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4. Thyristor Converters for DC drives

4.1 Introduction
The thyristor is perhaps the most efficient and most robust power semiconductor switch. It is a
four-layer (3-junction) semiconductor device which can be turned on by a small gate current of a
few milliamps. The voltage (and current) ratings of thyristors range from a few tens to several
thousands of Volts (Amperes). The thyristor, however, does not gate turn-off capability, at least
in its conventional form. (The gate turn-off thyristor exists, however, they require special drive
gate drive requirements). Once the thyristor is on its anode-cathode voltage is about 1.5 volts
and it can then be turned off by removing the gate current and by removing the anode current by
some auxiliary or the so-called natural means. The term natural, also called natural
commutation refers to thyristor operation in circuits with AC supply. In these circuits, the
anode current attempts to reverse due to the nature of the supply voltage, which is of alternate
polarity. The thyristor has very good reverse blocking capability, and also very good forward
blocking capability when its gate current is kept at zero. In this course we will only consider
conventional, naturally commutated thyristor converters.

4.2 The single-phase half-wave converter driven S. E. dc motor

Consider the simple single-phase, half-wave converter of figure 4.1 in which the thyristor gate is
triggered by a firing control circuit which is synchronised to the ac supply. The supply voltage to
the converter is single-phase ac at 50 Hz, at a voltage which is appropriate for developing the
required maximum dc voltage at the motor load. The peak of the ac supply voltage is Vmax. The
control voltage vc selects the angle of the trigger pulse (called the firing angle, ) in each ac
(positive) half-cycle.

Ra

vc

La

ia

Vmax sint

F
C
C

ea

va

Figure 4.1(a)
Vo

Va

Ea

ia

Va

Vs

Figure 4.1(b)
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In the above thyristor converter circuit, the gate trigger current pulse is issued at and the
armature current starts to build up through the RL circuit of the armature against its back emf.
The armature current reaches a peak and comes down to zero at angle , which is well before the
next trigger pulse due to the combined effect of the negative input voltage of the supply and the
back emf of the armature, ea. The armature current may sustain beyond angle due to the
inductance La of the armature. From to , the output voltage of the converter is the same as the
ac input voltage, which is now negative. This is because the thyristor is ON during this time.
Thereafter, the armature current remains zero and the the back emf of the motor voltage appears
across the output terminals of the converter. These waveforms are shown in figure 4.1(b). The
varaibles Va, Ea and Ia are the dc (average) levels of the respective quantities, but these vary with
time, as the control voltage vc (or ) varies. For a firing angle , the dc output voltage of the
converter is given by
Va =

1
2

Vmax sin t d( t ) =

Vmax
( cos cos )
2

(4.1)

Note that for this converter, the output voltage waveform across the motor has one voltage pulse
per input ac cycle and that the motor voltage and current waveforms have significant ripples.
These are undesirable, since they cause extra machine losses, torque ripple, noise and input ac
current distortion.

Analysis of armature current

In the following analysis, we assume that the armature current starts from zero in each cycle,
i.e., discontinuous conduction is assumed. For t ,

vs = Vmax sin t = Ra ia + La

dia
+ ea
dt

(4.2)

We constant speed of operation, and steady load, so that the back emf Ea is constant. The
solution for the armature current ia is
ia =

Ra

Vmax
Ra2 + ( La )

t
E
sin ( t ) a + Ae La
Ra

(4.3)

L
where = tan1 a is the power factor angle of the RL circuit.
Ra
We have assumed that ia starts to flow at , however, note that if the back emf Ea > Vmax sin ,
E
then conduction will start at b = sin1 a
Vmax
inequality applies.

