CHAPTER 6

DEVICE PROTECTION
6.1 INTRODUCTION

Due to the reverse recovery process of power devices and switching actions in the presence of circuit inductances,
voltage transients occur in the converter circuits. Even in carefully designed circuits, short-circuit fault conditions may
exist, resulting in an excessive current flow through the devices. The heat produced by losses in a semi-conductor
device must be dissipated sufficiently and effectively to operate the device within its upper temperature limit. The
reliable operation of a converter would require ensuring that at all times the circuit conditions do not exceed the ratings
of the power devices, by providing protection against over-voltage, over-current and overheating. In practice, the
power devices are protected from:
 Thermal runaway by heat sinks.
 High dv/dt and di/dt by snubbers.
 Reverse recovery transients.
 Supply and load-side transients.
 Fault conditions by means of fuses.

6.2 COOLING AND HEAT SINKS

Due to on-state and switching losses, heat is generated within the power device. This heat must be
transferred from the device to a cooling medium to maintain the operating junction temperature within the
specified range. Although this heat transfer can be accomplished by conduction, convection, radiation,
natural or forced-air, convection cooling is commonly used in industrial applications. The heat must
flow from the device to the case and then to the heat sink in the cooling medium. If PD is the average power
loss in the device, then the electrical analogue of a device mounted on a heat sink, is as shown in Figure 6.1.
The junction temperature of a device (TJ) is given by:

TJ  PD ( R JC  RCS  R SA )  TA

Where:

R JC  the thermal resistance from junction to case in ˚C/W or K/W

RCS  thermal resistance from case to sink in ˚C/W

R SA  thermal resistance from sink to ambient air in ˚C/W

TA  the ambient temperature in ˚C

Take note:

 TJ max  125 ˚C for an industrial application (production/money is involved)

 TJ max  90 ˚C for a military application and aerospace application (life depends on the health of the
component/system)
 TJ max  150 ˚C for a commercial application such as a hobbyist’s of DIY enthusiast’s equipment (it
does not really matter if it is now or tomorrow – money or life is not involved.
 TJ TC TS

PD

TA

Figure 6.1: Electrical analogue of a heat flow path

The less it matters, the higher the maximum temperature. If life is involved, the lower the maximum
temperature. R JC and RCS are normally specified by the manufacturer of the power device. Once the
device power loss PD is known, the required thermal resistance of the heat sink can be calculated for a known
ambient temperature TA. The next step is to choose a heat sink type and size it so that it would meet the
thermal resistance requirement. A wide variety of extruded aluminium heat sinks are commercially available.
They have cooling fins to increase their heat transfer capability.

In forced cooling, the thermal resistance decreases with the air velocity. However, above a certain velocity,
the reduction in thermal resistance is not significant.

The contact area between the device and the heat sink is extremely important to minimize the thermal
resistance between the case and sink. The surfaces should be flat, smooth and free of dirt, corrosion and
surface oxides. Silicone greases are normally applied to improve the heat transfer capability and to minimize
the formation of oxides and corrosion. The device must be mounted properly with the correct mounting
pressure between the mating surfaces.

Proper installation procedures are usually recommended by device manufacturers. In the case of stud-
mounted devices, excessive mounting torques may cause mechanical damage to the silicon wafer and the
stud and nut should not be greased or lubricated because the lubrication increases the tension on the stud.

The device may be cooled by heat pipes partially filled with a low-vapour pressure liquid. The device is
mounted on one side of the pipe and a condensing mechanism (or heat sink) on the other side. The heat
produced by the device vaporizes the liquid and the vapour then flows to the condensing end where it
condenses and the liquid returns to the heat source. The device may be some distance from the heat sink.

In high-power applications, the devices are more effectively cooled by liquids, normally oil or water. Water-
cooling is very efficient and approximately three times more effective than oil cooling. However, it is
necessary to use distilled water to minimize corrosion and antifreeze must be added to avoid freezing. Oil is
flammable. Oil cooling, which may be restricted to some applications, provides good insulation and eliminates
the problems of corrosion and freezing. Heat pipes and liquid-cooled heat sinks are commercially available.

The thermal impedance of a power device is very small and as a result the junction temperature of the device
varies with the instantaneous power loss. The instantaneous junction temperature must always be
maintained lower than the acceptable value. If the cooling medium fails in practical systems, the temperature
rise of the heat sinks is normally used to switch off the power converters, especially in high-power
applications.

