L7-L9 DC Chopper

Power Electronic Devices and Circuits (EEL-209)

Department of Electrical Engineering, Indian Institute of Technology Delhi,

Hauz Khas, New Delhi-10016, India- 110016email: bsingh@ee.iitd.ac.in

Ph.:011-2659-1045

By: Prof. Bhim Singh

1

DC-DC Converters

Choppers

Block Diagram of DC-DC Converters

3

Block Diagram of a DC-DC Converters

DC-DC Converters

4

Choppers Isolated DC-DC

Converters

Non- Isolated DC-DC

Converters

5

DC-DC Converters

Current Source Converter

Voltage Source Converter

Chopper Non Isolated Converter

Isolated Converter

Step Up Class B Chopper

Step Down Class A Chopper

Two Quadrant Class D Chopper

Two Quadrant Class C Chopper

Multi Phase Chopper

Four Quadrant Class E Chopper

Boost Converter Buck Converter

Buck Half Bridge

Converter

Buck Boost Converter

Boost Full Bridge

Converter

Buck Full Bridge

Converter

SEPIC ConverterCuk Converter

ZETA Converter

Boost Forward Converter

Buck Forward Converter

Buck Half Bridge

Converter

Fluyback Converter

Boost Full Bridge

Converter

Buck Full Bridge

Converter

Boost Push Pull Converter

Buck Push Pull Converter

SEPIC ConverterCuk Converter

ZETA Converter

Choppers

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Choppers

7

DC choppers are normally used to convertconstant dc voltage to variable dc voltages for realizing adc current source. It is considered a dc converterequivalent similar to ac transformer with a variable turnsratios.

The output dc voltage is varied through a solidslate switch in pulse width modulation PWM control atconstant or variable frequency switching.

Applications of DC-Choppers

8

• DC motors control in locomotives,• Subway cars,• Battery operated pallet trucks,• Folk lift trucks,• Electric boats,• Submarines,• Battery operated fans and small fans in cooling• Computers,• Toys etc.

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DC MOTORS CONTROL IN LOCOMOTIVES

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SUBWAY CARS

11

FOLK LIFT TRUCKS

12

ELECTRIC BOATS

13

SUBMARINES

14

BATTERY OPERATED FANS

15

COMPUTERS

16

BATTERY OPERATED TOYS

Solid state switches used for choppers

17

• Thyristors with forced commutation (Obsolete)

• GTO• (higher power rating and below 1kHz frequency)

• IGBT• (medium power rating and up-to 20kHz frequency)

• MOSFET• (small power rating and up-to 100kHz frequency)

Types of Chopper

• Step down class A chopper

• Step up class B chopper

• Two quadrant class C chopper

• Two quadrant class D chopper

• Four quadrant class E chopper

• Multi phase chopper18

Circuit Topologiesof Choppers

Circuit Topologies of Chopper

20

Step down chopper

o DV DV=

Circuit Topologies of Chopper

D0

VV(1 D)

=−

21

Step up chopper

Circuit Topologies of Chopper

22

Step up chopper for regenerative braking of DC motor

Circuit Topologies of Chopper

D

D

23

Two quadrant class C chopper

Circuit Topologies of Chopper

24

Two quadrant class D chopper

Circuit Topologies of Chopper

25

Four quadrant class E chopper

Circuit Topologies of Chopper

0 1 2 DV (D D )V= +26

Multi-phase chopper

Circuit Topologies of Chopper

Multi-phase chopper in DC Motor Speed Control

Quadrant Operation of Choppers

Chopper Class A Class B Class C Class D Class E

Voltage polarity Positive Positive Positive

Positive &

Negative

Positive &

Negative

Currentpolarity Positive Negative

Positive&

NegativePositive

Positive &

Negative

Quadrantof

operationI II I & II I & IV

I, II, III &IV

28

Mathematical Analysisof Choppers

29

Step Down Chopper

30

For continuous conduction, the chopper output voltage is independent ofoutput current.

It has two operating modes. In the first mode, solid state switch (S) is onand diode (D) is off and current flows from source to load.

Mode 1

Step Down Chopper cont..

31Mode 2

During second mode, solid switch is off and load current flows throughfree-wheeling diode (D).

In majority of loads, load or circuit time constant (Ta = L/R) is quite highcompared to chopping period (T) resulting in linear rise of load currentwith time and linear fall in second mode of operation.

