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Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus ni(Ti) = Nd = 10 14 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 Ti !-! 1 300 Solving for Ti using Eg = 1.1 eV, k = 1.4x10 -23 [1/°K] yields Ti = 262 °C or 535 °K. 19-2. N-side resistivity rn = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014) = 43.5 ohm-cm P-side resistivity rp = 1 q!mp!Na = 1 (1.6x10-19)(500)(1018) = 0.013 ohm-cm 19-3. Material is n-type with Nd = 10 13 cm-3 >> ni = 10 10 cm-3. Hence use approximate formulas given in Chapter 19. n = Nd = 10 13 cm-3 ; p = n 2 i Nd = 1020 1013 = 107 cm-3 19-4. po = n 2 i [300] Nd ; 2po = n 2 i [300!+!T] Nd 2 n 2 i [300] = n 2 i [300 + T] ; 2x1010 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 T!-! 1 300 Solving for T yields T = q!Eg!300 (q!Eg!-!k!300!ln(2)) = 305.2 °K DT = 305.2 - 300 = 5.2 °K. 19-5. I1 = Is exp( q!V1 k!T ; 10 I1 = Is exp( q!V1!+!dV k!T ) ; dV = k!T q ln(10) = 60 mV 19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias. Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus ni(Ti) = Nd = 10 14 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 Ti !-! 1 300 Solving for Ti using Eg = 1.1 eV, k = 1.4x10 -23 [1/°K] yields Ti = 262 °C or 535 °K. 19-2. N-side resistivity rn = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014) = 43.5 ohm-cm P-side resistivity rp = 1 q!mp!Na = 1 (1.6x10-19)(500)(1018) = 0.013 ohm-cm 19-3. Material is n-type with Nd = 10 13 cm-3 >> ni = 10 10 cm-3. Hence use approximate formulas given in Chapter 19. n = Nd = 10 13 cm-3 ; p = n 2 i Nd = 1020 1013 = 107 cm-3 19-4. po = n 2 i [300] Nd ; 2po = n 2 i [300!+!T] Nd 2 n 2 i [300] = n 2 i [300 + T] ; 2x1010 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 T!-! 1 300 Solving for T yields T = q!Eg!300 (q!Eg!-!k!300!ln(2)) = 305.2 °K DT = 305.2 - 300 = 5.2 °K. 19-5. I1 = Is exp( q!V1 k!T ; 10 I1 = Is exp( q!V1!+!dV k!T ) ; dV = k!T q ln(10) = 60 mV 19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias. Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus ni(Ti) = Nd = 10 14 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 Ti !-! 1 300 Solving for Ti using Eg = 1.1 eV, k = 1.4x10 -23 [1/°K] yields Ti = 262 °C or 535 °K. 19-2. N-side resistivity rn = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014) = 43.5 ohm-cm P-side resistivity rp = 1 q!mp!Na = 1 (1.6x10-19)(500)(1018) = 0.013 ohm-cm 19-3. Material is n-type with Nd = 10 13 cm-3 >> ni = 10 10 cm-3. Hence use approximate formulas given in Chapter 19. n = Nd = 10 13 cm-3 ; p = n 2 i Nd = 1020 1013 = 107 cm-3 19-4. po = n 2 i [300] Nd ; 2po = n 2 i [300!+!T] Nd 2 n 2 i [300] = n 2 i [300 + T] ; 2x1010 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 T!-! 1 300 Solving for T yields T = q!Eg!300 (q!Eg!-!k!300!ln(2)) = 305.2 °K DT = 305.2 - 300 = 5.2 °K. 19-5. I1 = Is exp( q!V1 k!T ; 10 I1 = Is exp( q!V1!+!dV k!T ) ; dV = k!T q ln(10) = 60 mV 19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias. Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus ni(Ti) = Nd = 10 14 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 Ti !-! 1 300 Solving for Ti using Eg = 1.1 eV, k = 1.4x10 -23 [1/°K] yields Ti = 262 °C or 535 °K. 19-2. N-side resistivity rn = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014) = 43.5 ohm-cm P-side resistivity rp = 1 q!mp!Na = 1 (1.6x10-19)(500)(1018) = 0.013 ohm-cm 19-3. Material is n-type with Nd = 10 13 cm-3 >> ni = 10 10 cm-3. Hence use approximate formulas given in Chapter 19. n = Nd = 10 13 cm-3 ; p = n 2 i Nd = 1020 1013 = 107 cm-3 19-4. po = n 2 i [300] Nd ; 2po = n 2 i [300!+!T] Nd 2 n 2 i [300] = n 2 i [300 + T] ; 2x1010 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 T!-! 1 300 Solving for T yields T = q!Eg!300 (q!Eg!-!k!300!ln(2)) = 305.2 °K DT = 305.2 - 300 = 5.2 °K. 19-5. I1 = Is exp( q!V1 k!T ; 10 I1 = Is exp( q!V1!+!dV k!T ) ; dV = k!T q ln(10) = 60 mV 19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias. Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus ni(Ti) = Nd = 10 14 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 Ti !-! 1 300 Solving for Ti using Eg = 1.1 eV, k = 1.4x10 -23 [1/°K] yields Ti = 262 °C or 535 °K. 19-2. N-side resistivity rn = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014) = 43.5 ohm-cm P-side resistivity rp = 1 q!mp!Na = 1 (1.6x10-19)(500)(1018) = 0.013 ohm-cm 19-3. Material is n-type with Nd = 10 13 cm-3 >> ni = 10 10 cm-3. Hence use approximate formulas given in Chapter 19. n = Nd = 10 13 cm-3 ; p = n 2 i Nd = 1020 1013 = 107 cm-3 19-4. po = n 2 i [300] Nd ; 2po = n 2 i [300!+!T] Nd 2 n 2 i [300] = n 2 i [300 + T] ; 2x1010 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 T!-! 1 300 Solving for T yields T = q!Eg!300 (q!Eg!-!k!300!ln(2)) = 305.2 °K DT = 305.2 - 300 = 5.2 °K. 19-5. I1 = Is exp( q!V1 k!T ; 10 I1 = Is exp( q!V1!+!dV k!T ) ; dV = k!T q ln(10) = 60 mV 19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias. Chapter 19 Problem Solutions 19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus ni(Ti) = Nd = 10 14 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 Ti !-! 1 300 Solving for Ti using Eg = 1.1 eV, k = 1.4x10 -23 [1/°K] yields Ti = 262 °C or 535 °K. 19-2. N-side resistivity rn = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014) = 43.5 ohm-cm P-side resistivity rp = 1 q!mp!Na = 1 (1.6x10-19)(500)(1018) = 0.013 ohm-cm 19-3. Materialis n-type with Nd = 10 13 cm-3 >> ni = 10 10 cm-3. Hence use approximate formulas given in Chapter 19. n = Nd = 10 13 cm-3 ; p = n 2 i Nd = 1020 1013 = 107 cm-3 19-4. po = n 2 i [300] Nd ; 2po = n 2 i [300!+!T] Nd 2 n 2 i [300] = n 2 i [300 + T] ; 2x1010 = 1010 exp Î Í È ˚ ˙ ˘ -! q!Eg 2k !ÓÌ Ï ˛ ˝ ¸1 T!-! 1 300 Solving for T yields T = q!Eg!300 (q!Eg!-!k!300!ln(2)) = 305.2 °K DT = 305.2 - 300 = 5.2 °K. 19-5. I1 = Is exp( q!V1 k!T ; 10 I1 = Is exp( q!V1!+!dV k!T ) ; dV = k!T q ln(10) = 60 mV 19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias. xn(0) + xp(0) = Wo = ! 2!e!fc!(Na!+!Nd) q!Na!Nd (1) fc = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! Na!Nd n 2 i = 0.026 ln Î Í È ˚ ˙ ˘ ! 1014!1015 1020 = 0.54 eV Conservation of charge: q Na xp = q Nd xn (2) Solving (1) and (2) simultaneously gives using the numerical values given in the problem statement gives: Wo = 2.8 microns ; xn(0) = 2.55 microns ; xp(0) = 0.25 microns (b) Electric field profile triangular-shaped as shown in Fig. 19-9b. Maximum electric at zero bias given by Emax = 2!fc Wo = (2)!(0.54) (2.8x10-4) = 3,900 V/cm (c) From part a) fc = 0.54 eV (d) C(V) A = e Wo 1!+! V fc ; C(V) = space-charge capacitance at reverse voltage V. C(0) A = (11.7)(8.9x10-14) 2.8x10-4 = 3.7x10-9 F/cm2 C(50) A = (11.7)(8.9x10-14) 2.8x10-4 1!+! 50 0.54 = 3.8x10-10 F/cm2 (e) I = Is exp( qV kT ) ; exp( qV kT ) = exp ( 0.7 0.026 ) = 5x10 11 Is = q n 2 i ÎÍ Í È ˚ ˙ ˙ ˘ ! Dnt Na!t !+! Dpt Nd!t A = (1.6x10-19)(1020) ÎÍ Í È ˚ ˙ ˙ ˘ ! (38)(10-6) (1015)(10-6) !