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Power Electronic Devices

Power Supply Systems. Electrical Energy Conversion and Power Systems . Universidad de Oviedo. Power Electronic Devices. Semester 1 . Lecturer: Javier Sebastián. Outline. Review of the physical principles of operation of semiconductor devices.

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Power Electronic Devices

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  1. Power Supply Systems Electrical Energy Conversion and Power Systems Universidad de Oviedo Power Electronic Devices Semester 1 Lecturer: Javier Sebastián

  2. Outline • Review of the physical principles of operation of semiconductor devices. • Thermal management in power semiconductor devices. • Power diodes. • Power MOSFETs. • The IGBT. • High-power, low-frequency semiconductor devices (thyristors).

  3. Electrical Energy Conversion and Power Systems Universidad de Oviedo Lesson 2 - Thermal management in power semiconductor devices Semester 1 - Power Electronics Devices

  4. Outline • Basic concepts and calculation about thermal management in power semiconductor devices. • The main topics to be addressed in this lesson are the following: • Introduction. • Thermal resistances and electric equivalent circuits. • Heat sinks for power electronic devices. • Thermal calculations in transient operation.

  5. Main ideas about heat transfer • Heat transfer occurs through three mechanisms: • Radiation. • Conduction. • Convection. Only significant for space applications. Significant for general power applications. • Conduction: the heat is transferred by the vibratory motion of atoms or molecules. • Convection: the heat is transferred by mass movement of a fluid. It could be natural convection (without a fan) or forced convection (with a fan). • In both cases, the heat transfer process can be approached using equivalent electric circuits. • However, a detailed study of the heat transfer mechanisms is more complex and it is beyond the scope of this course.

  6. Simple static thermal model (I) 2 1 T1 T2 A • Simple model for heat conduction: • Q12_cond = (T1 – T2)/Rth12_cond , where: • Q12_cond = rate of heat energy transferred from 1 to 2 due to conduction. • T1 = temperature at 1. T2 = temperature at 2. • Rth12_cond = thermal resistance between 1 and 2 due to conduction. Q12_cond Q12_conv • Simple model for convection: • Q12_conv = h(T1,T2,n)·A·(T1 – T2) , where: • Q12_conv = rate of heat energy transferred from 1 to 2 due to convection. • h = film coefficient of heat transfer. • n = fluid velocity. • A = cross-sectional area.

  7. Simple static thermal model (II) Q12_cond = (T1 – T2)/Rth12_cond Q12_conv = h(T1,T2,n)·A·(T1 – T2) 2 1 T1 T2 • Over the temperature ranges of interest (from -40o C +100 o C) h is fairly constant for a given value of n. Therefore: • Q12_conv = h(n)·A·(T1 – T2) = (T1 – T2)/Rth12_conv(n),where: Rth12_conv (n) = 1/[h(n)·A] • The total rate of heat energy transferred from 1 to 2 will be: • Q12 =Q12_cond + Q12_convand therefore: • Q12 =(T1 – T2)/Rth12(n) where: • Rth12(n) = 1/[1/Rth12_cond + 1/Rth12_conv(n)] is the thermal resistance between 1 and 2. Q12_cond Q12_conv Q12

  8. Simple static thermal model (III) Q12 =(T1 – T2)/Rth12(n) T1 T2 • We will use the notation “Rth12” (or “Rq12”) for the thermal resistance in natural and forced convection. Its value will depend on the fluid (air) velocity. • We can re-write the above mentioned equation replacing the rate of heat energy transferred from 1 to 2 with the power transferred from 1 to 2: • P12 =(T1 – T2)/Rth12 • Now, we can establish a direct relationship between this equation and Ohm’s law: • P12 =(T1 – T2)/Rth12 i12 =(V1 – V2)/R12 V1 V2 1 2 Q12 1 2 R12 i12 • RthÞ R • ΔT Þ ΔV • P Þ i Thermal world Electric world

  9. Simple static thermal model (IV) • Electric resistance: R = rl·/A,where: • r= electric resistivity. • l = length. • A = cross-sectional area. • Thermal resistance: Rth = rthl·/A ,where: • rth= thermal resistivity. A A T1 V1 T2 V2 r 1 1 2 2 Q12 i12 rth l l

  10. Thermal model for a power semiconductor device (I) • Typical mechanical structure used for mounting a power semiconductor device • Interface • (e.g., mica) • Bonding wire • Heat sink • Semiconductor junction • Case Si Interface Header (Cu, Al or Ti)