The packing orientation of printed circuit boards and semi-conductor heat sinks within a cabinet or enclosure
should minimize restrictions in the direction of airflow. Since heat rises, vertical panels, boards and fins are
desirable for natural convection.
Thermal compounds are to be used when mounting power devices on heat sinks since this reduces the
thermal resistivity. There are two types of thermal compounds available: Thermal paste and thermal grease.
Thermal paste has three times the thermal conductivity of thermal grease. Thermal paste fills the microscopic
voids between the joining surfaces and thus maximizes the heat transfer. Care must be exercised when
applying thermal paste. If it is too thick, depending on the environment, it will collect contaminants over a
period of time. This may cause a decrease in insulation resistance or intermittent shorts between heat sink
and the device.

Worked example 1

Two identical MOSFET’s that share current equally are mounted on a common heat sink. The thermal
resistances are as follows: Junction to case: 1.2 ˚C/W; case to heat sink: 0.3 ˚C/W; heat sink to ambient
air: 0.75 ˚C/W. The maximum allowed junction temperature is 125 ˚C. Ambient air temperature is 35 ˚C.

1.1 Sketch the thermal equivalent circuit for the combination.

PD PD
TJ TJ

1.2˚C/W 1.2˚C/W
TC TC

0.3˚C/W 0.3˚C/W
TS

RSA 0.75˚C/W

TA

Figure 6.2: Heat flow diagram for two devices on a common heat sink

1.2 Determine the maximum power that can be safely dissipated by each device.

TJ  PD ( R JC  RCS  2 R SA )  TA

TJ  TA 125  35
PD    30 W
( R JC  RCS  2 RSA ) (1.2  0.3  2  0.75 )

1.3 Determine the heat sink temperature.

TS  PD (2 R SA )  TA  30(2(0.75))  35  80 o C

1.4 If one MOSFET would be removed, determine the maximum power that the remaining one could
dissipate before exceeding the maximum junction temperature.

TJ  TA 125  35
PD    40 W
( R JC  RCS  RSA ) (1.2  0.3  0.75)
1.5 Determine the maximum heat sink temperature with only one MOSFET.

TS  PD ( RSA )  TA  30(0.75)
40  35  65 o C

1.6 Determine the maximum current through the single MOSFET if the drain to source resistance is equal
to 14 m.

P 40
ID    53.45 A
RDSON 14 103

Exercise 6.1

Two identical MOSFET’s that share current equally, are mounted on a common heat sink. The thermal
resistances are as follows: Junction to case: 1.2 ˚C/W; case to heat sink: 0.3 ˚C/W; heat sink to ambient
air: 0.75 ˚C/W; the maximum allowed junction temperature is 125 ˚C. Ambient air temperature is 25 ˚C.

6.1.1 Sketch the thermal equivalent circuit for the combination.

6.1.2 Determine the maximum power that can be safely dissipated by each device.

6.1.3 Determine the heat sink temperature.

6.1.4 If one MOSFET would be removed, determine the maximum power that the remaining one could
dissipate before exceeding the maximum junction temperature.

6.1.5 Determine the maximum heat sink temperature with only one MOSFET.

6.1.6 Determine the maximum current through the single MOSFET if the drain to source resistance is equal
to 14 m.

Answers: (33.33 W ; 75 ˚C ; 44.44 W ; 58.33 ˚C ; 56.34 A)

Worked example 2

An IBGT has a VCEsat = 2.2 V when a current of 6 A flows through it. It is used in an industrial application
with an insulator between it and a heat sink. Silicone grease is applied and this halves the heat resistance
of the insulator from 1 K/W to 0.5 K/W. The maximum ambient temperature is 35 ˚C. The IGBT thermal
resistance from junction to case is 2.1 K/W.

2.1 Calculate the power dissipated in the IGBT.

PD  VCEsat  I C  2.2  6  13.2 W

2.2 Calculate the thermal resistance rating of the heat sink required.

TJ  PD ( R JC  RCS  R SA )  TA

TJ  TA
 RSA   ( R JC  RCS )
PD

With silicone grease: RCS  0.5 K/W or C/W

o
 TJ  TA 125  35
 RSA   ( R JC  RCS )   (2.1  0.5)  4.22 o C/W
PD 13.2

2.3 If a technician replaces the IGBT, but neglects to apply silicone grease to the insulator, calculate the
maximum current that the device would now be able to carry before exceeding the maximum junction
temperature for an industrial application. Assume that VCEsat is still the same.

