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Power Electronics

CHOPPERS

DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM

2018 Dr. Francis M. Fernandez

Choppers

DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM 2

It is a DC-DC converter

It can be step up chopper or step down chopper

ChopperDCin DCout

Step-down Chopper (Buck converter)


S

OFFON

ONSO V

TT

TVVoutput,Averege

Switching is done at chopping frequency, typically 5 kHz to 25 kHz

SCR, GTO, BJT, MOSFET or IGBT can be used as switch

MOSFET and IGBT are preferred as switch for common applications since special commutation circuits are not required.

OFFON

ON

TT

Tcycleduty

toequal iswhich , theis

SS

OFFON

ONSRMSO VV

TT

TVV

22

)(

LOAD

Step-up Chopper (Boost Converter)


1

1output,Averege SO VV

T

Tie

TT

Tcycleduty ON

OFFON

ON

, toequal iswhich , theis

IS

VO

TON TOFF

IO

Buck-Boost Converter


• Output polarity is different when compared to buck or boost converter

• For 0 < α < 0.5 step down operation is achieved

• For 0.5 < α < 1.0 step up operation is achieved

Averege output,1

O SV V

T

Tcycleduty ON , theis

Control Strategies


Time Ratio Control

PWM Control(Frequency is constant and pulse width is controlled)

Variable Frequency Control(Pulse width is constant and frequency is controlled)

Current Limit Control(Switching is done at upper limit and lower limit of output current)

IO

VO

TON TOFF

IO(MIN)

IO(MAX)

Example

A DC chopper connected to a 120 V source supplies to an inductive loadof 30 mH in series with a 6 Ω resistance. A freewheeling diode isconnected across the load. The load current varies between 10A and12A. Determine the duty cycle of chopper.

7DEPT. OF ELECTRICAL ENGINEERING, COLLEGE OF ENGINEERING TRIVANDRUM

10 12Average value of load current 11 A

2

120Maximum value of load current 20 A

6

11Average value of ouput voltage 120 66 V

20

66Duty Cycle 0.55

120ON O

ON OFF S

T V

T T V

Example

A step-up chopper is used to generate 220 V from 100 V dc source. Theblocking period of switch is 80μs. Compute the required pulse width.


1Output voltage,

1O SV V

Duty cycle, = ON

ON OFF

T

T T

1220 100

1

then, 0.5454

0.5454 = 80

ON

ON

T

T = 96 μsONT

IS

VO

TON TOFF

IO

Chopper Classification


Type -> Class A Class B Class C Class D Class E

Quadrants

Output Polarity

Voltage: +Current: +

Voltage: +Current: -

Voltage: +Current: ±

Voltage: ±Current: +

Voltage: ±Current: ±

Typical Application

DC Motor operation

Regenerative braking of DC motor

Both motoring

and regenerative

braking of DC motor

Both motoring

and regenerative

braking of DC motor

Reversible regenerative motor drive

IO

VO

Type A Chopper


IO

VO

LOAD

Average value of both voltage and current are positive

This chopper is also called step down chopper

Freewheeling diode FD maintains the current when switch S is off

Power flow is always from Source to Load

Type B Chopper


The load should have a battery source or it may be a motor which has back emf

Average value of voltage is positive but current is negative

This chopper is also called step up chopper

Power flow is from Load to Source

dt

diLEVO ,voltageLoad

VO may be more than VS and make diode D forward biased

Type C Chopper


IO

S2

D1

VS

VO

L

E

S1 D2

Average value of voltage is positive but current may be positive or negative

Switches S1 and S2 should not be turned on at the same time

Power flow may be in either direction

Used for motoring and regenerative braking of dc motors

Is a combination of Type A and Type B Choppers

Type C operating modes


Type A operation: Type B operation:

IO

VO

Type C Waveforms


IO

S2

D1

VS

VO

L

E

S1 D2

Depending on duty cycle, the average value of load current IO may be either positive or negative