Step Down Chopper

32

The chopper output average voltage is as:VO = Vin Ton / (Ton + Toff) = Vin D … (1)

Where, Ton is on time, Toff is off time and

(Ton + Toff) = T =(1/f) …(2)

T is total on chopping period. f is switching or chopping frequency. D is duty ratio = Ton/T. … (3)

Step Down Chopper (CCM)

cont..

33

Waveforms for continuous conduction

mode

Step Down Chopper in mode 1

34

The load current (ion) for first mode may be achieved from:

Vin = R ion + L (dion/dt) + E …(4)At initial condition this current ion (t = 0) = Imin. The solution of eqngives the load current as:

ion(t) = Imin e-(tR/L) + (Vin-E/R) (1- e-(tR/L)) …(5)This first mode is for 0<t<Ton (=DT), and at the end of this mode,the load current is as:

ion (t = Ton= DT) = Imax …(6)

Step Down Chopper in mode 2

35

The load current in second mode may be achieved from the voltcurrent equation as:

0 = R ioff + L (dioff/dt) + E …(7)With initial current ioff ( t=0) = Imax and resetting the time origin (i.e.t=0), the current (ioff) in second mode can be expressed as:

ioff (t) = Imax e – (tR/L) – (E/R) (I – e (– tR/L)) …(8)This second mode is only for 0 ≤ t ≤ Toff = {(1 – D) T}. At the endof second mode, the load current is as:

ioff (t =Toff) = Imin …(9)

Step Down Chopper cont..

36

The relations between minimum and maximum currents fromeqns. can be written as:

Imax = Imin e– DT/Ta + {(Vin – E) /R} (1 – e DT/Ta) …(10)And

Imin = Imax e – (1 – D) T/Ta – (E/R) (1 – e – (1-D) T/Ta) …(11)Where Ta is time constant = L/R.Solving eqns. Imin and Imax are as:

Imin =(Vin/R) {(e –Dx– 1) / (ex – 1)} - (E/R) …(12)Where x=T/Ta = TR/L = ratio of chopping period to circuit timeconstant.

Imax = (Vin/R) {e – Dx – 1) / (e – x – 1)} - (E/R) …(13)

37

ΔI/(Vin/R)

The peak to peak load current ripple may be computed as.

ΔI = Imax – Imin = (Vin/R) (1 – e – Kx + e– x – e–(1 – D)x)/(1-e-x) …(14)

Step Down Chopper cont..

Step Down Chopper cont..

38

Conclusions from the above figure

1. The ripple current ΔI reduces to zero as D →0 and D →1.2. Differentiation of equation for ΔI reveals that the maximum

ripple current ΔI occurs at D = ½3. The longer the load L /R time constant, Ta, the lower the output

ripple current ΔI .4. The higher the switching frequency, 1/T, the lower the output

ripple.

Step Down Chopper cont..

The maximum current ripple can be obtained as:D(ΔI)/dD = 0

It results in e-Dx – e–(1–D)x = 0 or –D =-(1-D) or D=0.5.

Thus the maximum current ripple is at D=0.5 and it is:

ΔImax = (Vin/R) Tanh(R/4fL) =(Vin/R) (1 – e –0.5T/Ta)/(1+e–0.5T/Ta ) …(15)

For the assumption 4fL>>R, Tanhψ ≈ ψ and maximum ripplecurrent may be approximately as:ΔImax=Vin/(4fL) …(16)

Step Down Chopper cont..

40

For continuous conduction, various relations can bederived as:

Iin = DIo

from power equalityPin = Vin Iin = Po = VoIin

Input effective resistance is as:Rin=Vin/Iin=R/D

Inductor average current ILav= IoInductor minimum current ILmin = IminInductor maximum current ILmax = Imax

Step Down Chopper cont..

41

Inductor rms current is as:ILrms = [Imin

2 + {(Imax – Imin)2/3} + Imin (Imax – Imin)}Input source average current Iinav=DIoInput source minimum current Iinmin = 0Input source maximum current Iinmax = ImaxSolid state switch (S) average current Isav = DIoSolid state switch (S) maximum current Ismax = ImaxSwitch rms currentIsrms=D0.5[Imin

2+{Imax-Imin}2/3]+Imin(Imax-Imin)0.5

Switch average voltage Vsav=(1-D)VinSwitch maximum voltage Vsmax=VinSwitch rms voltage Vsrms=Vin(1-D)0.5

Step Down Chopper cont..

