review dc switched pwm single analysis dc … section 2...review of dc motors • control of a...

Section 2 - DC Motor Drives

• Review of DC motors and characteristics• Switched‐mode PWM converters.• Single‐ and three‐phase thyristor converter circuits.

• Analysis of converter and DC motor circuits.• Effects of discontinuous conduction on drive.

1ELEC4613 – Section 2 – DCM Drive F. Rahman/Aug 2019


• Review of DC motors and characteristics (elec3105)

• Switched‐mode PWM converters.

• Single‐ and three‐phase thyristor converter circuits.

• Analysis of converter and DC motor circuits.

• Effects of discontinuous conduction on drive.


Review of DC Motors

• Control of a separately excited DC motor is very straightforward, via ia and if, which are de-coupled from each other. The commutator-brush assembly provides for this simplicity.

• AC machines strive to emulate such control via machine-model based controllers which are rather complex.

• The commutator-brush has many limitations and maintenance issues.


Working Principle

Field is either from electro or permanent magnets The field circuit is stationary - in the stator The armature carries conductors in slots and rotates with the rotor


DC Motor Electric Circuits

• For large DC machines, the field is from electro‐magnets, there are two circuits which can both be controlled

– Field circuit– Armature circuit

• For small DC machines (< 20kW), the field is from permanent magnets, there is

– No field circuit; allows no field control– Only armature current control


The Field Circuit

ff f f f

div R i L

dt

ff

f

VI A

R

The air-gap B field (in Tesla) is constant for a constant fieldcurrent If, as is the flux per pole, f, in Weber. For a linearmagnetic circuit, .

When field excitation is constant,

Vf

if

Lf

Rf

If

Bf

f f fK I 6

ELEC4613 – Section 2 – DCM Drive F. Rahman/Aug 2019

Separately Excited DC Machine EquationsTorque

Back emf

KT in Nm/A = KE in V/rad/sec.

'em r a T f a T aT 2N lrBi k i K i

'a r m E f m E me 2N lrB k K

va ea

Ra La

vf

ia RfLfif

aa a a a a

div R i L edt

f

f f f f

div R i L

dt


In the steady‐state,

a a a aV R I E '

a a E f mR I K

a a a a a am ' '

E f E f f

V I R V I RK K K I

em ema ' '

T f T f f

T TIK K K I

a a

m em' 2' 'E f f E T f f

V R TK K I K K K I

+ m ,rad/sec

Va

m

V a

Tem

m

-Tem

-m

Va

-Va8

Steady‐state Torque‐Speed Characteristic with Variable va

m emA BT

Steady‐state Torque‐Speed Characteristic with Variable Va

9

a aem T a T

a

T a E T m

a

T a T E m

a a

V ET K I KR

K V K KR

K V K KR R

Here' '

'

T T f T f f

E f E

K k k k I

k K

in SI units. (rad/s, Nm, V, A).Tstall is the stall torque when rated Va is applied i.e., Va = Varated

ω-axis intercept Va

Constant Slope


T‐ω Characteristic with Variable Φ (Va=Vao, ω> ωb)

10

= 0.3 pu

= 0.5 pu

= 1 pu

m , rad/sec

T e, N m

,ao a a ratedb

E

V R IK

ao am em' ' ' 2

E E T

V R TK K K

m 0 'stall T ao aT K V R

eT 0 'm ao EV K

Here,

CONST' 'T EK ,K

Stall torque is proportional to field

No-load speed is inversely proportional to field


Va, If and Ia Boundaries

T, Nm

m, Rad/sec

T, Nm

m, Rad/sec

Field Control

ArmatureVoltage Control

Envelop for max

arm. current

Va, Rated

Ia, Rated

If, min

11

CONSTT


A

BD

C

Example 1: Steady‐state torque‐speed characteristic

12

An application requires a continuous torque of 0.86 Nm (7.6 lb-in) at a speed of 2750 RPM. The peak torque required for acceleration is 6.25 Nm (56.8 lb-in). Will M-3358-C work in this application with Va = 100V and 50V?

M-3358-CPoint A (0.75 Nm / 2750 rpm) is in the continuous operation area (OA).

Point B (6.25 Nm) is in the intermittent OA.

0.183 2750 3.14 / 30 0.14 4.953.4V

a E m a aV K R I

0.183 V/rad/s, 1.4 OhmE aK R

DC motor ratings:Vao = 100 V, Nr = 5000 rpm,Ia,rated = 4.9 A, Trated = 0.86 Nm.