  11. Thermal model for a power semiconductor device (II) • Example: an electronic device in TO-3 package over an HS02 heat sink • Semiconductor package (Case) TO-3 • Heat sink • Insulating mica for TO-3 • (Interface) HS02

  12. Thermal model for a power semiconductor device (III) • Examples of final assembly of electronic devices in TO-3 package over heat sinks

  13. Thermal model for a power semiconductor device (IV) • Thermal resistances without a heat sink Junction (J) Ambient (A) Case (C) Basic equations: TC = TA + RthCA·P TJ = TC + RthJC·P Therefore: TJ = TA + (RthJC + RthCA)·P RthCA Printing Circuit Board (PCB) RthJC RthCA RthJC TJ TC TA J A C P TA 0 oC

  14. Thermal model for a power semiconductor device (V) • Thermal resistances with a heat sink (I) Spacer (Mica plate) Ambient (A) RthCA Junction (J) RthHA Case (C) RthCH Heat sink (H) RthJC RthJC RthCA TA TC TJ C A J H RthCH TA RthHA P 0 oC

  15. Thermal model for a power semiconductor device (VI) • Thermal resistances with a heat sink (II) RthCA RthJC Basic equations: TC = TA + [RthCA·(RthHA + RthCH)/(RthCA+ RthHA + RthCH)]·P TJ = TC + RthJC·P Therefore: TJ = TA + [RthJC + RthCA·(RthHA + RthCH)/(RthCA+ RthHA + RthCH)]·P However, many times RthCA >> (RthHA + RthCH),and therefore: TJ» TA + (RthJC + RthHA + RthCH)·P TA TC TJ C A J RthHA RthCH H TA 0 oC P

  16. Thermal model for a power semiconductor device (VII) • Thermal resistances with a heat sink (III) TH TJ TC TA C H A J RthHA RthCH RthJC P Basic equations: TH = TA + RthHA·P TC = TH + RthCH·P TJ = TC + RthJC·P Therefore: TJ= TA + (RthJC + RthCH + RthHA)·P • Main issue in thermal management: • The junction temperature must be below the limit specified by the manufacturer. • For power silicon devices, this limit is about 150-200 oC. TA 0 oC

  17. Thermal resistance junction to case, RthJC (I) • Its value depends on the device. • Examples corresponding to different devices in TO-3: • MJ15003, NPN low-frequency power transistor for audio applications • LM350, adjustable voltage regulator • 2N3055, NPN low-frequency power transistor for audio applications

  18. IF(AV) = 5A, VRRM = 1200V Thermal resistance junction to case, RthJC (II) • The same device has different value of RthJC for different packages. • Examples corresponding to two devices:

  19. Thermal resistance case to ambient, RthCA • Its value depends on the case. • Manufacturers give information about RthJA = RthJC+ RthCA. • Therefore, RthCA = RthJA - RthJC. • This thermal resistance is important only for relative low-power devices. • IRF150, N-Channel power MOSFET in TO-3 package • IRF540, N-Channel power MOSFET in TO-220 package TO-220

  20. Other examples of packages for power semiconductor devices TO-247 D-56 SOT-227

  21. Thermal resistance case to heat sink, RthCH • Its value depends on the interface material between semiconductor and heat sink. • Examples of thermal pads for TO-3 package: Basedonsilicone: SP400-0.009-00-05 Based on mica: Rth=0.3 oC/W)

  22. Thermal resistance heat sink to ambient, RthHA (I) • Its value depends on the heat sink dimensions and shape and also on the convection mode (either natural or forced). • Examples of heat sinks for TO-3 package: UP-T03-CB HP1-TO3-CB HS02

  23. Thermal resistance heat sink to ambient, RthHA (II) • Examples of heat sinks for general purpose

  24. Thermal resistance heat sink to ambient, RthHA (III) • Heat sink profiles (I)

  25. Thermal resistance heat sink to ambient, RthHA (IV) • Heat sink profiles (II)

  26. Thermal resistance heat sink to ambient, RthHA (V) • Calculations with heat sink profiles (I) Thermal resistance for 15 cm and natural convection Thermal resistance for 15 cm and air speed of 2m/s (forced convection) Rth_10cm = 1.32·Rth_15cm Rth_1m/s = 1.41·Rth_2m/s