42

Diode average current IDav=(1-D)Io

Diode maximum current IDmax=Imax

Diode rms current Isrms=√(1-D)[Imax

2+{Imax-Imin}2/3]+Imax(Imax-Imin)] Diode average voltage VDav=-DVin

Diode maximum voltage VDmax=-Vin

Diode rms voltage VDrms=Vin D0.5

Step Down Chopper(Discontinuous Current Mode)

43

For low values of duty ratio especially at low voltage output and high value of battery voltage, E (or back emf), the load current may be discontinuous.

44

Vin/Vo=1.25

Vin/Vo=2.0

Vin/Vo=5.0

Step-Down DC-DC Converter: Limits of Cont./Discont. Conduction

Step Down Chopper (DCM) cont..

45

Waveforms for discontinuous

conduction mode

Step Down Chopper (DCM)

46

For the case of discontinuous current, Imin=0 and ion(t)={(Vin-E)/R}(1-e-t/Ta) …(17a)

Current flowing during the off stateioff(t)={(0-E)/R}(1-e-t/Ta) …(17b)

Moreover, for 0<t<Tcioff(t=Tc)=0,

which results in: Tc=(L/R) ln{1+(RImax)/E} …(18)

At t=DT, current ion(t) is as:ion(t)=Imax={(Vin-E)/R}(1-e-Dx) …(19)

After substituting for Imax, critical time Tc is as:Tc=(L/R) ln{1+{(Vin-E)/E}(1-e-x)] …(20)

Step Down Chopper (DCM) cont…

47

However, for boundary condition of continuous current for Imin ≥ 0,

[{(eDx-1)/(ex-1)}-(E/Vin)]≥0 …(21)

This provides the value of ratio (E/Vin)=Z as:

Z=(E/Vin)≤(eDx-1)/(ex-1) …(22)

Step Down ChopperFilters

48

To improve the performance of the chopper at inputsource and output load, a set of filters is used.The input filter is normally L–C filter andOutput filter is generally an inductor (L), which maybe a separate inductor or part of the load orcombination of external inductor and inductance of theload.The objective of output filter is to make an ideal ripplefree dc current in the load or at least continuouscurrent

Step Down ChopperFilters

49

Step Down ChopperOut-Put Filters

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Normally, the design of this output filter inductor is made for aspecified current ripple in the load current, which is alsodependent on chopper frequency f. The maximum current rippleoccurs at duty ratio D = 0.5 and

…(23)

Where x=T/Ta, TR/L=R/(Lf); y=∆ImaxR/Vin .

Eqn.(9.43) may be modified as:

T/Ta=2 ln{(1+y)/(1-y)} …(24)

L=(R/f)/ [2 ln{(1+y)/(1-y)}] …(25)

( ) ( )0.5 1 / 1xe y y− = − +

Step Down ChopperIn-Put Filter

51

It is observed that without input filter supply current is same aschopper switch current which varies from zero (during turn off) tomaximum load current (during on period of the chopper). Thistype of fluctuating current from dc supply, which may causeundesirable effects of voltage fluctuations, EMI and RFI, supplyvoltage distortion, additional heating etc.

Step Down ChopperIn-Put Filter

52

The nth harmonic current in the dc supply is as:Isn=[(Xc/n)/{(nXL)-(Xc/n)}]ICHn …(26)

Where, XL=2лfL, Xc=1/(2лfC), n is order of harmonic.Isn=ICHn/{(2лnf)2LC-1}= ICHn/{(nf/fr)2-1} …(27)

Where fr = resonant frequency of the filter = 1/{2л√(LC)}.These two frequencies f and fr must be different to avoidresonance, which results in large oscillation of supply voltage.Normally f is selected more than twice of fr In such case supplyharmonic current may be approximated to:

Isn={fr/(nf)}2ICHn …(28)

Step Down ChopperIn-Put Filter

53

For an ideal chopper with perfect filters on both input source andoutput load shown in Fig. 9.3, the following relations simplifiedas:

Voltage ration Vo/Vin=D …(29)

Current ratio Io/Iin=1/D …(30)

Average switch current Isav = DIo …(31)

RMS switch current Isrms=Io √D …(32)

Average diode current IDav = Io(1–D) …(33)

RMS diode current IDrms=Io√D …(34)

Step Up Chopper

54

Class B chopper is used to step up variable voltage to constant voltage

Mode 1 Mode2

Step Up Chopper

55

Waveforms in CCM Mode of operation

Step Up Chopper (CCM)

56

Average input voltage (Vin) can be expressed in terms of outputvoltage (Vo) as:

Vin=Vo Toff/(Ton+Toff)=Vo(1-D) …(35)

Alternately, a boost inductor used in step up chopper does notstore any energy under steady state condition in a PWM period. Itmeans the energy, which it stores in on period same it releasesduring off period. Therefore, the energy stored in input inductorduring on period must equal to off period considering current inthe inductor remains constant (I) as:

VinITon=(Vo-Vin)IToff …(36)

Step Up Chopper (CCM) cont..

