ; At t = t ri , V GS = V GS(th) + V d g m R D ; Solving for t ri = t = R

fv yields t ri = t fv = ln V GG -V GS(th) V GG -V GS(th) V d g m R D During t rv = t fi , v G (t) = 0 and solution is v GS (t) = V GS,Io e -t/ . At t = t rv , v GS (t) = V GS(th) . Solving for t rv yields t rv = ln V GS(th) + V d g m R

D V GS(th) . Invert equation for t ri to find C gd . Result is C gd = t ri ln V GG -V GS(th) V GG -V GS(th) V d g m R D -R G C gs 1 R G (1+g m R D ) C gd = 3x10 -8 ln

15-4 15-4-1 -5x10 -9 1 5(1+25) = 2.3x10 -9 F = 2.3 nF Solving for switching times in circuit with R D = 150 ohms. = [10 -9 + 2.3x10 -9 {1 + 150}][100] = 35 s t ri = 3.5x10 -5 ln 15-4 15-4-2 = 7 s ; t rv = 3.5x10 -5 ln 4+2 4 = 14 s 22-4. Waveforms for v DS (t) and i D (t) shown in previous problem. Power dissipation in MOSFET given by = [E on + E sw ] f s ; f s = 1 T ; E on = [I

D ] 2 r DS,on (T j ) T 2 ; dv GS dt + v GS = V G (t) ; G [C gs + C gd {1 + g m R D }] ; During t ri = t fv interval, V G (t) = V GG . Solution is v GS (t) = V GG + {V GS(th) - V GG } e -t/ ; At t = t ri , V GS = V GS(th) + = R

V d g m R D ; Solving for t ri = t fv yields t ri = t fv = ln V GG -V GS(th) V GG -V GS(th) V d g m R D During t rv = t fi , v G (t) = 0 and solution is v GS (t) = V GS,Io e -t/ . At t = t rv , v GS (t) = V GS(th) . Solving for t rv yields t rv = ln

V GS(th) + V d g m R D V GS(th) . Invert equation for t ri to find C gd . Result is C gd = t ri ln V GG -V GS(th) V GG -V GS(th) V d g m R D -R G C gs 1 R G (1+g m R D ) C gd

= 3x10 -8 ln 15-4 15-4-1 -5x10 -9 1 5(1+25) = 2.3x10 -9 F = 2.3 nF Solving for switching times in circuit with R D = 150 ohms. = [10 -9 + 2.3x10 -9 {1 + 150}][100] = 35 s t ri = 3.5x10 -5 ln 15-4 15-4-2 = 7 s ; t rv = 3.5x10 -5 ln 4+2 4 = 14 s 22-4. Waveforms for v DS (t) and i D (t) shown in previous problem. Power dissipation in MOSFET given by = [E on + E sw

] f s ; f s = 1 T ; E on = [I D ] 2 r DS,on (T j ) T 2 ; I D = V d R D = 300 150 = 2 A ; r DS,on (T j ) = 2 1+ T j -25 150 = 2 0.833+ T j 150 E on = (4)(2) 0.833+ T j 150 1 2f

s = {3.32 + 0.027 T j } 1 f s E sw = 1 T 0 t ri V d I D (1t t ri )( t t ri )dt 1 T

+

0 t fi V d I D (1t t fi )( t t fi )dt = V d I D 6 [t ri + t fi

] E sw = (300)(2) 6 [7x10 -6 + 14x10 -6 ] = 2.1x10 -3 joules = {3.32 + 0.027 T j } 1 f s f s + 2.1x10 -3 f s = {3.32 + 0.027 T j } +[ 2.1x10 -3 ][10 4 ] = 24.3 + 0.027 T j B B B B B 0 5 10 15 20 25 30 0 10 20 30 40 50 60 70 80 90 100 P MOSFET Watts Temperature [ K] 22-5. V on = on-state voltage of three MOSFETs in parallel = I o r eff

I D = V d R D = 300 150 = 2 A ; r DS,on (T j ) = 2 1+ T j -25 150 = 2 0.833+ T j 150 E on = (4)(2) 0.833+ T j 150 1 2f s = {3.32 + 0.027 T j } 1 f s E sw = 1 T 0 t ri V d I

D (1t t ri )( t t ri )dt 1 T

+

0 t fi V d I D (1t t fi )( t t fi )dt = V d I D 6 [t ri + t fi ] E sw = (300)(2) 6 [7x10 -6 + 14x10 -6 ] = 2.1x10 -3 joules = {3.32 + 0.027 T j } 1 f s

f s + 2.1x10 -3 f s = {3.32 + 0.027 T j } +[ 2.1x10 -3 ][10 4 ] = 24.3 + 0.027 T j B B B B B 0 5 10 15 20 25 30 0 10 20 30 40 50 60 70 80 90 100 P MOSFET Watts Temperature [ K] 22-5. V on = on-state voltage of three MOSFETs in parallel = I o r eff r eff = r 1 r 2 r 3 r 1 r 2 +r 2 r 3 +r 3 r 1

; r 1 etc. = on-state resistance of MOSFET #1 etc. r 1 (T j ) = r 1 (25 C) 1+0.8 T j -25 100 ; r 1 (105 C) = (1.64) r 1 (25 C) etc. r 1 (105 C) = 2.95 W ; r 2 (105 C) = 3.28 W ; r 3 (105 C) = 3.61 W r eff (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, P i = V on 2 2r i = I o 2 r eff 2 2r i ; Assume a 50% duty cycle and ignore switching losses. P 1 = (5) 2 (1.09) 2

(2)(2.95) = 5 W ; P 2 = (5) 2 (1.09) 2 (2)(3.28) = 4.5 W ; P 3 = (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 fas ter switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned o n before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks V d volts in the off-state and conducts I o amps in the on-state. r eff = r 1 r 2 r 3 r 1 r 2 +r 2 r 3 +r 3 r 1 ; r 1 etc. = on-state resistance of MOSFET #1 etc. r 1 (T j ) = r 1

(25 C) 1+0.8 T j -25 100 ; r 1 (105 C) = (1.64) r 1 (25 C) etc. r 1 (105 C) = 2.95 W ; r 2 (105 C) = 3.28 W ; r 3 (105 C) = 3.61 W r eff (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, P i = V on 2 2r i = I o 2 r eff 2 2r i ; Assume a 50% duty cycle and ignore switching losses. P 1 = (5) 2 (1.09) 2 (2)(2.95) = 5 W ; P 2 = (5) 2 (1.09) 2 (2)(3.28)

= 4.5 W ; P 3 = (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 fas ter switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned o n before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks V d volts in the off-state and conducts I o amps in the on-state. v = v DS CE i D i C v GS v BE V DS,on V CE,on I o I ; r < 1 o (1 - r )I o 22-7. BV DSS 1.3x10 17 N drift = 750 volts ; N drift = 1.7x10 14 cm -3 W drift (10 -5 )(750) = 75 microns ;

W d,body = protrusion of drain depletion layer into body region W drift N drift N body = (75)(1.7x10 14 ) 5x10 16 0.3 microns Even though body-source junction is shorted, there is a depletion layer associat ed with it which is contained entirely on the body side of the junction. This must be inclu ded in the estimate of the required length of the body region. W s,body 2e f c qN a,body ; f c = kT q ln N a N d n i 2 ; v = v DS CE i D i C v GS v BE V DS,on V CE,on

I o I ; r < 1 o (1 - r )I o 22-7. BV DSS 1.3x10 17 N drift = 750 volts ; N drift = 1.7x10 14 cm -3 W drift (10 -5 )(750) = 75 microns ; W d,body = protrusion of drain depletion layer into body region W drift N drift N body = (75)(1.7x10 14 ) 5x10 16 0.3 microns Even though body-source junction is shorted, there is a depletion layer associat ed with it which is contained entirely on the body side of the junction. This must be inclu ded in the estimate of the required length of the body region. W s,body 2e f c qN a,body ; f c = kT q ln

N a N d n i 2 ; f c = 0.026 ln (10 19 )(5x10 16 ) (10 20 ) = 0.94 W s,body (2)(11.7)(8.9x10 -14 )(0.94) (1.6x10 -19 )(5x10 16 ) 0.16 microns In order to avoid reach-through , W body > W d,body + W s,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = C gd dv GD dt gd dv DS dt ; v DS v GD >> v C

GS BJT will turn on if R body C gd dv DS dt = 0.7 V dv DS dt > 0.7 R body C gd will turn on the BJT. 22-9. V GS,max = (0.67) E BD t ox = (0.67) (5x10 6 ) (5x10 -6 ) = 16.7 volts 22-10. a) i D = m n C ox NW cell (v GS -V GS(th) ) 2L C ox = e t ox = (11.7)(8.9x10 -14 ) 10 -5

= 1.04x10 -7 F/cm 2 N = 2i D L m n C ox W cell (v GS -V GS(th) ) N = (2)(100)(10 -4 ) (1500)(1.04x10 -7 )(2x10 -3 )(15-4) 5,800 cells b) I cell = 100 5800 = 17 milliamps 22-11. V on = 4 volts = I on R on = (10 A) R on ; R on = 0.4 ohms R on = W drift qm n N d A : W drift = 10

-5 BV DSS = (10 -5 )(800) = 80 microns f c = 0.026 ln (10 19 )(5x10 16 ) (10 20 ) = 0.94 W s,body (2)(11.7)(8.9x10 -14 )(0.94) (1.6x10 -19 )(5x10 16 ) 0.16 microns In order to avoid reach-through , W body > W d,body + W s,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = C gd dv GD dt gd dv DS dt ; v DS v GD >> v GS BJT will turn on if R body C

