SECTION 8

DC DRIVES

INTRODUCTION

A major area of power electronic application is the dc and ac motor drives. A varietyof power switching circuits are employed for the drive and control of dc and ac motors. In thissection we see the basic defining equations of the different types of dc motors, their outputcharacteristics under conventional excitation and the various possible methods of their controlwith power electronic controllers.

SEPARATELY EXCITED DC MOTOR

The bulk of the motors used with power converters are either separately excited dc mo-tors (for constant torque and constant horse power applications in manufacturing industries),

and the series excited dc motors (for traction applications).

Figure 1 shows the schematic diagram of a separately excited dc motor. The excitationfield and the armature field are normal to each other. The field current determines the excita-tion flux in the machine. The armature current determines the armature flux in the ma-Φe Φachine. The defining equations of the motor are

field circuit equationve = Re ie + Led ied t

armature circuit equationva = Ra ia + Lad iad t

+ ea

mechanical system equationTg = TL + Bω + J dωdt

back emf equation ea = k1ieω

torque equationTg = k2iei a

Comparing the back emf and the torque equation, we get , because the electrical inputk1 = k2 will be equal to the mechanical output . The system equations may be put in theeai a Tgω

following form.

τed ied t

= veRe

− i e

τad iad t

= vaRa

− i a − eaRa

τmd ωd t

= −TLB

− ω +TgB

E6 201 Power Electronics

EE 185 IISc

T g

TLRaLa

eV a

ia

ie

Re

LeVeJ, B

ω

Fig. 1 Separately Excited DC Motor

and are defined as the field electrical time constant, armature electrical timeτe, τa τmconstant and the load mechanical time constant of the system. The independent inputs to thesystem are and . The output states of the system are and . The field circuitve, va TL ie, ia ωequation is seen to be an independent equation. The armature circuit equation and themechanical system equations are seen to be coupled to each other through the back emf andtorque relationships. For any set of inputs and , the above set of differentialve, va TLequations may be integrated to obtain the dynamic performance of the machine.

Steady state relationships

Under steady state, the above dynamic relationships reduce to

Ve = IeRe

Va = kIe Ω + Ia Ra

Tg = TL + BΩ

Armature control

In the armature control of the separately excited dc motor, the field current is kept con-stant at the rated value . The armature voltage applied to the machine is controlled. UnderIesuch a control, we may combine the above armature and mechanical system equations to obtainthe following.

kIeVaRa

− k2Ie2 Ω

Ra= TL + B Ω

a Va − b Ω = TLThe above equation defines the relationship between load torque and motor speed for anygiven armature input voltage ( ) under a given excitation ( ). The output characteristics (Va Ie

) of the armature controlled separately excited dc motor is shown in Fig. 2.Ω vs TLField control

The field control operation is employed for speeds above rated speed ( ). InVa/kIethis range of control the armature voltage is maintained at and the field current is reducedVabelow the rated value of . The steady state equation is thenIe

kIeVaRa

− k2Ie2 Ω

Ra= TL + B Ω

The nature of the output characteristics is the same as that of armature control except that themaximum torque that can now be delivered is limited to approximately (where is thekIeIa Iarated armature current and is the control current in the excitation). As the field is weakenedIemore, the peak torque capability proportionately reduces, the speed proportionately increases,

Va1

Va2

Va3

N1

N2

N3 Va1 > Va2 > Va3

Constant Torque Envelope

kIeIa

T

N

Fig. 2 Speed Control through Armature Voltage Control

and the output power remains constant. The torque speed curve in the field controlled region isshown in Fig. 3.

The armature control region is referred to as the constant torque operation and the fieldcontrolled region is referred to as the constant horse power operation.

Dynamic model of the separately excited dc motor

Under constant excitation the machine model is linear and the defining equations are

diadtdωdt

=

− 1τa −kIe

LakIeJ

−BJ

iaω

+

1La0

va +

0

−1J

TL

The characteristic equation of the system is

s2 + s

1τa + 1

τm +

1

τa τm +k2 Ie

2

La J

= 0

= natural frequency Ωo =

1

τa τm +k2 Ie

2

La J

= damping δ =

1τa + 1

τm

12 Ω o

We may qualitatively see that the motor response is a second order response. It could be underor overdamped. The response is faster for strong field excitation. The damping is low at strongfield excitation.