57

Solving it, the relation between input and output voltages is as:

Vo=Vin(Ton+Toff)/Toff=VinT/(T-Ton)=Vin/(1-D) …(37)

For deriving current relationship, it is considered that an inputpractical inductor (L) has a small definite resistance (R).Assuming an ideal filter at the output high voltage (Vo), a volt-current equation during turn on period is as:

L(dion/dt)+Rion=Vin …(38)

With initial condition ion= Imin at t =0.

Step Up Chopper (CCM) cont..

58

With initial condition ion= Imin at t =0.The volt-current equation during turn off period is as:

Vo-Vin=Rioff+L(dioff/dt) …(39)

With the initial condition ioff= Imax at t=0. …(40)

Imax and Imin maximum and minimum currents. These maximum and minimum currents values are to be same under steady state condition in different chopping cycles. The solution of eqns. is with proper initial conditions as:Imax=(Vin/R)(e-Ton/Ta)+Imine-Ton/Ta …(41)Where Ta= L/R.

Imin={(Vin-Vo)/R}(1-e-Toff/Ta)+Imaxe-Toff/Ta …(42)

Step Up Chopper (CCM)

59

Further solving these equations the Imax and Imin are as:Imax=(Vin/R)-(Vo/R){(e-Ton/Ta-e-T/Ta)/(1- e-T/Ta)} …(43)Imin=(Vin/R)-(Vo/R){(1-e-Toff/Ta)/(1- e-T/Ta)} …(44)

The peak to peak current ripple is as:∆I=Imax-Imin=(Vo/R){(1+ e-T/Ta)-(e-DT/Ta+e-(1-D)T/Ta)}/ (1- e-T/Ta) …(45)

The condition for maximum current ripple can be derived in thesimilar manner and it also appears at D = 0.5.The average input current can be derived as:

Iinav=(Vin-Vsav)/R={Vin-(1-D)Vo}/R …(46)

Step Up ChopperDiscontinuous Current Mode

60

The input current drawn from dc source (Vin) of low voltage maybecome zero before the end of the off period of the chopper. Theboundary condition between continuous and discontinuous currentmay be achieved for Imin=0, which results in as:

(Vin/Vo)={(1-e-Toff/Ta)/(1- e-T/Ta)} …(47)

However, DCM may be avoided either by increasing the value ofinductance or frequency of chopping or both.

61

Vin/Vo=0.25

Vin/Vo=0.5

Vin/Vo=0.8

Step-Up DC-DC Converter: Limits ofCont./Discont. Conduction

Step Up ChopperFilters

62

• Normally a high value of an inductor (Li) is used in theinput, which is low voltage side to reduce the currentripple to a specified value.

• A C-L filter is used on output high voltage side to haveripple free current and ripple free voltage.

Step Up ChopperFilters

63

The following relations can be derived as:Iin/Io=1/(1-D)Average input current Iinav=Io/(1-D)Switch average current Isav=IoD/(1-D)Switch average voltage Vsav=Vin=Vo(1-D)Switch peak voltage Vsmax=VoDiode average current IDav=IoDiode maximum current IDmax= ImaxDiode average voltage VDav=DVoDiode peak voltage VDmax=Vo

64

Class B chopper

Mode 1 Mode 2

continuous inductor current Discontinuous inductor current

65

R io+ L (dio/dt) =E …(48)

During the switch on-period, when vo = 0

Which yields

io(t) = Imin e-(t/Ta) + E/R (1- e-(t/Ta)) (0≤ t≤Ton) …(49)

During the switch off-period, when vo = Vs

E= R io + L (dio/dt) + Vs …(50)