Without silicone grease: RCS  1 K/W or o C/W

TJ  PD ( R JC  RCS  R SA )  TA

TJ  TA 125  35
PD    12.3 W
( R JC  RCS  RSA ) (2.1  1  4.22)

But PD  VCEsat  I C

PD 12.3
 IC    5.59 A
VCEsat 2.2

Exercise 6.2

A MOSFET is mounted on a heat sink. The thermal resistances are as follows: Junction to case: 1.2 ˚C/W;
case to heat sink: 0.3 ˚C/W; heat sink to ambient air: 0.75 ˚C/W. The maximum allowed junction
temperature is 120 ˚C. Ambient air temperature is 35 ˚C.

6.2.1 Sketch the thermal equivalent circuit.

6.2.2 Determine the maximum power that can be safely dissipated by the device.

6.2.3 Determine the heat sink temperature.

6.2.4 Determine the maximum current through the MOSFET if the drain to source resistance is equal to 15
m. Answers: (37.78 W ; 63.34 ˚C ; 50.19 A)

Worked example 3

Specify the thermal resistance ( R SA ) of a heat sink for a transistor if the ambient temperature is 40 ˚C, the
total power dissipated is 50 W, R JS  1 K/W and the application is industrial related.

TJ = 125 ˚C TC TS

PD = 50 W

TA = 40 ˚C

Figure 6.3: Heat flow diagram for worked example 3

 TJ  PD ( R JC  RCS  R SA )  TA

TJ  TA 125  40
 RSA   ( R JC  RCS )   1  0.7 o C/W
PD 50

Exercise 6.3

A power device application has a junction to ambient air thermal resistance of 25 K/W. Assume an acceptable
ambient temperature in an enclosure of 40 ˚C and determine the maximum power that can safely be
dissipated in the following environments: Industry; Aerospace and Hobbyist.

Answers: (3.4 W; 2 W; 4.4 W)

Worked example 4

A MOSFET has an rDS of 10 m. The MOSFET is mounted on a heat sink. The thermal resistance from
junction to ambient air is 3 K/W. Determine the maximum current that the MOSFET can handle if it is used
in an industrial application. The ambient air is 40 ˚C.

TJ  125 o C

TJ  TA
PD  I D2 rDS 
R JA

TJ  TA 125  40
 ID    53.23 A
rDS R JA 10 103  3

Worked example 4

A three-phase inverter supplies an AC load. The inverter’s two MOSFET groups are each connected to
common heat sinks which are isolated from the structure on which they are mounted by means of
insulators. Each MOSFET has a thermal resistance from junction–to-case of 0.5 ˚C/W. Between case
and sink it is 0.5 ˚C/W (if thermal compound is correctly applied and adequate mounting pressure is
maintained between the MOSFET and the heat sink). The highest ambient temperature measured in an
enclosure in the factory has been 45 ˚C and the lowest 15 ˚C. The average power dissipated in each
MOSFET is 30 W.

4.1 Sketch a labeled thermal equivalent circuit of the setup. See Figure 6.4.

4.2 Specify the thermal resistance of a heat sink for each group of this inverter.

TJ  PD ( R JS  3RSA )  TA

TJ  TA
 R JS 125  45  1
PD
R SA   30  0.556 C/W
3 3
4.3 Calculate the maximum heat sink temperature.

Ts  PD (3RSA )  TA  30  3  0.556    45  95 C

4.4 Calculate the heat sink temperature if the ambient temperature is 15 ˚C.

Ts  PD (3RSA )  TA  30  3  0.556    15  65 C

PD PD PD
TJ TJ TJ

0.5˚C/W 0.5˚C/W 0.5˚C/W

TC TC TC

0.5˚C/W 0.5˚C/W 0.5˚C/W

TS

? ˚C/W

TA

Figure 6.4: Thermal equivalent circuit for worked example 4

Worked example 5

A three-phase zig-zag rectifier supplies 100 A to a DC load. The diodes are connected to a common
heat sink which is isolated from the structure on which it is mounted by means of insulators. Each diode
has a forward volt drop of 1.45 V when it is conducting 100 A. Each diode has a thermal resistance from
junction–to-case of 0.5 K/W and between case and sink it is 0.5 K/W if thermal compound is correctly
applied and adequate mounting pressure is maintained between the diodes and the heat sink. The
highest ambient temperature measured in an enclosure in the factory has been 323 K and the lowest
300 K.

5.1 Sketch labeled waveforms for the current flowing through the diodes.

Figure 6.5: Diode currents of a zig-zag rectifier

5.2 Estimate the average power dissipated in each diode. Include calculations in order to show that
you did not guess to arrive at an answer! (Answer must be rounded to the nearest integer).