Control strategies for separately excited dc motor

The range of armature control is below base speed and that of field control is abovebase speed. Below base speed, the excitation is kept constant at , and the armature voltageIe

is varied in the range of . When the rated armature voltage is reached, a furtherva ±Vaspeed increase is possible by lowering the field current below . This strategy of control isIeshown in Fig. 4. Usually during acceleration and deceleration (braking) the motor will be oper-ated in torque control mode and during running under speed control. The reversal of speed ispossible either by reversing the armature voltage or by reversing the field current. However thesecond method of speed reversal (by field reversal) is seldom used, because it requires thede-excitation and re-excitation of the field. This process is slow on account of the large field

E6 201 Power Electronics

EE 187 IISc

Fig. 3 Speed Control through Field Current Control

Ie1, Va1

Va2

Va3

N1

N2

N3

Va1 > Va2 > Va3

kIeIa

T

N

Ie3

Ie2, Va1

Constant PowerEnvelope

circuit time constant. Further the armature current will have to be blocked during the time offield reversal causing discontinuity in the torque generated. For these reasons, four quadrantdrives are invariably equipped with an armature voltage source of bipolar voltage and currentcapability. The advantage is that the field excitation is unchanged and the response could befaster.

When several motors are fed from a common armature voltage source, as in many con-tinuous manufacturing industries, trim adjustment of the field is employed to closely match thespeeds of several drives operating sequentially.

SERIES EXCITED DC MOTOR

Figure 5 shows the schematic diagram of a series excited dc motor. On account of theiroperating characteristics the series motors are well suited for traction application. The excita-tion field and the armature field are both determined by the armature current. The de-Φe Φa

fining equations of the motor are

T

N

Field Control

Field Control

Speed Control

Torque Control

Arm

atur

e C

ontr

ol

DRIVEBRAKE

DRIVE BRAKE

Fig. 4 Overall Control Strategy for DC Motor Control

Fig. 5 Series Excited DC Motor

T g

TLRaLa

V a

iaRe Le

J, B

ω

ea

armature circuit equationva = R ia + Lad iad t

+ ea

; R = Re + Ra L = Le + La

mechanical system equationTg = TL + Bω + J d ωd t

back emf equation ea = kia ω

torque equationTg = kia2

The system equations may be put in the following form.

τad iad t

= vaR

− ia − eaR

τmd wd t

= −TLB

− ω +TgB

and are defined as the electrical and the mechanical time constants of the system. Theτa τmindependent inputs to the system are and . The output states of the system are and .va TL ia ωThe armature circuit equation and the mechanical system equations are seen to be coupled toeach other through the back emf and torque relationships. For any set of inputs , and , theva TLabove set of differential equations may be integrated to obtain the dynamic performance of themachine.

Steady state relationship

Under steady state,

Va = R Ia + kIa ω

Tg = TL + Bω

Speed control

Usually the machine is controlled by controlling the armature voltage (Field controlmay be incorporated if field diverter resistors are used). The mechanical and the electrical sys-tem equations may then be combined to get the following steady state relationship.

k

VaR + kΩ

2= TL + B Ω

The torque speed characteristics of the motor is shown in Fig. 6. The speed torque curve issteep at light loads. This is a desirable feature for traction applications as well as multi motordrives. Though field weakening is possible, the range of control in restricted. The effect of thesystem parameters on the speed torque curve is shown qualitatively in Fig. 7.

E6 201 Power Electronics

EE 189 IISc

N

T

Va1

Va2Va3

Va1>Va2>Va3

Fig. 6 Fig. 7

N

TIncreasing R

Increasing VaIncreasing eΦIncreasing

B

Dynamic model of the series excited dc motor

The series machine model is nonlinear. This may be put in the form of

d iad t

= va − R ia − kia ω

d ωd t

= −TLJ

− B ωJ

+ kia2

The system being nonlinear, the solution may be found through numerical integration. Or alinear small signal model may be found close to any desired operating point.

Control strategy of a series excited dc motor

By controlling the armature voltage, the motor may be driven at any desired speed(provided the load is not zero). In order to reverse the torque, a reversal of field winding isnecessary. The machine can thus be driven in the forward or reverse direction by reversing thefield. However, operation in the second and fourth quadrant will have to be through a differentcontrol strategy. This may be appreciated by seeing the electrical equation of the machine. At

, the effective resistance of the machine becomes zero. In other words, the backΩ = − R/k

ia

e, R ia

Field Resistance Line

Excitation Characteristicsia

R L

ω

Fig. 8

emf counters the resistive drop in the armature and the machine may operate without inputvoltage (self excitation). The self excited braking operation is shown in Fig. 8. For closervacontrol usually the series machine is reconnected as a separately excited machine for third andfourth quadrant operation.

