2. DC Motor Control

2.2.2.3 In separately excited DC motor (RLE) load

The basic circuit for a single phase full converter fed separately excited dc motor

is shown in figure 2.30.The armature voltage of the motor is controlled by a single phase

full converter and the field circuit is fed from the ac supply through a diode bridge

rectifier. For conducting the field control the uncontrolled rectifier is replaced by

controlled rectifier depends on the field ratings. The motor current cannot reverse due to

the thyrsitors in the converters.

Figure 2.30 Basic circuit for full converter fed separately excited motor.

The single phase full converters drives are used for low and medium horsepower

applications. Such drives have poor speed regulation on open loop firing angle control.

However, with armature voltage or speed feedback, good regulation can be achieved.

From the previous chapter, we know that the average speed

…………………… (2.7)

So the speed of the motor can vary, either by varying the terminal voltage or

armature current or the file current. The figure 2.30 shows the speed control of separately

excited motor by varying the voltage.

Figure 2.31 Single phase controlled rectifier fed separately excited motor

The figure 2.31 shows a single phase fully controlled rectifier supplying a dc

separately excited motor. The armature circuit of the motor is replaced by the armature

resistance Ra, armature inductance La and the back emf Eg

As discussed earlier, for positive half cycle at angle , the thyristors T1, T2

receives firing pulses. At the instant ( + ), the thyristors T3 and T4 receives the power.

The positive and the negative half cycle current flow path as like shown in figure 2.27

and 2.28 respectively. Due to the inductance value, there is a small change in the time

intervals 0to and to ( + ) in wave forms as discussed earlier, the output may be

continuous and discontinuous. Next section we deeply study about the modes of

operation.

2.2.2.4 Modes of operation

The single phase full converter with motor load has two modes of operation. They

are

1. Motoring mode

(a) Discontinuous conduction mode

(b) Continuous conduction mode

2. Regenerative braking mode

a) Discontinuous conduction mode

b) Continuous conduction mode

1. Motoring mode

The condition for making the full converter with RLE load as motoring mode is firing

angle should be less than 90 degree. So the net armature terminal voltage or full converter

output voltage is positive and relatively high compare to the back emf Eg. So the current

flowing through the armature is

= Positive…………………… (2.8)

Figure 2.32 Equivalent circuit of motoring mode

So the machine is rotate in the forward direction as shown in figure 2.32. This

mode is called motoring mode. The wave forms are shown in figure 2.33.

In motoring mode the following points are useful to know the operation.

1. In the motoring mode, the armature current flows through the source, and either

through the thyristors T1,T2 or through the pair T3,T4.When the pairs T1,T2

and T3,T4 conduct, the output voltage of the converter Va is Vs(Supply

voltage).When none of the thyristor pair conducts Ia = 0 and Va = E.

2. When Ia>0 at the instant of firing a thyristor pair, then the biasing of thyristor

pairs will be decided by the source voltage only. If the source voltage provides a

positive bias, thyristors will turn on even when the source voltage is less than E.

Since the property of thyristor to remains it in on state is, the anode current

above the latching current.

3. When Ia is zero at the instant of firing a thyristor pair, then the biasing on

thyristor pairs will be decided by the difference of the source voltage and the

back emf. That means (Vs-Eg) and Vs>Eg.

(a)Discontinuous conduction mode in motoring

In discontinuous conduction mode, current starts flowing with turn on of

thyristors T1 and T2 at the firing angle .Motor gets connected to the source and its

terminal voltage equal Vs. After the instant , the armature current falls to zero at ,

before to reach the instant of next thyristor pair(T3,T4) triggered ( + ).So due to the

absence of current, T1 and T2 turn-off. From the instant , the motor terminal voltage is

now equal to its back emf Eg. When thyristors T3 and T4 are fired at ( + ), next cycle

of the motor terminal voltage Va starts. The output voltage and the current wave forms

are shown in the figure 2.33.