+! (13)(10-6) (1014)(10-6) (2) 2 Is = 6.7x10 -14 A ; I = (6.7x10-14 )(5x1011) = 34 mA 19-7. Resistance R = r!L A ; L A = 0.02 0.01 = 2 cm -1 At 25 °C, Nd = 10 14 >> ni so r = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014!) = 41.7 W-cm R(25 °C) = (41.7)(2) = 83.4 ohms At 250 °C (523 °K), ni[523] = 10 10 exp Î Í È ˚ ˙ ˘ -! (1.6x10-19)!(1.1) (2)(1.4x10-23) ! Ó Ì Ï ˛ ˝ ¸1 523!-! 1 300 = (1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 10 14. Thus we should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields no = Nd 2 Î Í È ˚ ˙ ˘ 1!+! 1!+! 4!n 2 i N 2 d and po = n 2 i no . Putting in numerical values yields no = 1014 2 ÎÍ Í È ˚ ˙ ˙ ˘ 1!+! 1!+! (4)(7.6x1013)2! (1014)2 = 1.4x1014 and po = 5.8x1027 1014 = 5.8x1013 Assuming temperature-independent mobilities (not a valid assumption but no other information is given in text or the problem statement), resistance is R(250 °C) = r(250!° C)!!L A ; r(250 °C) ≈ 1 q!mn!no!+!q!mp!po = 1 (1.6x10-19)(1500)(1.4x1014!)!+!(1.6x10-19)(500)(5.8x1013!) = 26.2 W-cm ; R(250 °C) ≈ (26.2)(2) = 52.4 ohms 19-8. BVBD = e!(Na!+!Nd)!E 2 BD 2!q!Na!Nd = (11.7)(8.9x1014)(1015!+!1014)(3x105)2 (2)(1.6x10-19)(1015)(1014) = 3,340 volts 19-7. Resistance R = r!L A ; L A = 0.02 0.01 = 2 cm -1 At 25 °C, Nd = 10 14 >> ni so r = 1 q!mn!Nd = 1 (1.6x10-19)(1500)(1014!) = 41.7 W-cm R(25 °C) = (41.7)(2) = 83.4 ohms At 250 °C (523 °K), ni[523] = 10 10 exp Î Í È ˚ ˙ ˘ -! (1.6x10-19)!(1.1) (2)(1.4x10-23) ! Ó Ì Ï ˛ ˝ ¸1 523!-! 1 300 = (1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 10 14. Thus we should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields no = Nd 2 Î Í È ˚ ˙ ˘ 1!+! 1!+! 4!n 2 i N 2 d and po = n 2 i no . Putting in numerical values yields no = 1014 2 ÎÍ Í È ˚ ˙ ˙ ˘ 1!+! 1!+! (4)(7.6x1013)2! (1014)2 = 1.4x1014 and po = 5.8x1027 1014 = 5.8x1013 Assuming temperature-independent mobilities (not a valid assumption but no other information is given in text or the problem statement), resistance is R(250 °C) = r(250!° C)!!L A ; r(250 °C) ≈ 1 q!mn!no!+!q!mp!po = 1 (1.6x10-19)(1500)(1.4x1014!)!+!(1.6x10-19)(500)(5.8x1013!) = 26.2 W-cm ; R(250 °C) ≈ (26.2)(2) = 52.4 ohms 19-8. BVBD = e!(Na!+!Nd)!E 2 BD 2!q!Na!Nd = (11.7)(8.9x1014)(1015!+!1014)(3x105)2 (2)(1.6x10-19)(1015)(1014) = 3,340 volts 19-9. E 2 max = E 2 BD ≈ 4!f 2 c!BVBD W 2 o!fc ; Eq. (19-13); or W 2 o !fc = 4!BVBD E 2 BD W2(BVBD) = W 2 o!BVBD !fc ; Eq. (19-11) ; Inserting W 2 o !fc = 4!BVBD E 2 BD and taking the square root yields W (BVBD) ≈ 2!BVBD E 2 BD. 19-10. Lp = Dp!t = (13)(10 -6) = 36 microns ; Ln = Dn!t = (39)(10 -6) = 62 microns 19-11. Assume a one-sided step junction with Na >> Nd I1 = q n 2 i A Dp!t1 Nd!t1 exp( q!V k!T ) ; I2 = q n 2 i A Dp!t2 Nd!t2 exp( q!V k!T ) I2 I1 = 2 = t1 t2 ; Thus 4 t2 = t1 19-12. s = q mp p + q mn n ; np = n 2 i ; Combining yeilds s = q mp n 2 i n + q mn n ds dn = 0 = - q mp n 2 i n2 + q mn ; Solving for n yields n = ni mp mn and p = ni mn mp p = 1010 1500 500 = 1.7x10 10 cm-3; n = 1010 500 1500 = 6x10 9 cm-3; Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10 -19)(1010) (500)(1500) = 2.8x10-6 mhos-cm 19-9. E 2 max = E 2 BD ≈ 4!f 2 c!BVBD W 2 o!fc ; Eq. (19-13); or W 2 o !fc = 4!BVBD E 2 BD W2(BVBD) = W 2 o!BVBD !fc ; Eq. (19-11) ; Inserting W 2 o !fc = 4!BVBD E 2 BD and taking the square root yields W (BVBD) ≈ 2!BVBD E 2 BD. 19-10. Lp = Dp!t = (13)(10 -6) = 36 microns ; Ln = Dn!t = (39)(10 -6) = 62 microns 19-11. Assume a one-sided step junction with Na >> Nd I1 = q n 2 i A Dp!t1 Nd!t1 exp( q!V k!T ) ; I2 = q n 2 i A Dp!t2 Nd!t2 exp( q!V k!T ) I2 I1 = 2 = t1 t2 ; Thus 4 t2 = t1 19-12. s = q mp p + q mn n ; np = n 2 i ; Combining yeilds s = q mp n 2 i n + q mn n ds dn = 0 = - q mp n 2 i n2 + q mn ; Solving for n yields n = ni mp mn and p = ni mn mp p = 1010 1500 500 = 1.7x10 10 cm-3; n = 1010 500 1500 = 6x10 9 cm-3; Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10 -19)(1010) (500)(1500) = 2.8x10-6 mhos-cm 19-9. E 2 max = E 2 BD ≈ 4!f 2 c!BVBD W 2 o!fc ; Eq. (19-13); or W 2 o !fc = 4!BVBD E 2 BD W2(BVBD) = W 2 o!BVBD !fc ; Eq. (19-11) ; Inserting W 2 o !fc = 4!BVBD E 2 BD and taking the square root yields W (BVBD) ≈ 2!BVBD E 2 BD. 19-10.Lp = Dp!t = (13)(10 -6) = 36 microns ; Ln = Dn!t = (39)(10 -6) = 62 microns 19-11. Assume a one-sided step junction with Na >> Nd I1 = q n 2 i A Dp!t1 Nd!t1 exp( q!V k!T ) ; I2 = q n 2 i A Dp!t2 Nd!t2 exp( q!V k!T ) I2 I1 = 2 = t1 t2 ; Thus 4 t2 = t1 19-12. s = q mp p + q mn n ; np = n 2 i ; Combining yeilds s = q mp n 2 i n + q mn n ds dn = 0 = - q mp n 2 i n2 + q mn ; Solving for n yields n = ni mp mn and p = ni mn mp p = 1010 1500 500 = 1.7x10 10 cm-3; n = 1010 500 1500 = 6x10 9 cm-3; Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10 -19)(1010) (500)(1500) = 2.8x10-6 mhos-cm 19-9. E 2 max = E 2 BD ≈ 4!f 2 c!BVBD W 2 o!fc ; Eq. (19-13); or W 2 o !fc = 4!BVBD E 2 BD W2(BVBD) = W 2 o!BVBD !fc ; Eq. (19-11) ; Inserting W 2 o !fc = 4!BVBD E 2 BD and taking the square root yields W (BVBD) ≈ 2!BVBD E 2 BD. 19-10. Lp = Dp!t = (13)(10 -6) = 36 microns ; Ln = Dn!t = (39)(10 -6) = 62 microns 19-11. Assume a one-sided step junction with Na >> Nd I1 = q n 2 i A Dp!t1 Nd!t1 exp( q!V k!T ) ; I2 = q n 2 i A Dp!t2 Nd!t2 exp( q!V k!T ) I2 I1 = 2 = t1 t2 ; Thus 4 t2 = t1 19-12. s = q mp p + q mn n ; np = n 2 i ; Combining yeilds s = q mp n 2 i n + q mn n ds dn = 0 = - q mp n 2 i n2 + q mn ; Solving for n yields n = ni mp mn and p = ni mn mp p = 1010 1500 500 = 1.7x10 10 cm-3; n = 1010 500 1500 = 6x10 9 cm-3; Thus minimum conductivity realized when silicon is slightly p-type. Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn . Putting in numerical values smin = (2)(1.6x10 -19)(1010) (500)(1500) = 2.8x10-6 mhos-cm Chapter 20 Problem Solutions 20-1. Nd = 1.3x1017 BVBD! = 1.3x1017 2500 = 5x10 13 cm-3 ; W(2500 V) = (10-5)(2500) = 250 microns 20-2. Drift region length of 50 microns is much less than the 250 microns found in the previous problem (20-1) for the same drift region doping density. Hence this must be a punch- through structure and Eq. (20-9) applies. BVBD = (2x10 5)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2 (2)(11.7)(8.9x10-14) = 900 V 20-3. Von = Vj + Vdrift ; Vj = k!T q ln ÎÍ È ˚ ˙˘! I Is ; For one-sided step junction Is = q!A!n 2 i !Lp !Nd!to ; Evaluating Is yields (1.6x10-19)(2)(1010)2! (13)(2x10-6) !(5x1013)(2x10-6) = 1.6x10-9 A Vd = K1 I + K2 (I) 2/3 Eq. (20-16) with I = forward bias current through the diode. K1 = Wd q!mo!A!nb = 5x10-3 (1.6x10-19)(900)(2)(1017) = 1.7x10-4 K2 = 3 W 4 d q2!m 3 o!n 2 b!A2!to = 3 (5x10-3)4 (1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6) = 7.5x10-4 Chapter 20 Problem Solutions 20-1. Nd = 1.3x1017 BVBD! = 1.3x1017 2500 = 5x10 13 cm-3 ; W(2500 V) = (10-5)(2500) = 250 microns 20-2. Drift region length of 50 microns is much less than the 250 microns found in the previous problem (20-1) for the same drift region doping density. Hence this must be a punch- through structure and Eq. (20-9) applies. BVBD = (2x10 5)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2 (2)(11.7)(8.9x10-14) = 900 V 20-3. Von = Vj + Vdrift ; Vj = k!T q ln ÎÍ È ˚ ˙˘! I Is ; For one-sided step junction Is = q!A!n 2 i !Lp !Nd!to ; Evaluating Is yields (1.6x10-19)(2)(1010)2! (13)(2x10-6) !