Which yields

io (t) = Imax e – (t/Ta) – ((E-Vs)/R) (I – e (– t/Ta)) (0≤ t≤Tc) …(51)

m a x

m in

(5 2 )1

1 (5 3 )1

TTa a

sT

T a

T To nT a

sT

T a

tTT

VER R

VER R

e eIe

eIe

+

− −

−

−

−

−= −

−

−= −

−

66

Two-quadrant dc chopper - Q I and Q II

Mode 1 Mode 2

67

Waveforms for quadrant 1 Waveforms for quadrant 2

68

Four possible output modes can occur, depending on therelative polarity of the maximum and minimum currentsshown in figure as shown above

Imin > 0, Imax > 0 and Io > 0When the minimum current (hence average output current) isgreater than zero, the chopper is active in the first-quadrant. Theswitch T2 and diode D2 do not conduct during any portion of theoperating period.Imin < 0, Imax > 0 and Io > 0When the minimum current is negative but the maximum positivecurrent is larger in absolute magnitude, then for a highly inductiveload, the average output current is greater than zero, and thechopper operates in the first-quadrant. If the load is not highlyinductive the boundary is determined by the average output currentIo > 0. The various circuit waveforms are shown in figure (a).

69

Imin < 0, Imax > 0 and Io < 0For a highly inductive load, if the magnitude of the negativepeak is greater than the positive maximum, the average is lessthan zero and the chopper is operating in the regenerative mode,quadrant II. If the load is not highly inductive the boundary isdetermined by the average output current Io < 0.

Imin < 0, Imax < 0 and Io < 0When the maximum current and the average current are bothnegative, the chopper is operational in the second-quadrant.Since the load current never goes positive, switch T1 and diodeD1 never conduct, as shown in figure

70

T1 and T4 forming a +Vs path T1 and D4 (or T4 andD1) forming a zerovoltage loop

D1 and D4 creating a -Vspath

Two-quadrant dc chopper - Q 1 and Q IV

71

Operation in quadrant I Operation in quadrant IV

72

T1 and T4 VsT1 and D4 0T1 and T4 VsT4 and D1 0 (not T1 and D4 again)T1 and T4 VsT1 and D4 0, etc.

D1 and D4 -Vs (that is T1 and T4 off)T1 and D4 0D1 and D4 -VsT4 and D1 0 (not T1 and D4 again)D1 and D4 -VsT1 and D4 0, etc.

In regeneration mode

There are two types of switchingMultilevel switching is when both switches are controlled independently to

give all three output voltage states (three levels), namely ±Vs and 0V.Bipolar switching (or two level switching) is when both switches operate in

unison, where they turn on together and off together. Only two voltage outputstates (hence the term bipolar), are possible, +Vs and – Vs.

73

Four quadrant chopper

74

Multilevel (three-level) output voltage(a) with Vo> 0 and Io > 0

Multilevel (three-level) output voltage(b) with Vo < 0 and Io < 0

75

The output voltage switches between + Vs and – Vs and the relativeduration of each state depends on the magnitude of the modulation indexD.If D = 0 then T1 and T4 never turn-on since T2 and T3 conductcontinuously which impresses – Vs across the load.At the other extreme, if D = 1 then T1 and T4 are on continuously and Vsis impressed across the load.If D = ½ then T1 and T4 are turned on for half of the period T, while T2and T3 are on for the remaining half of the period. The output voltage is– Vs for half of the time and + Vs for the remaining half of any period.The average output voltage is therefore zero, but disadvantageously, theoutput current needlessly ripples about zero (with an average value ofzero).The chopper output voltage is defined in terms of the triangle voltagereference level vΔ by• vΔ > δ, vo = -Vs• vΔ < δ, vo = +Vs

76

Bipolar (two-level) output voltage(c) with Vo > 0 and Io > 0

Bipolar (two-level) output voltage(d) with Vo < 0 and Io < 0

Multi-phase Chopper

77

Multiphase chopper is a chopper, which has several identicalchoppers operated in parallel and phase shifted from each other.

These may also be configured in different quadrant operation.They offer the advantages of reduced ripple current and an

increase in ripple frequency resulting in reduction in supplyharmonic current.

Multi-phase Chopper

78

They are preferred in large power rating, when either use severalsolid state devices in parallel or use multiphase chopper, whichavoids the problems of matching of dynamic and staticcharacteristics of several devices in single chopper.

Each chopper in P-phase chopper is operated at same frequencyand duty cycle.The start of switching is phase shifted by (T/P) seconds and(360°/P) in terms of angle.Effect of phase shift on current waveforms can be observed as anincrease in ripple frequencies .