IL 100
I Drms    57.7 A
3 3

VD 1.45
VDrms    0.84 V
3 3

PD  I DrmsVDrms  57.7  0.84  49 W

Can you think of another way to calculate the power dissipated per device?

5.3 Sketch a labeled thermal equivalent circuit of the setup.

PD PD PD
TJ TJ TJ

0.5˚C/W 0.5˚C/W 0.5˚C/W

TC TC TC

0.5˚C/W 0.5˚C/W 0.5˚C/W

TS

? ˚C/W

TA

Figure 6.6: Thermal equivalent circuit for worked example 5

5.4 Specify the thermal resistance of a heat sink for this rectifier.

The maximum ambient temperature has to be converted from Kelvin to Celcius:

TA  323  273  50 C

TJ  TA
 R JS 125  50  1
PD 49
R SA    0.18 C/W
3 3
6.3 SNUBBERS

Two types of snubbers are presented in this section, i.e. turn-off snubbers and turn-on snubbers. Snubbers
delay the rate of rise of the voltage or current in a device in order to keep the transients within the safe
limits of the device and can also be used to minimise the power dissipated during the switching times by
means of wave shaping.

6.3.1 MATHEMATICAL MODELLING OF A TURN-OFF SNUBBER

iT iC
C
I
V
R

I
0 t1 t2

Figure 6.4: Turn-off snubber and waveforms

Assume a linear fall of current in the transistor circuit of Figure 6.4 at turn-off and also that the current in
the load (I) is essentially constant from one cycle to the next. The flywheel diode will switch on when the
voltage over it goes from reverse bias to forward bias. Considering an ideal diode, this will be when the
transistor voltage ( vT ) equals the supply voltage ( V ). From the graph in Figure 6.4 this can be seen as
happening at time t1 .
Derive an expression for the snubber capacitor C that will delay the rise of voltage over the transistor to its
off-state value of V in a time of t1.

From the graph the transistor current is given by:

t 
iT  I    I ……………… (6.1)
 t2 

Applying KCL to the circuit:

I  iT  iC

Thus up to t1 :
iC  I  iT ……………… (6.2)

At time t1 , vT  V , therefore at t1 the charge q on the capacitor is given by: q  CV  0 iC dt

t1

……………… (6.3)

Substitute Equations (6.1) and (6.2) into Equation (6.3):

t1
t t  CV  t 2  t12
CV   I  I    I dt
t1 t1
= I  0   dt    
 t1   t2  I  2t2 0 2t2
0

It12
C  ……………… (6.4)
2Vt2

The expression for the switching energy loss in terms of V, I, t1 and t2 is determined as follows:

The switching loss W by definition =   i  dt

There is thus a need for an expression for vT up to t1 .

From the expression for the instantaneous charge on a capacitor:

q
q  CvT  vT  ……………… (6.5)
C

Substitute Equation (6.3) into Equation (6.5):

vT 
iC dt
……………… (6.6)
C

Substitute Equation (6.4) into Equation (6.6):

vT 
iC dt
……………… (6.7)
2
It 1
2Vt2

Substitute Equations (6.1) and (6.2) into Equation (6.7):

2Vt2  t  2Vt2 t 2Vt t2

It12  
vT  I I  I  dt = 2
 I  dt  22  I
t2  It1 t2 It1 2t2

Vt2
vT  ……………… (6.8)
t12

Thus the switching loss is given by:

Vt 2  t   t 
W   vi dt  
t1
I  dt   V  I 
t2
I  I  dt
0 t12  t 2 
t1
 t 2 

 t1 t 2  t  t2  t  
  VI   2 1   dt   1   dt 
 0 t
1  t2  t1
 t2  
  t1 t 2 t 3 t2  t  
  VI   2  2 dt   1   dt 
 0 t
1 t1 t2 t1
 t2  

 t 3 t4  
t1
t2  
t2

  VI  2  2    t   
 3t1 4t1 t2 0  2t2 t1 

 t 3 t4   t2 t2 
  VI  1 2  12    t2  2  t1  1 
 3t1 4t1 t2   2t2 2t2  

 t 2t t2 
  VI  2  1  1 
 2 3 4t2 

But the energy stored in the capacitor is already lost, since it will be dissipated in resistor R when
the transistor is switched on again. Thus