(5x1013)(2x10-6) = 1.6x10-9 A Vd = K1 I + K2 (I) 2/3 Eq. (20-16) with I = forward bias current through the diode. K1 = Wd q!mo!A!nb = 5x10-3 (1.6x10-19)(900)(2)(1017) = 1.7x10-4 K2 = 3 W 4 d q2!m 3 o!n 2 b!A2!to = 3 (5x10-3)4 (1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6) = 7.5x10-4 Chapter 20 Problem Solutions 20-1. Nd = 1.3x1017 BVBD! = 1.3x1017 2500 = 5x10 13 cm-3 ; W(2500 V) = (10-5)(2500) = 250 microns 20-2. Drift region length of 50 microns is much less than the 250 microns found in the previous problem (20-1) for the same drift region doping density. Hence this must be a punch- through structure and Eq. (20-9) applies. BVBD = (2x10 5)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2 (2)(11.7)(8.9x10-14) = 900 V 20-3. Von = Vj + Vdrift ; Vj = k!T q ln ÎÍ È ˚ ˙˘! I Is ; For one-sided step junction Is = q!A!n 2 i !Lp !Nd!to ; Evaluating Is yields (1.6x10-19)(2)(1010)2! (13)(2x10-6) !(5x1013)(2x10-6) = 1.6x10-9 A Vd = K1 I + K2 (I) 2/3 Eq. (20-16) with I = forward bias current through the diode. K1 = Wd q!mo!A!nb = 5x10-3 (1.6x10-19)(900)(2)(1017) = 1.7x10-4 K2 = 3 W 4 d q2!m 3 o!n 2 b!A2!to = 3 (5x10-3)4 (1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6) = 7.5x10-4 I Vj Vdrift Von 0 A 0 V 0 V 0 V 1 0.53 0.001 0.53 10 0.59 0.005 0.59 100 0.65 0.033 0.68 1000 0.71 0.25 0.96 3000 0.74 0.67 1.41 • • • • • 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1 10 100 1000 10000 Von in volts Forward current in amperes 20-4. a) Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r L A ; r = 1 q!mn!Nd r = 1 (1.6x10-19)(1500)(5x1013) = 85 ohm-cm ; L A = 5x10-3 2 = 2.5x10 -3 Rdrift = (85)(2.5x10 -3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds Von(t) = (0.21)(2.5x10 8 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds Von(4 ms) = (5.3x10 7)(4x10-6) = 212 volts b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x10 7 t] Von(t) = {0.21[1 - 2.5x10 7 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds I Vj Vdrift Von 0 A 0 V 0 V 0 V 1 0.53 0.001 0.53 10 0.59 0.005 0.59 100 0.65 0.033 0.68 1000 0.71 0.25 0.96 3000 0.74 0.67 1.41 • • • • • 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1 10 100 1000 10000 Von in volts Forward current in amperes 20-4. a) Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r L A ; r = 1 q!mn!Nd r = 1 (1.6x10-19)(1500)(5x1013) = 85 ohm-cm ; L A = 5x10-3 2 = 2.5x10 -3 Rdrift = (85)(2.5x10 -3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds Von(t) = (0.21)(2.5x10 8 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds Von(4 ms) = (5.3x10 7)(4x10-6) = 212 volts b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x10 7 t] Von(t) = {0.21[1 - 2.5x10 7 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds 0 50 100 150 200 250 0 0.5 1 1.5 2 2.5 3 3.5 4 Von in volts With carrier injection No carrier injection 20-5. toff = trr + t3 = trr + IF diR/dt = trr + 2000 2.5x108 = trr + 8 ms trr = 2!t!IF diR/dt ; t = 4x10-12(BVBD) 2 = 4x10-12(2000)2 = 16 ms trr = (2)(1.6x10-5)(2x103) 2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms 20-6. Assume a non-punch-through structure for the Schottky diode. Nd = 1.3x1017 BVBD = 1.3x1017 150 = 8.7x10 14 cm-3 Wd = 10 -5 BVBD = (10 -5) (150) = 15 microns 20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1 q!mn!Nd L A A = 2x10-3 !(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2 0 50 100 150 200 250 0 0.5 1 1.5 2 2.5 3 3.5 4 Von in volts With carrier injection Nocarrier injection 20-5. toff = trr + t3 = trr + IF diR/dt = trr + 2000 2.5x108 = trr + 8 ms trr = 2!t!IF diR/dt ; t = 4x10-12(BVBD) 2 = 4x10-12(2000)2 = 16 ms trr = (2)(1.6x10-5)(2x103) 2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms 20-6. Assume a non-punch-through structure for the Schottky diode. Nd = 1.3x1017 BVBD = 1.3x1017 150 = 8.7x10 14 cm-3 Wd = 10 -5 BVBD = (10 -5) (150) = 15 microns 20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1 q!mn!Nd L A A = 2x10-3 !(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2 0 50 100 150 200 250 0 0.5 1 1.5 2 2.5 3 3.5 4 Von in volts With carrier injection No carrier injection 20-5. toff = trr + t3 = trr + IF diR/dt = trr + 2000 2.5x108 = trr + 8 ms trr = 2!t!IF diR/dt ; t = 4x10-12(BVBD) 2 = 4x10-12(2000)2 = 16 ms trr = (2)(1.6x10-5)(2x103) 2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms 20-6. Assume a non-punch-through structure for the Schottky diode. Nd = 1.3x1017 BVBD = 1.3x1017 150 = 8.7x10 14 cm-3 Wd = 10 -5 BVBD = (10 -5) (150) = 15 microns 20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1 q!mn!Nd L A A = 2x10-3 !(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2 0 50 100 150 200 250 0 0.5 1 1.5 2 2.5 3 3.5 4 Von in volts With carrier injection No carrier injection 20-5. toff = trr + t3 = trr + IF diR/dt = trr + 2000 2.5x108 = trr + 8 ms trr = 2!t!IF diR/dt ; t = 4x10-12(BVBD) 2 = 4x10-12(2000)2 = 16 ms trr = (2)(1.6x10-5)(2x103) 2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms 20-6. Assume a non-punch-through structure for the Schottky diode. Nd = 1.3x1017 BVBD = 1.3x1017 150 = 8.7x10 14 cm-3 Wd = 10 -5 BVBD = (10 -5) (150) = 15 microns 20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1 q!mn!Nd L A A = 2x10-3 !(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2 20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] Î Í Í È ˚ ˙ ˙ ˘2!e q!W 2 d = [(2x105)(2x10-3) - 300] (2)(11.7)(8.9x10-14) (1.6x10-19)(2x10-3)2 = Nd = 3.4x10 14 cm-3 20-9. R A npt = Wd(npt) q!mn!Nnpt ; Nnpt = Nd of non-punch-throuth (npt) diode R A pt = Wd(pt) q!mn!Npt ; Npt = Nd of punch-through (pt) diode Wd(npt) = e!EBD !q!Nnpt ; Derived from Eqs. (19-11), (19-12), and (19-13) Wd(pt) = e!EBD !q!Npt Î Í Í È ˚ ˙ ˙ ˘ 1!+_ ! 1!-! 2!q!Npt!BVBD e!E 2 BD 2!q!Npt!BVBD e!E 2 BD = q!Npt e!EBD Nnpt Nnpt 2!BVBD EBD = q!Nnpt e!EBD 2!BVBD EBD Npt Nnpt = 1 Wd(npt) Wd(npt) Npt Nnpt = x = Npt Nnpt ; Wd(pt) = Wd(npt) 1 x [ ]1!+_ ! 1!-!x If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10) Limit of 1 x [ ]1!+_ ! 1!-!x as x approaches is infinite for the plus root and 0.5 for the minus root. Hence the minus root is the correct choice to use. R A pt = Wd(npt)! 1 x![ ]1!-!! 1!-!x q!mn!Npt = Wd(npt)! 1 x![ ]1!-!! 1!-!x q!mn!Npt Nnpt !Nnpt = Wd(npt) !q!mn!Nnpt 1 x2 [1 - 1!-!x ] = R A npt 1 x2 [1 - 1!-!x ] d dx ÎÍ È ˚ ˙˘! 1 x2 !! 1!-!x! = 0 = -2 x3 [1 - 1!-!x ] + 1 2!x2! 1!-!x Solving for x yields x = 8 9 i.e. Npt = 8 9 Nnpt ; Wd(pt) = 0.75 Wd(npt) 20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] Î Í Í È ˚ ˙ ˙ ˘2!e q!W 2 d = [(2x105)(2x10-3) - 300] (2)(11.7)(8.9x10-14) (1.6x10-19)(2x10-3)2 = Nd = 3.4x10 14 cm-3 20-9. R A npt = Wd(npt) q!mn!Nnpt ; Nnpt = Nd of non-punch-throuth (npt) diode R A pt = Wd(pt) q!mn!Npt ; Npt = Nd of punch-through (pt) diode Wd(npt) = e!EBD !q!Nnpt ; Derived from Eqs. (19-11), (19-12), and (19-13) Wd(pt) = e!EBD !q!Npt Î Í Í È ˚ ˙ ˙ ˘ 1!+_ ! 1!-! 2!q!Npt!BVBD e!E 2 BD 2!q!Npt!BVBD e!E 2 BD = q!Npt e!EBD Nnpt Nnpt 2!BVBD EBD = q!Nnpt e!EBD 2!BVBD EBD Npt Nnpt = 1 Wd(npt) Wd(npt) Npt Nnpt = x = Npt Nnpt ; Wd(pt) = Wd(npt) 1 x [ ]1!