Figure 2.33.Discontinuous conduction mode (motoring mode)

Let us take the motor terminal voltage Va, in discontinuous mode the drive

operates in two intervals such as duty interval and zero interval.

In the duty intervals are the intervals in which the mote is connected in the source

and the terminal voltage Va = Vs (Supply voltage).The interval is .In this

interval, by using the basis of electric circuits, we can write the voltage equation as

, for …………………… (2.9)

In the zero intervals, there is no current flow in the load. So that the armature

current Ia = 0.

Therefore, , and Ia=0, for …………………… (2.10)

The equation (2.9) can be write as

Complimentary function (C.F)

The auxiliary equation of the above is

The auxiliary equation consists of only one real root

Therefore the complimentary function C.F =

Particular Integral P.I =

= = P.I1+P.I2

P.I1=

= =

=

P.I2 =

The Solution of this equation is =C.F +P.I

+ ………………………………….(2.11)

+ for ………. (2.12)

In the above solution consists of two components. One is due tot h ac

source , and other due to back emf .

Where, Z=

is the armature circuit time constant.

Between the intervals , the initial current , so at

, Substitute those values in equation 2.12.The equation 2.12 yields

………………………………… (2.13)

Substitute the value of 2.13 in equation 2.12

, for

……………………………………………………….. (2.14)

Between the same interval final time , , Substitute those values in

equation 2.14,We get

………………… (2.15)

From the equation 2.15, we can evaluate the value of by using numerical

methods.

Since the voltage drop across the armature inductance due to dc component of

armature current is zero

Therefore Va= Eg + IaRa…………………………………………………(2.16)

Where Va and Ia are respectively dc components of armature voltage and current

respectively.

From the waveforms 2.33, the armature voltage or output voltage of the full

converter is

………………………………….. (2.17)

Armature current consists of dc component Ia and harmonics. When flux constant,

only dc components produces steady torque. Harmonics produce alternating torque

components and the average value of torque is zero.

Compare equations 2.16 and 2.17 we get,

Eg + IaRa …………………………(2.18)

For constant flux,

Back emf Eg=K …………………………………. (2.19)

Torque T=KIa…………………………………….. (2.20)

Substitute the values of Eg and Ia from equation 2.19 and 2.20 respectively in

equation 2.18, Therefore equation 2.18 becomes

………………………(2.21)

From the figure 2.33, we have seen that

Boundary between continuous and discontinuous conduction is reached when

.Substitute the value in equation 2.21.So the speed at which the angle is

called critical speed .

Substitute the value in equation 2.15,we get

Therefore Eg=

Ie) K

Therefore the critical speed

……….. (2.22)

(b)Continuous conduction mode in motoring

In continuous conduction mode, positive current flows through the motor and

Thyristor pairs T3 and T4 are in conduction just before .At the instant gate pulses

given to the thyristor pairs T1 and T2.Conduction of T1 and T2 reverse biases the

thyristors T3 and T4and turns off them. So the continuous armature current Ia is flowing

to the motor. But it is not perfect dc, the motor torque fluctuates. The waveforms are

shown in below.

Figure 2.34 Motoring mode of full converter <90 (Continuous conduction)

From the figure 2.34, the average output voltage of the full converter or the

terminal voltage of the motor is

= ……………..………….. (2.23)

Compare equation 2.16 and 2.23 we get,

Eg + IaRa = …………………………………………… (2.24)

Substitute the values of Eg and Ia from equation 2.19 and 2.20 respectively in

equation 2.24, Therefore equation 2.24 becomes

……………………………………..(2.25)

Figure 2.35 Speed verses torque characteristics

Speed torque curves for the drive are shown in figure 2.35.When both pairs of

thyristor fail to fire, the armature current will be zero. The no load operation is obtained

when armature current Ia=0This will happen when the back emf greater than the supply

voltage through out the period for which firing pulses are present. Therefore, when firing

angle less than 90 degree, the back emf should be greater or equal to Vm and when firing

angle greater than 90 degree, back emf should be greater or equal to Vm .Therefore,

no load speeds are given by

, for

, for

When firing angle , Equation 2.23 gives the maximum average terminal

voltage as .This is the rated motor voltage. The ideal no load speed of the motor

when fed by a perfect direct voltage of rated value will be .The boundary between

the continuous and discontinuous conduction is shown in figure 2.35.So if degree

the output voltage is zero.