  27. Thermal resistance heat sink to ambient, RthHA (VI) • Calculations with heat sink profiles (II) Thermal resistance for 15 cm and natural convection Thermal resistance for 15 cm and air speed of 2m/s (forced convection) • In the case of natural convection, the value of Rth given by the manufacturer corresponds to the case of a difference of 75 oC between heat sink and ambient. • For other differences, the following plot must be used. Rth_50oC = 1.15·Rth_75oC

  28. Thermal resistance heat sink to ambient, RthHA (VII) • Examples Rth_50oC = 1.15·Rth_75oC Rth_10cm = 1.32·Rth_15cm Rth_1m/s = 1.41·Rth_2m/s • Rth for 10 cm at DT=50 oC and natural convection = 0.99·1.32·1.15 = 1.5 oC/W • Rth for 10 cm and 2 m/s air speed = 0.72·1.32= 0.95 oC/W • Rth for 15 cm and 1 m/s air speed = 0.72·1.41 = 1.01 oC/W • Rth for 10 cm and 1 m/s air speed = 0.72·1.32·1.41 = 1.34 oC/W 28

  29. Transient thermal model (I) • So far our discussion and models have been limited to systems in which both the energy being dissipated and the temperatures within the system are constant. • Our models do not represent the following situations: • Star-up processes, where dissipation may be constant, but temperatures are climbing. • Pulsed operation, where temperatures may be constant, but dissipation is not. • The latter situation is the most important one, since under such conditions the junction temperature can be much higher than the one predicted by static models. • Thermal inertia can be characterized by capacitors in the equivalent electric model. 2 1 • T2 > T1 Q21 T1 T2 Rth21 T1 T2 2 1 • Cth21 T1 P21 0 oC

  30. Transient thermal model (II) • Transient thermal model for a star-up process, P being constant: H C A J TA TC TH TJ RthHA RthCH RthJC • CthCH • CthJC • CthHA CthJC<< CthCH << CthHA P= constant • Under this condition, the junction temperature will be lower than the one predicted by static models, due to the fact that the total thermal impedance will be lower than the thermal resistance. TA 0 oC

  31. P tC tS Transient thermal model (III) • Transient thermal model for pulsed-power operation in steady-state (I) (after the start-up process) H C A J TA TC TH TJ CthJC<< CthCH << CthHA RthCH RthHA RthJC P = pulsed • CthHA • CthCH • CthJC • Pavg • TH and TC can be computed from Pavg, because RthHA·CthHA >> tS and RthCH·CthCH >> tS. Hence, the AC component of P can be removed and, therefore: • TH = TA + Pavg·RthHA and TC = TH + Pavg·RthCH • However, RthHA·CthHA may be lower than tS and, therefore, TJ cannot be computed from RthJC and Pavg. TA 0 oC Dutycycle: D = tC/tS

  32. P Ppeak tC tS Transient thermal model (IV) • Transient thermal model for pulsed-power operation in steady-state (II) (after the start-up process) Constant T C J TC TJ RthJC • CthJC • Tj_avg • Pavg P = pulsed • TC • T TJ_ wide_pulse= TC + Ppeak·RthJC • The maximum value of the actual temperature is not as low as TJ_average. • The maximum value of the actual temperature is not as high as TJ_wide_pulse. • How can we compute the maximum value of the actual temperature? ÞTransient thermal impedance. TC • Tj_actual TJ_avg = TC + Pavg·RthJC 0 oC

  33. tC tS Transient thermal model (V) • Concept of transient thermal impedance P Ppeak • We know that TJ_wide_pulse > TJ_max > TJ_avg. • We will compute TJ_max considering Ppeak and an impedance lower than RthJC. • This impedance is called transient thermal impedance, ZthJC(t). • Its value depends on the duty cycle and on the switching frequency. • The final equation to compute TJ_max is: • TJ_max = TC + Ppeak·ZthJC(t) • Pavg TJ_ wide_pulse • Tj_max • T TJ_ avg • Tj_actual TJ_ wide_pulse= TC + Ppeak·RthJC TJ_avg = TC + Pavg·RthJC

  34. Transient thermal model (VI) • Example of transient thermal impedance (I) TJ_max = TC + Ppeak·ZthJC(t)

  35. Transient thermal model (VII) • Example of transient thermal impedance (II) 0.43 0.00004 t1 = D·t2 = D/fS = 0.2/5 kHz = 0.00004 sec fs = 5 kHz D =0.2 ZthJC = 0.43 oC/W 25 W TJ_max - TC = 25 W· 0.43 oC/W = 10.75 oC

  36. Transient thermal model (VIII) • Relationship between transient thermal impedance and thermal resistance ZthJC() ®RthJC

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