. Thus should be replaced by b if the above

At t = , ia = 0, thus

0=

Ra

Vmax
Ra2 + ( La )

E
sin ( ) a + Ae La
Ra

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Ra
Vmax
Ea La

sin ( )
A=
e
2
Ra
Ra2 + ( La )

And ia =

Vmax
Ra2 + ( La )

(4.5)

Ra
Ea
Vmax
Ea La ( t )

sin ( t )
sin ( )
(4.6)
e
Ra R 2 + ( L )2
Ra
a
a

At t = , ia = 0, thus,

0=

Vmax
Ra2 + ( La )

Ra
Ea
Vmax
Ea La ( )
sin ( )
sin ( )

e
Ra R 2 + ( L )2
Ra
a
a

(4.7)

This transcedental equation needs to be solved iteratively to obtain b, which will then
completely define the voltage waveform across motor terminals. The average armature voltage
Va and current Ia can then be found. From these, the torque-speed (T-) characteristics of the
motor for different and load torque T can be found.
Furthermore, from the armature voltage and current waveforms, we can also find the ripple
voltages and currents of the motor. These anayses are quite complex and are best carried out on
a computer.

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4.3 Single-phase fully-controlled thyristor bridge converter driven S. E. dc

motor with continuous conduction of ia.
va

ip

Ra

T3

T1

ia
La

Va

Vmaxsint

Ea
T4

T2

T1&T2

Vs

T3&T4

T1&T2

Ea

va
= 45 o

ia

va

=145o

ia

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When conduction of armature current is continuous, a pair of thyristors conduct at all times.
With each tiggering, the armature currents transfers to a pair of incoming thyristors. If the ac
source has neglible inductance, then this transfer takes place immediately upon triggering.
The dc voltage across the motor is given by,

Va =

Vmax sin td( t )

2Vmax

cos

(4.8)

where Vmax is the peak line-line voltage of the ac supply to the converter. This converter operates
in two quadrants, Q1 and Q4, as indicated in the figures that follow. With continuous
conduction, this converter is able to develop negative voltage across it but can not supply
negative current.

Vd

Firing angle,
Vd
Q1

Q4

Id

Since the voltage waveform across the motor is fully determined by the ac supply voltage and
the firing angle, the voltage ripple components are easily found from Fourier analysis as:

an =

Vmax cos ( n 1) cos ( n + 1)

2
n+1
n1

bn =

Vmax
2

sin ( n 1) sin ( n + 1)

n+1
n1

vn = an2 + bn2

(4.9)

for n = 2, 4, 6, ..

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The amplitude of the ripple voltage reduces as the ripple harmonic number n increases. The
lowest ripple order is n = 2 for this converter and its amplitude is the highest (the dominant
ripple). From the Fourier coefficients of the ripple voltage, it can be shown that the ripple
voltage across the motor is the highest when = 90
For each ripple voltage component, the ripple current is found by dividing the rms ripple voltage
by the AC impedance of the armature. Thus
in =

vn

R + ( n L2a )
2
a

sin n ( t )

ia = I a +

where I a =

in

n= 2 ,4 ,6 ,.....

In =

where Vn =

(4.10)

Va Ea
Ra

(4.11)

Vn
R + ( n La )
2
a

(4.12)

v n
.
2

The total RMS ripple armature voltage and current are:

and

VaRMS = V22 + V42 + V62 +

(4.13)

2
2
I aRMS = I a2 + I a22 + I a4
+ I a6
+

(4.14)

I RMS
, called Form Factor of the armature current, is an important measure of motor
Ia
heating. Ideally, it should be as close to unity as practicable.

The ratio

Yet another aspect of the this type of converter is the fact that the input current waveform to the
thyristor converter is delayed from the input ac voltage by the firing angle . This implies a drop
in the operating power factor of the converter when the motor is driven at lower than rated
speed. In the limit, at near zero speed, when the firing angle is near 90, the input power factor is
also nearly zero. This is a very undesirable aspect of thyristor converters.
Note that input power factor at which the converter operates is given somewhat differently from
purely sinusoidal ac circuits for which the Power Factor is given by the cosine of the angle of the
input current and voltage. From basic definition of power factor,
Input Power Factor = Output Power/Input RMS VA
Thus, IPF =

Va I a
V I cos
I
= RMS 1
= 1 cos
VRMS I RMS
VRMS I RMS
I RMS

(4.15)

when converter losses are neglected. Here I1 is the RMS value of the input current to the
I
converter. The ratio 1 is known as the distortion facttor of the input current to the converter.
I RMS
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Analysis of armature current with continuous conduction

