switch losses

7/28/2019 Switch Losses

http://slidepdf.com/reader/full/switch-losses 1/24

Presentation Summary

•

Types of switch Losses

•Switch Losses under different operating conditions

•Numerical example on switching losses

• Turn-on and turn-off snubbers

•Heat sink design

•Two numerical examples on heat sink design



Losses in a power electronic device

• Switching losses:

Losses encountered during switch-on or switch-off transients. These losses are

proportional to switching frequency.

• Conduction losses:

Losses encountered by the device when it is ON or conducting .

• Off state losses:

Losses encountered by the device when it is OFF. These losses are often neglected.



Case I: Resistive load switching

The switch resistance is assumed to vary between R on and R off

(R on ≈ 0 and R off ∞ assuming ideal switch).

DC

R

is

vsV bus

+

- Switch



Switching Profile: Resistive load

(1) During switch on:

= 0

= 1− 0

=1

6

(2) When switch is turned off:

= 0

= 1 −

0=

1

6

V bus

Im

time

vs

is

t = 0 t =ton t = 0 t =toff

vs , is

Ton

Toff

Toff



Switching Profile: Resistive load (2)

If the switch operates at frequency , the total switching loss (power loss) is given by:

( + ) =1

6Vbus ( + )

Conduction loss = (

+)2 ∙ ∙ ; =

; = + ; =

1

Off state loss = (

+)2 ∙ ∙ (1 − ); 1− = 1−

=

Assuming off state loss to be negligible, the total loss is given by

Total Loss =16Vbus ( + ) + (

+)2 ∙ ∙ .

Note: & are negligible with respect to on or off .



Case II: Inductive load switching (constant

current)

•Relays, motors are examples of inductive loads.

•Inductive load is represented as follows.

•To simplify analysis constant current is assumed.

timet= 0 t =ton t= 0 t =toff

Vbus

Im

In practice these times may be

considerable

is

vs

vs , is

D onD off



Switching Profile: Inductive load

Switch on:

•Initially the diode is carrying the full current I.

• Once the switch is turned on, its resistance keeps falling at a very rapid rate and

current builds up in the switch till it takes all the current away from the diode.

•As long as the diode conducts, the voltage across it is zero and the switch

carries all the voltage.

• Once the diode switches off and the switch resistance falls at a less

rapid rate till it reaches zero.

• Although the initial time required for the current build up is shown negligible

for simplicity, in actual case it may be considerable.



Switching Profile: Inductive load (2)

Switch off:

•Initially the switch is carrying the full current I.

•The switch resistance keeps building up at a very rapid rate till the voltage across

the switch reaches V bus.

•Then the diode turns on.

•Subsequently the switch resistance increases at a slower rate till all the current is

taken over by the diode.

•Although the initial time required for the voltage build up across the switch is

shown negligible for simplicity, in actual case it may be considerable.



Switching Profile: Inductive load(3)

•During switch on:

•During switch off:

The total switching loss is given by :

The conduction loss for MOSFET will be:



Switching Profile: Inductive load(4)In case of IGBT/BJT, the conduction loss is given by :

The off state loss will be:

or

Therefore, total loss (neglecting off-state losses):

(MOSFET)

(IGBT/BJT)

or



Case III: RL type of load (Variable current)

(Continuous Current)

Assumption (S is BJT):

(1) Current in the load is continuous.

Total power loss=

T ON

iload

T s

t

Imax

Imin

V bus

D

S

L

R

i load



Case III: RL type of load (Variable current)

(Discontinuous Current)

(2) Current in the load is discontinuous:

Total power loss=

T ON

iload

t

Imax

T s



Example of loss calculation

Find the total loss for the circuit shown below:

,

V bus

ImD

S

1. Switching loss:



Example of loss calculation (contd.)

2. Conduction loss:

3. Off state loss:

(negligible).

Therefore, the total loss (neglecting off state loss) = 12.49 W.

In case, where switching losses are high we need to use snubber circuits

(to reduce switching losses).



Snubber Circuit

Suppose in the earlier circuit kHz f s 250 ?

What will be the switching losses?

Following the earlier calculation it is

A solution is to use a snubber circuit. Snubbers reduce overlap between switch voltageand current during switch-on and switch-off , reducing switching losses. Another

solution is to use soft-switching .

( + ) =1

2Vbus + =

30 ∗ 302

∗ 50+60 ∗ 10−9 ∗ 250 ∗ 103

= 11.24W



Turn-on and Turn-off snubbers

Turn-on snubber (inductor, L) : Reduces the rate of rise of current through the

switch when it is turned on.