SEPARATELY EXCITED DC MOTOR

We can put together the power converter circuit and the load (in this case a separatelyexcited dc motor) and study the system.

Single quadrant chopper

Figure 1 shows the separately excited dc motor driven through a dc to dc converter(chopper). The switch may be realised out of a number of possible devices (BJT and a diode,SCR and a diode, or MOSFET and a diode). The switch is operated as shown in Fig. 2. The

Fig. 1 Chopper Driven Separately Excited DC Motor

T g

TLRaLa

eV a

ia

ie

Re

LeVeJ, B

ωTon

Tof f

Vg

switching period is ( ). During the motor draws energy from the source.Ts Ton + Toff Ton

During armature current freewheels and no energy is drawn from the source. The voltageToffand the current waveforms are shown in Fig. 3. The system equations are

During on time:

vg = Ra ia + Ladiad t

+ kω

kia = Bω + J d ωd t

+ TL

During off time:

0 = Ra ia + Lad iad t

+ kω

kia = Bω + J d ωd t

+ TL

The system equations may be averaged as

vg d = Ra ia + Lad iadt

+ kω

k ia = B ω + J d ωd t

+ TL

= duty ratiod = TonToff

The machine's speed or torque may be controlled by control of the duty ratio . Thedsystem equations may be written in the following state space form.

d iad td ωd t

=

−RaLa

− kLa

kJ

−BJ

i aω

+

vgdLa

−TLJ

The inputs are: Control input d

E6 201 Power Electronics

EE 191 IISc

Fig. 2 Switch States

Ton Tof f t

Ton Tof f t

Vgt

i a

ig

t

t

Fig. 3 Switch States and Voltage and Currents in the Chopper

Power input vg

Load input TL

The system states are: Armature current iaRotor speed ω

The control objectives are: Speed control or Torque control through control of duty ratio d

Desired transfer functions: or for Torque or Current controlTgd

iad

for speed control ωd

The block diagram of the motor is shown in Fig. 4. The small signal system representation is

d^ia

d t

d^ω

d t

−RaLa

− kLa

kJ

−BJ

^ia^ω

+

VgLa0

^d +

DLa0

^vg +

0

−1J

^tL

From this the different transfer functions may be evaluated for each of the inputs while the otherinputs are made zero.

= torque control transfer function ; = load disturbance transfer function

^ia^d

^ω^tL

= speed control transfer function ;^ω^d

= power disturbance transfer function ;ω^vg

Discontinuous conduction

The above converter when realised for single quadrant operation with an IGBT switchand a diode is as shown in Fig. 5. This is a single quadrant chopper since the output current andvoltage are both restricted to be positive. The power flow can be only in one direction - fromsource to load. When such a chopper is used to drive a separately excited dc motor, and forconditions when the motor current becomes negative, the chopper cannot absorb this negative

Ra

1/La K

K

1/J

B

ω

ΤLd

Vg

Fig. 4 Block Diagram of Chopper Driven Separately Excited DC Motor

current. Then the load current becomes discontinuous. Such a mode of operation is called thediscontinuous current mode and is shown in Fig. 6. The system defining equations are

During on time ( ):d Ts

vg = Ra ia + Lad iad t

+ kω

k ia = B ω + J d ωd t

+ TL

During off time ( ):d2 Ts

0 = Ra ia + Lad iad t

+ kω

k ia = B ω + J d ωd t

+ TL

During idle time [ ]:(1 − d − d2)Ts

0 = Ra ia + Lad iad t

+ kω

0 = B ω + J d ωd t

+ TL

The system equations may be averaged as

vg d = Ra (d + d2) ia + Lad iad t

+ k (d + d2) ω

k (d + d2) ia = B ω + J d ωd t

+ TL

= duty ratiod = TonTs

E6 201 Power Electronics

EE 193 IISc

Vg

Ra Lai a

D

S

Fig. 5 Separately Excited DC Motor Driven by a Single Quadrant Chopper

ia

t

T s

e a

t

dTs d2Ts

Vg

Fig. 6 Armature Current and Armature Voltage in Discontinuous Conduction

Notice that is a dependent variable. The machines speed or torque may be controlled byd2control of the duty ratio d. The system equations may be written in the following state spaceform.