If Va = 50 V, the maximum speed at full load (0.86Nm) is 2375 rpm;The peak toque is only achieved at lower speed.

Series excited DC machine

13

Va Ea

Rf Lf Ra La Ia = If

Field circuit Armature circuit

A+

A-

F-

F+

'a a f a a a f a E f mV R R I E R R I K

'a f a E f a mR R I K K I

m, rad/sec

T, Nm

T, Nm

m, rad/sec

Va increases

Va increases

'em ema a f E f m' '

t f t f

T TV R R K KK K K K

eT K ELEC4613 – Section 2 – DCM Drive F. Rahman/Aug 2019

' 2em T f aT K K i

Shunt excited DC machine

14

Ra

La

Ea

IaRf

Lf Va

If

'a a a E f mV R I K

a a a a a am ' ' '

E f E f E f

V R I V R IK K K

f f f f a fK I K V R e fa '

t f a

T RI

K K V

2f a f

m e' ' ' 2 2E f E t f a

R R RT

K K K K K V

If

Bf


Losses in DC machine

15

shaft devT T

out out

in out losses

P P 100%P P P

Field loss (shunt)

Armature copper loss

Commutator-brush losses (contact drop

short-circuiting)

Mechanical losses

Windage and frictions

Core losses (hysteresis,

eddy current)

in a aP V I

2f f fP I R

bP

2a a aP I R

out m shaftP Tdev a a m devP E I T


Example 2: Losses in DC machine

16

The parameters of Kollmorgen U9D-B servo DC motor are given in the following table. What is the developed torque of the motor at rated output power?

Performance specifications Symbol Unit value

Rated power output P Watts 133

Rated continuous current I Amps 8.64

Back EMF constant KE V/kRPM 6

Torque constant KT N-cm/Amp 5.7Rated continuous torque T N-cm 42.4

Rated speed N RPM 3000

N cm

shaft out mT P133 3000 3.14 / 3042.4

N cm >

dev T

shaft

T K I5.7 8.6449.2 T

W >

dev

out

P EI6 3 8.64155.5 P

V/rad/s =

ET

K 6 1000 3.14 305.73 K



• Review of DC motors and characteristics


• Single‐ and three‐phase thyristor converter

circuits.




Converters for DC motor drives

Power Converter Armature

vc iF

Power Supply

va ea

Field Power

Converter

Power Supply

Two types of converters for DC motor drive:

1. PWM converters for low power DC motors (of several KW)2. Thyristor converters for medium and large DC motors

Armature voltage control up to rated Va; field control above base speed.


PWM switching pulses

P W M e c D

P o w er E lec tro n ic C o n v erte r

v a

Comparator output: High for TON ; Low for TOFF

vtri vc

t

t

19

a,avrg s

c tri s

V DVe / V V

SWsignals


PWM DC-DC converter in continuous conduction mode (CCM)

Vs

Ra

La

Ea

Ia

D

T

20

Vs

Ra

La

Ea

Ia

D

Ton0 ~ T

on ST ~ T

Vs

Ra

La

Ea

Ia

D

T

SV av

0

0

ai

minImaxI

on ST DTt

taE

ST

a SV DV

off ST 1 D T

00

SV av

0

0

ai

minImaxI

on ST DTt

taE

ST

a SV DV

off ST 1 D T

aI

PWM DC-DC converter in continuous conduction mode (CCM)

offs on s

s

TT T DTs s

a a s ss s s0 0 DT

1 1 V DTV v ( t )dt V dt 0dt DVT T T

ia

Ton =

DTs

Toff =

(1-D)Ts

IaImax

Imin

Vs Va

Ea

va

t

t

Ts

0

0

Vs

Ra

La

Ea

Ia

D

T


2 2 D

Sn a S

0 0

V1 1a v cos n t d t V cos n t d t sin 2n Dn

2

Sn a

0

V1b v sin n t d t 1 cos 2n Dn

2 2 sn an n n

2Vˆc V a b sin n Dn

From Fourier analysis

22

0a n n

n 1

av a cos n t b sin n t2

sT

0a a s

s 0

a 1V v ( t )dt DV2 T

Ripple amplitude, not RMS


• The ripple voltage is maximum for D = 0.5.

fs 3fs2fs0

Va

4fs

sD T2

aRM S s s0s

1V V dt D VT

2 2 2 2 2

a a1 a2 a3 a4V V V V V ........

where a1 a2 a3a1 a2 a3

ˆ ˆ ˆV V VV ; V ; V ;2 2 2

…..

