Giải bài tập đtcs mohamed power electric

Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses RG > 15 - = 30 ohms -10 (4x10 )(5x108) Estimate of RG at MOSFET turn-off: 100 VGG- + + 20 5x108 V/sec > (RG)(4x10-10) Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimize turn-off times 15 + = 115 ohms RG > (4x10-10)(5x108) Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG= 15 V and RG > 115 ohms 28-2 a Circuit diagram shown below + R G1 D - o = 200 A Tsw V = 1000 V d R I f G2 Qs RG2 1000 When FCT is off we need VKG = (1.25) 40 = 31.25 V = (1000) R G1 + RG2 Now R G1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2 RG2 = 31.25 kW and RG1 = 969 kW Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses RG > 15 - = 30 ohms -10 (4x10 )(5x108) Estimate of RG at MOSFET turn-off: 100 VGG- + + 20 5x108 V/sec > (RG)(4x10-10) Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimize turn-off times 15 + = 115 ohms RG > (4x10-10)(5x108) Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG= 15 V and RG > 115 ohms 28-2 a Circuit diagram shown below + R G1 D R o = 200 A Tsw V = 1000 V d - I f G2 Qs RG2 1000 When FCT is off we need VKG = (1.25) 40 = 31.25 V = (1000) R G1 + RG2 Now R G1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2 RG2 = 31.25 kW and RG1 = 969 kW b MOSFET characteristics - large on-state current capability - low Ron - low BVDSS (BVDSS of 50-100 V should work) Chapter 29 Problem Solutions 29-1 Assume Ts = 120 °C and T a = 20 °C 0.12 ; A in m2 ; Eq (29-18) Rq,rad ≈ A Èd ˘ Í vert˙ 1/4 Rq,conv ≈ (1.34)(A) Ỵ DT ˚ ; Eq (29-20) Rq,conv ≈ 0.24 1/4 for DT = 100 °C A [dvert] A cube having a side of length dvert has a surface area A = [dvert]2 Rq,conv ≈ 0.4 [dvert]1.75 0.02 ; Rq,rad ≈ [dvert]2 Rq,conv Rq,rad Rq,sa = net surface-to-ambient thermal resistance = R q,conv + Rq,rad Rq,sa = 0.04 [dvert]1.75 + [dvert]2 Heat Sink # Volume [m]3 7.6x10-5 10-4 1.8x10-4 2x10-4 3x10-4 dv = (vol.)1/3 [m] 0.042 0.046 0.057 0.058 0.067 A = [dv]2 [m]2 0.011 0.013 0.019 0.002 0.027 dv1.75 0.004 0.046 0.0066 0.0069 0.0088 dv2 0.0018 0.0021 0.0032 0.0034 0.0045 Rq,sa [°C/W] 5.3 4.5 3.1 2.9 2.3 Rq,sa (measured) 3.2 2.3 2.2 2.1 1.7 Heat Sink # 10 11 12 Volume [m]3 4.4x10-4 6.810-4 6.1x10-4 6.3x10-4 7x10-4 1.4x10-3 dv = (vol.)1/3 [m] 0.076 0.088 0.085 0.086 0.088 0.11 A = [dv]2 [m]2 0.034 0.046 0.043 0.044 0.047 0.072 dv1.75 0.011 0.014 0.013 0.014 0.014 0.021 dv2 0.0058 0.0078 0.0071 0.0073 0.0078 0.012 Rq,sa [°C/W] 1.8 1.4 1.5 1.4 1.3 0.9 Rq,sa (measured) 1.3 1.3 1.25 1.2 0.8 0.65 Heat sink #9 is relatively large and cubical in shape with only a few cooling fins Heat sink #9 is small and flat with much more surface area compared to its volume Large surface-to-volume ratios give smaller values of Rq,sa 29-2 Rq,conv ≈ 0.24 1/4 for DT = 100 °C ; From problem 29-1 A [dvert] Rq,conv ≈ 24 [d vert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2 dvert Rq,conv cm 76 °C/W cm 113 °C/W 12 cm 141 °C/W 20 cm 160 °C/W Volume [m]3 4.4x10-4 6.810-4 6.1x10-4 6.3x10-4 7x10-4 1.4x10-3 dv = (vol.)1/3 [m] 0.076 0.088 0.085 0.086 0.088 0.11 A = [dv]2 [m]2 0.034 0.046 0.043 0.044 0.047 0.072 dv1.75 0.011 0.014 0.013 0.014 0.014 0.021 dv2 0.0058 0.0078 0.0071 0.0073 0.0078 0.012 Rq,sa [°C/W] 1.8 1.4 1.5 1.4 1.3 0.9 Rq,sa (measured) 1.3 1.3 