d^i a

d t

d^ω

d t

=

−Ra(d+d2)

La−

k(d+d2)La

k(d+d2 )J

−BJ

^ia^ω

+

vgdLa

−TLJ

The inputs are: Control input d Power input vg

Load input TLThe system states are: Armature current ia

Rotor speed ωThe control objectives are: Speed control or Torque control through control of duty ratio

Desired transfer functions: or for Torque or Current controlTgd

iad

for Speed control ωd

Steady state solution:

If the motor resistance is neglected, then the steady state equations areRa D Vg = (D + D2 ) k Ω

Define ; Then ; M = k ΩVg

M = DD + D2

D2 =D (1−M)

MFrom the discontinuous current waveform shown in Fig. 6,

;k Ip (D + D2)

2= B Ω + TL

Ip =(Vg − k Ω) D Ts

La=

Vg (1−M) D TsLa

;where(D + D2) = 2 La B

k2Ts

1D (1−M)

[M + M∗] M∗ =

k TLB Vg

Define K = 2 La B

k2Ts

; DM

= KD (1−M) [M + M∗] D = K M (M+M∗)

(1−M)

; ;D2 = 1M

− 1 Ω =M Vg

KIav =

(B Ω + TL)k

M =−K M∗ + (K M∗)2 + 4 K D2

2 K

At the border of discontinuous conduction will be satisfied by both the equationsM

; D =Kcrit M (M+M∗)

(1−M)= M Kcrit =

D (1−D)(D + M∗)

From this the criterion for operation on the border of continuous and discontinuous conductionis found as shown in Fig. 7.

operation in continuous conductionK = 2 La B

k2Ts≥ Kcrit

operation in discontinuous conductionK = 2 La B

k2Ts≤ Kcrit

Dynamic model in DCM

The motor equations (when is negligible) areRa

d iad t

= − k d ωLa

−k d2 ω

La+

d vgLa

d ωd t

= k d iaJ

+k d2 ia

J−

tLJ

− B ωJ

Steady State

; ; Ω =D Vg

k (D + D2)D2 =

D Vg (1−M)k Ω

I =BΩ + TL

k(D + D2 )

Dynamic equations (small signal)

d^ia

d t

d^ω

d t

=

0 −D+D2

k

LaD+D2

k

J−B

J

^ia^ω

+

Vg − kΩLakIaJ

^d

+

−k ΩLa

k IaJ

^d2 +

0

−1J

^tL

Relationship between and d d2

E6 201 Power Electronics

EE 195 IISc

0 1D

Kcrit K

CCM

DCM

Fig. 7 Continuous and Discontinuous Conduction Operation

Vg D

k Ω− D = D2

^d2 =

Vg − k Ω

k Ω

^d + D

k Ω^

vg −D Vg

k Ω2^ω

substituting for in the dynamic equation, we get^

d2

d^ia

d t= 0

d^ω

d t=

(D + D2 ) k

J

^ia + k Ia

J(^d +

^d2 ) − B

J^ω − 1

J^tL

Relationship between and and ^ia

^d

^d2

Ia =Vg − kΩ

D Ts

2 La

^ia = Ts

2 La

Vg − k Ω

^d + D

^vg − k D

^ω

D + D2

=

D Vgk Ω

^

d2 +^d = D

k Ω^

vg +Vgk Ω

^d −

D Vg

k Ω2^ω

on substitution for all the dependent variables,

d^ω

d t= ^ω

−B

J− D2B

K M2 J

+

^d

2 B Vg (1−M)k M K J

+ ^

vg

D2B (2−M)k K J M

Notice that the system order has reduced in DCM.

Two quadrant chopper

The chopper may be made into a two quadrant chopper by simply making the switchesin the chopper bidirectional. Such a two quadrant chopper is shown in Fig. 8. The current andvoltage waveform when power is flowing from the motor to the source is shown in Fig. 9. The

two quadrant chopper is capable of driving and braking the motor in one direction.