The drive operates in quadrant I or forward motoring and reverses regenerative

braking or quadrant IV. In the equation 2.25.When firing angle less than 90 degree, the

power is fed from source to load. That means rectifying mode. So the speed is positive.

The polarities are shown in figure 2.32.

2. Regenerative braking mode

The condition for making the full converter with RLE load as regenerative braking

mode is firing angle should be greater than 90 degree. So the net armature terminal

voltage or full converter output voltage is negative and the magnitude of back emf is

greater than magnitude of full converter output.. So the current flowing through the

armature is negative. This operation is called inversion. So the power is fed from dc to ac

source.

= Negative…………………… (2.26)

Figure2.36. Equivalent circuit of regenerative braking mode

So the machine is rotate in the reverse direction as shown in figure 2.36. This

mode is called regenerative braking mode. The wave forms are shown in figure 2.37.In

regenerative braking the armature current should be reversed. But it is not possible in the

full converter, because the controlling element in which the converter is used as a

unidirectional device. So by changing the polarity of the back emf we can make

regenerative braking without change in the direction of armature current. The reversal of

the motor emf with respect to rectifier terminal can be done by any of the following

changes.

1. An active load coupled to a motor shaft may drive it in the reverse direction.

This gives reverse regeneration. In this case no changes are required in the

armature connection with respect to the rectifier terminals.

2. The field current may be reversed, with the motor running in the forward

direction. This gives forward regeneration. In this case also no changes are

required in the armature connection.

3. The motor armature connection may be reversed with respect to the rectifier

output terminals, with the motor till running in the forward direction. This will

give forward regeneration.

In regenerative braking mode the following points are useful to know the

operation.

1. The regenerative braking operation is obtained by reversing the back emf by

any of those three methods and adjusting so that the average terminal

voltage of the converter Va is negative. Thus the modes of operation drawn

for (+Eg) are for motoring and those drawn for (Eg) are for regeneration.

2. The rate of change of the armature current is given by the following

.According to the equation, if the armature resistance

drop is neglected, the rate of change of he current will be positive when Va

>Eg; otherwise it will be negative.

The mathematical proof is same as the expression which we have seen already. The

wave forms for discontinuous conduction and continuous conduction are shown in figure

2.37 and 2.38 respectively.

Figure 2.37.Discontinuous conduction mode (regenerative braking mode)

Figure 2.38.Continuous conduction mode (regenerative braking mode)

The characteristics curve armature voltage Va verses the firing angle is shown in

figure 2.39

Figure 2.39 Output voltage Verses firing angle for full converter with RLE load

Example1

The speed of a 10hp, 230V, 1200rpm separately excited dc motor is controlled by a

single phase full converter. The rated motor armature current is 38A and the armature

resistance Ra=0.3Ohm.The ac supply voltage is 260V.The motor voltage constant

V/rpm. Assume that sufficient inductance is present in the armature circuit

to make the motor current continuous and ripple free.

1. Rectifier operation(Motoring action)

For a firing angle 30degree and rated motor current, calculate

a) The motor torque

b) The speed of the motor

c) The supply power factor

2. Inverter operation(regenerating action)

The polarity of the motor back emf Eb is reversed, say by reversing the

field excitation. Calculate

a) The firing angle to keep the motor current at its rated value

b) The power fed back to the supply

Solution

V/rpm.