+_ ! 1!-!x If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10) Limit of 1 x [ ]1!+_ ! 1!-!x as x approaches is infinite for the plus root and 0.5 for the minus root. Hence the minus root is the correct choice to use. R A pt = Wd(npt)! 1 x![ ]1!-!! 1!-!x q!mn!Npt = Wd(npt)! 1 x![ ]1!-!! 1!-!x q!mn!Npt Nnpt !Nnpt = Wd(npt) !q!mn!Nnpt 1 x2 [1 - 1!-!x ] = R A npt 1 x2 [1 - 1!-!x ] d dx ÎÍ È ˚ ˙˘! 1 x2 !! 1!-!x! = 0 = -2 x3 [1 - 1!-!x ] + 1 2!x2! 1!-!x Solving for x yields x = 8 9 i.e. Npt = 8 9 Nnpt ; Wd(pt) = 0.75 Wd(npt) R A npt = 0.84 R A npt 20-10. Cjo = eA Wd(0) ; Area A determined by on-state voltage and current. Depletion-layer width Wd(0) set by breakdown voltage. Wd(0) same for both diodes. Wd(150) = Wd(0) 1!+! 150 0.7 = (10 -5)(150) = 15 mm ; Wd(0) = 15 (150)/(0.7) ≈ 1 mm Schottky diode area ; Vdrift = 2 volts = Wd(150) q!mn!Nd!ASchottky IF Drift region doping density Nd same for both diodes. Nd = 1.3x1017 150 = 8.7x1014 cm-3 ASchottky = (1.5x10-3)(300) (2)(1.6x10-19)(1500)(8.7x1014) = 1.07 cm2 PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF) 2/3 ; Eq. (20-16) K1 = Wd(150) q!mo!Apn!nb = 1.5x10-3 (1.6x10-19)(900)(Apn)(10 17) = 10-4 Apn K2 = 3 W 4 d(150) q2!m 3 o!n 2 b!A 2 pn!to = 3 (1.5x10-3)4 (1.6x10-19)2!(900)3!(1017)2!(A 2 pn)!(2x10-6) K2 = 2.4x10 -4 (Apn) -0.67 ; 2 volts = 10-4 Apn (300) + 2.4x10-4 (Apn) -0.67 (300)0.67 Apn = 0.015 + 0.00154 (Apn) 0.333 ; Solve by successive approximation ; Apn ≈ 0.017 cm 2 Schottky diode Cjo = (11.7)(8.9x10-14)(1.07) 10-4 ≈ 11 nF = 0.011 mF R A npt = 0.84 R A npt 20-10. Cjo = eA Wd(0) ; Area A determined by on-state voltage and current. Depletion-layer width Wd(0) set by breakdown voltage. Wd(0) same for both diodes. Wd(150) = Wd(0) 1!+! 150 0.7 = (10 -5)(150) = 15 mm ; Wd(0) = 15 (150)/(0.7) ≈ 1 mm Schottky diode area ; Vdrift = 2 volts = Wd(150) q!mn!Nd!ASchottky IF Drift region doping density Nd same for both diodes. Nd = 1.3x1017 150 = 8.7x1014 cm-3 ASchottky = (1.5x10-3)(300) (2)(1.6x10-19)(1500)(8.7x1014) = 1.07 cm2 PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF) 2/3 ; Eq. (20-16) K1 = Wd(150) q!mo!Apn!nb = 1.5x10-3 (1.6x10-19)(900)(Apn)(10 17) = 10-4 Apn K2 = 3 W 4 d(150) q2!m 3 o!n 2 b!A 2 pn!to = 3 (1.5x10-3)4 (1.6x10-19)2!(900)3!(1017)2!(A 2 pn)!(2x10-6) K2 = 2.4x10 -4 (Apn) -0.67 ; 2 volts = 10-4 Apn (300) + 2.4x10-4 (Apn) -0.67 (300)0.67 Apn = 0.015 + 0.00154 (Apn) 0.333 ; Solve by successive approximation ; Apn ≈ 0.017 cm 2 Schottky diode Cjo = (11.7)(8.9x10-14)(1.07) 10-4≈ 11 nF = 0.011 mF PN junction diode Cjo = (11.7)(8.9x10-14)(0.017) 10-4 ≈ 180 pF 20-11. BVcyl BVp = 2 r2 (1 + 1 r ) ln (1 + 1 r ) - 2r • • • • • • • • • 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 1 10 BVcyl / BVp r = R/(2W n) 20-12. BVcyl BVp = 950 1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = R 2!Wn R ≈ 12 Wn PN junction diode Cjo = (11.7)(8.9x10-14)(0.017) 10-4 ≈ 180 pF 20-11. BVcyl BVp = 2 r2 (1 + 1 r ) ln (1 + 1 r ) - 2r • • • • • • • • • 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 1 10 BVcyl / BVp r = R/(2W n) 20-12. BVcyl BVp = 950 1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = R 2!Wn R ≈ 12 Wn PN junction diode Cjo = (11.7)(8.9x10-14)(0.017) 10-4 ≈ 180 pF 20-11. BVcyl BVp = 2 r2 (1 + 1 r ) ln (1 + 1 r ) - 2r • • • • • • • • • 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.1 1 10 BVcyl / BVp r = R/(2W n) 20-12. BVcyl BVp = 950 1000 = 0.95 ; From graph in problem 20-11, r ≈ 6 = R 2!Wn R ≈ 12 Wn Chapter 21 Problem Solutions 21-1. NPN BJT ; BVCEO = BVCBO b1/4 ; PNP BJT ; BVCEO = BVCBO b1/6 B * * * * * * * * * * * * * * * * * J • • • • • •••• • • • • • ••• 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 10 100 BVCEO BVCBO PNP NPN 21-2. The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b. When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur. 21-3. a) Idealized BJT current and voltage waveforms in step-down converter Chapter 21 Problem Solutions 21-1. NPN BJT ; BVCEO = BVCBO b1/4 ; PNP BJT ; BVCEO = BVCBO b1/6 B * * * * * * * * * * * * * * * * * J • • • • • •••• • • • • • ••• 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 10 100 BVCEO BVCBO PNP NPN 21-2. The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b. When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur. 21-3. a) Idealized BJT current and voltage waveforms in step-down converter Chapter 21 Problem Solutions 21-1. NPN BJT ; BVCEO = BVCBO b1/4 ; PNP BJT ; BVCEO = BVCBO b1/6 B * * * * * * * * * * * * * * * * * J • • • • • •••• • • • • • ••• 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 10 100 BVCEO BVCBO PNP NPN 21-2. The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b. When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur. 21-3. a) Idealized BJT current and voltage waveforms in step-down converter T/2 I oVd V (t) CEi (t)C t d,off t rv t fit d,on t ri t fv BJT power dissipation Pc = 1 T ıÛ 0 T !vCE(t)!iC(t)!dt = {Eon + Esw }fs ; fs = 1 T Eon = VCE,on Io { T 2 + td,off - td,on} ≈ (2)(40) 1 2!fs = 40 fs Joules Esw = Eri + Efv + Erv + Efi Eri = Vd!Io!tri 2 = (100)(40)(2x10-7) 2 = 4x10 -4 Joules Efv = Vd!Io!tfv 2 = (100)(40)(1x10-7) 2 = 2x10 -4 Joules Erv = Vd!Io!trv 2 = (100)(40)(1x10-7) 2 = 2x10 -4 Joules Efi = Vd!Io!tfi 2 = (100)(40)(2x10-7) 2 = 4x10 -4 Joules Esw = 1.2x10 -3 Joules ; Pc = [ 40 fs + 1.2x10-3] fs = 40 + 1.2x10 -3 fs 40 80 0 33 kHz fs [watts] Pc b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25 1 = 125 watts = 40 + 1.2x10 -3 fs,max fs,max = (125!-!40) 1.2x10-3 = 71 kHz 21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4 (125!-!25) = 4x10 -3 110 - 50 = Rqja {40 + 1.2x10 -3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit. 21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC = q!Dn!Na!A !Wb = (1.6x10-19)(38)(1016)(1) (3x10-4) = 200 A 21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation. 21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base- side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias. 40 80 0 33 kHz fs [watts] Pc b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25 1 = 125 watts = 40 + 1.2x10 -3 fs,max fs,max = (125!-!40) 1.2x10-3 = 71 kHz 21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4 (125!-!25) = 4x10 -3 110 - 50 = Rqja {40 + 1.2x10 -3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit. 21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC = q!Dn!Na!A !Wb = (1.6x10-19)(38)(1016)(1) (3x10-4) = 200 A 21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation. 21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base- side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias. 40 80 0 33 kHz fs [watts] Pc b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25 1 = 125 watts = 40+ 1.2x10 -3 fs,max fs,max = (125!-!40) 1.2x10-3 = 71 kHz 21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4 (125!-!25) = 4x10 -3 110 - 50 = Rqja {40 + 1.2x10 -3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit. 21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC = q!Dn!Na!A !Wb = (1.6x10-19)(38)(1016)(1) (3x10-4) = 200 A 21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation. 