C gd dv DS dt = 0.7 V dv DS dt > 0.7 R body C gd will turn on the BJT. 22-9. V GS,max = (0.67) E BD t ox = (0.67) (5x10 6 ) (5x10 -6 ) = 16.7 volts 22-10. a) i D = m n C ox NW cell (v GS -V GS(th) ) 2L C ox = e t ox = (11.7)(8.9x10 -14 ) 10 -5 = 1.04x10 -7 F/cm

2 N = 2i D L m n C ox W cell (v GS -V GS(th) ) N = (2)(100)(10 -4 ) (1500)(1.04x10 -7 )(2x10 -3 )(15-4) 5,800 cells b) I cell = 100 5800 = 17 milliamps 22-11. V on = 4 volts = I on R on = (10 A) R on ; R on = 0.4 ohms R on = W drift qm n N d A : W drift = 10 -5 BV DSS

= (10 -5 )(800) = 80 microns f c = 0.026 ln (10 19 )(5x10 16 ) (10 20 ) = 0.94 W s,body (2)(11.7)(8.9x10 -14 )(0.94) (1.6x10 -19 )(5x10 16 ) 0.16 microns In order to avoid reach-through , W body > W d,body + W s,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = C gd dv GD dt gd dv DS dt ; v DS v GD >> v GS BJT will turn on if R body C gd C

dv DS dt = 0.7 V dv DS dt > 0.7 R body C gd will turn on the BJT. 22-9. V GS,max = (0.67) E BD t ox = (0.67) (5x10 6 ) (5x10 -6 ) = 16.7 volts 22-10. a) i D = m n C ox NW cell (v GS -V GS(th) ) 2L C ox = e t ox = (11.7)(8.9x10 -14 ) 10 -5 = 1.04x10 -7 F/cm 2 N = 2i

D L m n C ox W cell (v GS -V GS(th) ) N = (2)(100)(10 -4 ) (1500)(1.04x10 -7 )(2x10 -3 )(15-4) 5,800 cells b) I cell = 100 5800 = 17 milliamps 22-11. V on = 4 volts = I on R on = (10 A) R on ; R on = 0.4 ohms R on = W drift qm n N d A : W drift = 10 -5 BV DSS = (10 -5 )(800) = 80 microns

f c = 0.026 ln (10 19 )(5x10 16 ) (10 20 ) = 0.94 W s,body (2)(11.7)(8.9x10 -14 )(0.94) (1.6x10 -19 )(5x10 16 ) 0.16 microns In order to avoid reach-through , W body > W d,body + W s,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = C gd dv GD dt gd dv DS dt ; v DS v GD >> v GS BJT will turn on if R body C gd dv DS dt C

= 0.7 V dv DS dt > 0.7 R body C gd will turn on the BJT. 22-9. V GS,max = (0.67) E BD t ox = (0.67) (5x10 6 ) (5x10 -6 ) = 16.7 volts 22-10. a) i D = m n C ox NW cell (v GS -V GS(th) ) 2L C ox = e t ox = (11.7)(8.9x10 -14 ) 10 -5 = 1.04x10 -7 F/cm 2 N = 2i D L m

n C ox W cell (v GS -V GS(th) ) N = (2)(100)(10 -4 ) (1500)(1.04x10 -7 )(2x10 -3 )(15-4) 5,800 cells b) I cell = 100 5800 = 17 milliamps 22-11. V on = 4 volts = I on R on = (10 A) R on ; R on = 0.4 ohms R on = W drift qm n N d A : W drift = 10 -5 BV DSS = (10 -5 )(800) = 80 microns f c = 0.026 ln

(10 19 )(5x10 16 ) (10 20 ) = 0.94 W s,body (2)(11.7)(8.9x10 -14 )(0.94) (1.6x10 -19 )(5x10 16 ) 0.16 microns In order to avoid reach-through , W body > W d,body + W s,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = C gd dv GD dt gd dv DS dt ; v DS v GD >> v GS BJT will turn on if R body C gd dv DS dt = 0.7 V dv C

DS dt > 0.7 R body C gd will turn on the BJT. 22-9. V GS,max = (0.67) E BD t ox = (0.67) (5x10 6 ) (5x10 -6 ) = 16.7 volts 22-10. a) i D = m n C ox NW cell (v GS -V GS(th) ) 2L C ox = e t ox = (11.7)(8.9x10 -14 ) 10 -5 = 1.04x10 -7 F/cm 2 N = 2i D L m n C ox

W cell (v GS -V GS(th) ) N = (2)(100)(10 -4 ) (1500)(1.04x10 -7 )(2x10 -3 )(15-4) 5,800 cells b) I cell = 100 5800 = 17 milliamps 22-11. V on = 4 volts = I on R on = (10 A) R on ; R on = 0.4 ohms R on = W drift qm n N d A : W drift = 10 -5 BV DSS = (10 -5 )(800) = 80 microns N d = 1.3x10 17 BV

DSS = 1.3x10 17 800 1.6x10 14 cm -3 A = 8x10 -3 (1.6x10 -19 )(1500)(1.6x10 14 )(0.4) 0.5 cm 2 10A 0.5cm 2 = 20 A cm 2 BV DSS = 150 V Check for excessive power dissipation. P allowed = T j,max -T a R q ,j-a = 150-50 1 = 100 watts ; P dissipated = [E on + E sw ] f s E on f s = I o 2 r DS(on) 2 = (100) 2 (0.01) 2 = 50 watts E sw = V d I o 2 [t ri + t fi + t rv

+t fv ] = (100)(100) 2 [(2)(5x10 -8 ) + (2)(2x10 -7 )] E sw = 2.5x10 -3 joules ; E sw f s = (2.5x10 -3 )(3x10 4 ) = 75 watts P dissipated = 50 + 75 = 125 watts > P allowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14. Gate current which charges/discharges C gs and C gd during turn-on and turn-off is approximately constant during the four time intervals. However during the curren t rise and fall times the voltages V GS and V GD change by only a few tens of volts. However during the voltage rise and fall times V GD changes by approximately V d which is much larger than a few tens of volts. Thus we have: N d = 1.3x10 17 BV DSS = 1.3x10 17 800 1.6x10 14

cm -3 A = 8x10 -3 (1.6x10 -19 )(1500)(1.6x10 14 )(0.4) 0.5 cm 2 10A 0.5cm 2 = 20 A cm 2 BV DSS = 150 V

Check for excessive power dissipation. P allowed = T j,max -T a R q ,j-a = 150-50 1 = 100 watts ; P dissipated = [E on + E sw ] f s E on f s = I o 2 r DS(on) 2 = (100) 2 (0.01) 2 = 50 watts E sw = V d I o 2 [t ri + t fi + t rv +t fv ] = (100)(100) 2 [(2)(5x10 -8

) + (2)(2x10 -7 )] E sw = 2.5x10 -3 joules ; E sw f s = (2.5x10 -3 )(3x10 4 ) = 75 watts P dissipated = 50 + 75 = 125 watts > P allowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14. Gate current which charges/discharges C gs and C gd during turn-on and turn-off is approximately constant during the four time intervals. However during the curren t rise and fall times the voltages V GS and V GD change by only a few tens of volts. However during the voltage rise and fall times V GD changes by approximately V d which is much larger than a few tens of volts. Thus we have: N d = 1.3x10 17 BV DSS = 1.3x10 17 800 1.6x10 14 cm -3 A = 8x10 -3 (1.6x10 -19

)(1500)(1.6x10 14 )(0.4) 0.5 cm 2 10A 0.5cm 2 = 20 A cm 2 BV DSS = 150 V Check for excessive power dissipation. P allowed = T j,max -T

a R q ,j-a = 150-50 1 = 100 watts ; P dissipated = [E on + E sw ] f s E on f s = I o 2 r DS(on) 2 = (100) 2 (0.01) 2 = 50 watts E sw = V d I o 2 [t ri + t fi + t rv +t fv ] = (100)(100) 2 [(2)(5x10 -8 ) + (2)(2x10 -7 )] E sw = 2.5x10 -3

joules ; E sw f s = (2.5x10 -3 )(3x10 4 ) = 75 watts P dissipated = 50 + 75 = 125 watts > P allowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14. Gate current which charges/discharges C gs and C gd during turn-on and turn-off is approximately constant during the four time intervals. However during the curren t rise and fall times the voltages V GS and V GD change by only a few tens of volts. However during the voltage rise and fall times V GD changes by approximately V d which is much larger than a few tens of volts. Thus we have: N d = 1.3x10 17 BV DSS = 1.3x10 17 800 1.6x10 14 cm -3 A = 8x10 -3 (1.6x10 -19 )(1500)(1.6x10 14 )(0.4) 0.5 cm 2 10A 0.5cm

2 = 20 A cm 2 BV DSS = 150 V Check for excessive power dissipation. P allowed = T j,max -T a R q ,j-a = 150-50 1 = 100 watts ; P

dissipated = [E on + E sw ] f s E on f s = I o 2 r DS(on) 2 = (100) 2 (0.01) 2 = 50 watts E sw = V d I o 2 [t ri + t fi + t rv +t fv ] = (100)(100) 2 [(2)(5x10 -8 ) + (2)(2x10 -7 )] E sw = 2.5x10 -3 joules ; E sw f s = (2.5x10 -3 )(3x10