WTotal  Wcap  W

where

1 1  It 2  2 t2
Wcap  CV 2   1 V  VI 1
2 2  2Vt2  4t2

Thus

t 2t t2  t2 t 2t t2 
WTotal  VI  2  1  1   VI 1  VI  2  1  1 
 2 3 4t2  4t2  2 3 2t2 

d
For a minimum loss WTotal   0
dt1

d   t2 2t1 t12    2 t1 
  VI           0
dt1   2 3 2t2    3 t2 
2
 t1  t2
3

Worked example

A transistor that has a switch-off time of 3 μs carries a current of 60 A. The supply voltage is 120 V.
Determine a turn-off snubber capacitor for the transistor. Design the capacitor for a minimum loss in the
transistor.

t2   3 106   2 μs
2 2
For a minimum loss t1 
3 3

It 2  60   2 10  6 2

C  1   0.333 μF
2Vt2 2 120   3 106 
6.3.2 MATHEMATICAL MODELLING OF A TURN-ON SNUBBER

L V
I

T

I
0 t1 t2

Figure 6.5: Turn-on snubber and waveforms

Assume a linear fall of voltage in the transistor circuit of Figure 6.5 at turn-on and also that the current in
the load (I) is essentially constant from one cycle to the next. The flywheel diode will switch off when the
current through it has been reduced to zero. This will be when the transistor current equals the load current.
From the graph in Figure 6.5 this can be seen as happening at time t1 .

Derive an expression for the inductance L that will delay the rise of current to its on-state value of I in a time
of t1 .
From the graph the transistor voltage is given by:

t 
vT  V    V ……………… (6.9)
 t2 

Applying KVL to the circuit remembering that the freewheeling diode is ON up until t1 :

V  vT  vL

Thus up to t1 :

vL  V  vT ……………… (6.10)

At time t1 , iT  I , therefore at t1 the magnetic flux of the inductor is given by:   LI  0 vL dt

t1

……………… (6.11)

Substitute Equations (6.9) and (6.10) into Equation (6.11):

t1
t t1  t  LI  t 2  t12
LI   V  V    V dt  V    dt 
t1
  
 t1   1 V  2t2 0 2t2
0 0 t
 Vt12
L ……………… (6.12)
2 It2

The expression for the switching energy loss in terms of V, I, t1 and t2 is determined as follows:

The switching loss W by definition =   i  dt

There is thus a need for an expression for iT up to t1 .

From the expression for the instantaneous voltage over an inductor:

diT
vL  L ……………… (6.13)
dt

1
L
 iT  vL dt ……………… (6.14)

Substitute Equation (6.12) into Equation (6.14):

2 It2
 iT 
Vt12  vL dt ……………… (6.15)

Substitute Equations (6.9) and (6.10) into Equation (6.15):

2 It2 t  2 It  t2 
iT  2
  V  V  V   dt  22  V  
Vt1  t2  Vt1  2t2 

t2
 iT  I ……………… (6.16)
t12

Thus the switching loss is given by:

It 2  t  t2  t 
W   vi dt  
t1
2 
V  V  dt   I  V  V  dt
0 t1  t2  t1
 t2 

 t1 t 2  t  t2  t  
  VI   2 1   dt   1   dt 
 0 t1  t2  t1
 t2  

 t1 t 2 t 3 t2  t  
  VI   2  2 dt   1   dt 
 0 t
1 t1 t2 t1
 t2  

 t 3 t4  
t1
t2  
t2

  VI  2  2    t   
 3t1 4t1 t2 0  2t2 t1 

 t 3 t4   t2 t 2 
  VI  1 2  12    t2  2  t1  1  
 3t1 4t1 t2   2t2 2t2  
  t 2t t2 
  VI  2  1  1 
 2 3 4t2 

But the energy stored in the inductor is already lost, since the current will drop to zero when load is
switched off again. Thus

WTotal  Wind  W

Where:

1 2 1  Vt12  2 t2
Wind  LI    I  VI 1
2 2  2 It2  4t2

Thus:

t 2t t2  t2 t 2t t2 
WTotal  VI  2  1  1   VI 1  VI  2  1  1 
 2 3 4t2  4t2  2 3 2t2 

d
For a minimum loss  ETotal   0
dt1

d   t2 2t1 t12    2 t1 
  VI           0
dt1   2 3 2t2    3 t2 
2
 t1  t2
3

Worked example

A transistor that has a switch-off time of 3 μs carries a current of 60 A. The supply voltage is 120 V.
Determine a turn-on snubber inductor for the transistor. Design the inductor for a minimum loss in the
transistor.

t2   3 106   2 μs
2 2
For a minimum loss t1 
3 3

Vt 2 120   2 10 
6 2

L 1   1.333 μH
2 It2 2  60   3 106 