• The DC voltage Va develops ia, torque and useful output power.• The ripple voltages cause ripple currents in the armature and

additional loss in the machine. 23

sD T2

aRM S s s0s

1V V dt D VT

2 2 2 2

acRMS a a1 a2 a3V V V V V ........

• The ripple voltage is maximum for D = 0.5.• The DC voltage Va develops Ia, torque and useful output power.• The ripple voltages cause ripple currents in the armature circuit, resulting in

additional loss in the machine.24

Exercise 1: Max. ripple voltage in PWM DC-DC converter in CCM

• Find the RMS value of the AC voltage across the armature and the duty cycle when the RMS ripple voltage is maximum?

2 2acRM S aRM S a SV V V V D( 1 D)

acRM SS

dV 1 2 D V 0dD 2 D( 1 D)

acRM S SV 0.5V

D 0.5


Analysis of ia at constant speed. Continuous Conduction Mode (CCM)

During 0 t Tona

s a a a adiV R i L Edt

25

SV av

0

0

ai

minImaxI

on ST DTt

taE

ST

a SV DV

off ST 1 D T

Vs

Ra

La

Ea

Ia

D

T

Particular solution: s aa1

a

V EiR

Homogeneous solution:a

a

R tL

a2 1 2i C C e

s a a1 aV R i E

a2a a2 a

di0 R i Ldt

1 a1t C i

1 2 a mint 0 C C I a a

a a

R Rt tL Ls a

a amina

V Ei 1 e I e

R

Exercise 2: Analysis of ia at constant speed. (CCM, DTS t TS)

At t = DTs, ia = Imax, s a s aDT / DT /s a

a max a mina

V EI 1 e I e

R

aa a a a

di0 R i L Edt'

on st' t T t DT where

a at'/ t '/aa a max

a

Ei 1 e I eR

(13)

26

During DTS t Ts,

Vs

Ra

La

Ea

Ia

D

T

SV av

0

0

ai

minImaxI

on ST DTt

taE

ST

a SV DV

off ST 1 D T

aI

1 a a

1 2 a max

t' C E Rt' 0 C C I

and , the electrical time constantaa

a

LR

Iamin occurs at t’ = Ts – DTs =(1 – D)Ts

s a s a( 1 D )T / ( 1 D )T /aa min a max

a

EI 1 e I eR

(16)

From 13 and 16

s a

s a

DT /s a

a max T /a a

V E1 eIR R1 e

s a

s a

DT /s a

a min T /a a

V Ee 1IR Re 1

S a S a

S a S a

DT / DT /S S

a ripple amax amin T / T /a a

V V1 e e 1I I IR R1 e e 1

(17)

(18)

(19)

27

s a s as a DT / DT /a max a min

a

V EI 1 e I e

R

(13)


T- characteristic with CCM

a s a a aV DV R I E 'a E f m s a aE K DV I R

s aa

a

DV EIR

s a a s a a

m ' 'E f E f f

DV I R DV I RK K K I

Disc

min Ia or Tem

Ea or m

D = 1.0

D = 0.5

D = 0.25

D = 0.75

028

Boundary of CCM and DCMImin = 0


PWM DC-DC converter indiscontinuous conduction mode (DCM)

Vs

Ra

La

Ea

Ia

D

T

29

ia=0


ia

Ton = DTs

Ia

VsEava

Ts

t

Toff =(1-D)Ts

0

a sv VDuring 0 t DTs

During DTs t t va = 0

During t t Ts va = Ea

sT

a a s a0s s

t1V v dt DV 1 ET T

s sDT T

2 2 2 2aRMS s a s a

0 ts s

t1V V dt E dt DV 1 ET T

s a

ns

tV Ea sin 2 nD sin 2 nn n T

s an

s

2 ntV Eb 1 cos 2 nD 1 cos ;n n T

2 2n an n n

ˆc V a b

30

Analysis of ia at constant speed. Discontinuous Conduction Mode (DCM)

During 0 t DTs at /s aa

a

V Ei 1 eR

s aDT /s aa max

a

V EI 1 eR

During freewheeling (i.e., diode conducting)

a at'/ t'/aa a max

a

Ei 1 e I eR

s a

s a s a

( t DT )/a

a

DT / ( t DT )/s a

a

E 1 eR

V E 1 e eR

s a s aDT / DT /s aa

a

V Et ln e 1 1 eE

31

The armature current becomes zero at t , given by

(2.2.33)

s a

s a

D' T /a

T /s

E e 1V e 1

For a given speed (Ea), the boundary between CCM and DCM (whenIamin = 0) occurs for a duty cycle D’ for which ia = 0 at Ts. Thus

D > D’ implies operation in CCM (Imin > 0).

