Robust Power Semiconductor Systems 2
Exercise 1
Tutor:
Ruben Schnitzler (ruben.schnitzler @ilh.uni-stuttgart.de)
SoSe 2023
Task 1 Content related questions
1. Give 4 different KPI (key performace indicators) for power converters. Give one example to
improve each KPI.
2. Give the different influences on the efficiency of a power converter and order them by their typical
magnitude of influence.
3. Explain the different kinds of heat transport.
4. Give the equivalences between electrical and thermal domain.
5. What is the dominating thermal resistance in power converters?
FTIM
Task 2 Layout of a buck converter
The high power buck convertor of Fig. 1 operates with MOSFETs M1 and M2 in parallel due to the high
current requirement of this application. The operating conditions are Uin = 400 V, IL = 170 A,
TA = 40 °C and D = 0,35. The two MOSFETs are of type STY145N65M5 from ST packaged in a
TO-247 package (see Fig. 2). They operate at a switching frequency of 10 kHz with a 10 Ω gate resistor.
For cooling the packaged power semiconductors are mounted on a common aluminum heat sink with a
Thermal Interface Material (TIM) in between for electrical isolation. The thermal conductivity kTIM of
the TIM silicone pad is 0,67 W/(mK) with a thickness of dTIM = 0,1 mm.
Bond wire
MOSFET
Case
Pin
PCB
Figure 1. Buck Converter
Leadframe
Figure 2. TO-247 Package (Infineon)
Your task is to select a fitting heat sink for the MOSFETS so that they operate within their specified
temperature range.
1. Draw the equivalent thermal Cauer circuit of the system. How does the Cauer circuit differ from the
Forster circuit?
2. Simplify the previous circuit for steady state and specify the system temperature equations for junction
temperatures Tj1 and Tj2.
maxT > max Rosim
3. Determine the symmetrical worst-case power losses of the MOSFETs. Assume a very small current
ripple over the choke L. Use the datasheet values for on resistance RDS,on and switching energies Eon/off.
-
4. Determine the maximum value of thermal resistance for the heat sink.
5. Select a heat sink from the catalog. Make sure the weight of the system is kept to a minimum.
Task 3 Layout of a heat sink
Calculate with a density of
sAl = 2419 kJ/(m³K) for aluminum.
ρAl = 2,7 g/cm³
and
a
volumetric
heat
capacity
of
1. Estimate the thermal capacitance of the heat sink selected in task 1.
2. Compute the transient junction temperature step response for a symmetrical power loss step. Neglect
upts the heat sink capacitance. Calculate the time t0 when 90% of the steady
all thermal capacitances besides
state temperature is reached.
↑
Alutronic catalog for standard heat sink profiles:
R = 0 086
.
HS1 : 190mm
R = 0 086
.
= mi = 18
3
86kg/m
190 mm
·
=
.
.
3772kg
H Se = 250mm
My = 11.02
R = 0 086
2
=
.
.
kg Im
,
250mm
753kg
HS3 : 100mm
R = 0 086
=m =
3
.
=
ST STY145N65M5 Datasheet:
17 25kg/me
= 1
.
,
100mm
725kg
PR719
http://www.st.com/content/ccc/resource/technical/document/datasheet/50/d4/cb/13/82/05/4d/4
9/DM00066266.pdf/files/DM00066266.pdf/jcr:content/translations/en.DM00066266.pdf
1,
a
KPI for power converters , one example to
4
[W/m3]
power density
,
-increase
frequency
Thermal allowance
-
-
>
↓
utilization
of soft
↑ I
:
1 &
-
can
be reduced
Switching
To
-
by
WBG Seminat a
I
v
z
v
=
L
&=SISN US : Hard sWitching
soft switching
.
cost and time to market
1
Virtual prototyping
>
C
,
-
↳
>
-
d
simulation
Reduced time
of
and thermal
losses
consumption
layout
. Robustness andReliabilitasing
a,
b
,
DUT power switch
Transformators losses
:
1
ii
c,
passive
losses
Peod
:
Eddy
,
,
iii ,
:
·
,
psw leakage losses
,
currents
proximity losses
core
capacitances
Inductances
PCB related losses
:
~
heatsink
Smaller
Conduction lossy
Efficiency /
b.
switching
the
improve
losses
>
electronics
di Control
i
:
Drivers
,
ii)
conduction And
Thermal
a:
:
b . Thermal
convection
Thermal
energy
:
:
heat
=
Your
c
,
integrated DC-DC Supplies
radiation
:
h
=
Transport of heat via fluids
=
R
.
length aarea
,
,
(w)
power
R
(12)
Voltage
(v)
Resistance
proportional factor
K:
Temperature Gradient
Zh
=
Electical domain
(k/w)
at
#
Z
S V
.
rocume
gases
energy into electromagnetic
Bolzma constant
=
Lth
of
+*
-
or
A (t-Tamb
conversion
Thermal domain
d :
2- XT)
Prad =_ CA(Tpu -Tamb)
Stefan
Rh
.