Vmax sin t = Ra ia + La

Solving

dia
+ Ea for t +
dt
Ra

t
V
E
ia = max sin ( t ) a Ae La
Z
Ra

where Z = Ra2 + ( La )

(4.16)

L
and = tan 1 a .
Ra

(4.17)

In the steady-state, the armature current falls to its minimum value at t = . Thus, by setting
I a min = ia ( ) = ia ( + )

(4.18)

we can find that,

I a min

RaL

Vmax
e a + 1 Ea
= ia ( ) =
sin ( ) Ra
R

Za
a
e La 1

(4.19)

Note that for this converter, we have assumed that the voltage applied to armature circuit for
each half cycle of the AC cycle is the same and that the current waveform repeats in every half
cycle. The solution for ia for this converter is found by replacing the constant A in equation
(4.16) by Iamin form (4.19).
Once the solution of ia is found, the average value of the armature current, Ia, is found from

Ia =

ia ( t ) d ( t )

(4.20)

The AC components of the armature current can be found from Fourier analysis of the solution
of ia. This will be a tedius task (refer to pages 198-199 Shepherd, Hulley and Liang for Fourier
coefficients of ia). It is easier to perform the task by using Fourier analysis of the voltage ripples
and AC circuit analysis, as given by equations (4.9 - 4.12).
Analysis of discontinuous conduction

The armature current can become discontinuous at light load and also for certain operating
conditions of firing angles and motor parameters. Conduction becomes just discontinuous when
ia falls to zero at t = n + and extinction angle is equal to .+. Thus, using the condition
that ia = 0 for t = ,
Thus
Ra

V R
E
R
E
( t )
i = max a sin( t ) a + a a sin( ) e La
Ra Z
Vmax
Vmax Z

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The current falls to zero at which is obtained by equating the above to zero. Thus
Eb
Vmax

= e R / L
Eb
cos sin( )
Vmax
cos sin( )

e R / L

(4.22)

The extinction angle is found by solving the above transcendental equation. Vd can then be
calculated. The boundary between continuous and discontinuous conduction (the dotted
semicircle) is found by setting = + in equation 4.22.
By solving for for a given firing angle and back emf Ea, the output voltage waveform across
the armature and the current waveform through it are determined. The average Va and Ia are then
obtained. During discontinuous conduction, the voltage at the converter output terminals rises to
the back emaf Ea, and as a consequence Va rises with fall of Ia. Note that when conduction is
continuous, the converter output voltage is independent of the load current; the converter
essentially behaving as an ideal dc source. This is no longer true when armature current
discontinuous, the converter voltage rising sharply with fall of load, or falling sharply with rise
of load current, as shown in figure below.
Va

= 0
= 60

Ia

= 150
= 180

From the solution of ia, the average Ia can be found, thus giving the developed torque T. The
coordinates of the operating point for the chosen and are thus found. By repeating the above
calculations for a given firing angle and a range of speed , the torque-speed charactersitic of
the drive for a given firing angle can be found. The T - characteristic of a seprately excited
dc motor for a range of firing angles are indicated in the figure below. Note that during
continuous conduction in quadrant 1, the back emf of the armature Ea is lower than Va, by the
IaRa drop which is small. This voltage drop, divided by KE, is the droop in speed, which depends
on Ia and hence load torque. During discontinuous conduction, the change of speed with a
change of load toque can be rather high, implying a poor speed regulation with load. The rise in
Va with discontinuous conduction can be considered to be a loss of converter voltage gain
dVa/d or rather dVa/dEc, where Ec is the control voltage to the firing control circuit. This loss of
gain (usually it is a severe loss of gain) makes the converter response time to change of
operating conditions unacceptably slow.

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A high inductance in the armature circuit (or a large a) reduces the likelyhood of discontinuous
conduction, so that the minimum required inductance to ensure continuous conduction for the
minimum expected load may have to be found.