S1 ONS2 OFF

S1 OFFS2 ON

Type D Chopper


IO

VS

VO

L E

S1 D2

S2D1 Is a combination of two Type B Choppers

Type D Operation


IO IO

Direction of load current IO

is the same

Type D Waveforms


TON > TOFF TON < TOFF

Average voltage is positive and output current is also positive; so first quadrant operation

Average voltage is negative and output current is positive; so fourth quadrant operation

Type E Chopper


Is a combination of two Type C Choppers

IO

S2

D1

VS

VO

L

E

S1 D2

Type E Chopper – in Q1


First Quadrant Operation

Common Condition

Operation Current flow Action

S2, S3 off, S4 is kept on

S1 is turned on through S1, L, E and S4 Energy stored in Inductor L

S1 is turned off through D2, L, E and S4 Freewheeling current flows



Second Quadrant Operation

Common Condition


S1, S3 and S4 are kept off

S2 is turned on through L, S2, D4 and E Energy stored in Inductor L

S2 is turned off through D1, L, E and D4 Current fed back to source



Third Quadrant Operation

Common Condition


S1, S4 off, S2 is kept on

S3 is turned on through S3, E, L and S2 Energy stored in Inductor L

S3 is turned off through L, S2, D4 and E Freewheeling current flows

Polarity of E is reversed here



Fourth Quadrant Operation

Common Condition


S1, S2 and S3 are kept off

S4 is turned on through S4, D2, L and E Energy stored in Inductor L

S4 is turned off through D3, E, L and D2 Current fed back to source

Polarity of E is reversed here

Switching Regulators


iC

FDVS L

OAD

L

Control

VOV1iL

iS

iL

V1

VS

VO

TON TOFF

Switching regulators are used to convert an unregulated DC voltage to regulated DC voltage

Buck, Boost and Buck-boost type are available

Efficient compared to linear regulators

Buck type regulator

Filter Design


LOAD

Voltage across inductor, L

div L

dt

2 1S DC

ON ON

I I IV V L L

T T

ON

S DC

L IT

V V

1 2DC

OFF OFF

I I IV L L

T T

OFF

DC

L IT

V

S DC ON DC OFFV V T V T

IL L

ONT T 1OFFT T

DC SV V DC

S

V

V

During rise time of iL

During fall time of iL

Equating ΔI from the two expressions

If α is duty cycle

ON OFFT T T

ON

ON OFF

T

T T


1ON OFF

DC S DC S

S DC DC DC S DC DC S DC

T T Tf

V L I V V L I V L IL I L I

V V V V V V V V V

ON

S DC

L IT

V V

OFF

DC

L IT

V

1DC S DC S S S S

S S

V V V V V V VI

V f L V f L f L

L C Oi i i

If load ripple current is assumed to be zero L Ci i

1

. S

DC S DC

V L Iie

f V V V

Applying KCL

1SVIf L


2

0

1Peak-to-peak ripple voltage of capacitor,

4 8 8

T

C

I I T Iv dt

C C C f

1SVIf L

2

(1 )

8S

C

Vv

f LC

Substituting the value of ��

Average capacitor current for a duration ,2 4

C

T II

Ci

Example

A buck regulator has an input voltage of 120V, and the average output voltage is 70V. Average load current is 25A and the switching frequency of 25 kHz. The peak to peak ripple current in inductor is limited to 1A and the peak-to-peak ripple voltage at the output is 2V. Find the value of inductor and capacitor.


1SVIf L

2

(1 )

8S

C

Vv

f LC

Applicable equations:

Solution


70Duty cycle, 0.5833

120DC

S

V

V

3

120 0.5833 1 0.58331.0

25 10 L

1.17 mHL

2

3 3

120 0.5833(1 0.5833)2.0

8 25 10 1.17 10 C

2.49 μFC

1SVIf L

2

(1 )

8S

C

Vv

f LC

Typical MOSFETS


Typical IGBTs


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