V.Sec / radian

=1.74V.Sec/radian

(a) The motor torque T=Ke Ia=KIa [Since flux is constant]

=

(b) Given the motor current is continuous

Therefore the armature voltage from the full converter

= =

Therefore Eg=Va-IaRa

=202.82-(38 0.3)

=191.42V

Therefore Volts

Therefore speed rpm [Since flux Constant]

= 1051.8rpm

We may calculate the value of speed as follows

In continuous conduction speed

=110.01rad/Sec

=

=1051 rpm

(c) If the motor current is constant and ripple free, the rms supply current is same as

load current

Is=Irms=Ia=38Amps

Supply volt-ampere is =VsIs

=260 38

=9880VA

If the losses are neglected the real power or power from the supply

Ps=VaIa

=202.82 38

=7707.2W

Therefore supply power factor =

=0.78

2. At the time of regenerating action

The polarity of the motor back emf Eb is reversed,

(a) Therefore -Eg=Va- IaRa

Ie) -191.42=Va-(38 0.3)

Therefore Va= -180.02Volts

The firing angle to keep the motor current at its rated value is

Va= = -180.02

Therefore

(b) Power from the dc machine Pg=EgIa=191.42 38=7273.96Watts

Power lost by the armature resistance = =38 0.3=433.2Watts

Therefore,

The power fed back to the supply Ps=Generated power-Losses or (VaIa)

=7273.96-433.2=6840.76Watts

Three phase full converter fed dc drive

The most widely used dc drive control is the three phase fully controlled or six

pulse bridge rectifier fed dc separately excited drive as shown in figure 2.40. Thyristors

are fired with a phase difference of 60 degree. Each thyristor conducts for 120 degree and

only two thyristors conduct at a time –one is positive group and one negative group.

Thyristors are commutated by natural commutation, in positive cross over point positive

group thyristors are commutated and in negative cross over point negative group

thyristors are commutated.

Figure2.40 Three phase fully controlled rectifier fed Dc drive

What ever may be the load, it is making continuous and discontinues conduction.

If the firing angle of the thyristors is less than 60 degree it is producing continuous

conduction. If is greater than 60 degree, it is producing discontinuous conduction. For

our convenience, the firing angle for the thyristors are measured from cross over points.

The conducting sequence and the output voltage and current waveforms in motoring

operation are shown in figure 2.41.Figure 2.42 shows the waveforms for continuous

conduction in braking operation of three phase full converter with RLE load.

General operation

At the instant , Thyristor T6 is already conducting and thyristor T1 is

turn on and duration for each thyristor remains in on state is 120 degree. During the

interval ( ) to ( ) thyristors T1 and T6 conduct and the line to line voltage VRY

appears across the motor terminals. Since the VR voltage is positively high and the VY

voltage is negatively high.

At , thyristor T2 is fired and Thyristor T6 is reverse biased .So during

the interval to ,thyristors T1 and T2 conduct and the line to line voltage VRB

appears across the load. Since the VR voltage is positively high and the VB voltage is

negatively high. So the conducting sequence for three phase full converter is (T1,T2),

(T2,T3),(T3,T4),(T4,T5),(T5,T6) and (T6,T1).

Figure2.41. Wave form for 3-phase full converter with R or RL or RLE load at

=30 Degree Motoring operation (Continuous conduction)

Figure2.42. Wave form for 3-phase full converter with RLE load at

=120 Degree braking operation (Continuous conduction)

During the discontinuous conduction, if there is no conduction in the circuit, the

back emf is considered in RLE load, as shown in figure 2.43.

Figure 2.43.Three phase full converter with RLE load (Discontinuous conduction)

The discontinuous conduction is not significant for control the dc motor. So

consider for continuous conduction the armature output voltage

Va= = ………………………………….

(2.27)

According to the expression which studied earlier in single phase full converter

with RLE load, the speed is

………………………………………(2.28)

The output voltage verses the firing angle pf the full converter is same as shown

in figure 2.39.It is also working in first and fourth quadrant.