= 0

= 60

(rad/sec)

T, Nm

= 150
= 170

The critcal armature inductance

The minimum inducatance, Lamin, required in the armatude circuit to prevent discontinuous
conduction at a given speed and load can be found by equating Iamin = 0 in equation 4.19. Since,
for just discontinuous conduction, = + , the condition for minimum Lamin is given by

Ra
R + ( La )
2
a

RaL

e a + 1

> a
sin ( )
Ra
L
Vmax
e a 1

(4.23)

Effect of source inductance

The flat VaIa characteristics of the converter with continuous conduction at any firing angle ,
are also idealised. In reality, the dc output voltage of a phase-controlled thyristor converter falls
linealy with load current because of the source inductance in the AC supply (mainly the leakage
inductance of the supply transformer). A phenomenon called commutation overlap is responsible
for this droop. It will not be covered in this course. For a converter with a given source
inductance, the droop of the converter output dc voltage with load current may simply be
represented by an equivalent dc source reactance Lx. Thus, for the single-phase fully-controlled
converter, the dc voltage to the motor for a firing angle may be given by

Va =

2Vmax

cos Lx I a

(4.24)

where Lx represents the voltage droop factor of the converter due to the commutation overlap
mentioned above.

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The effect of source inductance is therefore responsible for a further speed droop that what is
indicated in the figure above. With AC side source inductance and commutation overlap, the
speed of the motor is thus given by
2Vmax

cos Lx I a I a Ra
KE

rad/sec

(4.25)

when conduction is continuous.

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Single-phase half-controlled thyristor bridge converter

V m a x sin t

Ia

La

Ra

Ea

A half controlled bridge converter has two thyristors and two diodes in the bridge arms, and a
addtionally a commutating diode across the motor terminal to allow the thryristor current to
commutate into when the AC supply voltage reverses polarity. The commutating, or
freewheeling diode is required to prevent unconrolled operation of a bridge arm. The presence of
the freewheeling diode circulates (freewheels) the armature current and prevents the load voltage
from becoming negative.
Va =

1
2

Vmax

Vmax sin ( t )d ( t )

(1 + cos )

when the conduction is continuous.

(4.26)

Vs

iL

Vo

iT 1

iD f

is

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Va (V) or
(rad/secc)

Ia (A) or T (Nm)

Vd

Firing angle,

The torque-speed characteristic of a separately excited DC motor is given by

Vmax

( 1 + cos ) ( Lx + Ra ) T / KT
KE

rad/sec

(4.27)

where continuous conduction has been assumed.

The half controlled converter has the desirable attribute that it presents a higher input power
factor than the fully controlled converter. This is because the lagging current of the source
current is contrained to flow locally through the motor circuit by the freewheeling diode. Its
output voltage and current ripples are also lower and the armature current is likely to become
disconitnuos, so that the crtical inductance required for continuous conduction is not as large.
The gate drive and converter circuit costs are also somewhat cheaper. It however operates in one
quadrant only, so regenerative braking of the drive is not possible.

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Three-phase, fully-controlled thyristor bridge converter

vo
van
ia
vbn

T1

T3

iL

T5

ib

Ra

Va

ic

La

+
vcn

T4

T6

T2

Ea V

For larger DC motor drives, three-phase thyristor converter circuits are preferred due their better
output ripple and input harmonic performance.
Assuming continuous conduction,
1
Vd =
/3
=

2
+
3

3Vmax l l

Vmax l l sin td( t )

[4.28]

cos

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This converter operates in quadrants 1 and 4, developing both positive and negative polarity dc
output voltage. For firing angles 0 o 90 o , the converter operates in quadrant 1 and
for 90 o 180 o , the operation is in quadrant 4. Operation in quadrant 4 is of course possible
only when the load includes an active dc source, able to source power into the ac circuit.