21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base- side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias. 40 80 0 33 kHz fs [watts] Pc b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25 1 = 125 watts = 40 + 1.2x10 -3 fs,max fs,max = (125!-!40) 1.2x10-3 = 71 kHz 21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4 (125!-!25) = 4x10 -3 110 - 50 = Rqja {40 + 1.2x10 -3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit. 21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC = q!Dn!Na!A !Wb = (1.6x10-19)(38)(1016)(1) (3x10-4) = 200 A 21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation. 21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base- side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias. 40 80 0 33 kHz fs [watts] Pc b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25 1 = 125 watts = 40 + 1.2x10 -3 fs,max fs,max = (125!-!40) 1.2x10-3 = 71 kHz 21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4 (125!-!25) = 4x10 -3 110 - 50 = Rqja {40 + 1.2x10 -3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit. 21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT. IC = q!Dn!Na!A !Wb = (1.6x10-19)(38)(1016)(1) (3x10-4) = 200 A 21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation. 21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base- side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below. Estimate of base-emitter base-side protusion = WEB,depl at zero bias. Wo,EB = 2!e!fc!(NaB!+!NdE) !q!NaB!NdE ; fcE = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!NdE n 2 i fcE = 0.26 ln Î Í È ˚ ˙ ˘ ! 1034 1020 = 0.84 V Wo,EB = (2)(11.7)(8.9x10-14)(0.84)(1019!+!1015) (1.6x10-19)(1019)(1015) = 0.33 microns Estimate of collector-base base-side protusion of WCB,depl. CB depletion layer thickness W(V) = Wo,CB 1!+! V fc = xp + xn ; xp = protrusion of CB depletion layer into p-type base region. xp = W(V) 11 using xp Na = xn Nd (charge neutrality). Wo,CB = 2!e!fcC!(NaB!+!NdC) !q!NaB!NdC ; fcC = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!NdC n 2 i fcC = 0.26 ln Î Í È ˚ ˙ ˘ ! 1029 1020 = 0.54 V Wo,CB = (2)(11.7)(8.9x10-14)(0.54)(1014!+!1015) (1.6x10-19)(1014)(1015) = 2.8 microns {(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+! V 0.54 : Solving for V yields V = 59 volts. Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V. 21-8. BVEBO = 10 V = 1.3x1017 !NaB ; NaB = acceptor doping density in base = 1.3x10 16 cm-3 BVCBO = b 1/4 BVCEO = (5) 1/4(1000) = ≈ 1500 Volts ≈ 1.3x1017 !NdC NdC = collector drift region donor density = 8.7x10 13 cn -3 Wo,EB = 2!e!fc!(NaB!+!NdE) !q!NaB!NdE ; fcE = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!NdE n 2 i fcE = 0.26 ln Î Í È ˚ ˙ ˘ ! 1034 1020 = 0.84 V Wo,EB = (2)(11.7)(8.9x10-14)(0.84)(1019!+!1015) (1.6x10-19)(1019)(1015) = 0.33 microns Estimate of collector-base base-side protusion of WCB,depl. CB depletion layer thickness W(V) = Wo,CB 1!+! V fc = xp + xn ; xp = protrusion of CB depletion layer into p-type base region. xp = W(V) 11 using xp Na = xn Nd (charge neutrality). Wo,CB = 2!e!fcC!(NaB!+!NdC) !q!NaB!NdC ; fcC = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!NdC n 2 i fcC = 0.26 ln Î Í È ˚ ˙ ˘ ! 1029 1020 = 0.54 V Wo,CB = (2)(11.7)(8.9x10-14)(0.54)(1014!+!1015) (1.6x10-19)(1014)(1015) = 2.8 microns {(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+! V 0.54 : Solving for V yields V = 59 volts. Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V. 21-8. BVEBO = 10 V = 1.3x1017 !NaB ; NaB = acceptor doping density in base = 1.3x10 16 cm-3 BVCBO = b 1/4 BVCEO = (5) 1/4(1000) = ≈ 1500 Volts ≈ 1.3x1017 !NdC NdC = collector drift region donor density = 8.7x10 13 cn -3 Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO) Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10 -5)(1500) = 150 mm xp(BVCBO) = xn(BVCBO)!NdC NaB = xn(BVCBO) 8.7x1013 1.3x1016 = xn(BVCBO) 6.7x10 -3 xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm Hence xp(BVCBO) = (1.5x10 -2 cm)(6.7x10-3) = 1 micron = Wbase 21-9. Beta = 150 = bD bM + bD + bM =20 bM + 20 + bM bM = 150!-!20 21 = 6.2 21-10. Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active. 10 I B,M B,MI 0.02 W 0.8 V 0.6 V 0.8 V - + - + - + C B E 10 IB,M + IB,M = 100 A IB,M = IC,D = 9.1A IC,M = 91 A VCE,D = 0.2 +(.02)(9.1) = 0.382 V VCE,M = VCE,D + 0.8 V = 1.18 v But a saturated main BJT with IC,M = 91 A going through 0.02 ohms generates a voltage drop of 1.8 V which is > 1.18 V. Hence main BJT must be saturated. Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO) Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10 -5)(1500) = 150 mm xp(BVCBO) = xn(BVCBO)!NdC NaB = xn(BVCBO) 8.7x1013 1.3x1016 = xn(BVCBO) 6.7x10 -3 xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm Hence xp(BVCBO) = (1.5x10 -2 cm)(6.7x10-3) = 1 micron = Wbase 21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM bM = 150!-!20 21 = 6.2 21-10. Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active. 10 I B,M B,MI 0.02 W 0.8 V 0.6 V 0.8 V - + - + - + C B E 10 IB,M + IB,M = 100 A IB,M = IC,D = 9.1A IC,M = 91 A VCE,D = 0.2 +(.02)(9.1) = 0.382 V VCE,M = VCE,D + 0.8 V = 1.18 v But a saturated main BJT with IC,M = 91 A going through 0.02 ohms generates a voltage drop of 1.8 V which is > 1.18 V. Hence main BJT must be saturated. Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO) Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10 -5)(1500) = 150 mm xp(BVCBO) = xn(BVCBO)!NdC NaB = xn(BVCBO) 8.7x1013 1.3x1016 = xn(BVCBO) 6.7x10 -3 xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm Hence xp(BVCBO) = (1.5x10 -2 cm)(6.7x10-3) = 1 micron = Wbase 21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM bM = 150!-!20 21 = 6.2 21-10. Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active. 10 I B,M B,MI 0.02 W 0.8 V 0.6 V 0.8 V - + - + - + C B E 10 IB,M + IB,M = 100 A IB,M = IC,D = 9.1A IC,M = 91 A VCE,D = 0.2 +(.02)(9.1) = 0.382 V VCE,M = VCE,D + 0.8 V = 1.18 v But a saturated main BJT with IC,M = 91 A going through 0.02 ohms generates a voltage drop of 1.8 V which is > 1.18 V. Hence main BJT must be saturated. B,MI 0.02 W 0.8 V 0.6 V 0.8 V - + - + - + C B E 0.02 W 0.6 V - + I C,M Neglecting IB,D IC,M + IC,D = 100 A (.02)IC,M -0.6 = (.02) I C,D +0.2 IC,D = 30 A ; IC,M = 70 A VCE,M = 0.2 + (70)(.02) = 1.6 V PDarl = VCE,M [IC,M + IC,D] PDarl = (1.6 V)(100 A) = 160 W 21-11. CEBO = e!AE WEBO ; CCBO = e!AC WCBO WEBO = zero-bias emitter-base depletion layer thickness WCBO = zero-bias collector-base depletion layer thickness WEBO = 2!e!fcE!(NaB!+!NdE) !q!NaB!NdE ; fcE = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!NdE n 2 i fcE = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(1016) 1020 = 0.89 V ; WEBO = (2)(11.7)(8.9x10-14)(0.89)(1019!+!1016) (1.6x10-19)(1019)(1016) = 0.34 microns CEBO = (11.7)(8.9x10-14)(0.3) 3.4x10-5 = 9.2 nF WCBO = 2!e!fcC!(NaB!