I
:
TIM
2
I ask
:
Thermal
:
interface
material
⑪
Cross Section
pi2
PT1
package
TO -247
↓
Ric
Tj1
↑
PTr
RTIM
& I-P7
⑨
-
SC
I
Ric
Til
-
-1
PTz 1 Se
↑
I
Phi
I
ICTIM
Ground reference to ambient temperature
Forster Model
Cault
-T
T1
paysical properties
TrioTe
Same
R
viorth
(Simulating)
②
Ric
RTIM
-
-
-
Rih
Til
V
Ric + R + M
·
~
-
7
Pro
I
Ras
Tia
↳ TA
Rih
ThS
O
_
-
as
TA
-
·
Tas
I
① Tas
:
5.1- Ths
=
=
Ras
(Pr P2)
P2)
+ TA
Rus (P1 +
=
#
Tjn =
Ta
-
+
Rih Pr
=
Rjh Pu
.
+
Imsert(a)
Tas
Rjh Pr + Ras (P1 + Pr)
+
Ta
+
TA
.
Similarly ; +i
&
IL
Esk
=
17 OA
......
-
=
Brond
=
.
P2
Ras(P1 P2)
+
+
max +, max
- - - - -
RDTON
I mean
>
O
Forms
Ria
conduction DOT
parallel
-
:=&
rus
RDS m
,
.
Ifirms
=
Ros on
,
.
(E)? D
=
Eload VP
1502
Tj max =
=
,
Ra = 10-
At
Eon
=
Ptot
2 2
(Ea
2 2
=
.
26
=
Rds
⑮
&
Eogy)
10KHE GMF
=
GOW
=
.
Jsu /Eon
=
%
4mo
.
215
+
(25)
, on
datashee
is
=
Psw + Pland
=
12m-
.
.
Egg
·
fsw
=
=
(Figure
4m5
=
Psw
=>
=
1158c
Rds on
(Tjimax)
Rdsion
>
-
,
Eog) Rosion ans ED
+
+
·
LA 126 8W F Each MOSFET
D
.
RdSon·
F
60W + 0 35 26
.
.
4me
.
·
.
&
Pr
=
Pz
=
=>
> Tj
Pc
Ras
Table 3
Ric
=
+
X
: 0
.
2 Rhs
Pc #TA
·
Timax
=
-
Pa
2
datasheet
+
Tsimax-TA
<
Rih
.C
Rjh
=
-
RTM
T
Rtim
214W
=
201
m
.
↑
.
-
41mm2
=
0
.
495
K
A = 15 3 mm 19 7mm
.
.
=
.
301
·
41mm
-
[datasheet (E D))
,
=>
Rjh =
Ras
=
0
.
2 + 0
.
495
=
Tsimax-TA
-
2
.
Pa
0
.
695k/W
Rih
_
=
190-06
0
.
086k/W
a
2
↑ above
k3
.
Capacitance Chil
2419KT/mk L =SKS
Thermal
:
S
.
=
V
=
S
=
.
. 2
2
=
Ct / LS) > Cth TIM
,
+
CthiseC
Neglect CRITIM and Cth sa
,
Rih/z
- - - - -
te =
I
#
,
S
↑
2
Polt) O
Re
Rizt
h
Pc(t)
=
h(+)
2
.
.
Pc(t)
7
zu-
Lapape
:
Pr(p)
=
t
Tj(p) -TA
=>
Tj1(t)
2"[(Eth(p)
=
Resh)
+
Rh
=
.
Pr(p)]
·
Rip]
+
Tj(t) = RihP((H) + 2 RaPi( - i
T;
Th = Rh
(to)
2
=>
Ch
.
Rih
=
Tj(to)
=
P2
,
0
TA- Bil
-
086k/W
2
+
Rh
.
Bh
.
Pr
PL
.
exp( =)
=>
,
=
.
to
=
.
.
39
.
=
.
1 545
.
·
F)
KT1k
=
+
in
132 87s
.
Pr[1-exp(i(]
1
-
E
+
TA
expli
- (t)-TA
2
Rh
.
Rio
-
P2
(
I (1-04115004-0691
%
-Y
= to = - 132 87s
-50
TA
+
Ch
<h
2
=
=>
P(p)
·
2()h +)
=
With Tth
(th(p) +h)
=
·
In
(1- Tj(to)
90
-
2
.
Ja- Rich
Rh
.
PL
.
PC