Multi-phase Chopper

79

Multi-phase Chopper

80

The number of choppers are to be on simultaneously atany instant of time, is dependant on the value of dutycycle D.For 0 < D<(1/P), there is an interval when only onechopper switch is on and all other phase choppers arefree wheeling.During this period net load current rising to maximumvalue Imax. Followed by this, there is an interval whenall chopper switches are off and all phase’s diodes arefreewheeling. During this period, net load current isdecaying towards minimum value Imin.

Multi-phase Chopper

81

In this fashion, it can be generalized that for any value ofD, {(Q-1)/P} < D < (Q/P), where Q is a positive integerless than or equal to P, there is two modes of conductionas:Q choppers are on and load current is increasingtowards Imax.Only (Q-1) choppers are on and load current decreasesto Imin.

Where P is no of phases while Q is no of phasesconducting

Multi-phase Chopper

82

For writing basic volt-current relations, it is considered that Qphases are conducting and (P-Q) phases freewheeling diodes areconducting. For such condition, these equations are as:

Ri1+L(di1/dt)+RLio+LL(dio/dt)=Vin …(54)

Ri2+L(di2/dt)+RLio+LL(dio/dt)=Vin …(55)

RiQ+L(diQ/dt)+RLio+LL(dio/dt)=Vin …(56)

RiQ+1+L(diQ+1/dt)+RLio+LL(dio/dt)=Vin …(57)

RiP+L(diP/dt)+RLio+LL(dio/dt)=Vin …(58)

Multi-phase Chopper

83

Eqns. (54)-(58) can be lumped together as:R1io+L1(dio/dt)=V1 …(59)

Where R1=R+PRL, L1=L+PLL, and V1=PVin …(60)

Eqn. (59) is applicable to the condition of increasing io, when Qchopper switches are on. Considering this period Ti of increasingcurrent as:

Ti=DT-{(Q-1)/P}T=T{D-(Q-1)/P} …(61)

Considering reference time t=0 at the instant of rising current withinitial condition as:

io=Imin at t=0 …(62)

Multi-phase Chopper

84

Based on this eqn. (62), the solution of eqn. (59) is as:io=Imine-t/Ta+(V1/R1) (1-e-t/Ta) …(63)

Where equivalent time constant is as:Ta=(L1/R1) …(64)

The maximum current Imax occurs at t=Ti, therefore:Imax=(V1/R1)(1-e-Ti/Ta)+Imine-Ti/Ta …(65)

During the period of the decay of load current io, (Q-1) switches areon and other phase choppers are freewheeling. The duration of thisperiod is from turn off switch 1 to the turn on (Q+1) switch, which isas:

Td=(Q/P)T-DT=T{(Q/P)-D} …(66)

Multi-phase Chopper

85

During decay of current, only (Q-1) equations will have Vin onRHS and remaining are having zero. By adding all equations, theresulting equation is as:

R1i0+L1(di0/dt)=V2 …(67)Where V2=(Q-1) Vin …(68)

Considering a new reference for the period of decay of current,the initial condition for eqn. (67) is as:

I0=Imax at t=0 …(69)Solving the eqn. (67) during decay of current is as:

i0=Imax e-t/Ta+(V2/R1)(1-e-t/Ta) …(70)

This current decay to minimum current Imin at t=Td, as:Imin = Imax e –Td/Ta – (V2/R1) (1 – e – Td/Ta)

Multi-phase Chopper

86

Solving eqns. (66) and (70) for Imin and Imax as:

Imax = (Vin/R1) {Q-(e – Ti/Ta – e – T/PTa)/(1- e – T/PTa)} …(71)

Imin=(Vin/R1) {Q-(1-e – Td/Ta)/(1- e – T/PTa)} …(72)The relation between P and Q can be written as:

(Q-1)/P<D<(Q/P) …(73)

(T/P) is the duration of the output current ripple, which is as:(T/P)=Ti+Td …(74)

Multi-phase Chopper

87

Average value of output voltage and current are as:V0=DVin-R(I0/P)=I0RL …(75)I0=DVin/(RL+R/P) …(76)V0=DVinRL/(RL+R/P) …(77)

For the load having battery or back emf (E), eqns.(71) and (72)may be modified similar to single equivalent chopper as:

Imax=(-PE/R1)+(Vin/R1){Q -(e – Ti/Ta – e – T/PTa)/(1- e – T/PTa)} …(78)Imin=(-PE/R1)+(Vin/R1){Q-(1-e – Td/Ta)/(1- e – T/PTa)} …(79)