4 ) = 75 watts P dissipated = 50 + 75 = 125 watts > P allowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14. Gate current which charges/discharges C gs and C gd during turn-on and turn-off is approximately constant during the four time intervals. However during the curren t rise and fall times the voltages V GS and V GD change by only a few tens of volts. However during the voltage rise and fall times V GD changes by approximately V d which is much larger than a few tens of volts. Thus we have: Current rise/fall times proportional to [C gs + C gd ] V GG I G Voltage rise/fall times proportional to C gd V d I G C gs roughly the same size as C gd and V d >> V GG Hence voltage switching times much greater than current switching times. Chapter 23 Problem Solutions 23-1. v s (t) = 2 V s sin( t) ; i L

(t) = v s (t) R L ; = 1 2

I BJT (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. Chapter 25 Problem Solutions 25-1. R on (MOS) A proprotional to 1 m majority ; m n = 3 m p ; Hence R on (p-channel) A = 3 R on (n-channel) A R on (IGBT) A proportional to 1

d n(m n +m p ) ; d n = excess carrier density d n = d p so p-channel IGBTs have the same R on as n-channel IGBTs 25-2. Turn-off waveforms of short versus long lifetime IGBTs i D long lifetime short lifetime I (long) BJT I (short) BJT t 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 I BJT (long) > I BJT (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. 25-3. 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 DS1 V > V DS1 DS2 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 inc reases as v DS

increases. This is base width modulation and it results in a lower output resis tance r o (steeper slope in the active region of the i D -v DS characteristics). 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 Depletion layer Drain of IGBT N + V DS Effective base width independent of V DS Depletion encroaches into the N layer but the advance is halted at moderate v DS values by the N + buffer layer. The PNP base width becomes constant and so the effective resistance r o remains large. 25-4. One dimensional model of n-channel IGBT P P + N N + Source Drain 10 17 10 14 10 19 10

19 25 m m N + 10 19 Reverse blocking junction is the P + - N + junction because of body-source short. BV RB 1.3x10 17 10 19 < 1 volt. No reverse blocking capability. Forward breakdown - limited by P - N junction. 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 Depletion layer Drain of IGBT N + V DS Effective base width independent of V DS Depletion encroaches into the N layer but the advance is halted at moderate v DS values by the N + buffer layer. The PNP base width becomes constant and so the effective resistance r o remains large. 25-4. One dimensional model of n-channel IGBT P P +

N N + Source Drain 10 17 10 14 10 19 10 19 25 m m N + 10 19 Reverse blocking junction is the P + - N + junction because of body-source short. BV RB 1.3x10 17 10 19 < 1 volt. No reverse blocking capability. Forward breakdown - limited by P - N junction. BV FB = 1.3x10 17 10 14 1300 volts ; But W depl (1300 V) = (10 -5 )(1300) = 130 microns 130 microns > 25 micron drift region length. Hence forward blocking limited by punch-through. BV FB = (2x10 5 )(2.5x10 -3 ) (1.6x10 -19 )(10 14

)(2.5x10 -3 ) 2 (2)(11.7)(8.9x10 -14 ) = 453 volts 25-5. IGBT current - I on,IGBT ; V on (IGBT) = V j + I on,IGBT R on,IGBT Assume V j 0.8 V ; Exact value not critical to an approximate estimate of I on,IGBT . I on,IGBT 3-0.8 R on,IGBT ; R on,IGBT W d q(m n +m p )n b A W d = (10 -5 )(750) = 75 m m ; R on,IGBT = 7.5x10 -3 (1.6x10 -19 )(900)(10 16 )(2) = 2.6x10 -3 W

I on,IGBT 2.2 2.6x10 -3 850 amps MOSFET current - I on,MOS ; V on (MOS) = I on,MOS R on,MOS I on,MOS = V on (MOS) R on,MOS ; R on,MOS = W d qm n N d A ; W d = 75 m m N d = 1.3x10 17 750 = 1.7x10 14 cm -3 R on,MOS = 7.5x10 -3 (1.6x10 -19 )(1.5x10 3 )(1.7x10 14 )(2) = 0.09 ohms I

on,MOS = 3 0.09 = 33 amps 25-6. V on (PT) = V j,PT + I on,PT R on,PT = V j,NPT + I on,NPT R on,NPT I on,PT I on,NPT R on,NPT R on,PT since V j,NPT V j,PT BV FB = 1.3x10 17 10 14 1300 volts ; But W depl (1300 V) = (10 -5 )(1300) = 130 microns 130 microns > 25 micron drift region length. Hence forward blocking limited by punch-through. BV FB = (2x10 5 )(2.5x10 -3 ) (1.6x10 -19 )(10 14 )(2.5x10

-3 ) 2 (2)(11.7)(8.9x10 -14 ) = 453 volts 25-5. IGBT current - I on,IGBT ; V on (IGBT) = V j + I on,IGBT R on,IGBT Assume V j 0.8 V ; Exact value not critical to an approximate estimate of I on,IGBT . I on,IGBT 3-0.8 R on,IGBT ; R on,IGBT W d q(m n +m p )n b A W d = (10 -5 )(750) = 75 m m ; R on,IGBT = 7.5x10 -3 (1.6x10 -19 )(900)(10 16 )(2) = 2.6x10 -3 W I

on,IGBT 2.2 2.6x10 -3 850 amps MOSFET current - I on,MOS ; V on (MOS) = I on,MOS R on,MOS I on,MOS = V on (MOS) R on,MOS ; R on,MOS = W d qm n N d A ; W d = 75 m m N d = 1.3x10 17 750 = 1.7x10 14 cm -3 R on,MOS = 7.5x10 -3 (1.6x10 -19 )(1.5x10 3 )(1.7x10 14 )(2) = 0.09 ohms I on,MOS

= 3 0.09 = 33 amps 25-6. V on (PT) = V j,PT + I on,PT R on,PT = V j,NPT + I on,NPT R on,NPT I on,PT I on,NPT R on,NPT R on,PT since V j,NPT V j,PT BV FB = 1.3x10 17 10 14 1300 volts ; But W depl (1300 V) = (10 -5 )(1300) = 130 microns 130 microns > 25 micron drift region length. Hence forward blocking limited by punch-through. BV FB = (2x10 5 )(2.5x10 -3 ) (1.6x10 -19 )(10 14 )(2.5x10 -3

) 2 (2)(11.7)(8.9x10 -14 ) = 453 volts 25-5. IGBT current - I on,IGBT ; V on (IGBT) = V j + I on,IGBT R on,IGBT Assume V j 0.8 V ; Exact value not critical to an approximate estimate of I on,IGBT . I on,IGBT 3-0.8 R on,IGBT ; R on,IGBT W d q(m n +m p )n b A W d = (10 -5 )(750) = 75 m m ; R on,IGBT = 7.5x10 -3 (1.6x10 -19 )(900)(10 16 )(2) = 2.6x10 -3 W I on,IGBT

2.2 2.6x10 -3 850 amps MOSFET current - I on,MOS ; V on (MOS) = I on,MOS R on,MOS I on,MOS = V on (MOS) R on,MOS ; R on,MOS = W d qm n N d A ; W d = 75 m m N d = 1.3x10 17 750 = 1.7x10 14 cm -3 R on,MOS = 7.5x10 -3 (1.6x10 -19 )(1.5x10 3 )(1.7x10 14 )(2) = 0.09 ohms I on,MOS =

3 0.09 = 33 amps 25-6. V on (PT) = V j,PT + I on,PT R on,PT = V j,NPT + I on,NPT R on,NPT I on,PT I on,NPT R on,NPT R on,PT since V j,NPT V j,PT

R on,NPT R on,PT = W d,NPT W d,PT 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Hence I on,PT I on,NPT 2 25-7. C v d T = d Q ; d Q = P V d t ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = d t. P = I

ov 2 R on ; I ov = VC v d T d tR on ; V A W drift R on = W drift q(m n +m p )n b A ; ov = q(m n +m p )n b A 2 C v d T d t

I

I ov = (1.6x10 -19 )(900)(10 16 )(0.5) 2 (1.75)(100) (10 -5 ) 2.5x10 3 amps Estimate is overly optimistic because it ignores any other ohmic losses in the d

evice such as channel resistance or resistance of the heavily doped source and drain diffus ions. 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 C gd and C gs 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 con ductivity modulation of the drift region to significantly reduce the specific on-state res istance. 25-9. Check dv DS dt at turn-off ; dv DS dt = V d t rv ; t rv < 0.75 microseconds dv DS dt > V d t off = 700 3x10 -7 = 2330 V/m s > 800 V/m s limit. Device is overstressed by an overly large dv DS dt .

R on,NPT R on,PT =

W d,NPT W d,PT 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Hence I on,PT I on,NPT 2 25-7. C v d T = d Q ; d Q = P V d t ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = d t. P = I ov 2 R on ; I ov = VC v d T d tR on ; V A W drift R on = W drift q(m n +m p )n b A ; ov = q(m n +m p )n b A 2 C v

I

d T d t I ov = (1.6x10 -19 )(900)(10 16 )(0.5) 2 (1.75)(100) (10 -5 ) 2.5x10 3 amps Estimate is overly optimistic because it ignores any other ohmic losses in the d evice such as channel resistance or resistance of the heavily doped source and drain diffus ions. 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 C gd and C gs 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 con ductivity modulation of the drift region to significantly reduce the specific on-state res istance. 25-9. Check dv DS dt at turn-off ; dv DS dt = V d t rv ; t rv < 0.75 microseconds dv DS dt > V d

t off = 700 3x10 -7 = 2330 V/m s > 800 V/m s limit. Device is overstressed by an overly large dv DS dt .