+!NdC) !q!NaB!NdC ; fcC = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!!NdC n 2 i fcC = 0.026 ln Î Í È ˚ ˙ ˘ ! (1.3x1016)(8.7x1013) 1020 = 0.6 V ; B,MI 0.02 W 0.8 V 0.6 V 0.8 V - + - + - + C B E 0.02 W 0.6 V - + I C,M Neglecting IB,D IC,M + IC,D = 100 A (.02)IC,M -0.6 = (.02) I C,D +0.2 IC,D = 30 A ; IC,M = 70 A VCE,M = 0.2 + (70)(.02) = 1.6 V PDarl = VCE,M [IC,M + IC,D] PDarl = (1.6 V)(100 A) = 160 W 21-11. CEBO = e!AE WEBO ; CCBO = e!AC WCBO WEBO = zero-bias emitter-base depletion layer thickness WCBO = zero-bias collector-base depletion layer thickness WEBO = 2!e!fcE!(NaB!+!NdE) !q!NaB!NdE ; fcE = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!NdE n 2 i fcE = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(1016) 1020 = 0.89 V ; WEBO = (2)(11.7)(8.9x10-14)(0.89)(1019!+!1016) (1.6x10-19)(1019)(1016) = 0.34 microns CEBO = (11.7)(8.9x10-14)(0.3) 3.4x10-5 = 9.2 nF WCBO = 2!e!fcC!(NaB!+!NdC) !q!NaB!NdC ; fcC = k!T q ln Î Í Í È ˚ ˙ ˙ ˘ ! NaB!!NdC n 2 i fcC = 0.026 ln Î Í È ˚ ˙ ˘ ! (1.3x1016)(8.7x1013) 1020 = 0.6 V ; WCBO = (2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013) (1.6x10-19)(1.3x1016)(8.7x1013) = 2.1 microns CCBO = (11.7)(8.9x10-14)(3) 2.1x10-4 = 14.8 nF 21-12. Equivalent circuit for turn-on delay time, td,on, calculation. + - 10 W CCB CBE VBEVin + - -8 V 8 V Vin t VBE(t) = 8 - 16 exp Î ÍÈ ˚ ˙˘-t !t ; t = (10 W)(CBE + CCB) At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln Î ÍÈ ˚ ˙˘16 7.3 The space-charge capacitances are nonlinear functions of the voltages across them. Need to find an average value for each of the two capacitors. During the td,on interval, the voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. Hence CCB will be given by CCB = CCBO ! 1!+! VCB fcC = 1.5x10-8 1!+! 100 0.6 = 1.2 nF The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find the average value of CBE. WCBO = (2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013) (1.6x10-19)(1.3x1016)(8.7x1013) = 2.1 microns CCBO = (11.7)(8.9x10-14)(3) 2.1x10-4 = 14.8 nF 21-12. Equivalent circuit for turn-on delay time, td,on, calculation. + - 10 W CCB CBE VBEVin + - -8 V 8 V Vin t VBE(t) = 8 - 16 exp Î ÍÈ ˚ ˙˘-t !t ; t = (10 W)(CBE + CCB) At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln Î ÍÈ ˚ ˙˘16 7.3 The space-charge capacitances are nonlinear functions of the voltages across them. Need to find an average value for each of the two capacitors. During the td,on interval, the voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. Hence CCB will be given by CCB = CCBO ! 1!+! VCB fcC = 1.5x10-8 1!+! 100 0.6 = 1.2 nF The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find the average value of CBE. CBE = ıÙ ÙÛ 8 0 ! CEBO ! 1!+! VEB fcE !dVEB ıÛ 8 0 dVEB = CEBO!fcE (-8) Î Í È ˚ ˙ ˘ 1!+! 0 fcE!!-!! 1!+! 8 fcE CBE = (9.2x10-9)(0.89) 8 [ 1!+! 8 0.89 - 1] = 2.2 nF td,on = (10) [2.2x10 -9 + 1.2x10-9] ln ÎÍ È ˚ ˙˘16 7.3 = (3.4x10 -8)(0.78) ≈ 27 nanoseconds Chapter 22 Problem Solutions 22-1. The capacitance of the gate-source terminals can be modeled as two capacitors connected electrically in series. Cox is the capacitance of the oxide layer and is a constant independent of vGS. Cdepl(vGS) isthe capacitance of the depletion layer which increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gate- source capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), the depletion layer thickness becomes constant because the formation of the inversion layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS (any additional increase in vGS is dropped across the oxide layer). Thus both components of Cgs are constant for vGS > VGS(th). 22-2. a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we need the numerical values of the various waveform parameters. The voltage and current amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) = 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the various switching times. Chapter 22 Problem Solutions 22-1. The capacitance of the gate-source terminals can be modeled as two capacitors connected electrically in series. Cox is the capacitance of the oxide layer and is a constant independent of vGS. Cdepl(vGS) is the capacitance of the depletion layer which increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gate- source capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), the depletion layer thickness becomes constant because the formation of the inversion layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS (any additional increase in vGS is dropped across the oxide layer). Thus both components of Cgs are constant for vGS > VGS(th). 22-2. a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we need the numerical values of the various waveform parameters. The voltage and current amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) = 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the various switching times. T/2 I oVd V (t) CE i (t) C t d,off t rv t fit d,on t ri t fv -VGG 0 VGS(th) VGS,Io VGG V (t)GS td,on estimate - Use equivalent circuit of Fig. 22-12a. Governing equation is dvGS dt + vGS RG(Cgs!+!Cgd) = VGG RG(Cgs!+!Cgd) ; Boundary condition vGS(0) = - VGG Solution is vGS(t) = VGG - 2 VGG e -t/t ; t = RG(Cgs + Cgd) ; At t = td,on , vGS = VGS(th). Solving for td,on yields td,on = RG(Cgs + Cgd) ln Î Í È ˚ ˙ ˘ ! 2!VGG !VGG!-!VGS(th) td,on = (50) (1.15x10 -9) ln ÎÍ È ˚ ˙˘! (2)(15) (15!-!4) = 58 ns tri estimate - Use equivalent circuit of Fig. 22-12b. vGS(t) still given by governing equation given above in td,on estimate. Changing time origin to when vGS = VGS(th) yields; vGS(t) = VGG + [VGG - VGS(th)] e -t/t . The drain current is given by iD(t) = Cgd d(Vd!-!vGS) dt + gm[vGS(t) - VGs(th)] ; gm = 10 7!-!4 = 3.3 mhos At t = tri, iD = Io. Substituting vGS(t) into iD(t) and solving for tri yields tri = RG(Cgs + Cgd) ln Î Í È ˚ ˙ ˘ ! (VGG!-!VGS(th)){gm!+ ! Cgd RG(Cgs!+!Cgd) } gm(VGG!-!VGS(th))!-!Io tri = (50)(1.15x10 -9) ln Î Í Í È ˚ ˙ ˙ ˘ ! (15!-!4)(3.3!+! 1.5x10-10 (50)(1.15x10-9) ) (3.3)(15!-!4)!-!10 = 21 ns tfv estimate - Use equivalent circuit of Fig. 22-12c. vGS approximately constant at VGS,Io = Io gm + VGS(th) during this interval. Governing equation is Cgd dvDS dt = - Î Í È ˚ ˙ ˘ ! VGG!-!VGS(th)!-! Io gm RG with vDS(0) = Vd. Solution is given by vDS(t) = Vd - Î Í È ˚ ˙ ˘ VGG!-!VGS(th)!-! Io gm t RGCgd ; At t = tfv, vDS = 0. Solving for tfv yields tfv = RG!Cgd!Vd VGG!-!VGS(th)!-! Io gm = (50)(1.5x10-10) 300 (15!-!4!-!3) = 300 ns td,off estimate - use equivalent circuit of Fig. 22-12d with the input voltagge VGG reversed. vGS(t) = - VGG + 2 VGG e -t/t ; At t = td,off, vGS = VGS,Io. Solving for td,off td,off = RG(Cgs + Cgd) ln Î Í È ˚ ˙ ˘ ! 2!VGG VGG!+!VGS(th)!+! Io gm = (50)(1.15x10-9) ln Î Í Í È ˚ ˙ ˙ ˘ ! (2)(15) 10 3.3!+!4!+!15 = 18 ns trv estimate - Use equivalent circuit of Fig. 22-12c with the input voltage VGG reversed. vGS approximately constant at VGS,Io as in previous of tfv. Governing equation is Cgd d{vDS!-!VGS,Io} dt = VGG!+!VGS,Io RG with vDS(0) = 0. Solution given by vDS(t) = VGG!+!VGS,Io RG!Cgd t . At t = trv , vDS = Vd . Solving for trv yields trv = Vd!RG!Cgd VGG!+!VGS,Io = (300)(50)(1.5x10-10) (15!