Multi-phase Chopper (DCM)

88

For a boundary condition between CCM and DCM,keeping Imin=0, it results in as:

(E/Vin)= {Q-(1-e – Ti/Ta)/(1- e – T/PTa)}/P …(80)

Numerical Examples

89

Numerical Example -11. In a battery-operated car, the battery voltage is 144 V. It

employs a dc motor with armature resistance of 0.25 ohmsthrough step up dc chopper. During regenerative braking,when car is moving down hill at some speed its back-emf is120 V and braking motor current is 15A. Calculate (a) theduty ratio, (b) value of peak, rms, and average current in thebattery, and (c) values of peak, rms, and average voltage andcurrent of the solid-state switch and diode

90

Solution

91

a aV E R IV 120 3.75 116.25

= −= − =

B a tVV

(1 D )1 1 6 .2 51 4 41 D

=−

=−

Dutyratio,D .19=

inpeak a

inaverage a

inrms a

Battery ParametersI I 15A

I (1 D)I 12.15A

I (1 D)I 13.5A

= =

= − =

= − =

Dpeak in

Daverage in

Drms in

V V 144V

V DV 27.36V

V DV 62.76V

= − = −

= − = −

= =

92

Speak a

Saverage a

Srms a

I I 15A

I DI 2.85A

I DI 6.54A

= =

= =

= =

Switch voltage and currents

Speak in

Saverage in

Srms in

V V 144V

V (1 D)V 116.64V

V (1 D)V 129.6V

= =

= − =

= − =

Diode currents and voltages

Dpeak a

Daverage a

Drms a

I I 15A

I (1 D)I 12.15A

I (1 D)I 13.5A

= =

= − =

= − =

Numerical Example- 2

2. A first quadrant dc chopper feeding to a dc motor the out putcurrent is constant due to large inductance of motor. (a) Showthat max. input current ripple takes place at D=0.5, (b) designan input filter with switching frequency is 400 Hz, armaturecurrent 100 amp if I1rms is 10% of average dc current letavailable capacitors are of 1µF and can take 5amp current

93

Solution

94

inaverage a

inrms a

2 2 2inripple a a a

2inripple a

2inripple

I DI

I DI

I (( DI ) (DI ) ) I D D

d I I (1 2D)dDd I 0

dD1 2D 0D 0.5

=

=

= − = −

= −

=

− ==

ain inave

inave a

ain1rms

I4i I (sin t sin 3 t ...........)2n

I I D 100*.5 50AI4 1I ( ) 45.1733A

2 1 2

= + ω + ω +Π

= = =

= × =Π ×

95

But input ac current not to surpass 10% of dc current

in 1 rm s in a v e

Ci in 1 rm s

L C

C

L C

L C

LC 1 in 1 rm s

L C

C 1

1 0I I 5 A1 0 0

XI ( ) IX X

X5 ( ) 4 5X X

X 1 0 XXI ( ) I

X XI 5 0 A

= =

=−

= ×−

=

=−

=

But maximum capacitor current is 5 amp. Therefore 10cap can be connected in parallel

equivalent

L C

C 10 F

X 10XL 0.15821mH

= μ

==

Numerical Example -3

3. A dc step down chopper controlled electric train is fedfrom a 1750 V dc source. The minimum effective ON timesolid state switching device is 30 μsec. If the chopperfrequency is 1.5 kHz, what is the minimum possibleoutput voltage?

SolutionVin = 1750VTon = DT = 30µsecf = 1.5 kHzT = 1/f = 0.667msecD = Ton/T = 0.045Vo = DVin = 0.045*1750 = 78.75V

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Numerical Example - 4

4. A battery powered, chopper fed dc motor driven pallettruck is running at constant speed. The battery voltage is112V and motor back-emf is 60 V. The chopper frequencyis 2 kHz and duty ratio is 80%. The motor circuitresistance is 0.8 ohms and inductance is 0.25 mH.Calculate (a) peak, average and rms motor currents, and(b) peak to peak value of the current ripple of the motor.