Example 5

The speed of 125hp, 600V, 1800rpm, and separately excited dc motor is controlled by

a three phase full converter. The converter is operated from a three phase, 480 V, 60Hz

supply. The rated armature current of the motor is 165A.The motor Parameters are

Ra=0.0874 Ohm, La=6.5mH and Ka =0.33V/rpm.The converter and ac supply are

considered to be ideal.

1. Find no-load speeds at firing angles 0 and 30 degree. Assume that at no load, the

armature current is 10% of the rated current and is continuous

2. Find the firing angle to obtain rated speed of 1800rpm at rated motor current.

Compute the supply power factor

3. Compute the speed regulation for the firing angle obtained in part2.

Solution

The output voltage of the full converter Va=

1. At = ,Va =648.55 Volts

Given =16.5

Therefore Eg = Va-IaRa = 648.55

= 646.6 Volts

At no-load, the speed rpm

At = ,Va = 648.55 =561.2 Volts

Therefore Eg = Va-IaRa = 561.2

= 559.8 Volts

At no-load, the speed

2. We know that, speed

But, Speed

Therefore, Eg = Volts

Motor terminal voltage at rated current is Va= Eg+IaRa

= Volts

Therefore,

Cos

At full load condition, the ripples in the motor current are very less. Therefore the supply

current or source current IR (Phase current) is a square wave of amplitude 165 Amps and

width 120 degree.

Therefore, rms value of the supply current is

=

Ie) IR = 134.64 Amps

The supply volt- ampere

= 111,885.8 VA

Consider, there are no losses in the converter. Therefore the supply power to the

converter is same as the power input to the motor.

Hence, the supply power to the converter is

Watts

Therefore, the supply power factor is

3. Speed regulation

At the firing angle the output voltage Va = 608.4 Volts

At full load condition motor current is 165 Amps and the speed is 1800rpm.If the load

is thrown off keeping the firing angle the same at the motor current decreases

to 16.5 Amps. Therefore

At the firing angle the no load speed rpm

Therefore the percentage of speed regulation is

=

=

Example 6

A 220 V,1500rpm dc motor has the armature resistance and inductance of 2 Ohm

and 28.36mH,respectively.It is controlled by a three phase fully controlled rectifier from

an ac source operating at 50Hz.Calculate the ac source voltage required to get the rated

voltage across the motor terminals when operating in continuous conduction

Solution

At

The output voltage of the full converter Va=

The rated voltage Va=220 Volts

Volts (Phase voltage)

Example 7

A 220V, 1500rpm, 50Amps separately excited motor with armature resistance of 0.5

Ohm, is fed from a three phase fully controlled rectifier. Available ac source has a line

voltage of 440Volts,50Hz.A star–delta connected transformer is used to feed the armature

so that motor terminal voltage equals rated voltage when converter firing angle is zero

1. Calculate transformer turns ratio?

2. Determine the value of firing angle when:

a. Motor is running at 1200rpm and rated

torque

b. When motor is running ( ) rpm and

twice the rated torque.

Assume continuous conduction

Solution

Given in the transformer, primary is star connected and the secondary is delta

connected. So in the secondary side the phase and line voltages are same. But in primary

side the phase voltage is equal to times of line voltage.

a) For three phase full converter in continuous conduction the input voltage to

the motor

For rated terminal voltage Va =220, Therefore,

Volts

Rms converter input line voltage = Volts (Secondary voltage)

Therefore turns ratio = .

b) (a) Motoring is running at 1200rpm

But the back emf at 1500rpm Eg=Va-IaRa

Volts

Therefore, the back emf at 1200rpm Volts

The terminal voltage at the same speed Va1=Eg1+ IaRa

= Volts

The output voltage of the full converter

Therefore, = =0.8227 [Since Va=Va1]