To Motor

Vg

Fig. 8 Bidirectional Chopper

V a Ton Tof f

Vg

i a

t

Fig. 9 Chopper Voltage and Current During Braking Operation

Four quadrant chopper

A four quadrant chopper is capable of driving and braking the motor in both the direc-tions. A four quadrant chopper made up of IGBTs is shown in Fig. 10. The power circuit does

not show the snubber elements. It may be seen that the four quadrant chopper consists of twonumbers of two quadrant choppers with the load connected between the two choppers in differ-ential fashion. The switch on/off timing signals are shown in Fig. 11.

SERIES EXCITED DC MOTOR

We can develop the chopper driven dc series motor model as follows.

E6 201 Power Electronics

EE 197 IISc

Fig. 10 Four Quadrant Chopper

Vg To Motor

P

P

N

N

t Va < 0

t Va = 0

t Va > 0

P

N

P

N

P

N

Fig. 11 Switch Enable Signals for Four Quadrant Chopper

Fig. 1 Chopper Driven Series Excited Motor

Vg

Lai aTof f

Ton

Le

V a

ig

J, Bω

T g

TL

Single quadrant chopper

Figure 1 shows the series excited dc motor driven through a dc to dc converter(chopper). The switch is realised using SCRs. In practice the dc series motors are used fortraction application where the voltage, current, and power are high (100s of kW). The devicesused are usually GTOs or SCRs. The chopper operation is as before. The defining equationsare

During on time

Vg = Ra ia + L d iad t

+ k ia ω

k ia2 = B ω + J d ω

d t+ TL

During off time

0 = Ra ia + L d iad t

+ k ia ω

k ia2 = B ω + J d ω

d t+ TL

The system equations may be averaged as

Vg d = Ra ia + L d iad t

+ k ia ω

k ia2 = B ω + J d ω

d t+ TL

= duty ratiod = TonTs

The machines speed or torque may be controlled by control of the duty ratio .d

The inputs are: Control input Power input d Vg Load input TL

The system states are: Current Speed ia ω

The control objectives are: Speed Control or Torque Control through duty ratio d

Desired transfer functions: or for torque controlTgd

iad

for speed controlωd

Fig. 2 Block Diagram of Chopper Driven Series Excited DC Motor

ω

Ra

1/La

K

K 1/J

B

ΤLd

Vg

The system equations are nonlinear. The block diagram representation of the system equationsis shown in Fig. 2. The dynamic solution can be obtained only by numerical simulation. Thesmall signal and steady state model may be readily obtained.

Dynamic model

The small signal dynamic model of the chopper driven series motor may be obtained to be asfollows.

d^ia

d t

d^ω

d t

=

−R+kΩL

−kIaL

2kIaJ

−BJ

^ia^ω

+

VgL0

^d +

DL0

^vg

+

0

−1J

^tL

From this the different transfer functions may be evaluated for each of the inputs while the otherinputs are being zero.

= torque control transfer function ; = power disturbance function

^ia^d

^ω^vg

= speed control transfer function ; = load disturbance transfer function^ω^d

^ω^tL

The dc series motor unlike the separately excited motor does not operate in the discontinuousconduction mode.

Ramp

V control

Ton (min)

Ton (max)

Ton

M

A

Fig. 4 forced Commutated Chopper Controller

Steady state solution:

If the motor resistance is neglected, then the steady state equations areRaD Vg = k Ω Ia

E6 201 Power Electronics

EE 199 IISc

D

Le La

M

AC

Vg

L

Fig. 3 A Series Motor Driven by a Force Commutated SCR Chopper

k Ia2 = B Ω + TL

A single quadrant chopper controlling a dc series motor along with the commutation circuit isshown in Fig. 3. A possible control scheme for the chopper devices is shown in Fig. 4.

Two quadrant chopper

The motor when operated in the second quadrant the torque has to be negative. Thismay be obtained by the dynamic braking scheme shown in Fig. 5. During braking the chopper is

off and the armature and the field are shorted through the braking resistor . Notice that theRbfield current direction is maintained to be the same as in running. The mechanical energy storedin the moving system is transferred to the electrical circuit and dissipated in the resistor . InRbdrives where the braking is more frequent (as in suburban trains), one may not like to dissipate

Le

Run

Brake

Vg

Fig. 5 Dynamic Braking of Series Excited Motor

Fig. 6 Regenerative Braking Chopper for DC Series Excited Motor

Vg

START

ACCELERATE

RUN

Fig. 7 Series & Parallel Connection of DC Series Motors in Traction Application

the stored energy during braking. It will be economical to recover the energy in the moving sys-tem and feed it to the source. Such a braking operation is called regenerative braking. Ascheme to achieve the same is shown in Fig. 6.