CCM

Ia or Tem

Ea or m

D = 1.0

D = 0.5

D = 0.25

D = 0.75

D = 0.00

DCM

32

Additional inductance in the armature may be required to reduce thepower loss due to ripple current in the armature and to prevent DCMoperation. The required minimum inductance Lamin for CCM can befrom 2.2.33 or 2.2.34.

T-w characteristics with discontinuous conduction

(2.2.34)


Determine the boundary on motor characteristics (Matlab)

Ea

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40

20

40

60

80

100

120

140

160

180

200

220

Ia

For a given substitute into

= for boundary

is a function of DCM

Characteristics

a

a a a a

s

a a

a a

D tt V

I V E / Rt T

I EE ,I

2.2.33


a sv VDuring 0 t DTs

During DTs t t va = 0

During t t Ts va = Ea

s s sT DT T

a a s a s a0 0 ts s s

t1 1V v dt V dt E dt DV 1 ET T T

s sDT T

2 2 2 2aRMS s a s a

0 ts s

t1V V dt E dt DV 1 ET T

s an

s

tV Ea sin 2 nD sin 2 nn n T

s an

s

2 ntV Eb 1 cos 2 nD 1 cos ;n n T

2 2n an n n

ˆc V a b


Armature current ripple via Fourier analysis

an

n 22a n a

V / 2I

R L

2 2 2 2aRMS a 1 2 3I I I I I .......

The input power to the armature, ignoring other losses,

sT2

in a a aRMS a a a0s

1P v i dt I R E IT

35

DC power input to the motor, neglecting core losses,2

in,dc a a a aP I R E I

Why is there no Ea in the equation?

Only DC component contributes to the developed power.


Operation in quadrants 1 & 2

Ra La

Ea

IaVs

T

D

+

Vs

Ra

La

Ea

Ia

D

T

Q1

Q2

36

Regenerative PWM DC-DC converter

Ts 0

Va

Ea

Ton = DTs Toff = (1-D)Ts

D is ON

ia

Imin

Imax

D is OFFt

Ia

0

Vsva

t

Ra La Ea

Ia Vs

T

D

+


Analysis of ia in Q2 at constant speed

During 0 £ t £ Ton

Note: Diode D is on during DTs, switch T is on during (1 - D)Ts.a

s a a a adiV E L i Rdt

During t - Ton £ t £ Ts, aa a a a

di0 R i L Edt

s a

s a

DT /s a

a max T /a a

V E1 eIR R1 e

s a

s a

DT /s a

a min T /a a

V Ee 1IR Re 1

a ripple a max a minI I I 38

The differential equations are the same as those in Q1


T1

T2

D1

D2

Ra La

Ea

Ia Vs

is

Ia or Tem

Ea or mD = 1.0

D = 0.5

D = 0.25

D = 0.75

0

Q1Q2

T1

T 2

is

V sv a

ia Ia

39

T1

T2D1 D1D2

The straight-line T-ω characteristics of Q1 CCM extends into Q2 with the same slopes and intercept

T1

T2D4

Ra LaIa Vs

D2T4

T3 D3D1

4Q PWM DC-DC converter drive

40

Q1

Q3

Q2

Q4+ia, +T-ia, -T

+va, + m

-va, - m

FB FM

RBRM

• Unipolar: (Vs/0; 0/-Vs)Only one switch is controlled for a given polarity of the output voltage, the current freewheels through the other switch and a diode during Toff ;• Bipolar (±Vs)Two diagonal switches are always switched together.