1.25 1.2 0.8 0.65 Heat sink #9 is relatively large and cubical in shape with only a few cooling fins Heat sink #9 is small and flat with much more surface area compared to its volume Large surface-to-volume ratios give smaller values of Rq,sa 29-2 Rq,conv ≈ 0.24 1/4 for DT = 100 °C ; From problem 29-1 A [dvert] Rq,conv ≈ 24 [d vert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2 dvert Rq,conv cm 76 °C/W cm 113 °C/W 12 cm 141 °C/W 20 cm 160 °C/W B 160 B 140 120 R q ,conv °C/W B 100 80 B 60 40 20 0 10 12 14 16 18 20 dvert [cm] 29-3 Rq,conv ≈ (1.34)(A) Èd ˘ Í vert˙ 1/4 ; Eq (29-20) Ỵ DT ˚ Rq,conv ≈ 353 [DT]-.25 A = 10 cm2 and dvert = cm DT Rq,conv 60 °C 127 °C/W 80 °C 118 °C/W 100 °C 112 °C/W 120 °C 107 °C/W B 160 B 140 120 R q ,conv °C/W B 100 80 B 60 40 20 0 10 12 14 16 18 20 dvert [cm] 29-3 Rq,conv ≈ (1.34)(A) Èd ˘ Í vert˙ 1/4 ; Eq (29-20) Ỵ DT ˚ Rq,conv ≈ 353 [DT]-.25 A = 10 cm2 and dvert = cm DT Rq,conv 60 °C 127 °C/W 80 °C 118 °C/W 100 °C 112 °C/W 120 °C 107 °C/W 140 120 B B B B 100 R q ,conv °C/W 80 60 40 20 60 70 80 90 100 110 120 DT [°C] 29-4 Rq,rad = DT Ê Ê Ts ˆ4 Ê Ta ˆ4 ˆ ÁÁ ˜ Á ˜ ˜ 5.1 A Ë Ë100¯ - Ë100¯ ¯ 120 - Ta(° C) Rq,rad = 196 È ÈTa(° K)˘4 ˘˙ Í Í ˙ Ỵ239 - Ỵ 100 ˚ ˚ ; Eq (29-17) ; A = 10 cm2 and Ts = 120 °C Ta Rq,rad °C 128 °C/W 10 °C 123 °C/W 20 °C 119 °C/W 40 °C 110 °C/W 140 120 B B B B 100 R q ,conv °C/W 80 60 40 20 60 70 80 90 100 110 120 DT [°C] 29-4 Rq,rad = DT Ê Ê Ts ˆ4 Ê Ta ˆ4 ˆ ÁÁ ˜ Á ˜ ˜ 5.1 A Ë Ë100¯ - Ë100¯ ¯ 120 - Ta(° C) Rq,rad = 196 È ÈTa(° K)˘4 ˘˙ Í Í ˙ Ỵ239 - Ỵ 100 ˚ ˚ ; Eq (29-17) ; A = 10 cm2 and Ts = 120 °C Ta Rq,rad °C 128 °C/W 10 °C 123 °C/W 20 °C 119 °C/W 40 °C 110 °C/W 140 120 R B B B B 100 q ,rad 80 [ °C / W ] 60 40 20 0 10 15 20 25 30 35 T [°C ] a 29-5 Rq,rad = DT Ê Ê Ts ˆ4 Ê Ta ˆ4 ˆ ÁÁ ˜ Á ˜ ˜ 5.1 A Ë Ë100¯ - Ë100¯ ¯ Ts(° C) - 40 Rq,rad = 196 ÈÈT (° K)˘4 ˘˙ ÍÍ s ˙ ỴỴ 100 ˚ - 96˚ ; Eq (29-17) ; A = 10 cm2 and Ta= 40 °C Ts Rq,rad 80 °C 114 °C/W 100 °C 120 °C/W 120 °C 110 °C/W 140 °C 101 °C/W 40 140 120 R B B B B 100 q ,rad 80 [ °C / W ] 60 40 20 0 10 15 20 25 30 35 T [°C ] a 29-5 Rq,rad = DT Ê Ê Ts ˆ4 Ê Ta ˆ4 ˆ ÁÁ ˜ Á ˜ ˜ 5.1 A Ë Ë100¯ - Ë100¯ ¯ Ts(° C) - 40 Rq,rad = 196 ÈÈT (° K)˘4 ˘˙ ÍÍ s ˙ ỴỴ 100 ˚ - 96˚ ; Eq (29-17) ; A = 10 cm2 and Ta= 40 °C Ts Rq,rad 80 °C 114 °C/W 100 °C 120 °C/W 120 °C 110 °C/W 140 °C 101 °C/W 40 120 B B B B 100 R q,rad 80 [ °C / W ] 60 40 20 80 90 100 T 29-6 PMOSFET,max = 150 ° C - 50 ° C ° C/W 110 s 120 = 100 W ; 100 W = 50 + 10-3 fs PMOSFET = 50 + 10-3 • 2.5x104 = 75 W Rq,ja = Rq,jc + Rq,ca = 150 ° C - 35 ° C 75 W Rq,ca = 1.53 - 1.00 = 0.53 °C/W 140 [ °C ] Solving for fs yields fs = 50 kHz 29-7 130 = 1.53 °C/W 120 B B B B 100 R q,rad 80 [ °C / W ] 60 40 20 80 90 100 T 29-6 PMOSFET,max = 150 ° C - 50 ° C ° C/W 110 s 120 = 100 W ; 100 W = 50 + 10-3 fs PMOSFET = 50 + 10-3 • 2.5x104 = 75 W Rq,ja = Rq,jc + Rq,ca = 150 ° C - 35 ° C 75 W Rq,ca = 1.53 - 1.00 = 0.53 °C/W 140 [ °C ] Solving for fs yields fs = 50 kHz 29-7 130 = 1.53 °C/W 120 B B B B 100 R q,rad 80 [ °C / W ] 60 40 20 80 90 100 T 29-6 PMOSFET,max = 150 ° C - 50 ° C ° C/W 110 s 120 = 100 W ; 100 W = 50 + 10-3 fs PMOSFET = 50 + 10-3 • 2.5x104 = 75 W Rq,ja = Rq,jc + Rq,ca = 150 ° C - 35 ° C 75 W Rq,ca = 1.53 - 1.00 = 0.53 °C/W 140 [ °C ] Solving for fs yields fs = 50 kHz 29-7 130 = 1.53 °C/W