Four quadrant chopper

Usually it is not required in series motor drives to obtain fast multiquadrant transitions.Therefore usually four quadrant operation is achieved by relay change over circuits.

Multiphase chopper

Traction application invariably uses multiple motor drives. In such cases several lowerrated choppers may be used in place of a single higher rated chopper. The individual choppersmay be operated with a phase shift from each other with obvious advantages. The arrangementsare shown in Figs 7 and 8.

CLOSED LOOP CONTROL OF MOTOR DRIVES

Control Equations of the DC motor

In this section we see the closed loop control of the separately excited dc motor. Thestate space model of the separately excited motor driven from an armature supply is given as

E6 201 Power Electronics

EE 201 IISc

ACCELERATE

RUN

START

Fig. 8 Multiphase Chopper Applications

diadtdωdt

=

−RaLa

− kLa

kJ

−BJ

iaω

+

GLa0

vc +

0

−1J

tL

where is the gain of the power converter used for the armature power source.G•X = A X + b

^vc + h

^tL

ω = c X = [0 1] X

The control of the motor is done through the control of the control input . The control transfervcfunction is

G(s) =^ω (s)^

vc (s)= c [sI − A]−1b

; ; = k GBRa(1+sτe)(1+sτm)+k2

τe = La

Raτm = J

B

The transfer function is of order 2When the motor is lossless ( ), the system poles are on the imaginaryB = 0 = Raaxis.When the losses are very high ( ), the system poles are real and theBRa >> k2

same as the electrical and mechanical frequencies of the motor.When the electrical time constant is very much less than the mechanical timeconstant of the motor ( ), then the system is effectively a first orderτe << τm

system with the single pole at equal to .τm1τmBRa

k2+BRaIn general if the loss is low and the time constants and are not wellτe τmseparated, then the motor is an underdamped second order system with the natural

frequency at .k2+BRaLaJ

Fig. 1

Lossless Motor

Fig. 2

Highloss Motor

Fig. 3

Fig. 4

Underdamped Motor

k/ LaJ

G

log ω

log ω log ω

log ω

G

G

G

ωe

ωm

ωm1 ωo

ωe >> ω m

These different cases are shown in Figs 1 to 4. Practical machines will have a response asgiven in Figs 3 or 4. For those in Fig. 3, a proportional controller may be used. For those inFig. 4, a lead-lag controller may be used

Proportional Controller

Consider case shown in Fig. 3 above. The motor transfer function is

G(s) = K1 + sτm1

; τm1 = τmBRa

BRa+k2

We may introduce a proportional controller with a gain as shown in Fig. 5. TheHc = Kcclosed loop transfer function is

; Gcl(s) =

KKc1 + sτm1

1 + KKc1 + sτm1

= KKc1 + sτm1 + KKc

≈ 11+sτ

τ =τm1KKc

The closed loop speed response for step change in speed will be

ω = ω ∗

1 − e− t

τ

The closed loop current transfer function is

Gi(s) =

BKKc(1 + sτm)

k1 + sτm1

1 + KKc1 + sτm1

≈ Bk

(1 + sτm)1 + sτ

The closed loop response in current to step change in speed reference is given by

ia(t) = ω∗ Bk

1 − e− t

τ

+ ω∗ B

k

τmτ

e− t

τ

The first part eventually leads to the steady state current equal to . The sec-Ia Bω∗/kond part is a transient current and can be several times the steady state current (on account ofthe multiplier ). In typical systems will be more than 10 and so this type of con-τm/τ τm/τtroller will result in large current transients which stress the motor windings enormously. So asa rule this kind of speed controllers are not employed in motor control. Instead the motor

E6 201 Power Electronics

EE 203 IISc

Kc k

TLω∗ ω

Fig. 5 Block Diagram of a DC Drive with Proportional Controller

1B

11 + sτm

KBk

1 + sτm1 + sτm1

current is controlled in closed loop and the speed control loop is realised as an outer loop.Such a controller is as shown in Fig. 6.