T1

T2D4

Ra LaIa Vs

D2T4

T3 D3D1

4Q PWM DC-DC converter drive (Unipolar)

41

T1

T2D4

Ra LaIaVs

D2T4

T3 D3D1

• Unipolar (Va>0)

During Ton During Toff

SV av

0

0

ai

onTt

t

ST

aI

offT

T1&T2 T1&T2T2&D4


T1

T2D4

Ra LaIa Vs

D2T4

T3 D3D1

4Q PWM DC-DC converter drive (Unipolar)

42

T1

T2D4

Ra LaIaVs

D2T4

T3 D3D1

• Unipolar (Va<0)

During Ton During Toff

SV av0

0ai

onT

t

t

offT


43

Exercise 3: 4-Q DC-DC converter drive

Sketch va and ia in bipolar switching mode, indicating the conduction paths of ia through the switches and diodes.

T1

T2 D4

Ra La Ia Vs

D2T4

T3 D3D1

T1

T2 D4

Ra La Ia Vs

D2T4

T3 D3D1

T1

T2D4

Ra LaIaVs

D2T4

T3 D3D1

T1

T2D4

Ra LaIaVs

D2T4

T3 D3D1

SV av

0

0

ai

onTt

t

offT

SV SV av0

0ai

onT

t

t

offT

SV

Unipolar versus bipolar switching in 4-quadrant converter (SW signals, ripple) The PWM switching frequency is selected from the following considerations:1. for the switching frequency, 2fsLa >> Ra (from current ripple consideration)

2. High switching frequency reduces the current ripple and motor losses. It alsoavoids discontinuous conduction.

3. fs should be much higher than the speed control bandwidth. Thus fs > 10×speedcontrol bandwidth.

4. fs should be higher than any significant resonant frequencies

5. fs should be sufficiently high to avoid audible noise (> 5kHz)

6. Too high switching frequency will result in excessive switching losses in theswitching devices (transistors).

7. Too high switching frequency limits the range of output and introduces offsetinto the power converter input-output characteristics. At high switchingfrequencies the finite delay times of gate switching circuits and dead-times fordevice protection may become comparable to the switching period.

Switching scheme and PWM switching frequency



• Review of DC motors and characteristics


• Single‐ and three‐phase thyristor converter

circuits.




Thyristor converter drive for DC motor

• A controlled diode, turned on by gate current pulse when forward biased.• It continues to conduct while the voltage across it is not reversed, even

when the current into the gate stops. • It will be turned off when the anode current falls to zero.


Single phase half-wave Thyristor AC-DC converter

F C C

vc

LaRa

eaVmax sint

va

ia

47

vs va

ia

Ea

Va

0adidt

0adidt

= 45

S a a av E i R 0


as max a a a a

div V sin t R i L edt

a

a

R tLmax a

a 22 aa a

V Ei sin t Ae ;RR L

1 a

a

Ltan

R

At t = , ia = 0,

a

a

RLmax a

22 aa a

V E0 sin AeRR L

a

a

RLmax a

22 aa a

V EA sin eRR L

a

a

R tLmax a max a

a 2 22 2a aa a a a

V E V Ei sin t sin eR RR L R L

48

2

max aa max a

V E1V V sin t d( t ) E d( t ) cos cos 22 2 2


At at , i 0, thus,

a

a

RLmax a max a

2 22 2a aa a a a

V E V E0 sin sin eR RR L R L

49

When is found, ia and va waveforms are completely known, and then Ia and Va can be determined.

Va, m

Ia, T

1

2


Single-phase fully-controlled thyristor bridge converter drive

50

Va

ip

Vmaxsint

T1

T2T4

T3 Ra

La

Ea

iais

T1 and T2 are triggeredva = vs , is = ia

T3 and T4 are triggeredva = -vs , is = -ia

Va

ip

Vmaxsint

T1

T2T4

T3 Ra

La

Ea

iaisVa

ip

Vmaxsint

T1

T2T4

T3 Ra

La

Ea

iais


Single-phase fully-controlled thyristor bridge converter drive

a max

max

1V V sin td( t )

2V cos

51

va 135 (Q 4)

va 45 (Q1)

-vs vs

-vsvs

-isis

max amax L'a aT

m ' 'E E

2V R2V cos Tcos I R KK K

52

Ia

Ea Va

1

Ia

Ea Va

1


Armature voltage and current ripples

maxn

cos n 1 cos n 1Va2 n 1 n 1

maxn

sin n 1 sin n 1Vb

2 n 1 n 1

2 2n n nv a b n = 2, 4, 6, ..... Only even-order harmonics

n

n n2 2a a

vi sin n t ;R n L

a a nn 2,4 ,6 ,.....

i I i ;

a a max aa

a a a

V E 2V EI cos

R R R

Ripples in armature current cause additional losses 2n aI R

1 an

a

n LtanR


n

n 22a a

VIR n L

n maxn

VV

2

2 2 2 2aRMS a 2 4 6I I I I I

aRMS

a

II

Armature Form Factor =

a a RMS 1 1 11

RMS RMS RMS RMS RMS

V I V I cos IIPF cosV I V I I

Converter input Power Factor (ideal converter) =

54

Note that 1 is the power factor angle associated with thefundamental (i.e., harmonic order n = 1) of input voltage andcurrent waveforms. It is largely determined by the firing angle .