R on,NPT R on,PT = W d,NPT W d,PT 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Hence I on,PT I on,NPT 2 25-7. C v d T = d Q ; d Q = P V d t ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = d t. P = I ov 2 R on ; I ov = VC v d T d tR on ; V A W drift R on = W drift

q(m n +m p )n b A ; ov = q(m n +m p )n b A 2 C v d T d t I ov = (1.6x10 -19 )(900)(10 16 )(0.5) 2 (1.75)(100) (10 -5 ) 2.5x10 3 amps Estimate is overly optimistic because it ignores any other ohmic losses in the d evice such as channel resistance or resistance of the heavily doped source and drain diffus ions. 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 C gd and C gs 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 con ductivity modulation of the drift region to significantly reduce the specific on-state res istance. 25-9. Check dv DS I

dt at turn-off ; dv DS dt = V d t rv ; t rv < 0.75 microseconds dv DS dt > V d t off = 700 3x10 -7 = 2330 V/m s > 800 V/m s limit. Device is overstressed by an overly large dv DS dt .

R on,NPT R on,PT = W d,NPT W d,PT 2 assuming doping level in PT drift region is much less than the doping level in the NPT drift region. Hence I on,PT I on,NPT 2 25-7. C v d T = d Q ; d Q = P V d t ; P = power dissipated in IGBT during overcurrent transient. V = volume in IGBT where power is dissipated.Duration of transient = d t. P = I

ov 2 R on ; I ov = VC v d T d tR on ; V A W drift R on = W drift q(m n +m p )n b A ; ov = q(m n +m p )n b A 2 C v d T d t

I

I ov = (1.6x10 -19 )(900)(10 16 )(0.5) 2 (1.75)(100) (10 -5 ) 2.5x10 3 amps Estimate is overly optimistic because it ignores any other ohmic losses in the d

evice such as channel resistance or resistance of the heavily doped source and drain diffus ions. 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 C gd and C gs 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 con ductivity modulation of the drift region to significantly reduce the specific on-state res istance. 25-9. Check dv DS dt at turn-off ; dv DS dt = V d t rv ; t rv < 0.75 microseconds dv DS dt > V d t off = 700 3x10 -7 = 2330 V/m s > 800 V/m s limit. Device is overstressed by an overly large dv DS dt . Check switching losses P sw = V d I o t ri

+t rv +t fi +t fv 2 f s P sw = (700)(100) (3x10 -7 +7.5x10 -7 ) 2 (2.5x10 4 ) = 875 W Allowable power loss = T j,max -T a R q ja = 150-25 0.5 = 250 watts Switching losses exceed allowable losses. Module is overstressed by too much power dissipation. Chapter 26 Problem Solutions 26-1. Equivalent circuit for JFET in active region. + C GD C GS r o m v GS v GS v DS + + Equivelent circuit for JFET Linearized I-V charac teristics in the blocking state. C

GD C GS v GS v DS + + 0 -V GS2 GS1 -V i D V DS1 V DS2 0 26-2. Drive circuit configuration + V DD R L 2 R 1 R V drive MOSFET off V DS = V KG = - V GK = V DD R 2 R 1 +R 2 Negative enough to insure that the FCT is off. MOSFET on V DS = V

KG = - V GK = 0 and FCT is on. Chapter 26 Problem Solutions 26-1. Equivalent circuit for JFET in active region. + C GD C GS r o m v GS v GS v DS + + Equivelent circuit for JFET teristics in the blocking state. C GD C GS v GS v DS + + 0 -V GS2 GS1 -V i D V DS1 V DS2 0 26-2. Drive circuit configuration + V DD R L

Linearized I-V charac

2 R 1 R V drive MOSFET off V DS = V KG = - V GK = V DD R 2 R 1 +R 2 Negative enough to insure that the FCT is off. MOSFET on V DS = V KG = - V GK = 0 and FCT is on. MOSFET characteirstics: - High current sinking capability - Low R on ; Low BV Dss 26-3. V drift = W d 2 (m n +m p )t ; BV BD = E BD W d 2 ; V

drift,GaAs = V drift,Si W d 2 (Si) (m n +m p )| Si t Si = W d 2 (GaAs) (m n +m p )| GaAs t GaAs t GaA t Si = (m n +m p )| Si (m n +m p )| GaAs

E BD (Si) E BD (GaAs)2 (m n + m

p )| Si = 2000 cm 2 /V-sec ; (m n + m p )| GaAs = 9000 cm 2 /V-sec E BD (Si) = 300 kV/cm ; BD (GaAs) = 400 kV/cm t GaA t Si = 2 9

E

3 42 = 0.125 ; GaAs has the shorter lifetime. 26-4. E BD = 10 7 V/cm ; BV BD = E BD t ox t ox = 10 3 10 7 = 10 -4 cm = 1 micron 26-5. I A,max = (10 5 )(1.5x10 -2 ) = 1500 amperes

26-6. P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in t he P-MCT and p-channel OFF-FET in the N-MCT. Hence I A,max = (3)(10 5 )(1.5x10 -2 ) = 4500 amperes 26-7. Assume an n-type drift region since m n > m p . MOSFET characteirstics: - High current sinking capability - Low R on ; Low BV Dss 26-3. V drift = W d 2 (m n +m p )t ; BV BD = E BD W d 2 ; V drift,GaAs = V drift,Si W d 2 (Si) (m n +m p )| Si t Si =

W d 2 (GaAs) (m n +m p )| GaAs t GaAs t GaA t Si = (m n +m p )| Si (m n +m p )| GaAs

E BD (Si) E BD (GaAs)2 (m n + m p )| Si = 2000 cm 2 /V-sec ; (m n + m p )| GaAs = 9000 cm 2 /V-sec E BD (Si) = 300 kV/cm

;

E

BD (GaAs) = 400 kV/cm t GaA t Si = 2 9

3 42 = 0.125 ; GaAs has the shorter lifetime. 26-4. E BD = 10 7 V/cm ; BV BD = E BD t ox t ox = 10 3 10 7 = 10 -4 cm = 1 micron 26-5. I A,max = (10 5 )(1.5x10 -2 ) = 1500 amperes 26-6. P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in t he P-MCT and p-channel OFF-FET in the N-MCT. Hence I A,max = (3)(10 5 )(1.5x10 -2 ) = 4500 amperes 26-7. Assume an n-type drift region since m n > m p .

MOSFET characteirstics: - High current sinking capability - Low R on ; Low BV Dss 26-3. V drift = W d 2 (m n +m p )t ; BV BD = E BD W d 2 ; V drift,GaAs = V drift,Si W d 2 (Si) (m n +m p )| Si t Si = W d 2 (GaAs) (m n +m p )| GaAs t GaAs t GaA t Si

= (m n +m p )| Si (m n +m p )| GaAs

E BD (Si) E BD (GaAs)2 (m n + m p )| Si = 2000 cm 2 /V-sec ; (m n + m p )| GaAs = 9000 cm 2 /V-sec E BD (Si) = 300 kV/cm ; BD (GaAs) = 400 kV/cm t GaA t Si = 2 9

E

3 42 = 0.125 ; GaAs has the shorter lifetime. 26-4. E

BD = 10 7 V/cm ; BV BD = E BD t ox t ox = 10 3 10 7 = 10 -4 cm = 1 micron 26-5. I A,max = (10 5 )(1.5x10 -2 ) = 1500 amperes 26-6. P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in t he P-MCT and p-channel OFF-FET in the N-MCT. Hence I A,max = (3)(10 5 )(1.5x10 -2 ) = 4500 amperes 26-7. Assume an n-type drift region since m n > m p . MOSFET characteirstics: - High current sinking capability - Low R on ; Low BV Dss 26-3. V drift = W d 2 (m n +m p )t

; BV BD = E BD W d 2 ; V drift,GaAs = V drift,Si W d 2 (Si) (m n +m p )| Si t Si = W d 2 (GaAs) (m n +m p )| GaAs t GaAs t GaA t Si = (m n +m p )| Si (m n +m p )| GaAs

E BD

(Si) E BD (GaAs) 2 (m n + m p )| Si = 2000 cm 2 /V-sec ; (m n + m p )| GaAs = 9000 cm 2 /V-sec E BD (Si) = 300 kV/cm ; BD (GaAs) = 400 kV/cm t GaA t Si = 2 9

E

3 42 = 0.125 ; GaAs has the shorter lifetime. 26-4. E BD = 10 7 V/cm ; BV BD = E BD t ox t ox = 10 3 10 7 = 10

-4 cm = 1 micron 26-5. I A,max = (10 5 )(1.5x10 -2 ) = 1500 amperes 26-6. P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in t he P-MCT and p-channel OFF-FET in the N-MCT. Hence I A,max = (3)(10 5 )(1.5x10 -2 ) = 4500 amperes 26-7. Assume an n-type drift region since m n > m p . MOSFET characteirstics: - High current sinking capability - Low R on ; Low BV Dss 26-3. V drift = W d 2 (m n +m p )t ; BV BD = E BD W d 2 ; V drift,GaAs = V drift,Si W d 2 (Si)

(m n +m p )| Si t Si = W d 2 (GaAs) (m n +m p )| GaAs t GaAs t GaA t Si = (m n +m p )| Si (m n +m p )| GaAs

