Power Electronic Devices

Semester 1

Lecturer: Javier Sebastián

Electrical Energy Conversion and Power Systems

Universidadde 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

Universidadde 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.

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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.

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1 2Simple static thermal model (I)

T1 T2A

Q12_cond

• 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.

Q12_conv

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1 2Simple static thermal model (II)

T1 T2Q12

Q12_cond = (T1 – T2)/Rth12_cond

Q12_conv = h(T1,T2,n)·A·(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_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.

Q12_cond

Q12_conv

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Simple static thermal model (III)

1 2

T1 T2Q12

Q12 = (T1 – T2)/Rth12(n)

• 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

Rth Þ R ΔT Þ ΔV P Þ i

R12 1 2

i12

V1 V2

Electric worldThermal world

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Simple static thermal model (IV)

1 2

V1 V2i12

• Electric resistance: R = r l·/A, where:

r = electric resistivity. l = length. A = cross-sectional area.

• Thermal resistance: Rth = rth l·/A , where:

rth = thermal resistivity.

1 2

T1 T2Q12

A

A

l

l

r

rth

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

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Thermal model for a power semiconductor device (I)

Si

Header (Cu, Al or Ti)Interface

Interface (e.g., mica)

Case

Semiconductor junction

Heat sink

Bonding wire

• Typical mechanical structure used for mounting a power semiconductor device

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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)

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Thermal model for a power semiconductor device (III)

• Examples of final assembly of electronic devices in TO-3 package over heat sinks

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Printing Circuit Board (PCB)

Thermal model for a power semiconductor device (IV)• Thermal resistances without a heat sink

Junction (J)

Case (C)Ambient (A)

RthCA

RthJC

A C J

P

TJTCTA

RthCA RthJC

TA

0 oC

Basic equations:TC = TA + RthCA·P

TJ = TC + RthJC·P

Therefore:TJ = TA + (RthJC + RthCA)·P

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Thermal model for a power semiconductor device (V)• Thermal resistances with a heat sink (I)

A

TA

C J

P

TJTC

RthCA RthJC

TA

0 oC

Heat sink (H)

RthCA

Junction (J)Case (C)Ambient (A)

RthJCRthCH

RthHA

H

Spacer(Mica plate)

RthCHRthHA

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Thermal model for a power semiconductor device (VI)• Thermal resistances with a heat sink (II)

A

TA

C J

P

TJTC

RthCA RthJC

TA

0 oC

HRthCHRthHA

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)]·PHowever, many times RthCA >> (RthHA + RthCH), and therefore:

TJ » TA + (RthJC + RthHA + RthCH)·P

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• Thermal resistances with a heat sink (III)

TA C J

P

TJTC

RthHA RthJC

TA

0 oC

H RthCHA

TH

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.

Thermal model for a power semiconductor device (VII)

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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

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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

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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 packageTO-220

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TO-247

SOT-227

Other examples of packages for power semiconductor devices

D-56

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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:

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 silicone: SP400-0.009-00-05

Rth=0.3 oC/W)

Based on mica:

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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:

Material Aluminum

Rth @ natural 9 °C/W

UP-T03-CB HP1-TO3-CB

Material Aluminum

Rth @ natural 5.4 °C/W

Material Aluminum

Rth @ natural 4.5 °C/W

HS02

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

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Thermal resistance heat sink to ambient, RthHA (III)

• Heat sink profiles (I)

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Thermal resistance heat sink to ambient, RthHA (IV)

• Heat sink profiles (II)

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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

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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.

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Rth_10cm = 1.32·Rth_15cm

Thermal resistance heat sink to ambient, RthHA (VII)

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Rth_1m/s = 1.41·Rth_2m/s

Rth_50oC = 1.15·Rth_75oC

• 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

• Examples

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1 2

T1 T2

T2 > T1

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.

T1

P21

Rth21

T1

0 oC

21

T2

Cth21Q21

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Transient thermal model (II)

• Transient thermal model for a star-up process, P being constant:

TAC J

P= constant

TJTCRthHA RthJC

TA

0 oC

H

RthCH

A

TH

CthHA CthCH

• 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.

CthJC

CthJC << CthCH << CthHA

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Transient thermal model (III)• Transient thermal model for pulsed-power operation in

steady-state (I) (after the start-up process)

• 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.

TAC J

P = pulsed

TJTCRthHA RthJCTA

0 oC

H

RthCH

A

TH

CthHA CthCH CthJC CthJC << CthCH << CthHA

P

tS

tC

Pavg

Duty cycle:D = tC/tS

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Transient thermal model (IV)

• 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? Þ Transient thermal impedance.

C J

P = pulsed

TJTC RthJC

TC

0 oC

CthJCP Ppeak

tS

tC

Pavg

• Transient thermal model for pulsed-power operation in steady-state (II) (after the start-up process)

Tj_avg

Tj_actual

T TJ_ wide_pulse = TC + Ppeak·RthJC

TJ_avg = TC + Pavg·RthJC

Constant T

TC

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Transient thermal model (V)

• 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)

• Concept of transient thermal impedance

TJ_ wide_pulse = TC + Ppeak·RthJC

TJ_avg = TC + Pavg·RthJC

P Ppeak

tS

tC

Pavg

Tj_actual

T TJ_ wide_pulse

TJ_ avg

Tj_max

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Transient thermal model (VI)

TJ_max = TC + Ppeak·ZthJC(t)• Example of transient thermal impedance (I)

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Transient thermal model (VII)

ZthJC = 0.43 oC/W

• Example of transient thermal impedance (II)

25 W

fs = 5 kHzD =0.2

t1 = D·t2 = D/fS = 0.2/5 kHz = 0.00004 sec

TJ_max - TC = 25 W· 0.43 oC/W = 10.75 oC

0.43

0.00004

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