CURRENT

All practical drives incorporate an inner current controller ( ) and an outer speedHicontroller ( ) as shown in Fig. 6. Consider the current loop as shown in Fig. 7.Hs

Gi(s) =^ia (s)^

vc (s)=

(1 + sτm) BGJLa

1 + s

Qω p+ s2

ωp2

BRa + k2

JLa

ωp2 =

BRa + k2

JLa

; Q =

ωp

RaLa

+ BJ

The transfer function is shown in Fig. 8. The phase of the transfer function is always inside±90° and hence the design of a closed loop compensator is simple. Either a proportional or aproportional plus integral (PI) current compensator may be used. Consider a PI compensator ofthe form

Hi(s) =1+ s

ωps

ωp

The loop gain is shown in Fig. 8. The closed loop bandwidth may be seen to be higherGiHithan and and so the closed loop gain in the range of will be practically unity.ωp ωm ω ≤ ω pThis is shown in Fig. 7b.

SPEED CONTROLLER

Fig. 6 Closed Loop Controller with Nested Control Loops

ωω∗Hs Hi Gi(s) k

TL1B

11 + sτm

I*I1

Fig. 7 Closed Loop Current Controller and Its Low Frequency Model

Hi Gi(s)I∗ I

Fig. 8 Motor Current Transfer Function

G

log ωωmωp

With the current controller , the current gain may be taken as 1, and the system isHi(s)simplified as shown in Fig. 10.

The speed transfer function is now first order

ω(s)I∗(s)

= kB (1 + sτm)

and is shown in Fig. 11. The design of closed loop compensator for speed is quite simple.

Select

Hs(s) = Ks1 + s

ω ms

ωmThe loop gain is shown in Fig. 12. The bandwidth of the overall system is .ωmkKs/B

E6 201 Power Electronics

EE 205 IISc

Fig. 9 Loopgain Transfer Function

GiHi

log ωωmωp

Fig. 10 Block Diagram of the Current Controlled DC Motor

1

Current Controller

ii* ωω∗Hs k

TL1B

11 + sτm

Fig. 11 Current Speed Transfer Function

G

log ωωm

Fig.12 Loopgain of the Speed Controlled Motor Drive

Gω(s)Hs(s)

log ωωmωmkKs/B

All practical motor drives (separately excited dc motor, series dc motor, ac motors) in-corporate the above type of speed control. The speed controller ( ) output may be clampedHsto and . Then during speed transients, the speed loop will not be effective and+Imax −Imaxthe motor will be controlled to draw constant current and deliver constant torque correspon-ding to . Thus the motor current will be within safe limits during transients also.± Imax

FEED FORWARD CONTROLLERS

A Simple Example

Consider the following circuit shown in Fig. 1. The circuit consists of an RLcircuit terminated to a current load. The control objective is to maintain the current through theinductor constant.

The circuit equation is

Ld idt

= u− (i − IL)R

i =u+ILR

(sL+R )The above system may be represented by the following block diagram shown in Fig. 2.

Following the classical method we may design a PI controller of the form

; Hi(s) = K 1+sTasTa

Ta = LR

The controller loop gain is

T = K1+sTa

Ta1R

11+sTa

= Ks L

Such a design will lead to a steady state error of zero and a control bandwidth of . Such aKL

controller is shown in Fig. 3. The response in current for step change in command is given by

; i = I∗

1 − e− t

τ

τ = L

K

u

i L

R I L

Fig. 1 A Simple RL Circuit

u

RI L

i

Fig.2 Block Diagram of the Circuit in Fig. 1

1R +s L

Fig. 3 Block Diagram with Classical PI Controller

I∗

i

K(1+ sTa)sTa

1R +s L

i

u

RIL

For faster response, the gain must be made large. The performance in the response ofKinductor current will be quite satisfactory for changes in the command input. However, if thereare disturbances in the load current , the performance is not quite satisfactory. It can beILshown that for step changes in the load current the response in the inductor current is givenIL

∗

by

i = IL∗

τ

Ta − τ

e− t

τ − e− t

Ta

This response consists of a slow component , and is not satisfactory. Further

e

− tTa

if either the resistance or the inductor is a function of the current through them then the sys-R Ltem equations become non-linear. In such a case a design based on the classical approach in-troduces further limitations by way of small signal validity. The following explains the strategyof feedforward control. The system equation is given by

L d id t

= u − v

If the desired response in current is required to be first order with a time constant of , theTisame may be expressed mathematically as