400

Va

200

-200
-400

Firing angle, , degrees

Va
Q1

Q4

Ia

From Fourier analysis, the out voltage harmonics for continuous conduction are given by
an =

Vmax l l cos ( n t )d t

2 sin ( n + 1) cos ( n + 1) 2 sin ( n 1) cos ( n 1)

3Vmax l l
6
6
=
+

n+1
n1

2 sin ( n + 1) sin ( n + 1) 2 sin ( n 1) sin ( n 1)

3Vmax l l
6
6
+
bn =

n+1
n1

[4.29]

[4.30]

where n = 6, 12, 18 etc,.

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The peak ripple voltage of a harmonic number 6n is given by

cn = an2 + bn2 =

3Vmax l l

1
1
2 cos 2
+

n + 1 n 1 ( n 1)( n + 1)

[4.31]

of the Three-phase bridge converters circuits have the following attributes, compared to the
single- phase bridge converter:
1. The output voltage waveform of the converter is smoother, having the lowest harmonic
order of six acopared to two for the single-phase bridge converter. The ripple voltage to
the motor have harmonic order or 6k where k is any positive integer. The armarure
current ripple has the same harmonic order. The ripple voltage and currents are also of
lower magnitude. The highest output ripple occurs for = 90.
2. The lower ripple in the armature current and the smoother voltage waveform calls for
smaller armature inductance which may be required to ensure continuous conduction.
3. The effective converter switching frequency is six times the supply frequency (300 Hz,
compared to 100 Hz for a single-phase bridge converter).
4. The input current waveform to the converter is more closer to being a sinusoid (i.e.,
better distortion factor), compared to the input current waveform for the single-phase
bridge converter. The harmonic order of the AC input line current is given by 6k 1,
compared to 2k 1 for the single-phase converter, where k is any positive integer. This
calls for reduced filter requirement at the input AC side.

Armature current for continuous conduction

The motor current can be exactly determined from the DC and harmonic currents, using DC and
AC circuit analysis techniques and then adding the currents for each voltage component.
Alternatively, for the interval + 30 o t + 90 o ,
Vmax l l sin ( t + 30 o ) = Ra ia + La

dia
+ Ea , assuming a constant back emf.
dt

Note that the output voltage wavefrom repeats every 60. Solving for ia,
Ra

t
V

E
ia ( t ) = max l l sin t + a + Ae La 6
Za
6

Ra

[4.32]

The constant of integration A is given by

A=

Vmax l l sin ( )

[4.33]

Ra

Z a e La 3 1

The minimum and maximum armature current values are given by:
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I a min =

Ra
La 6

sin ( ) e
Vmax l l
E V

sin + a + max l l
Za
3 Ra
Za
Ra

e La 3 1

[4.34]

From the complete solution of the armature current, the average and RMS armature current can
be determined. For the average armature current, the motor developed torque is found, for the
speed for which Ea was used in the calculation.
Note that if the conduction is continuous, which is readily shown by the sign of
3Vmax l l
Ia =

cos Ea
Ra

[4.35]

then the average armature current is easily found by the abobe equation (equation 4.35).

Torque-speed characteristic with continuous conduction

For a given firing angle a, the T- characteristics are given by,
3Vmax l l

cos I a Ra L3 x I a

[4.36]

KE

where L3x is the equivalent source reactance of the three-phase AC source.

= 0

= 60

(rad/sec)

T, Nm

= 150
= 170
Note that opration in quadrant 1 is for forward driving and operation in quadrant 4 is for reverse
(regenerative) braking. The droop in speed with load is partly due to the voltage drop in
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armature resistance and partly due to the voltage regulation characteristic of the converter as a
consequence of the AC side source inductance. The steeper drop in speed indicated by the
graphs to the left of the semicircle are due to dicontinuous conduction.