(A measure of motor heating)

Distortion factor


Analysis of ia in CCM

max sin aa a a a

diV t R i L Edt

t Solving for

a

a

R tLmax a

aa

V Ei sin t AeZ R

22a aZ R L

1tan a

a

LR

In the steady-state, armature current falls to its minimum value at t = , , …

min ( ) ( )a a aI i i

55

A


RaLa

maxRa

La

2V eA sinZ

e 1

a

a

a

a

RL

max aa min a R

aL

V Ee 1I i ( ) sinZ R

e 1

1 ( )a aI i t d t

maxor, ( / c s 2 o)a a a aa

a aI V E R V E

R R

Complicated

Simple

It is also a straight line for a certain firing angle.

'a a a

E

V I RK

The critical (required minimum) armature inductance

For operation at the boundary of CCM and DCM, .

The condition for minimum Lamin is given by

a

a

a

a

RL

a aR22 maxLa a

e 1R Esin

VR Le 1

57

Note that the critical (minimum required) armature inductance is found from this transcendental eqn. 2.2.36.

[2.3.36]


Exercise 4: Find in DCM and Va

ia falls to zero at t = .

58

a

a

R tLmax a

aa

V Ei sin t AeZ R

( ) ( ) 0a ai i

a

a

a

a

RLmax a

aRLmax a

a

V Esin AeZ R

V Esin AeZ R

a

R / L R / Lmaxa a a a

a

max

Ecos sin( )Ve eEcos sin( )V

aRcosZ

2.3.25


Exercise 4: Find in DCM and Va

59

a max a

max a

1V V sin t d( t ) E d( t )

V Ecos cos


ω T characteristics in DCM

T, Nm

= 60

= 0

= 150

= 170

m rad/sec

60

for a given Ea (2.3.25).

Substitute into

A series of ω-T curves for different firing angles can be drawn. At the boundary

DCMaV

aV

a a a aI V E R

a max a

max a

1V V sin t d( t ) E d( t )

V Ecos cos

DCM

a

R / L R / Lmax

a

max

Ecos sin( )Ve eEcos sin( )V

2.3.25

Effect of source inductance on speed

Ia

Va

ip

Vmaxsint

T1

T4 T2

T3Lsi

vsi

Ra

La

+ Ea

max2 2cos s aa

V L IV

max2 2cos sa a

mE

V L R I

K

61

A further droop in ωTCharacteristics (CCM)

voltage droop factor


62

va

µ

• The input current through Ls changes by 2Ia , the missing voltage is

• All four thyristors conduct during commutation overlap (µ)due to the source inductance;

max sin sdiV t Ldt

2a

a

I

max s s aIV sin t d t L di L I

max2 2cos s aa

V L IV

Effect of source inductance on speed

Single-phase half-controlled thyristor bridge converter drive

Vmax sint

Ea

Ra

La

Ia

T1 T2

D1 D2

Df

Ls a max1V V sin t d t

maxV 1 cos

V s

i a

V a

i T 1

i D f

i s

0 2

Ia or T

Va

or m

Q1

63

T - ω characteristic under HC drive

max sa T

mE

V L1 cos R T / K

K

64

• The input current through Ls changes by Ia due to thefreewheeling path.

0

1 1 aI smax s a

LV sin t d t L di I

max 1 cos s aa

V L IV


T - ω characteristic under HC drive

65

• HC has a higher IPF, because the lagging component of inputcurrent is freewheeled locally rather than fed back to input.

• Lower ripples in armature current and voltage and less likely to be DCM.

• Low cost.

• First quadrant only, because the freewheel diode prevents thearmature voltage to become negative.

