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

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).

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

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.

1 2 6 Simple static thermal model (I) T1T1 T2T2 A Q 12_cond Simple model for heat conduction: Q 12_cond = (T 1 – T 2 )/R th12_cond, where: Q 12_cond = rate of heat energy transferred from 1 to 2 due to conduction. T 1 = temperature at 1. T 2 = temperature at 2. R th12_cond = thermal resistance between 1 and 2 due to conduction. Simple model for convection: Q 12_conv = h(T 1,T 2, )·A·(T 1 – T 2 ), where: Q 12_conv = rate of heat energy transferred from 1 to 2 due to convection. h = film coefficient of heat transfer. = fluid velocity. A = cross-sectional area. Q 12_conv

1 2 7 Simple static thermal model (II) T1T1 T2T2 Q 12 Q 12_cond = (T 1 – T 2 )/R th12_cond Q 12_conv = h(T 1,T 2, )·A·(T 1 – T 2 ) Over the temperature ranges of interest (from -40 o C +100 o C) h is fairly constant for a given value of. Therefore: Q 12_conv = h( )·A·(T 1 – T 2 ) = (T 1 – T 2 )/R th12_conv ( ), where: R th12_conv ( ) = 1/[h( )·A] The total rate of heat energy transferred from 1 to 2 will be: Q 12 = Q 12_cond + Q 12_conv and therefore: Q 12 = (T 1 – T 2 )/R th12 ( ) where: R th12 ( ) = 1/[1/R th12_cond + 1/R th12_conv ( )] is the thermal resistance between 1 and 2. Q 12_cond Q 12_conv

8 Simple static thermal model (III) 1 2 T1T1 T2T2 Q 12 Q 12 = (T 1 – T 2 )/R th12 ( ) We will use the notation “R th12 ” (or “R  12 ”) 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: P 12 = (T 1 – T 2 )/R th12 Now, we can establish a direct relationship between this equation and Ohm’s law: P 12 = (T 1 – T 2 )/R th12 i 12 = (V 1 – V 2 )/R 12 R th  R ΔT  ΔV P  i R i 12 V1V1 V2V2 Electric world Thermal world

9 Simple static thermal model (IV) 1 2 V1V1 V2V2 i 12 Electric resistance: R =  l ·/A, where:  = electric resistivity. l = length. A = cross-sectional area. Thermal resistance: R th =  th l ·/A, where:  th = thermal resistivity. 1 2 T1T1 T2T2 Q 12 A A l l   th MaterialResistivity [ o C·cm/W] Still air3050 Mica150 Filled silicone grease130 Alumina (Al 2 O 3 )6.0 Berylia (BeO)1.0 Aluminum Nitride (AlN)0.64 Aluminum0.48 Copper0.25

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

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

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

Printing Circuit Board (PCB) 13 Thermal model for a power semiconductor device (IV) Thermal resistances without a heat sink Junction (J) Case (C) Ambient (A) R thCA R thJC A C J P TJTJ TCTC TATA R thCA R thJC TATA 0 o C Basic equations: T C = T A + R thCA ·P T J = T C + R thJC ·P Therefore: T J = T A + (R thJC + R thCA )·P

14 Thermal model for a power semiconductor device (V) Thermal resistances with a heat sink (I) A TATA C J P TJTJ TCTC R thCA R thJC TATA 0 o C Heat sink (H) R thCA Junction (J) Case (C) Ambient (A) R thJC R thCH R thHA H Spacer (Mica plate) R thCH R thHA

15 Thermal model for a power semiconductor device (VI) Thermal resistances with a heat sink (II) A TATA C J P TJTJ TCTC R thCA R thJC TATA 0 o C H R thCH R thHA Basic equations: T C = T A + [R thCA ·(R thHA + R thCH )/(R thCA + R thHA + R thCH )]·P T J = T C + R thJC ·P Therefore: T J = T A + [R thJC + R thCA ·(R thHA + R thCH )/(R thCA + R thHA + R thCH )]·P However, many times R thCA >> (R thHA + R thCH ), and therefore: T J  T A + (R thJC + R thHA + R thCH )·P

16 Thermal resistances with a heat sink (III) TATA C J P TJTJ TCTC R thHA R thJC TATA 0 o C H R thCH A THTH Basic equations: T H = T A + R thHA ·P T C = T H + R thCH ·P T J = T C + R thJC ·P Therefore: T J = T A + (R thJC + R thCH + R thHA )·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 o C. Thermal model for a power semiconductor device (VII)

17 Thermal resistance junction to case, R thJC (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 Thermal resistance junction to case, R thJC (II) The same device has different value of R thJC for different packages. Examples corresponding to two devices: I F(AV) = 5A, V RRM = 1200V

19 Thermal resistance case to ambient, R thCA Its value depends on the case. Manufacturers give information about R thJA = R thJC + R thCA. Therefore, R thCA = R thJA - R thJC. 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