The critical inductance

As for the single-phase converter, the armature current can become discontinuous, depending on
the load, firing angle and motor parameters. The boundary between continuous and
discontinuous conduction is found by equating the expression for Iamin (eqn 4.34) to zero. This is
indicated by the dottes semicircle in the T- plane in the above figure. The critical inductance
for just discontinuous conduction is given by

Ra
Ea
sin ( )
sin + + R

Za
3
Vmax l l
a

e La 3 1

[4.37]

Operation with discontinuous conduction

When the armature current becomes discontinuous, the motor terminal voltage rises to the level
of the back emf, instead of following the AC line-line voltages. The effect is a net increase in the
average voltage across the motor and hence speed. If the armature current falls to zero at angle
+ , where is the conduction angle, the solution for the armature current for this condition
operation is found by noting the ia at zero in a thyristor is triggered. Thus,
Ra

V
E

0 = max l l sin + a + Ae La 3
Za
3 Ra

[4.38]

From (4.32) and (4.38)

R
R

a t +
a t +
E
sin t + sin + e La 6 a 1 e La 6
3
3

Ra

[4.39]

V
ia = max l l
Za

Current ia becomes zero at + . Hence

R
R
a +
a +
Vmax l l
Ea

L
6

a
sin + + sin + e

1 e La 6
0 =
Za
3
3
R

[4.40]

The conduction angle can be found by solving equation 4.40 iteratively. The the average and
RMS armature current can be found from

Ia =

I aRMS =

+ / 6 +
+ / 6

i a d ( t )

+ / 6 +

+ / 6

[4.41]

ia2 d ( t )

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Once the conduction angle is determined, the average armature voltage Va is given by
Va =

+ / 6 + / 2

+ / 6

va d ( t ) )

[4.43]

3Vmax l l

3E

cos + cos + + + a

3
3

[4.44]

The average armature can be more easily from (4.44) once angle has been found.

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Half-controlled three-phase thyristor bridge converter

vo

+V D /2
v an
ia
v bn

T1

T3

ib

Df Vd

ic

v cn

iL

T5

D4

D6

Load
L

D2

V D /2

Assuming continuous conduction,

Vd =

1
2 / 3

Vmax l l sin td ( t )

3Vmax l l
(1 + cos )
2

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Note that this converter only operates in quadrant 1, because negative voltage across the dc rails
will initiate freewheeling of the load current through Df. It should be noted that freewheeling is
present when the firing angle is greater than 60. It should also be noted that the input current
waveform is not symmetrical, because the commutation of the load current through the
freewheeling diode, thereby relieving the input source of the lagging component of current. [The
converter input current waveform will include some even order harmonics]. As a result, the
input power factor of this converter is higher than the fully controlled converter and its output
voltage ripples are also smaller.
400

Va
200

Firing angle (degrees)

Va
Q1
Ia

As for the single-phase, half-controlled converter, the output putput voltage and current ripple
are smaller for for this converter, compared to the fully controlled converter. It also operates
with a better input displacement factor.

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Electric Drive Systems

The converter Voltage Gain

The firing control circuits of phase-controlled AC-DC converters normally includes a
conversion such that the firing angle made equivalent to cos-1(vc) where vc is the control
volatge to the firing angle controller. This is easily incorporated with the firing control circuit.
For the three-phase fully controlled bridge converter, the output DC voltage to the motor is then
given by
Va =

3Vmax l l

( )} =

cos cos1 vc

3Vmax l l

vc

Between the firing controller and motor terminals, the converter thus behaves as avoltage gain of
3Vmax l l
.

Firing
Control
Circuit

vc

cos-1()

4-Quadrant Converter
Single and three-phase fully-controlled thyristor converters described so far operate in two
quadrants, 1 and 4. In order to drive the motor in all four quadrants, two such converters are
needed, to be connected back to back as shown below.

Ra
Ia
La
Ea

1 + 2 = 180

Motor de-rating due to ripple current.

For a DC motor, its continuous current rating is given by the level of smmoth DC current it can
carry continuously. (Note that the RMS value of a constant DC value is the DC value itself).
When the motor is supplied from a converter in the steady state, the armature current includes a
DC value, which is resposible for the developed torque and some ripples which produce ripple
torque and heating due to the I2Ra loss. For such drives, DC motors have to be de-rated in the
ratio of the DC to RMS current the motor will need to carry at the rated load condition.
Lecture 5 7 Thyristor converters for dc drives

L5 - 21

F. Rahman/August 2002