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Electrical Energy Conversion and Power Systems Universidad de Oviedo Power Electronic Devices Semester 1 Power Supply Systems Lecturer: Javier Sebastián 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). 2 Electrical Energy Conversion and Power Systems Universidad de Oviedo Lesson 2 - Thermal management in power semiconductor devices Semester 1 - Power Electronics Devices 3 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. 4 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. 5 Simple static thermal model (I) T1 2 1 A Q12_cond Q12_conv T2 • 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. • 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. 6 Simple static thermal model (II) Q12_cond = (T1 – T2)/Rth12_cond Q12_conv = h(T1,T2,n)·A·(T1 – T2) T1 2 1 Q12_cond Q12_conv Q12 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_conv and 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. 7 Simple static thermal model (III) Q12 = (T1 – T2)/Rth12(n) T1 Q12 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. 1 V1 V2 • 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: 1 2 2 R12 i12 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 Thermal world Electric world Rth R ΔT ΔV P i 8 Simple static thermal model (IV) l • Electric resistance: R = r l·/A, where: r = electric resistivity. l = length. A = cross-sectional area. V1 i12 2 l rth = thermal resistivity. Material Resistivity [oC·cm/W] Still air 3050 Mica 150 Filled silicone grease 130 Alumina (Al2O3) 6.0 Berylia (BeO) 1.0 Aluminum Nitride (AlN) 0.64 Aluminum 0.48 Copper 0.25 r 1 • Thermal resistance: Rth = rth l·/A , where: V2 A T1 Q12 1 T2 A rth 2 9 Thermal model for a power semiconductor device (I) • Typical mechanical structure used for mounting a power semiconductor device Bonding wire Semiconductor junction Case Interface (e.g., mica) Si Interface Header (Cu, Al or Ti) Heat sink 10 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 HS02 Insulating mica for TO-3 (Interface) Heat sink 11 Thermal model for a power semiconductor device (III) • Examples of final assembly of electronic devices in TO-3 package over heat sinks 12 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 RthCA Printing Circuit Board (PCB) Therefore: TJ = TA + (RthJC + RthCA)·P TA A TA 0 oC RthCA TC C RthJC RthJC TJ J P 13 Thermal model for a power semiconductor device (V) • Thermal resistances with a heat sink (I) Ambient (A) Case (C) Junction (J) RthCA RthHA Heat sink (H) TA RthJC RthCH RthCA Spacer (Mica plate) RthJC TC C A TJ J H TA 0 oC RthCH RthHA P 14 Thermal model for a power semiconductor device (VI) • Thermal resistances with a heat sink (II) TA A RthCA TC C RthHA H RthJC TJ J RthCH TA 0 oC P 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 15 Thermal model for a power semiconductor device (VII) • Thermal resistances with a heat sink (III) TA A RthHA TA TH TC H C RthCH TJ RthJC J P 0 oC 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. 16 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 17 Thermal resistance junction to case, RthJC (II) • The same device has different value of RthJC for different packages. • Examples corresponding to two devices: IF(AV) = 5A, VRRM = 1200V 18 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 19 Other examples of packages for power semiconductor devices TO-247 SOT-227 D-56 20 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: Based on silicone: SP400-0.009-00-05 Description THERMAL PAD TO-3 .009" SP400 Material Silicone Based Thermal Conductivity 0.9 W/m-K Thermal Resistance 1.40 °C/W Thickness 0.229 mm Based on mica: Rth=0.3 oC/W) 21 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 Material Aluminum Material Aluminum Material Aluminum Rth @ natural 9 °C/W Rth @ natural 5.4 °C/W Rth @ natural 4.5 °C/W 22 Thermal resistance heat sink to ambient, RthHA (II) • Examples of heat sinks for general purpose 23 Thermal resistance heat sink to ambient, RthHA (III) • Heat sink profiles (I) 24 Thermal resistance heat sink to ambient, RthHA (IV) • Heat sink profiles (II) 25 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 26 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) Rth_50oC = 1.15·Rth_75oC • 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. 27 Thermal resistance heat sink to ambient, RthHA (VII) • Examples Rth_10cm = 1.32·Rth_15cm Rth_50oC = 1.15·Rth_75oC Rth_1m/s = 1.41·Rth_2m/s 28 • 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 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. 1 T1 T2 > T1 Rth21 T1 2 1 T2 2 Cth21 Q21 T2 T1 0 oC P21 29 Transient thermal model (II) • Transient thermal model for a star-up process, P being constant: TA A RthHA TA CthJC CthCH CthHA H C TH TC RthCH J RthJC TJ CthJC << CthCH << CthHA 0 oC 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. 30 Transient thermal model (III) • Transient thermal model for pulsed-power operation in steady-state (I) (after the start-up process) CthHA TA TA 0 oC A RthHA CthJC CthCH H C TH R thCH TC CthJC << CthCH << CthHA J RthJC TJ P = pulsed • 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. P Pavg tC tS Duty cycle: D = tC/tS 31 Transient thermal model (IV) • Transient thermal model for pulsed-power operation in steady-state (II) (after the start-up process) CthJC Constant T TC 0 oC C TC P J RthJC TJ P = pulsed Ppeak tC tS 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? Transient thermal impedance. Pavg Tj_avg Tj_actual TJ_avg = TC + Pavg·RthJC TC 32 Transient thermal model (V) • Concept of transient thermal impedance • 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) P Ppeak Pavg tC tS Tj_max T TJ_ wide_pulse TJ_ avg Tj_actual TJ_ wide_pulse = TC + Ppeak·RthJC TJ_avg = TC + Pavg·RthJC 33 Transient thermal model (VI) • Example of transient thermal impedance (I) TJ_max = TC + Ppeak·ZthJC(t) 34 Transient thermal model (VII) • Example of transient thermal impedance (II) 0.43 0.00004 fs = 5 kHz D =0.2 t1 = D·t2 = D/fS = 0.2/5 kHz = 0.00004 sec 25 W ZthJC = 0.43 oC/W TJ_max - TC = 25 W· 0.43 oC/W = 10.75 oC 35 Transient thermal model (VIII) • Relationship between transient thermal impedance and thermal resistance ZthJC() RthJC 36