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

21 Thermal resistance case to heat sink, R thCH Its value depends on the interface material between semiconductor and heat sink. Examples of thermal pads for TO-3 package: DescriptionTHERMAL PAD TO-3.009" SP400 MaterialSilicone Based Thermal Conductivity0.9 W/m-K Thermal Resistance1.40 °C/W Thickness0.229 mm Based on silicone: SP R th =0.3 o C/W) Based on mica:

22 Thermal resistance heat sink to ambient, R thHA (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: MaterialAluminum R natural 9 °C/W UP-T03-CB HP1-TO3-CB MaterialAluminum R natural 5.4 °C/W MaterialAluminum R natural 4.5 °C/W HS02

23 Thermal resistance heat sink to ambient, R thHA (II) Examples of heat sinks for general purpose

24 Thermal resistance heat sink to ambient, R thHA (III) Heat sink profiles (I)

25 Thermal resistance heat sink to ambient, R thHA (IV) Heat sink profiles (II)

26 Thermal resistance heat sink to ambient, R thHA (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) R th_10cm = 1.32·R th_15cm R th_1m/s = 1.41·R th_2m/s

27 Thermal resistance heat sink to ambient, R thHA (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) R th_50 o C = 1.15·R th_75 o C In the case of natural convection, the value of R th given by the manufacturer corresponds to the case of a difference of 75 o C between heat sink and ambient. For other differences, the following plot must be used.

R th_10cm = 1.32·R th_15cm 28 Thermal resistance heat sink to ambient, R thHA (VII) 28 R th_1m/s = 1.41·R th_2m/s R th_50 o C = 1.15·R th_75 o C R th for 10 cm at  T=50 o C and natural convection = 0.99·1.32·1.15 = 1.5 o C/W R th for 10 cm and 2 m/s air speed = 0.72·1.32= 0.95 o C/W R th for 15 cm and 1 m/s air speed = 0.72·1.41 = 1.01 o C/W R th for 10 cm and 1 m/s air speed = 0.72·1.32·1.41 = 1.34 o C/W Examples

1 2 T1T1 T2T2 T 2 > T 1 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. T1T1 P 21 R th21 T1T1 0 o C 2 1 T2T2 C th21 Q 21

30 Transient thermal model (II) Transient thermal model for a star-up process, P being constant: TATA C J P= constant TJTJ TCTC R thHA R thJC TATA 0 o C H R thCH A THTH C thHA C thCH 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. C thJC C thJC << C thCH << C thHA

31 Transient thermal model (III) Transient thermal model for pulsed-power operation in steady-state (I) (after the start-up process) T H and T C can be computed from P avg, because R thHA ·C thHA >> t S and R thCH ·C thCH >> t S. Hence, the AC component of P can be removed and, therefore: T H = T A + P avg ·R thHA and T C = T H + P avg ·R thCH However, R thHA ·C thHA may be lower than t S and, therefore, T J cannot be computed from R thJC and P avg. TATA C J P = pulsed TJTJ TCTC R thHA R thJC TATA 0 o C H R thCH A THTH C thHA C thCH C thJC C thJC << C thCH << C thHA P tStS tCtC P avg Duty cycle: D = t C /t S

32 Transient thermal model (IV) The maximum value of the actual temperature is not as low as T J_average. The maximum value of the actual temperature is not as high as T J_wide_pulse. How can we compute the maximum value of the actual temperature?  Transient thermal impedance. C J P = pulsed TJTJ TCTC R thJC TCTC 0 o C C thJC P P peak tStS tCtC P avg Transient thermal model for pulsed-power operation in steady-state (II) (after the start-up process) T j_avg T j_actual T T J_ wide_pulse = T C + P peak ·R thJC T J_avg = T C + P avg ·R thJC Constant T TCTC

33 Transient thermal model (V) We know that T J_wide_pulse > T J_max > T J_avg. We will compute T J_max considering P peak and an impedance lower than R thJC. This impedance is called transient thermal impedance, Z thJC (t). Its value depends on the duty cycle and on the switching frequency. The final equation to compute T J_max is: T J_max = T C + P peak ·Z thJC (t) Concept of transient thermal impedance T J_ wide_pulse = T C + P peak ·R thJC T J_avg = T C + P avg ·R thJC P P peak tStS tCtC P avg T j_actual T T J_ wide_pulse T J_ avg T j_max

34 Transient thermal model (VI) T J_max = T C + P peak ·Z thJC (t) Example of transient thermal impedance (I)

35 Transient thermal model (VII) Z thJC = 0.43 o C/W Example of transient thermal impedance (II) 25 W f s = 5 kHz D =0.2 t 1 = D·t 2 = D/f S = 0.2/5 kHz = sec T J_max - T C = 25 W· 0.43 o C/W = o C

36 Transient thermal model (VIII) Relationship between transient thermal impedance and thermal resistance Z thJC (  )  R thJC