E BD (Si) E BD (GaAs)2 (m n + m p )| Si = 2000 cm 2 /V-sec ; (m n + m

p )| GaAs = 9000 cm 2 /V-sec E BD (Si) = 300 kV/cm ; BD (GaAs) = 400 kV/cm t GaA t Si = 2 9

E

3 42 = 0.125 ; GaAs has the shorter lifetime. 26-4. E BD = 10 7 V/cm ; BV BD = E BD t ox t ox = 10 3 10 7 = 10 -4 cm = 1 micron 26-5. I A,max = (10 5 )(1.5x10 -2 ) = 1500 amperes 26-6. P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in t he P-MCT and p-channel OFF-FET in the N-MCT. Hence I A,max = (3)(10

5 )(1.5x10 -2 ) = 4500 amperes 26-7. Assume an n-type drift region since m n > m p . MOSFET characteirstics: - High current sinking capability - Low R on ; Low BV Dss 26-3. V drift = W d 2 (m n +m p )t ; BV BD = E BD W d 2 ; V drift,GaAs = V drift,Si W d 2 (Si) (m n +m p )| Si t Si = W d 2 (GaAs) (m n +m p

)| GaAs t GaAs t GaA t Si = (m n +m p )| Si (m n +m p )| GaAs

E BD (Si) E BD (GaAs)2 (m n + m p )| Si = 2000 cm 2 /V-sec ; (m n + m p )| GaAs = 9000 cm 2 /V-sec E BD (Si) = 300 kV/cm ; BD (GaAs) = 400 kV/cm t GaA t Si = 2

E

9

3 42 = 0.125 ; GaAs has the shorter lifetime. 26-4. E BD = 10 7 V/cm ; BV BD = E BD t ox t ox = 10 3 10 7 = 10 -4 cm = 1 micron 26-5. I A,max = (10 5 )(1.5x10 -2 ) = 1500 amperes 26-6. P-MCT fabricated in silicon can turn-off three times more current than an identical N-MCT due to the difference between the mobilities in the n-channel OFF-FET in t he P-MCT and p-channel OFF-FET in the N-MCT. Hence I A,max = (3)(10 5 )(1.5x10 -2 ) = 4500 amperes 26-7. Assume an n-type drift region since m n > m p . R drift = W d qm n N

d A ; drift,sp = R drift A = W d qm n N d R

Using Eq. (20-1) N d = e E BD 2 2qBV BD ; Using Eq. (20-3) d = 2BV BD E BD

W

Substituting into the expression for R drift,sp yields R drift,sp = 2BV BD E BD 1 qe 2qBV BD e E BD 2 = 4(BV BD ) 2 e m n (E BD ) 3

26-8. Silicon : R drift,sp = (4)(500) 2 (11.7)(1500)(8.9x10 -14 )(3x10 5 ) 3 = 0.024 ohms-cm 2 GaAs: R drift,sp = (4)(500) 2 (12.8)(8500)(8.9x10 -14 )(4x10 5 ) 3 = 0.0016 ohms-cm 2 6H-SiC: R drift,sp = (4)(500) 2 (10)(600)(8.9x10 -14 )(2x10 6 ) 3 = 2.3x10 -4 ohms-cm 2 Diamond: R drift,sp = (4)(500) 2 (5.5)(2200)(8.9x10 -14 )(10 7 ) 3 = 9.3x10 -7 ohms-cm 2 26-9. Diamond is the most suitable material for high temperature operation. It h as the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier dens

ity at any given temperature. This statement presumes that the phase change listed for diam ond in the table of material properties exceeds the sublimation temperature of SiC (180 0 C). 26-10. Eq. (20-1): N d = e E BD 2 2qBV BD For GaAs: N d = (12.8)(8.9x10 -14 )(4x10 5 ) 2 (2)(1.6x10 -19 )(BV BD ) = 5.7x10 17 BV BD For 6H-SiC: N d = (10)(8.9x10 -14 )(2x10 6 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.1x10 19 BV BD R drift = W d qm

n N d A ; drift,sp = R drift A = W d qm n N d R


W

Substituting into the expression for R drift,sp yields R drift,sp = 2BV BD E BD 1 qe 2qBV BD e E BD 2 = 4(BV BD ) 2 e m n (E BD

) 3 26-8. Silicon : R drift,sp = (4)(500) 2 (11.7)(1500)(8.9x10 -14 )(3x10 5 ) 3 = 0.024 ohms-cm 2 GaAs: R drift,sp = (4)(500) 2 (12.8)(8500)(8.9x10 -14 )(4x10 5 ) 3 = 0.0016 ohms-cm 2 6H-SiC: R drift,sp = (4)(500) 2 (10)(600)(8.9x10 -14 )(2x10 6 ) 3 = 2.3x10 -4 ohms-cm 2 Diamond: R drift,sp = (4)(500) 2 (5.5)(2200)(8.9x10 -14 )(10 7 ) 3 = 9.3x10 -7 ohms-cm 2 26-9. Diamond is the most suitable material for high temperature operation. It h

as the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier dens ity at any given temperature. This statement presumes that the phase change listed for diam ond in the table of material properties exceeds the sublimation temperature of SiC (180 0 C). 26-10. Eq. (20-1): N d = e E BD 2 2qBV BD For GaAs: N d = (12.8)(8.9x10 -14 )(4x10 5 ) 2 (2)(1.6x10 -19 )(BV BD ) = 5.7x10 17 BV BD For 6H-SiC: N d = (10)(8.9x10 -14 )(2x10 6 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.1x10 19 BV BD R drift = W

d qm n N d A ; drift,sp = R drift A = W d qm n N d R


W

Substituting into the expression for R drift,sp yields R drift,sp = 2BV BD E BD 1 qe 2qBV BD e E BD 2 = 4(BV BD ) 2 e m n

(E BD ) 3 26-8. Silicon : R drift,sp = (4)(500) 2 (11.7)(1500)(8.9x10 -14 )(3x10 5 ) 3 = 0.024 ohms-cm 2 GaAs: R drift,sp = (4)(500) 2 (12.8)(8500)(8.9x10 -14 )(4x10 5 ) 3 = 0.0016 ohms-cm 2 6H-SiC: R drift,sp = (4)(500) 2 (10)(600)(8.9x10 -14 )(2x10 6 ) 3 = 2.3x10 -4 ohms-cm 2 Diamond: R drift,sp = (4)(500) 2 (5.5)(2200)(8.9x10 -14 )(10 7 ) 3 = 9.3x10 -7 ohms-cm

2 26-9. Diamond is the most suitable material for high temperature operation. as the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier ity at any given temperature. This statement presumes that the phase change listed for ond in the table of material properties exceeds the sublimation temperature of SiC 0 C). 26-10. Eq. (20-1): N d = e E BD 2 2qBV BD For GaAs: N d = (12.8)(8.9x10 -14 )(4x10 5 ) 2 (2)(1.6x10 -19 )(BV BD ) = 5.7x10 17 BV BD For 6H-SiC: N d = (10)(8.9x10 -14 )(2x10 6 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.1x10 19 BV BD R drift

It h dens diam (180

= W d qm n N d A ; drift,sp = R drift A = W d qm n N d R


W

Substituting into the expression for R drift,sp yields R drift,sp = 2BV BD E BD 1 qe 2qBV BD e E BD 2 = 4(BV BD ) 2

e m n (E BD ) 3 26-8. Silicon : R drift,sp = (4)(500) 2 (11.7)(1500)(8.9x10 -14 )(3x10 5 ) 3 = 0.024 ohms-cm 2 GaAs: R drift,sp = (4)(500) 2 (12.8)(8500)(8.9x10 -14 )(4x10 5 ) 3 = 0.0016 ohms-cm 2 6H-SiC: R drift,sp = (4)(500) 2 (10)(600)(8.9x10 -14 )(2x10 6 ) 3 = 2.3x10 -4 ohms-cm 2 Diamond: R drift,sp = (4)(500) 2 (5.5)(2200)(8.9x10 -14 )(10 7 ) 3 = 9.3x10

-7 ohms-cm 2 26-9. Diamond is the most suitable material for high temperature operation. as the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier ity at any given temperature. This statement presumes that the phase change listed for ond in the table of material properties exceeds the sublimation temperature of SiC 0 C). 26-10. Eq. (20-1): N d = e E BD 2 2qBV BD For GaAs: N d = (12.8)(8.9x10 -14 )(4x10 5 ) 2 (2)(1.6x10 -19 )(BV BD ) = 5.7x10 17 BV BD For 6H-SiC: N d = (10)(8.9x10 -14 )(2x10 6 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.1x10 19 BV BD

It h dens diam (180

For diamond: N d = (5.5)(8.9x10 -14 )(10 7 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.5x10 20 BV BD Eq. (20-3): W d = 2BV BD E BD For GaAs: W d = 2BV BD 4x10 5 = 5x10 -6 BV BD [cm] For 6H-SiC: W d = 2BV BD 2x10 6 = 10 -6 BV BD [cm] For diamond: W d = 2BV BD 10 7 = 2x10

