Download Lesson 1 - Introduction

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