+!7) = 100 ns tfi estimate - use equivalent circuit of Fig. 22-12b with the input voltage VGG reversed. Governing equation the same as in previous calculation of tri. At t = 0, vGS(0) = VGS,Io. Solution in this caae is given by vGS(t) = - VGG + [VGS,Io + VGG] e -t/t ; At t = tfi, vGS = VGS(th). Solving for tfi tfi = RG(Cgs + Cgd) ln Î Í È ˚ ˙ ˘ ! VGG!+!VGS,Io VGG+!VGS(th) = (50)(1.15x10-9)ln ÎÍ È ˚ ˙˘! 15!+!7 15!+!4 = 9 ns b) Estimate the power dissipated in the MOSFET in the same manner as was done for the BJT in problem 21-3.Waveforms for the MOSFET are the same as for the BJT except for appropriate re-labeling of the currents and voltages. Eri = (0.5)(300)(10)(2.1x10 -8) = 3x10-5 Joules Efv = (0.5)(300)(10)(3x10 -7) = 4.5x10-4 Joules Eon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 V T >> td,on and td,off Eon = (10)(300)(0.5)(5x10 -5) = 1.25x10-3 Joules Erv = (0.5)(300)(10)(10 -7) = 1.5x10-4 Joules Efi = (0.5)(300)(10)(9x10 -9) = 1.5x10-5 Joules Pc = (1.95x10 -3)(2x104) = 39 watts 22-3. Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below. Eri = (0.5)(300)(10)(2.1x10 -8) = 3x10-5 Joules Efv = (0.5)(300)(10)(3x10 -7) = 4.5x10-4 Joules Eon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 V T >> td,on and td,off Eon = (10)(300)(0.5)(5x10 -5) = 1.25x10-3 Joules Erv = (0.5)(300)(10)(10 -7) = 1.5x10-4 Joules Efi = (0.5)(300)(10)(9x10 -9) = 1.5x10-5 Joules Pc = (1.95x10 -3)(2x104) = 39 watts 22-3. Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below. V (t)G VGG V (t)GS VGS(th) VGS,Io t t d,on t = t r i fv t = tf i rv td,off t VGG Vd V (t)DS I o i (t) D t Equivalent circuit during voltage and current rise and fall intervals: + - Cgs Cgd V (t)G R D V d g (V - V ) m GS GS(th) RG Governing equation using Miller capacitance approximation: dvGS dt + vGS t = !VG(t) t ; t = RG [Cgs + Cgd{1 + gmRD}] ; During tri = tfv interval, VG(t) = VGG. Solution is vGS(t) = VGG + {VGS(th) - VGG} e -t/t ; At t = tri, VGS = VGS(th) + Vd gmRD ; Solving for tri = tfv yields tri = tfv = t ln Î Í È ˚ ˙ ˘ ! VGG!-!VGS(th) VGG!-!VGS(th)!-!! Vd !gm!RD During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e -t/t . At t = trv, vGS(t) = VGS(th). Solving for trv yields trv = t ln Î Í È ˚ ˙ ˘ ! VGS(th)!+! Vd !gm!RD VGS(th) . Invert equation for tri to find Cgd. Result is Cgd = ÓÔ Ì ÔÏ Ô˛ ˝ Ô¸ tri !ln! Î Í È ˚ ˙ ˘ ! VGG!-!VGS(th) VGG!-!VGS(th)!-!! !Vd !gm!RD !!-!!RG!Cgs ÓÌ Ï ˛ ˝ ¸1 RG(1!+!gmRD) Cgd= ÓÔ Ì ÔÏ Ô˛ ˝ Ô¸3x10-8 !ln!ÎÍ È ˚ ˙˘! 15!-!4 15!-!4!-!1 !!-!!5x10-9 Ó Ì Ï ˛ ˝ ¸1 5(1!+!25) = 2.3x10 -9 F = 2.3 nF Solving for switching times in circuit with RD = 150 ohms. t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms tri = 3.5x10 -5 ln ÎÍ È ˚ ˙˘! 15!-!4 15!-!4!-!2 = 7 ms ; trv = 3.5x10 -5 ln ÎÍ È ˚ ˙˘! 4!+!2 4 = 14 ms 22-4. Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in MOSFET given by <PMOSFET> = [Eon + Esw] fs ; fs = 1 T ; Eon = [ID] 2 rDS,on(Tj) T 2 ; dvGS dt + vGS t = !VG(t) t ; t = RG [Cgs + Cgd{1 + gmRD}] ; During tri = tfv interval, VG(t) = VGG. Solution is vGS(t) = VGG + {VGS(th) - VGG} e -t/t ; At t = tri, VGS = VGS(th) + Vd gmRD ; Solving for tri = tfv yields tri = tfv = t ln Î Í È ˚ ˙ ˘ ! VGG!-!VGS(th) VGG!-!VGS(th)!-!! Vd !gm!RD During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e -t/t . At t = trv, vGS(t) = VGS(th). Solving for trv yields trv = t ln Î Í È ˚ ˙ ˘ ! VGS(th)!+! Vd !gm!RD VGS(th) . Invert equation for tri to find Cgd. Result is Cgd = ÓÔ Ì ÔÏ Ô˛ ˝ Ô¸ tri !ln! Î Í È ˚ ˙ ˘ ! VGG!-!VGS(th) VGG!-!VGS(th)!-!! !Vd !gm!RD !!-!!RG!Cgs ÓÌ Ï ˛ ˝ ¸1 RG(1!+!gmRD) Cgd = ÓÔ Ì ÔÏ Ô˛ ˝ Ô¸3x10-8 !ln!ÎÍ È ˚ ˙˘! 15!-!4 15!-!4!-!1 !!-!!5x10-9 Ó Ì Ï ˛ ˝ ¸1 5(1!+!25) = 2.3x10 -9 F = 2.3 nF Solving for switching times in circuit with RD = 150 ohms. t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms tri = 3.5x10 -5 ln ÎÍ È ˚ ˙˘! 15!-!4 15!-!4!-!2 = 7 ms ; trv = 3.5x10 -5 ln ÎÍ È ˚ ˙˘! 4!+!2 4 = 14 ms 22-4. Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in MOSFET given by <PMOSFET> = [Eon + Esw] fs ; fs = 1 T ; Eon = [ID] 2 rDS,on(Tj) T 2 ; ID = Vd RD = 300 150 = 2 A ; rDS,on(Tj) = 2 Î ÍÈ ˚ ˙˘1!+! Tj!-!25 150 = 2 Î ÍÈ ˚ ˙˘0.833!+! Tj 150 Eon = (4)(2) Î ÍÈ ˚ ˙˘0.833!+! Tj 150 1 2fs = {3.32 + 0.027 Tj} 1 fs Esw = 1 T ı ÙÛ 0 tri Vd!ID(1!-! t tri )( t tri )dt + 1 T ı ÙÛ 0 tfi Vd!ID(1!-! t tfi )( t tfi )dt = Vd!ID 6 [tri + tfi] Esw = (300)(2) 6 [7x10 -6 + 14x10-6] = 2.1x10-3 joules <PMOSFET> = {3.32 + 0.027 Tj} 1 fs fs + 2.1x10 -3 fs <PMOSFET> = {3.32 + 0.027 Tj} +[ 2.1x10 -3][104] = 24.3 + 0.027 Tj B B B B B 0 5 10 15 20 25 30 0 10 20 30 40 50 60 70 80 90 100 PMOSFET Watts Temperature [ °K] 22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff ID = Vd RD = 300 150 = 2 A ; rDS,on(Tj) = 2 Î ÍÈ ˚ ˙˘1!+! Tj!-!25 150 = 2 Î ÍÈ ˚ ˙˘0.833!+! Tj 150 Eon = (4)(2) Î ÍÈ ˚ ˙˘0.833!+! Tj 150 1 2fs = {3.32 + 0.027 Tj} 1 fs Esw = 1 T ı ÙÛ 0 tri Vd!ID(1!-! t tri )( t tri )dt + 1 T ı ÙÛ 0 tfi Vd!ID(1!-! t tfi )( t tfi )dt = Vd!ID 6 [tri + tfi] Esw = (300)(2) 6 [7x10 -6 + 14x10-6] = 2.1x10-3 joules <PMOSFET> = {3.32 + 0.027 Tj} 1 fs fs + 2.1x10 -3 fs <PMOSFET> = {3.32 + 0.027 Tj} +[ 2.1x10 -3][104] = 24.3 + 0.027 Tj B B B B B 0 5 10 15 20 25 30 0 10 20 30 40 50 60 70 80 90 100 PMOSFET Watts Temperature [ °K] 22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff reff = r1!r2!r3 r1!r2!+!r2!r3!+!r3!r1 ; r1 etc. = on-state resistance of MOSFET #1 etc. r1(Tj) = r1(25 °C) Î ÍÈ ˚ ˙˘1!+!0.8!! Tj!-!25 100 ; r1(105 °C) = (1.64) r1(25 °C) etc. r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W reff(105 °C) = (2.95)(3.28)(3.61) [(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms For the ith MOSFET, Pi = Von 2 2!ri = Io 2!reff 2 2!ri ; Assume a 50% duty cycle and ignore switching losses. P1 = (5)2(1.09)2 (2)(2.95) = 5 W ; P2 = (5)2(1.09)2 (2)(3.28) = 4.5 W ; P3 = (5)2(1.09)2 (2)(3.61) = 4.1 W 22-6. Hybrid switch would combine the low on-state losses of the BJT and the faster switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the on-state. reff = r1!r2!r3 r1!r2!+!r2!r3!+!r3!r1 ; r1 etc. = on-state resistance of MOSFET #1 etc. r1(Tj) = r1(25 °C) Î ÍÈ ˚ ˙˘1!+!0.8!! Tj!-!25 100 ; r1(105 °C) = (1.64) r1(25 °C) etc. r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W reff(105 °C) = (2.95)(3.28)(3.61) [(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms For the ith MOSFET, Pi = Von 2 2!ri = Io 2!reff 2 2!ri ; Assume a 50% duty cycle and ignore switching losses. P1 = (5)2(1.09)2 (2)(2.95) = 5 W ; P2 = (5)2(1.09)2 (2)(3.28) = 4.5 W ; P3 = (5)2(1.09)2 (2)(3.61) = 4.1 W 22-6. Hybrid switch would combine the low on-state losses of the BJT and the faster switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the on-state. v = vDS CE i D i C vGS vBE VDS,on VCE,on I o r I ; r < 1o (1 - r)I o 22-7. BVDSS ≈ 1.3x1017 Ndrift = 750 volts ; Ndrift = 1.7x10 14 cm-3 Wdrift ≈ (10 -5)(750) = 75 microns ; Wd,body = protrusion of drain depletion layer into body region ≈ Wdrift!Ndrift Nbody = (75)(1.7x1014) 5x1016 ≈ 0.3 microns Even though body-source junction is shorted, there is a depletion layer associated with it which is contained entirely on the body side of the junction. This must be included in the estimate of the required length