Turn-off (RCD) snubber: Takes away the energy in the turn-on snubber and the other

associated lead inductances and stores it in capacitor C at turn-off. This reduces the

rate of rise of voltage and voltage spikes across the switch at turn-off. At turn-on ,C

discharges through the switch. The peak discharge current is limited by the resistor, R.The RC time constant controls the discharge time. The diode bypasses the resistor at

turn-off. ( See Assignment problem 3.19).

V

Turn-on

snubber

Turn-off

snubber

L

D

R

C

Note: The snubbers reduce voltage current overlap and hence switching-

losses also.

time

Vbus

Im

Switching locus changed in presence of

turn-on and turn-off snubbers.

is

vs

vs , is



Heat Sinks

• Heats sinks are required to save devices from thermal destruction.

•Heat sinks dissipate the losses in the form of heat taking place in a power semiconductor.

•Usually, the device is fixed tightly to the heat sink by means of nuts and bolts.

• Good thermal contact is ensured between the device and the heat sink by means

of heat sink compound.



Heat sink calculation

Elements involved in a heat sink calculation

(i) Source of heat (power loss) = P

(ii) Final destination of heat (cooling medium usually ambient)

Let the temperature of the ambient be TA °C

(iii) Heat that is generated at the semiconductor junction be at a temperature

TJ °C(iv) Temperature of the semiconductor case TC °C

(v) Temperature of the heat sink = TS °C

(vi) R JC – thermal resistance between junction and case

(vii) R CS – thermal resistance between case and sink

(viii) R SA – thermal resistance between heat sink and ambient

k d



Heat Sink design

TC TS TATJ

R JC

R CS

R SA

P

Model (electrical equivalent of device connected to a heat sink)

From the above figure,

TJ – TC = R JC P (1)

TC – TS = R CS P (2)

TS – TA = R SA P (3)Adding the equations (1), (2) and (3)

TJ – TA = P (R JC + R CS + R SA)



Heat Sink design (2)

In most cases, TA is known.

R JC - given by device manufacturer.

R CS – also given by the device manufacturer.

R SA - depends upon heat sink, to be designed.

TJ – is also known (device manufacturer specifies).

Units involved:P – W; T - °C; R - °C/W.

Lower R SA means bigger heat sink. Lower R SA will keep the device cooler.

Forced convection by blowing air over heat sink reduces R SA.



Heat Sink Design Example I

The following data for a MOSFET is provided.

TJ = 150 °C; R JC = 0.8 °C/W; R CS = 0.2 °C/W.

If TA = 50 °C find a heat sink with suitable R SA if the MOSFET loss is 25 W.

Solution:

R SA = (TJ – TA)/ P – (R JC + R CS) = (150-50)/25 – (0.8+0.2)=3 °C/W



Heat Sink Design Example 2

V bus

ImD

M

Consider the above power circuit with current source type of load. The diode D and MOSFET M are mounted together on a heat sink.

Find R SA for the heat sink to be used. Given:

PD = 25 W; TJD(max) = 125 °C; R JCD = 1 °C/W; R CSD = 0.1 °C/W

PM = 40 W; TJM(max) = 140 °C; R JCM= 0.5 °C/W; R CSM = 0.1 °C/WTA = 50 °C



Solution to Example 2

TS TA

R SA

PD

TJD

PM

TJM

R JCD TCDR CSD

R JCM TCM

R CSM

(TJD – TS)/ PD = R JCD + R CSD = R JSD = 1+0.1=1.1°C/W (1)

(TJM – TS)/ PM = R JCM + R CSM =R JSM = 0.5+0.1=0.6 °C/W (2)

(TS-TA)/(PD+PM) = R SA (3)



Substituting TS from (3) in (1)TJD = TA + R SA. (PD+PM) +R JSD.PD =50+ R SA *65+ 1.1*25 ≤ 125 °C (4)

Substituting TS from (3) in (2)

TJM = TA + R SA. (PD+PM) + R JSM.PM = 50+ R SA *65+ 0.6*40 ≤ 140 °C (5)

From equation (4)

R SA ≤ 0.731 °C/W

From equation (5)

R SA ≤ 1.015 °C/W

Common sense dictates that the heat sink with the lower thermal resistance must be selected.

Thus a heat sink with R SA ≤ 0.731 °C/W should be chosen.

Solution to Example 2 (contd.)

switch losses

Documents