(b) At rpm the back emf Eg2= Volts

The terminal voltage at the same speed Va2=Eg2+ IaRa

= Volts

Therefore

Chopper fed DC drive

Principles of chopper operation

A chopper is nothing but a thyristor on/off switch that connects load to and

disconnects it from the supply and produces a chopped or variable load voltage from a

constant input supply voltage. The chopper is represented by an SCR inside a dotted

square. If we use the SCR as a switching device in dc supply, the commutation circuit is

needed. So that, for explaining the operation of dc chopper SCR is replaced by GTO as

shown in figure 2.43.The name of this type of chopper is step-down chopper

Figure 2.43.Basic configuration of chopper (step-down chopper)

The operation of the chopper is shown in figure 2.44.During the period “Ton”, or

when chopper is on, the supply terminals are connected to the load terminals. During the

interval of Toff or when chopper is in off, load current flows through the free wheeling

diode D, because of the energy delivering of load inductor and the load terminals are

shorted. A chopped dc voltage is thus produced at the load terminals as shown in

figure 2.44.So the average output voltage and current are positive.

Figure 2.44.Operation of step-down chopper

From the figure 2.44, we can see that, during the Ton time there is a output

voltage

Total time T=Ton+Toff

Therefore, dc output voltage

= VdcVdcT

Ton

……………………… (2.29)

Where, Duty cycle =

From the equation (2.29), the load voltage is controlled by the duty cycle of the

chopper. So the output voltage can be varied from one of the following ways such as

a) Time ratio control and

b) Current limit control

The time ratio-control is further classified as in to

1. Constant frequency system

2. Variable frequency system

Time ratio control

In the time ratio control, the value of or the ratio of the turn on time to the total

time is varied. This is effected in two ways .The are constant frequency operation

and variable frequency operation

Constant frequency system

We know that, the chopping frequency .Constant frequency scheme is

nothing but, the total time T is kept constant and one-time Ton is varied. Because of that,

remains the chopping frequency is constant. This may be called as pulse width

modulation. Figure 2.45 shows the wave forms of constant frequency scheme with less

on time and high on time.

Figure 2.45.Constatnt frequency scheme

Variable frequency system

In the variable frequency scheme, the chopping period T is varied, either by on

time is kept constant or off-time is kept constant. This may be called as frequency

modulation. Figure 2.46 illustrates the principle of frequency modulation. In that first

case is chopping period T is varied but on-time is kept constant. In the second case the

chopping period T is varied but off time is kept constant.

Figure 2.46.Variable frequency scheme

The constant frequency scheme is thus preferred scheme because of the

disadvantages of frequency modulation technique. The disadvantages such as

a) The frequency has to be varied over a wide range to provide the full output

voltage range. Filter design for variable frequency operation is difficult.

b) The possibility of interference with signaling and telephone line is greater.

c) The large off-time at low output voltage will make the current of a dc motor

load discontinuous.

Steady state analysis of time ratio control

The steady state analysis gives the motor speed-torque curves and the armature current

ripple. To analysis the system we have to take two intervals such as duty interval and free

wheeling interval.

Duty interval is nothing but the on time of the chopper. The time duration of the

interval is 0<t<Ton or 0<t< [Since ]

During this interval the voltage equation becomes

……………………………..(2.30)

Take the Laplace transform of equation (2.30)

Let Ia(0) = Ia1

Therefore

…………….(2.31)

By using partial fraction method

A=

Take the inverse Laplace transform of equation (2.31)

[Since armature circuit time constant

]

If the current at the end of the duty interval (Ton= ) is Ia2, then the above

equation becomes [Since ]

……………….(2.32)

Free wheeling interval is nothing but the off time of the chopper. The time

duration of the interval is Ton<t<T or <t< T

The voltage equation is

……………………………………..(2.33)

Where,

The initial current (at t’=0) is Ia2

Solution of the equation 2.33 is as like the solution of equation 2.30 is

………………………….(2.34)

In steady state, the value of Ia, at the end of the chopping cycle should be the

same as at the beginning of the cycle. Thus, the value of Ia for t’=T-Ton= will be

Ia1.Subsititue the value in equation 2.34.