of the body region. Ws,body ≈ 2!e!fc q!Na,body ; fc = k!T q ln ÎÍ Í È ˚ ˙ ˙ ˘ ! Na!Nd ni 2 ; v = vDS CE i D i C vGS vBE VDS,on VCE,on I o r I ; r < 1o (1 - r)I o 22-7. BVDSS ≈ 1.3x1017 Ndrift = 750 volts ; Ndrift = 1.7x10 14 cm-3 Wdrift ≈ (10 -5)(750) = 75 microns ; Wd,body = protrusion of drain depletion layer into body region ≈ Wdrift!Ndrift Nbody = (75)(1.7x1014) 5x1016 ≈ 0.3 microns Even though body-source junction is shorted, there is a depletion layer associated with it which is contained entirely on the body side of the junction. This must be included in the estimate of the required length of the body region. Ws,body ≈ 2!e!fc q!Na,body ; fc = k!T q ln ÎÍ Í È ˚ ˙ ˙ ˘ ! Na!Nd ni 2 ; fc = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(5x1016) (1020) = 0.94 Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94) (1.6x10-19)(5x1016) ≈ 0.16 microns In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = Cgd dvGD dt ≈ Cgd dvDS dt ; vDS ≈ vGD>> vGS BJT will turn on if Rbody Cgd dvDS dt = 0.7 V dvDS dt > 0.7 Rbody!Cgd will turn on the BJT. 22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10 6) (5x10-6) = 16.7 volts 22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L Cox = e tox = (11.7)(8.9x10-14) 10-5 = 1.04x10-7 F/cm2 N = 2!iD!Lmn!Cox!Wcell!(vGS!-!VGS(th)) N = !(2)(100)(10-4) (1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells b) Icell = 100 5800 = 17 milliamps 22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms Ron = Wdrift q!mn!Nd!A : Wdrift = 10 -5 BVDSS = (10 -5)(800) = 80 microns fc = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(5x1016) (1020) = 0.94 Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94) (1.6x10-19)(5x1016) ≈ 0.16 microns In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = Cgd dvGD dt ≈ Cgd dvDS dt ; vDS ≈ vGD>> vGS BJT will turn on if Rbody Cgd dvDS dt = 0.7 V dvDS dt > 0.7 Rbody!Cgd will turn on the BJT. 22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10 6) (5x10-6) = 16.7 volts 22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L Cox = e tox = (11.7)(8.9x10-14) 10-5 = 1.04x10-7 F/cm2 N = 2!iD!L mn!Cox!Wcell!(vGS!-!VGS(th)) N = !(2)(100)(10-4) (1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells b) Icell = 100 5800 = 17 milliamps 22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms Ron = Wdrift q!mn!Nd!A : Wdrift = 10 -5 BVDSS = (10 -5)(800) = 80 microns fc = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(5x1016) (1020) = 0.94 Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94) (1.6x10-19)(5x1016) ≈ 0.16 microns In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = Cgd dvGD dt ≈ Cgd dvDS dt ; vDS ≈ vGD>> vGS BJT will turn on if Rbody Cgd dvDS dt = 0.7 V dvDS dt > 0.7 Rbody!Cgd will turn on the BJT. 22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10 6) (5x10-6) = 16.7 volts 22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L Cox = e tox = (11.7)(8.9x10-14) 10-5 = 1.04x10-7 F/cm2 N = 2!iD!L mn!Cox!Wcell!(vGS!-!VGS(th)) N = !(2)(100)(10-4) (1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells b) Icell = 100 5800 = 17 milliamps 22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms Ron = Wdrift q!mn!Nd!A : Wdrift = 10 -5 BVDSS = (10 -5)(800) = 80 microns fc = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(5x1016) (1020) = 0.94 Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94) (1.6x10-19)(5x1016) ≈ 0.16 microns In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = Cgd dvGD dt ≈ Cgd dvDS dt ; vDS ≈ vGD>> vGS BJT will turn on if Rbody Cgd dvDS dt = 0.7 V dvDS dt > 0.7 Rbody!Cgd will turn on the BJT. 22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10 6) (5x10-6) = 16.7 volts 22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L Cox = e tox = (11.7)(8.9x10-14) 10-5 = 1.04x10-7 F/cm2 N = 2!iD!L mn!Cox!Wcell!(vGS!-!VGS(th)) N = !(2)(100)(10-4) (1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells b) Icell = 100 5800 = 17 milliamps 22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms Ron = Wdrift q!mn!Nd!A : Wdrift = 10 -5 BVDSS = (10 -5)(800) = 80 microns fc = 0.026 ln Î Í È ˚ ˙ ˘ ! (1019)(5x1016) (1020) = 0.94 Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94) (1.6x10-19)(5x1016) ≈ 0.16 microns In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns 22-8. Displacement current = Cgd dvGD dt ≈ Cgd dvDS dt ; vDS ≈ vGD>> vGS BJT will turn on if Rbody Cgd dvDS dt = 0.7 V dvDS dt > 0.7 Rbody!Cgd will turn on the BJT. 22-9. VGS,max = (0.67) EBD tox = (0.67) (5x10 6) (5x10-6) = 16.7 volts 22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th)) 2!L Cox = e tox = (11.7)(8.9x10-14) 10-5 = 1.04x10-7 F/cm2 N = 2!iD!L mn!Cox!Wcell!(vGS!-!VGS(th)) N = !(2)(100)(10-4) (1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells b) Icell = 100 5800 = 17 milliamps 22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms Ron = Wdrift q!mn!Nd!A : Wdrift = 10 -5 BVDSS = (10 -5)(800) = 80 microns Nd = 1.3x1017 BVDSS = 1.3x1017 800 ≈ 1.6x10 14 cm-3 A = 8x10-3 (1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2 10!A 0.5!cm2 = 20 A cm2 << the allowable maximum of 200 A cm2 , so estimate is alright. 22-12. Cgs ≈ Cox N Wcell L = (1.04x10 -7)(5.8x103)(2x10-3)(10-4) = 121 pF 22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first. VDS(turn-off) = Vd + L di dt = 100 + (10 -7) ÎÍ È ˚ ˙˘! 100 5x10-8 = 300 V > BVDSS = 150 V Check for excessive power dissipation. Pallowed = Tj,max!-!Ta Rq,j-a = 150!-!50 1 = 100 watts ; Pdissipated = [Eon + Esw] fs Eonfs = Io 2!rDS(on) 2 = (100)2(0.01) 2 = 50 watts Esw = Vd!Io 2 [tri + tfi + trv +tfv] = (100)(100) 2 [(2)(5x10 -8) + (2)(2x10-7)] Esw = 2.5x10 -3 joules ; Eswfs = (2.5x10 -3)(3x104) = 75 watts Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have: Nd = 1.3x1017 BVDSS = 1.3x1017 800 ≈ 1.6x10 14 cm-3 A = 8x10-3 (1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2 10!A 0.5!cm2 = 20 A cm2 << the allowable maximum of 200 A cm2 , so estimate is alright. 22-12. Cgs ≈ Cox N Wcell L = (1.04x10 -7)(5.8x103)(2x10-3)(10-4) = 121 pF 22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first. VDS(turn-off) = Vd + L di dt = 100 + (10 -7) ÎÍ È ˚ ˙˘! 100 5x10-8 = 300 V > BVDSS = 150 V Check for excessive power dissipation. Pallowed = Tj,max!-!Ta Rq,j-a = 150!-!50 1 = 100 watts ; Pdissipated = [Eon + Esw] fs Eonfs = Io 2!rDS(on) 2 = (100)2(0.01) 2 = 50 watts Esw = Vd!Io 2 [tri + tfi + trv +tfv] = (100)(100) 2 [(2)(5x10 -8) + (2)(2x10-7)] Esw = 2.5x10 -3 joules ; Eswfs = (2.5x10 -3)(3x104) = 75 watts Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts MOSFET overstressed by both overvoltages and excessive power dissipation. 22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have: Nd = 1.3x1017 BVDSS = 1.3x1017 800 ≈ 1.6x10 14 cm-3 A = 8x10-3 (1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2 10!A 0.5!cm2 = 20 A cm2 << the allowable maximum of 200 A cm2 , so estimate is alright. 22-12. Cgs ≈ Cox N Wcell L = (1.04x10 -7)(5.8x103)(2x10-3)(10-4) = 121 pF 22-13. Two overstress possibilities,
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