-7 BV BD [cm] 26-11. Use equations from problem 26-10. Material N d W d GaAs 2.9x10 15 cm -3 10 -2 cm 6H-SiC 5.5x10 16 cm -3 2x10 -3 cm Diamond 7.5x10 17 cm -3 4x10 -5 cm 26-12. T j = R q jc P diode + T case : R q jc = C. (k ) -1 k = thermal conductivity and C = constant Using silicon diode data: C = (T j -T case )k P diode = (150-50)(1.5) 200 = 0.75 cm -1 R

q jc (GaAs) = 0.75 0.5 = 1.5 C/W : R q jc (SiC) = 0.75 5 = 0.15C/W R q jc (diamond) = 0.75 20 = 0.038C/W T j (GaAs) = (1.5)(200) + 50 = 350 C : T j (SiC) = (0.15)(200) + 50 = 80 C For diamond: N d = (5.5)(8.9x10 -14 )(10 7 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.5x10 20 BV BD Eq. (20-3): W d = 2BV BD E BD For GaAs: W d = 2BV BD 4x10 5 = 5x10 -6 BV BD

[cm] For 6H-SiC: W d = 2BV BD 2x10 6 = 10 -6 BV BD [cm] For diamond: W d = 2BV BD 10 7 = 2x10 -7 BV BD [cm] 26-11. Use equations from problem 26-10. Material N d W d GaAs 2.9x10 15 cm -3 10 -2 cm 6H-SiC 5.5x10 16 cm -3 2x10 -3 cm Diamond 7.5x10 17 cm -3 4x10 -5 cm 26-12. T j = R q jc P diode

+ T case : R q jc = C. (k ) -1 k = thermal conductivity and C = constant Using silicon diode data: C = (T j -T case )k P diode = (150-50)(1.5) 200 = 0.75 cm -1 R q jc (GaAs) = 0.75 0.5 = 1.5 C/W : R q jc (SiC) = 0.75 5 = 0.15C/W R q jc (diamond) = 0.75 20 = 0.038C/W T j (GaAs) = (1.5)(200) + 50 = 350 C : T j (SiC) = (0.15)(200) + 50 = 80 C For diamond: N d = (5.5)(8.9x10 -14 )(10 7 ) 2 (2)(1.6x10 -19 )(BV BD ) = 1.5x10 20 BV

BD Eq. (20-3): W d = 2BV BD E BD For GaAs: W d = 2BV BD 4x10 5 = 5x10 -6 BV BD [cm] For 6H-SiC: W d = 2BV BD 2x10 6 = 10 -6 BV BD [cm] For diamond: W d = 2BV BD 10 7 = 2x10 -7 BV BD [cm] 26-11. Use equations from problem 26-10. Material N d W d GaAs 2.9x10 15 cm -3 10 -2 cm 6H-SiC

5.5x10 16 cm -3 2x10 -3 cm Diamond 7.5x10 17 cm -3 4x10 -5 cm 26-12. T j = R q jc P diode + T case : R q jc = C. (k ) -1 k = thermal conductivity and C = constant Using silicon diode data: C = (T j -T case )k P diode = (150-50)(1.5) 200 = 0.75 cm -1 R q jc (GaAs) = 0.75 0.5 = 1.5 C/W : R q jc (SiC) = 0.75 5 = 0.15C/W R q jc (diamond) = 0.75 20 = 0.038C/W T j

(GaAs) = (1.5)(200) + 50 = 350 C : T j (SiC) = (0.15)(200) + 50 = 80 C T j (diamond) = (0.038)(200) + 50 = 57.4 C 26-13. Pinch-off of the channel occurs when the depletion region of the gate-cha nnel (P + N ) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-s ource voltage of -V p . 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. W 2 = W o 1+ V p f c ; f c = kT q ln N a N d n i 2 ; W o = 2e f c qN d Solving for V p yields V p = f c

W 2W

o 2 c f c = 0.026 ln (10 19 )(2x10 14 ) 10 10 = 0.8 V W o = (2)(11.7)(8.9x10 -14 )(0.8) (1.6x10 -19 )(2x10 14 ) = 2.3 microns V p = (0.8) 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 fulle r 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 v arious geometrical factors that determine the resistance. Each cell is d centimeters de ep in the direction perpendicular to the plane (page) of the diagram. The gate-source volt age is set at zero. T j (diamond) = (0.038)(200) + 50 = 57.4 C 26-13. Pinch-off of the channel occurs when the depletion region of the gate-cha nnel (P + N ) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-s ource voltage of -V f

p . 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. W 2 = W o 1+ V p f c ; f c = kT q ln N a N d n i 2 ; W o = 2e f c qN d Solving for V p yields V p = f c

W 2W o2 c f c = 0.026 ln (10 19 )(2x10 14 ) 10 f

10 = 0.8 V W o = (2)(11.7)(8.9x10 -14 )(0.8) (1.6x10 -19 )(2x10 14 ) = 2.3 microns V p = (0.8) 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 fulle r 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 v arious geometrical factors that determine the resistance. Each cell is d centimeters de ep in the direction perpendicular to the plane (page) of the diagram. The gate-source volt age is set at zero. T j (diamond) = (0.038)(200) + 50 = 57.4 C 26-13. Pinch-off of the channel occurs when the depletion region of the gate-cha nnel (P + N ) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-s ource voltage of -V p . 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. W 2 = W o 1+ V p f c ; f c

= kT q

N a N d n i 2o

ln

; = 2e f c qN d

W

Solving for V p yields V p = f c

W 2W o2 c f c = 0.026 ln (10 19 )(2x10 14 ) 10 10 = 0.8 V W o = (2)(11.7)(8.9x10 -14 )(0.8) (1.6x10 -19 )(2x10 14 ) = 2.3 microns V f

p = (0.8) 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 fulle r 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 v arious geometrical factors that determine the resistance. Each cell is d centimeters de ep in the direction perpendicular to the plane (page) of the diagram. The gate-source volt age is set at zero. R d R t R c R s P + P + drain source P + P + W - 2W o W W + W g W + W /2 o g l c l gs W g l - W - W /2 gd o g R s = l gs qm n N

d dW = 10 -3 (1.6x10 -19 )(1500)(2x10 14 )(0.07)(10 -3 ) = 298 ohms R c = l c qm n N d d(W-2W o ) = 10 -3 (1.6x10 -19 )(1500)(2x10 14 )(0.07)(10 -3 -4.6x10 -4 ) = 552 ohms R t estimate. Treat the region of thickness W o + W g /2 as though it has an average width given by (W-2W o )+(W+W g ) 2 = W + W g /2 - W o . R t now approximately given by

R t = W o +W g /2 qm n N d d(W+W g /2-W o ) R t = (10 -3 +5x10 -4 ) (1.6x10 -19 )(1500)(2x10 14 )(0.07)(10 -3 +5x10 -4 -2.3x10 -4 ) = 351 ohms R d = (l gd -W o -W g /2) qm n N d d(W+W g ) = R d = (35x10

-4 -2.3x10 -4 -5x10 -4 ) (1.6x10 -19 )(1500)(2x10 14 )(0.07)(10 -3 +10 -3 ) = 412 ohms Total resistance of a single cell is R cell = R s + R c + R t + R d R cell = 298 + 552 + 351 + 412 = 1613 ohms There are N = 28 cells in parallel so the the net on-state resistance is R on = R cell N = 1613 28 = 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 l gd and doping N d 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 = 1.3x10 17 2x10

14 = 650 V ; W d > (10 -5 )(6.5x10 2 ) = 65 microns > 35 microns Hence this is a punch-through structure. Use Eq. (21-21) and E BD = 2x10 5 V/cm BV = (2x10 5 )(3.5x10 -3 ) (1.6x10 -19 )(2x10 14 )(3.5x10 -3 ) 2 (2)(11.7)(8.9x10 -14 ) = 700 - 189 BV = 511 V R cell = 298 + 552 + 351 + 412 = 1613 ohms There are N = 28 cells in parallel so the the net on-state resistance is R on = R cell N = 1613 28 = 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 l gd and doping N d 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 = 1.3x10 17 2x10 14 = 650 V ; W d > (10 -5 )(6.5x10 2 ) = 65 microns > 35 microns Hence this is a punch-through structure. Use Eq. (21-21) and E BD = 2x10 5 V/cm BV = (2x10 5 )(3.5x10 -3 ) (1.6x10 -19 )(2x10 14 )(3.5x10 -3 ) 2 (2)(11.7)(8.9x10 -14 ) = 700 - 189 BV = 511 V Chapter 27 Problem Solutions 27-1. a. During turn-off of the GTO, I o communtates linearly to C s . C s dv C dt = I o t t fi ; dv C dt = dv

AK dt = I o t C s t fi < 7 V/s Maximum dv AK dt occurs at t fi . Solving for C s yields C s > I o 5x10

dv AK dt -1 = 500 5x10 7 = 10 microfarads R s chosen on basis of limiting discharge current from C s to safe level when GTO turns on. I Cs,max = I AM - I o - I rr . Assume i rr = 0.2 I o . Then I Cs,max = 1000 - 500 -100 = 400 A

R s = 500 400 1.3 ohms Snubber recovery time = 2.3 R s C s = (2.3)(10 -5 )(1.3) = 30 microseconds. b. Power dissipated in snubber P Rs f sw C s V d 2 2 P Rs = (0.5)(10 3 )(10 -5 )(500) 2 = 1.25 kW 27-2. L s di A dt max = V d ; L s 500 3x10 8 1.7 microhenries. Voltage across GTO at turn-off = V d + I o R s : Assume I o