Therefore, Ia1= …………(2.35)

Solve the equations (2.32) and (2.35) we get

The current ripple

………(2.36)

The steady state average voltage drop across the inductance is zero, therefore

Va=Eg+IaRa……………………………………………..(2.37)

Where,Va and Ia aare the average values of the armature terminal voltage and

current respectively.

We know that Va=

Therefore equation (2.37) …………………………….(2.38)

The flux is constant, so the average motor torque depends only on the dc

component or average value of the armature current.

Therefore the motor torque Ta=KIa……………………………………… (2.39)

Substitute the value of Ia from the equation (2, 38) to (2.39)

But Eg=

Where is the motor speed in radian/Sec.

Therefore …………………………………………… (2.40)

In lossless chopper , therefore the speed

In the series motor

Current limit control

In the current limit control strategy, the chopper is switched ON and OFF so that

the current in the load is maintained between two limits such as upper limit I2 and lower

limit I1.When the current exceeds the upper limit, the chopper is switched off. During off

period, the load current freewheels and decrease exponential. When it reaches the lower

limit, the chopper is switched ON. Current limit control is possible either with constant

frequency or with constant Ton. The current limit control is used only when the load has

energy storage element especially inductor. Since the chopper operates between

prescribed current limits, discontinuity cannot occur. The difference between maximum

and the minimum output current limit decides the switching frequency. The ripples in the

load current can be reduced if the difference between the maximum and minimum limits

is less. Figure 2.47 illustrate the current limit control.

Figure 2.47.Current limit control

Steady state analysis for current limit control

In current limit control, chopper operates to control the armature current between

the prescribed limits Ia1 and Ia2. The chopper adjusts the value of and T such that the

current fluctuates between these prescribed limits. In CLC, we can calculate the speed

and torque as follows.

In the duty interval the equation of Ia is same as derived in time ratio control

Therefore from equation (2.32)

From the above equation

……………………………….(2.41)

From the equation (2.35)

Ia1=

From the above equation

………………………………..(2.42)

Adding equations (2.41) and (2.42) gives

…………………..(2.43)

If we get the value of T from equation (2.43), we can calculate the value of from

equation (2.41).Then we can calculate the value of armature current Ia and the torque Ta.

Since the durations and are small compared to the armature circuit

time constant, the variation of armature current Ia between Ia1 and Ia2 takes place along

the initial parts of the exponential curves, which can be approximated by straight line.

Therefore

The current limit changes the motor speed-torque characteristic from a constant

speed to a constant torque characteristic. This type of characteristic can be used for

constant current starting of a motor or for torque control of a motor driving a battery

operated vehicle. This type of characteristic is not suitable for driving most loads to the

problem of instability.

Example.8

The speed of a separately excited dc motor is controlled by a chopper as figure

2.43.The dc supply voltage is 120V, armature circuit resistance is Ra=0.5 Ohm, armature

circuit inductance is La=20mH, and motor constant is Ka =0.05V/rpm.The motor drives

a constant- torque load requiring an average armature current of 20A.Assume that motor

current is continuous .Determine:

1. The range of speed control

2. The range of the duty cycle.

Solution

We know that, the armature terminal voltage Va = Eg+IaRa

When Eg =0,the speed becomes zero

Therefore Va= Volts

In step-down chopper the output voltage Vo=Va=

Ie) 10 =

The speed becomes maximum when at Va=120V

Therefore Eg =120 Volts

Speed [Since flux is constant]

From that the range of speed is 0< rpm, and the range of duty cycle is

.

Example: 9

A 230V, 960 rpm and 200 Amps separately excited dc motor has an armature

resistance of 0.02 Ohm. The motor is fed from a chopper which provides both motoring

and braking operations. The source has a voltage of 230 Volts. Assume continuous

conduction.

1. Calculate duty ratio of chopper for motoring operation at rated torque and 350

rpm

2. Calculate duty ratio of chopper for braking operation at rated torque and 350 rpm

3. If maximum duty ratio of chopper is limited to 0.95 and maximum permissible

motor current is twice the rated, calculate maximum permissible motor speed

obtainable without field weakening and power fed to the source

4. If motor field is also controlled in 3rd part, calculate field current as a fraction of

its rated value for a speed of 1200rpm.