Solution(a) (i)Imax = (Vin/R){(1-e-(Ton/Ta))/( 1-e-(T/Ta))} – E/R

= (112/0.8){(1-e-(0.4/0.3125))/( 1-e-(0.5/0.3125))} – 60/0.8= 51.32A

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(ii) Imin = (Vin/R){(e(Ton/Ta)-1)/( e(T/Ta)-1)} – E/R= (112/0.8){(e(0.4/0.3125)-1)/( e(0.5/0.3125)-1)} – 60/0.8= 16.96A

(iii) Io = (Vo – E)/R = (DVin- E)/R = (0.8*112 – 60)/25 = 37A(iv) Iorms = {Imin2+(Imax - Imin)2/3+ Imin(Imax - Imin)}0.5

= {16.962+(51.32 – 16.96)2/3+16.96(51.32 – 16.96)}0.5= 35.55A

(b) ∆Io = (Imax - Imin) = 51.32 – 16.96 = 34.36A

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Numerical Example - 55. The four phase chopper is operating at 125 Hz and at a duty cycle of

0.55. These choppers are having the inductors of 5.0 mH with theirresistances 0.5 ohms. This chopper is feeding RLE load having R=5ohms and L=5 mH and E=120V. The dc supply voltage is 240 V.Calculate (a) ripple current in the load, (b) frequency of output ripplecurrent, (c) average voltage across load resistance, (d) average loadcurrent, (e) values of peak, rms, and average voltage and current ofthe solid state switch and diode, (f) value of E at which load currentwill become just discontinuous, and (g) value of duty cycle at whichload current will become just discontinuous with E=120 V.

SolutionTa = (L+Ls/N)/(R+Rs/N) = 1.22msecD1 = (Ta/T)ln[1+E/Vin*(e(T/Ta) – 1)]= (1.22/8)ln[1+(120/240)*(e(8/1.22) – 1)= 0.895

99

Here the actual duty ratio D becoming less than D1, therefore it becomes DCM operation.

Imax = ((Vin – E)/Req)*(1-e-( Ton/ Ta))= ((240– 120)/5.125)*(1-e-(4.4/1.22))= 22.78A

(a) ∆Io = Imax = 22.78A(b) frequency of o/p ripple current = 125Hz(c) Io = (Vo – E)/R

Vo = DVin+(1 – Tx/T)ETx = Ton+Ta*ln[1+((Vin – E)/E)*(1-e-( Ton/ Ta))]

= 7.77msecVo = 0.55*240+(1 – 7.77/8)*120 = 135.45V

Io = (135.45 - 120)/5.125 = 3.015A

Voltage across load resistance = IoR = 3.015*5 = 15.075A

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(d) Io = 3.015A(e) (i) Vspeak = Vin = 240V

(ii) Vsave = (1/T){Vin(Tx - Ton)+(Vin - E)(T - Tx)} = 104.55V(iii) Vsrms = {(1/T)[Vin

2(Tx - Ton)+(Vin - E)2(T - Tx)]}1/2 = 157.09V(iv) VDpeak = Vin = 240V(v) VDave = (1/T)(VinTon+E(T - Tx)) = 135.45V

(vi) VDrms = {(1/T)(Vin2Ton+E2(T - Tx))}1/2 = 179.15V

(vii) Ispeak = Imax/4 = 5.695A(viii) Isave = Imax*(Ton/(2T))/4 = 22.78*4.44/(2*8)/4 = 1.58A(ix) Isrms = Imax*√(Ton/(3T))/4 = 22.78*√(4.44/(3*8))/4 = 2.495A(x) IDpeak = Imax/4 = 5.695A

(xi) IDave = Imax*((Tx – Ton)/(2T))/4 = 1.2A(xii) IDrms = Imax*√((Tx – Ton)/(3T))/4 = 2.134A

(f) If D = 0.55D = (Ta/T)ln[1+E/Vin*(e(T/Ta) – 1)]0.55= (1.22/8)ln[1+(E/240)*(e(8/1.22) – 1)E = 12.22V

(g) D1 = 0.895 101

References

1. N. Mohan, T. M. Undeland and W. P. Robbins, “Power Electronics, Converter, Application and Design”, Second Edition, John Willey & Sons, 1995, New York, ISBN 9971-51-177-0.

2. M. H. Rashid, “Power Electronics, circuits, Devices andApplications”, Second Edition, Prentice-Hall, 1995, India,ISBN 81-203-0869-7.

3. B. W. Williams, “Power Electronics: Devices, Drivers andApplications”,Wiley,1987.NewYork,ISBN: 0470206969.

4. W. C. Lander, "Power Electronics", 3rd Edition, McGraw-Hill, 1993, New York, ISBN: 0077077148S.

5. Joseph Vithyathil, “Power Electronics principles andapplications”, Tata McGraw-Hill Edition.

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Thank you

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