R s = 0.2 V d R s = (0.2)(500) (500) = 0.2 ohms. 27-3. v Cs (t) = V d - V d cos(w o t) + V d C base C s sin(w o t) = V d + K sin(w o t - f ) v Cs,max = V d + K ; K sin(w o t - f ) = K sin(w o t) cos(f ) - Kcos(w o t) sin(f ) Chapter 27 Problem Solutions 27-1. a. During turn-off of the GTO, I o communtates linearly to C s . C s dv C dt = I o t t

fi ; dv C dt = dv AK dt = I o t C s t fi < 7 V/s Maximum dv AK dt occurs at t fi . Solving for C s yields C s > I o 5x10

dv AK dt -1 = 500 5x10 7 = 10 microfarads R s chosen on basis of limiting discharge current from C s to safe level when GTO turns on. I Cs,max = I AM - I o - I rr . Assume i

rr = 0.2 I o . Then I Cs,max = 1000 - 500 -100 = 400 A R s = 500 400 1.3 ohms Snubber recovery time = 2.3 R s C s = (2.3)(10 -5 )(1.3) = 30 microseconds. b. Power dissipated in snubber P Rs f sw C s V d 2 2 P Rs = (0.5)(10 3 )(10 -5 )(500) 2 = 1.25 kW 27-2. L s di A dt max = V d ; L s 500 3x10 8 1.7 microhenries. Voltage across GTO at turn-off = V d + I

o R s : Assume I o R s = 0.2 V d R s = (0.2)(500) (500) = 0.2 ohms. 27-3. v Cs (t) = V d - V d cos(w o t) + V d C base C s sin(w o t) = V d + K sin(w o t - f ) v Cs,max = V d + K ; K sin(w o t - f ) = K sin(w o t) cos(f ) - Kcos(w o t) sin(f ) Chapter 27 Problem Solutions 27-1. a. During turn-off of the GTO, I o communtates linearly to C s . C s dv

C dt = I o t t fi ; dv C dt = dv AK dt = I o t C s t fi < 7 V/s Maximum dv AK dt occurs at t fi . Solving for C s yields C s > I o 5x10

dv AK dt -1 = 500 5x10 7 = 10 microfarads R s chosen on basis of limiting discharge current from C s to safe level when GTO turns on. I Cs,max

= I AM - I o - I rr . Assume i rr = 0.2 I o . Then I Cs,max = 1000 - 500 -100 = 400 A R s = 500 400 1.3 ohms Snubber recovery time = 2.3 R s C s = (2.3)(10 -5 )(1.3) = 30 microseconds. b. Power dissipated in snubber P Rs f sw C s V d 2 2 P Rs = (0.5)(10 3 )(10 -5 )(500) 2 = 1.25 kW 27-2. L s di A dt max = V d ; L s

500 3x10 8 1.7 microhenries. Voltage across GTO at turn-off = V d + I o R s : Assume I o R s = 0.2 V d R s = (0.2)(500) (500) = 0.2 ohms. 27-3. v Cs (t) = V d - V d cos(w o t) + V d C base C s sin(w o t) = V d + K sin(w o t - f ) v Cs,max = V d + K ; K sin(w o t - f ) = K sin(w o t) cos(f ) - Kcos(w o t) sin(f ) K sin(w o t) cos(f ) - K cos(w

o t) sin(f ) = V d C base C s sin(w o t) - V d cos(w o t) K cos(f ) = V d C base C s and d ; [ ] Kcos(f ) 2 + [ ] Ksin(f ) 2 = K 2 = V d 2 C base C s + V d 2 v Cs,max = V d + K = V d + V d 1+ C base C s 27-4. a. Equivalent circuit after diode reverse recovery. K sin(f ) = V

L = R s C s 200 + i L i L (0 + ) = rr ; di R dt = di R dt = I rr t rr

10 m H

V

I During reverse recovery L

200 V

= 200 10 -5 = 2x10 7 A/sec ; I rr = (2x10 7 )(3x10 -7 ) = 6 A b. v Cs,max = 500 V = 200 + 200 1+ C base C s 1+ C base C s = 1.5 ; C base

C s = 1.5 1.25 C base = (10 -5 ) 6 2 (200) 2 = 9 nF C s = 9nF 1.25 7 nF K sin(w o t) cos(f ) - K cos(w o t) sin(f ) = V d C base C s sin(w o t) - V d cos(w o t) K cos(f ) = V d C base C s and d ; [ ] Kcos(f ) 2 + [ ] Ksin(f ) 2 = K 2 = V d 2 K sin(f ) = V

C base C s + V d 2 v Cs,max = V d + K = V d + V d 1+ C base C s 27-4. L = R s C s 200 + i L i L (0 + ) = rr ; di R dt = di R dt = I rr t rr = 200 10 -5 = 2x10 7 a. Equivalent circuit after diode reverse recovery.

10 m H

V

I During reverse recovery L

200 V

A/sec ; I rr = (2x10 7 )(3x10 -7 ) = 6 A b. v Cs,max = 500 V = 200 + 200 1+ C base C s 1+ C base C s = 1.5 ; C base C s = 1.5 1.25 C base = (10 -5 ) 6 2 (200) 2 = 9 nF C s = 9nF 1.25 7 nF 27-5. Use the circuit shown in problem 27-4. C s = C base = (10 -5 ) 6 2 (200) 2 = 9 nF R s = 1.3 R

base = (1.3) 200 6 = 43 ohms v Cs,max = (1.5)(200) = 300 V 27-6. P = W R f sw = L s I rr 2 +C s V in 2 2 f sw W R = (0.5)(10 -5 )(6) 2 + (0.5)(9x10 -9 )(200) 2 = 3.6x10 -4 Joules P = W R f sw = (3.6x10 -4 )(2x10 4 ) = 7.2 watts 27-7. a. BJT waveforms (t rv assumed to be zero for C s = 0) I o i Cs t

t f i t v CE V d C = 0 s C > 0 s 0 0 Power dissipation for C s = 0 is P c = V d I o t fi 2 f sw 27-5. Use the circuit shown in problem 27-4. C s = C base = (10 -5 ) 6 2 (200) 2 = 9 nF R s = 1.3 R base = (1.3) 200 6 = 43 ohms v Cs,max = (1.5)(200) = 300 V 27-6. P = W R f sw = L

s I rr 2 +C s V in 2 2 f sw W R = (0.5)(10 -5 )(6) 2 + (0.5)(9x10 -9 )(200) 2 = 3.6x10 -4 Joules P = W R f sw = (3.6x10 -4 )(2x10 4 ) = 7.2 watts 27-7. a. BJT waveforms (t rv assumed to be zero for C s = 0) I o i Cs t t f i t v CE V d C = 0 s C > 0 s 0 0 Power dissipation for C s = 0 is P

c = V d I o t fi 2 f sw 27-5. Use the circuit shown in problem 27-4. C s = C base = (10 -5 ) 6 2 (200) 2 = 9 nF R s = 1.3 R base = (1.3) 200 6 = 43 ohms v Cs,max = (1.5)(200) = 300 V 27-6. P = W R f sw = L s I rr 2 +C s V in 2 2 f sw W R = (0.5)(10 -5

)(6) 2 + (0.5)(9x10 -9 )(200) 2 = 3.6x10 -4 Joules P = W R f sw = (3.6x10 -4 )(2x10 4 ) = 7.2 watts 27-7. a. BJT waveforms (t rv assumed to be zero for C s = 0) I o i Cs t t f i t v CE V d C = 0 s C > 0 s 0 0 Power dissipation for C s = 0 is P c = V d I o t fi 2 f sw P c = (200)(25)(4x10 -7

) 2 (2x10 4 ) = 20 W Power dissipation for C s = C s1 P c = W c f sw ; W c = 0 t fi I o (1t t fi ) I o 2C s1 t 2 t fi dt = V d I o t fi 12 P c = (200)(25)(4x10 -7 )(2x10 4 ) 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 C s

being dissipated in the BJT, but also because the time to complete turn-on is extended as shown i n Fig. 27-14a. This extended duration of traversal of the active region also incre ases the turn-on losses. During the turn-on interval, the collector-emitter voltage is given by (assuming that the external circuit dominates the transient) C s1 dv CE dt = di C dt t - I rr ; v CE (t) = V d di C dt t 2 2C s1 rr t C s1 Seting the expression for v CE (t) equal to zero and solving for the time D T = t 2 - (t ri + t rr ) (see Fig. 27-14a) required for v CE to reach zero yields D T = I rr di I

C /dt

I rr di C /dt 2 + 2V d C s1 di C /dt

+

Note that D T = 0 if C s1 = 0 which is consistent with the assumption that the external circuit and not the BJT that dominates the turn-on transient. Extra ene rgy disspated in the BJT at turn-on due to C s1 is thus 0 D T v CE (t)i C (t)dt = V d I rr D T + V d 2 di C dt [I o +I rr ]I rr 2C s1 D T 2 -

(I o +3I rr ) 2C s1 di C dt D T 3 - di C dt 2 D T 4 8C s1 The increase in the BJT loss is 0 D T v CE (t)i C (t)dt f s where f s is the switching frequency. Numerical evaluation of D T gives D T = 0.29 m s (I rr = 10 A and C s1 = 25 nF). Evaluation of the loss gives 1.3x10 -3 f s = 26.5 watts 27-8. a. D s shorts out the snubber resistance during the BJT turn-off. Hence C s1 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 reco very

is shown below. R s s C I + r r di C dt v CE + v C + v (0 ) = V + C d v CE = C s R s dv C dt + v C ; ; C s dv C dt

t

= - I rr di C dt t Combining equations and solving for v CE (t) yields v CE (t) = V d - I rr R s -