Solution

We know that, the back emf Eg = Va-IaRa

= 230- Volts

1. Given speed 350 rpm and it is a motoring operation

Back emf at 350 rpm is Eg1= = 82.4 Volts

Terminal voltage of the motor at 350 rpm is Va1=Eg1+IaRa

= Volts

Therefore, duty cycle Volts

2. Given speed 350 rpm and it is a braking operation,

While braking the current direction will be reverse. Therefore

Terminal voltage of the motor at 350 rpm is Va1=Eg1- IaRa

Volts

Therefore, duty cycle Volts

3. Given the maximum duty ratio

Therefore maximum available voltage Vamax =

= =218.5 Volts

Given the armature current at the time is twice the rated current

The back emf Eg3 =Vamax +

= Volts

The maximum permissible motor speed = rpm

Here losses are not mentioned. So consider loss is negligible.

Therefore power fed to the source =

4. Given, the motor field is controlled by the method which is mentioned in

part3.Therefore back emf Eg3 = 226.5 Volts. Assume linear magnetic circuit.

Therefore back emf Eg3 will be inversely proportional to field current. Field

current as a ratio of its rated speed and the given speed.

That is equal to

Operation of four quadrant chopperThe four quadrant operation can be obtained by using the class E chopper as

shown in figure 2.48. The operation of this chopper can be explained by using the following quadrants. The simplified quadrant operation is shown in figure 2.49.

Figure2.48. Four quadrant class E choppersQuadrant – (1)

The first quadrant operation is called forward motoring. In this operation, the output voltage as well as the current is positive. In this mode chopper (CH4) is permanently ON and chopper (CH1) is made on and Off. Otherwise the time ratio control (TRC) is done by varying the duty cycle from minimum to maximum so as to provide the speed from minimum to maximum. When CH1 is off, output current (Io) freewheels through diode D2 and chopper CH4 gives zero output voltage. Both output voltage Vo and current Io is positive and Io varies from I1 to I2.In this mode, the chopper is act like a step-down (Type-A) chopper as shown in figure2.50.Quadrant – (2)

Let the motor rotate forward direction with positive armature current Io. The motor back emf Eg is less than Vo. Hence the current Io cannot reverse. The current Io can reverse if the speed of the motor increases. The motor curt can reverse if chopper CH4, CH1 are off and the duty cycle of chopper CH2 is varied by using time ratio control. Now the current Io reverses and free wheels through CH2,D4 and D1,D4 and returns the energy to the source Vdc as shown in figure .This applies a negative torque or braking action on the motor and its speed comes to zero in a short time and also the Io becomes zero. So in the second quadrant the output voltage is positive and the current is

negative. The wave forms for quadrant I and quadrant II are shown in figure 2.52.In this quadrant chopper is act like a step-up chopper.Quadrant – (3)

In this quadrant, Chopper CH2 is permanently on and the time ratio control is provided by CH3 from minimum to maximum so as to increase the speed in reverse direction. In this instant, the motor voltage and current Vo, Io both are reversed. The motor current is kept continuous due to free wheeling action of CH2, D4 as shown in figure2.53.So in this instant also, the chopper is act like a step down chopper, but the output voltage is negative.

Quadrant – (4)In this quadrant, the motor is rotating in reverse direction and its voltage is negative. The

motor current can be reversed only if the speed increased beyond normal. The motor

current can also be reversed if the chopper CH2, CH3 are off and duty cycle of CH4 is

varied

Now the current Io will be reversed that is positive direction, and it freewheels through

D2,D3 and Ch4D2 and return energy to the source Vdc as shown in figure 2.54.This

applies a negative torque or braking action in reverse rotation of the motor, which

decreases its speed and the motor comes to zero speed in a short time.