HC drive vs. FC drive:


Ra

LaVa

T1 T3 T5

T4 T6 T2

iaia

ibic

van

vbn

vcn

+ Ea

23

a maxl l3

maxl l

1V V sin td t/ 3

3V cos

Q1

Q4

Va

Ia

66

Three-phase fully-controlled thyristor bridge converter drive

V1n

V2n


FC Converter waveforms

67

T1 on, V1n=van;T3 on, V1n=vbn;T5 on, V1n=vcn;T4 on, V2n=van;T6 on, V2n=vbn;T2 on, V2n=vcn;Va=V1n-V2n;T6,T1 on, Va=vab;T1,T2 on, Va=vac;T2,T3 on, Va=vbc;T3,T4 on, Va=vba;T4,T5 on, Va=vca;T5,T6 on, Va=vcb.

FC Converter waveforms (Ls)

68

02aImaxl l

s s aV sin t d t L di L I

The voltage pulse is missing during µ:

maxl l sa a

mE

3V 3 Lcos R I

K

With CCM and negligible source inductance

maxl la a

mE

3V cos I R

K

With CCM and source inductance LS

maxl la

aa

3V cos EI ;

R


Output voltage ripples

maxl l

n

2sin n 1 cos n 1 2sin n 1 cos n 13V 6 6an 1 n 1

max l l

n

2 sin n 1 sin n 1 2 sin n 1 sin n 13V 6 6bn 1 n 1

2 2 max l l

n n max n n 2 23V 1 1 2cos 2c V a b

n 1 n 1n 1 n 1

for n = 6, 12, 18, .... Multiples of 6th harmonics


Critical inductance for continuous conduction

a

a

a aR

a max l lL 3

sinR Esin

Z 3 Ve 1

T, Nm

= 60

= 0

= 150

= 170

m rad/sec


The output waveforms of 3-ph converter are smoother.3-ph: multiples of 6th order;1-ph: multiples of 2nd order.

The lower current ripple calls for smaller inductance required for CCM.The effective converter switching frequency:

3-ph: 300 Hz;1-ph: 100 Hz.

The input current waveform has better distortion factor. This calls forreduced filter requirement at the input AC side.

3-ph: 6k ± 1;1-ph: 2k ± 1.

Complexity, cost and power handling capacity of 1-ph converter is lowerthan 3-ph converter.

72

Three-phase vs. single-phase converter drive


Half-controlled three-phase thyristor bridge converter driven DC motor

Ra

La

Vd

va

T1 T3 T5

D4D6 D2

iLia

ib

ic

van

vbn

vcnn

+VD/2

VD/2

Df

e a

maxl la

3VV 1 cos2

Firing angle

Va

Va

Ia

Q1


HC converter waveforms

74

2

3 3 maxl la maxl l maxl l

3

3V3V V sin t d t V sin t d t 1 cos2 3 2

1 2

21


Converter Voltage Gain (3-ph FC)

1max l l max l la c c

3V 3VV cos cos v v

cos-1 Firing

Control Circuit

vc va

75

max l l3V

Assuming CCM

The firing angle is made equivalent to the control voltage, vc , to the firing controller.

Between the firing controller and motor terminal, the converter behaves as a voltage gain of

1ccos v

maxl l3V


Four-Quadrant Converter

Ra

La

Ea

Ia

1 + 2 = 180 1 2

C1 C2

76

Assuming CCM

Suppressed half control mode: (either C1 or C2 is enabled)C1, α1: Q1 (Va1>0, Ia>0, α1<90) and Q4 (Va1<0, Ia>0, α1>90); C2 disabled. C2, α2: Q2 (Va2>0, Ia<0, α2>90) and Q3 (Va2<0, Ia<0, α2<90); C1 disabled.Crossover delay for a smooth transfer of current between motoring and generating.


Circulating current mode

77

Ra

La

Ea

Ia

1 2

C1 C2

1 + 2 = 180

• Both converters are operated together, 1 + 2 = 180.

• Due to the instantaneous output voltage differences of C1 and C2, a circulating current flows, which is normally limited by a center-tapped inductor.

• Zero transfer delay.


The armature current includes a DC value and ripples.DC value produces the developed torque Kt Ia; Ripples produce ripple torque and extra heating

Motor output power is proportional to Ia.

Copper loss is proportional to Iarms squared.

Because of the ripple current In, the motor is to be de-rated by the factor Ia/Iarms.

A 50kW DC motor is regarded as a 40kW machine if the ratio is 0.8, extra heating loss is 10kW;

21

nn aI R

2copper arms aP I R

dev a aP E I

Motor de-rating due to ripple current


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