I rr C s +R s di C dt t di C dt t 2 2C s At t = D T, v CE = 0 and turn-on is completed. D T = I rr +R s C s di C dt di C dt

I rr +R s C sdi C dt di C dt 2 + 2C

+

s di C dt [V d -I rr R s ] D T goes to zero when R s = V d I rr . hence there is no increase in power dissipation in the BJT due to the presence of C s . The increase in the BJT loss is 0 D T v CE (t)i C (t)dt f s where f s is the switching frequency. Numerical evaluation of D T gives D T = 0.29 m s (I rr = 10 A and C s1 = 25 nF). Evaluation of the loss gives 1.3x10 -3 f s = 26.5 watts 27-8. a. D s shorts out the snubber resistance during the BJT turn-off. Hence C s1 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 reco very is shown below. R s

s C I + r r di C dt v CE + v C + v (0 ) = V + C d v CE = C s R s dv C dt + v C ; ; C s dv C dt

t

= - I rr di C dt t Combining equations and solving for v CE (t) yields v CE (t) = V d - I rr R s I

rr C s +R s di C dt t di C dt t 2 2C s At t = D T, v CE = 0 and turn-on is completed. D T = I rr +R s C s di C dt di C dt

I rr +R s C sdi C dt di C dt 2 + 2C s di C

+

dt [V d -I rr R s ] D T goes to zero when R s = V d I rr . hence there is no increase in power dissipation in the BJT due to the presence of C s . 27-9. a. Proposed snubber circuit configuration shown below. 1 2 3 4 L s R s C s I o 2 V s 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 v s (t) goes through zero and becomes negative). L s R s C s 2 V s 3 or 4 1 or 2 I r r With 3 & 4 on, 1 & 2 are off, and effectively in parallel with the R s -C s snubber. same is true when 1 & 2 are on and 3 & 4 are off. Thus the R s -C s

snubber functions as a turn-off snubber. b. w L s = 0.05V s I a1 ; 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, L s di dt = 2 V s and di dt = I rr t rr . Solving for I rr yields I rr = 2w t rr I a1 0.05 ; C base = L s

I rr 2V s 2 = w I a1 t rr 2 0.05V s Smallest overvoltage occurs when C

s = C base . Putting in numerical values C s = 0.9933 I a1 m F R base = 2V s I rr = 0.05V s w t rr I a1 = (0.05)(230) (377)(10 -5 )I a1 = 3050 I a1 ohms R s,opt = 1.3 R base = 4000 I a1 ohms c. Peak line voltage = 2 (230) = 322 V Smallest overvoltage = (1.5)(322) = 483 V which occurs when C s = C base and R s = 1.3 R base . For I a1 = 100 A R s,opt = 1.3 R base =

4000 100 = 40 ohms ; C s = (0.0033)(100) = 0.33 m F 27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the dev ice turn-off so that the snubber capacitance is in parallel with the device. The unc harged capacitor delays the build-up of the large V d voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forw ard biased during turn-off, thus providing the shorting of the snubber resistor as r equired. 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 a re obtained when R s = 1.3 R base and C s = C base. Overvoltages are 70-% larger if R s is zero. Hence a diode in parallel with the resistor is not desirable. 27-11. a. Check dv DS dt at turn-off; dv DS dt = V d t rv ; t rv < 0.75 microseconds dv DS dt > 700 7.5x10 -7 = 930 V/m s > 800 V/m s limit so snubber is needed. Not enough information available to check on power dissipation or overcurrents. di dt

= I rr t rr . Solving for I rr yields I rr = 2w t rr I a1 0.05 ; C base = L s

I rr 2V s 2 = w I a1 t rr 2 0.05V s Smallest overvoltage occurs when C s = C base . Putting in numerical values C s = 0.9933 I a1 m F R base = 2V s I rr = 0.05V s w t rr

I a1 = (0.05)(230) (377)(10 -5 )I a1 = 3050 I a1 ohms R s,opt = 1.3 R base = 4000 I a1 ohms c. Peak line voltage = 2 (230) = 322 V Smallest overvoltage = (1.5)(322) = 483 V which occurs when C s = C base and R s = 1.3 R base . For I a1 = 100 A R s,opt = 1.3 R base = 4000 100 = 40 ohms ; C s = (0.0033)(100) = 0.33 m F 27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the dev ice turn-off so that the snubber capacitance is in parallel with the device. The unc harged capacitor delays the build-up of the large V d voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forw ard biased during turn-off, thus providing the shorting of the snubber resistor as r equired. 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 a re obtained when R

s = 1.3 R base and C s = C base. Overvoltages are 70-% larger if R s is zero. Hence a diode in parallel with the resistor is not desirable. 27-11. a. Check dv DS dt at turn-off; dv DS dt = V d t rv ; t rv < 0.75 microseconds dv DS dt > 700 7.5x10 -7 = 930 V/m s > 800 V/m s limit so snubber is needed. Not enough information available to check on power dissipation or overcurrents. di dt = I rr t rr . Solving for I rr yields I rr = 2w t rr I a1 0.05 ; C base = L s

I rr 2V s 2 = w I a1 t rr 2 0.05V s Smallest overvoltage occurs when C s = C base . Putting in numerical values C s = 0.9933 I a1 m F R base = 2V s I rr = 0.05V s w t rr I a1 = (0.05)(230) (377)(10 -5 )I a1 = 3050 I a1 ohms R s,opt = 1.3 R base = 4000 I a1

ohms c. Peak line voltage = 2 (230) = 322 V Smallest overvoltage = (1.5)(322) = 483 V which occurs when C s = C base and R s = 1.3 R base . For I a1 = 100 A R s,opt = 1.3 R base = 4000 100 = 40 ohms ; C s = (0.0033)(100) = 0.33 m F 27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the dev ice turn-off so that the snubber capacitance is in parallel with the device. The unc harged capacitor delays the build-up of the large V d voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forw ard biased during turn-off, thus providing the shorting of the snubber resistor as r equired. 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 a re obtained when R s = 1.3 R base and C s = C base. Overvoltages are 70-% larger if R s is zero. Hence a diode in parallel with the resistor is not desirable. 27-11. a. Check dv DS dt at turn-off; dv DS dt = V

d t rv ; t rv < 0.75 microseconds dv DS dt > 700 7.5x10 -7 = 930 V/m s > 800 V/m s limit so snubber is needed. Not enough information available to check on power dissipation or overcurrents. b. dv DS dt = I o C s = 400 V/m s ; C s = 100 4x10 8 = 0.25 m F Choose R s to limit total current I D to less than 150 A 150 A = 100 + 700 R s ; R s = 700 150-100 = 14 ohms Check snubber recovery time = 2.3 R s C s = (2.3)(14)(2.5x10 -7 ) = 8 m s Off time of the IGBT is 10 microseconds which is greater than the snubber recovery time. Hence choice of R s is fine. Chapater 28 Problem Solutions 28-1. Schematic of drive circuit shown below.

R G V GG+ V GGV DS 100 A 100 V + v DS (t) waveform same as in problem 22-2. During MOSFET turn-on dv DS dt = V d t fv < 500 V/m s From problem 22-2, V d t fv = V GG+ -V GSth I o g m R G C gd During MOSFET turn-off, dv DS dt = V d t rv < 500 V/m s From problem 23-2, V d t

rv

V GG+V GSth + I o g m R G C gdg m

=

= I o V GS -V GSth = 60 7-4 = 20 A/V Estimate of R G for MOSFET turn-on: 5x10 8 V/sec > V GG+ -4100 20 (R G )(4x10 -10 ) ; V GG+,min = V GSth + I o g m = 4 + 100 20 = 9 V Choose V

GG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses. R G > 15-9 (4x10 -10 )(5x10 8 ) = 30 ohms Estimate of R G at MOSFET turn-off: 5x10 8 V/sec > V GG+4+ 100 20 (R G )(4x10 -10 ) Choose V GG= -15 V to insure that MOSFET is held in off-state and to minimize turn-off times. R G > 15+9 (4x10 -10 )(5x10 8 ) = 115 ohms Satisfy both turn-on and turn-joff requirements by choosing V GG+ = V GG= 15 V and R G > 115 ohms. 28-2. a. Circuit diagram shown below. R G1 R G2 Q s T sw

D f + I o = 200 A V d = 1000 V When FCT is off we need V KG = (1.25) 1000 40 = 31.25 V = (1000) R G2 R G1 +R G2 Now R G1 + R G2 = 10 -6 ohms so 31.25 = (1000) (10 -6 ) R G2 R G2 = 31.25 kW and R G1 = 969 kW Choose V GG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses. R G > 15-9 (4x10 -10 )(5x10 8 ) = 30 ohms Estimate of R G at MOSFET turn-off: 5x10 8 V/sec > V GG-

+4+ 100 20 (R G )(4x10 -10 ) Choose V GG= -15 V to insure that MOSFET is held in off-state and to minimize turn-off times. R G > 15+9 (4x10 -10 )(5x10 8 ) = 115 ohms Satisfy both turn-on and turn-joff requirements by choosing V GG+ = V GG= 15 V and R G > 115 ohms. 28-2. a. Circuit diagram shown below. R G1 R G2 Q s T sw D f + I o = 200 A V d = 1000 V When FCT is off we need V KG = (1.25) 1000 40 = 31.25

Documents

53326496 Power Electronics Mohan 2nd Edition Solution