Tid id t

+ i = I∗

When the above two equations are combined, we get

u = LTi

(I∗ − i) + v

If the input is made up according to the above equation then the resulting response inucurrent will be first order with a time constant of . Such a controller is shown in Fig. 4. TheTicontroller is a simple proportional control (of gain ) followed by the additional feed for-L/Tiward term ( ).v

Advantages of feed forward control:The response is same for command tracking as well as disturbance rejection.The method works for non-linear systems also.Disadvantages of the feed forward control:

It requires extra signals to be sensed and fed forward (in this case ).vA knowledge of system parameter is needed ( in this example).L

Application of Feed Forward to Speed Control of DC Motor

The feed forward control is applicable to the speed controller. Consider the dc motorfor which a current controller has been realised. When the current bandwidth is large, the

E6 201 Power Electronics

EE 207 IISc

Fig. 4 Feedforward Current Control in a Simple RL Circuit

i L

R I L

I*

i

vLTi

current controller may be assumed to be a simple gain of magnitude , for frequencies lessGithan the bandwidth of the current controller. The motor may be represented by

I∗ Gi KT = md = J d ωd t

+ B ω + mL

The mechanical system is represented in Fig. 5.

Let the desired response be with a time constant of . This may be expressed mathematicallyTwas

Twd ωd t

+ ω = ω∗

We may combine this to the system mechanical equation to obtain the following relationship onthe current reference .I∗

I∗ = 1GiKT

mL + B ω + J

Tw[ω∗ − ω]

The above controller is shown in block diagram in Fig. 6. It consists of a proportional currentcontroller and the feed forward terms corresponding to the frictional torque term and theB ωload torque term .mL

In realising the above control apart from the speed signal for feedback, load torque mL also is required to be sensed. Further the quantities and are all also to be known.J, B, KT GiError in speed error gain will result in a difference in the response time, but will not causesteady state error in speed. Error in feed forward terms will result in an error in the steadystate response without causing any error in response time. This steady state error may be cor-rected by an integral controller operating across the proportional controller of speed error op-erating very slowly.

The following schematic in Fig. 7 shows a feed forward controller for the cur-rent control in dc motor.

B

I*

md

mLω

Fig. 5 Mechanica System of a Motor Drive

1J ∫KT Gi

ω∗

ωf b ωf b mL

I*

Fig. 6 Feedforward Speed Controller

1BTW

B 1GiKT

R

I *e

iRi

va

Fig. 7 Feedforward Current Controller

LaTa

SPEED CONTROL OF DC MOTOR

(Symmetrical Optimum Method)

In Section 1 it was seen that the speed controller may be realised through a PIcontroller around the current controlled dc motor. In such a case it was mentioned that thespeed amplifier PI time constant is chosen to be the same as the mechanical time constant of themotor. Even though this is a satisfactory method for small machines, it will result in inconven-iently large circuit elements for the speed amplifier in the case of large motors which havelarge mechanical time constant. In such applications a method known as the symmetrical opti-mum method has become popular.

Machine Model

The model of the current controlled dc machine is shown in Fig. 1

Closed Loop Controller

With such a system we may employ a PI speed controller. The closed loop control ofspeed with such a system is shown in Fig. 2.

The various parameters of the system are: Current Controller Gain.Ki: Torque Constant of the Motor.KT

: Gain of the Mechanical system .Km Km = 1

J

: Speed Measurement Gain.Ks

: Response Time Constant of the Current Controller (typically 2 to 5mS).Ti: Filter Time Constant of Speed Measurement (typically about 1 to 5mS).Ts

The design consists of evaluating the proportional time constant and the proportional gainTp, for a given phase margin of design specification and evaluate the correspondingKp Φm

bandwidth of the design.

E6 201 Power Electronics

EE 209 IISc

I*

mL

ω

ωfb

Fig. 1 Current Controlled DC Machine

Ki KT1 +s Ti

1J s

Ks1 +s Ti

Fig. 2 Current Controlled DC Machine

I*

mL

ω

ωfb

ω Ki KT1 +s Ti

KmJ s

Ks1 +s Ti

Kp (1 + sTp)s Tp

Design Method

Figure 3 shows the simplified equivalent block diagram of the system. The cur-rent controller time constant and the speed measurement blocks have been combined and repre-sented by their low frequency equivalent.

σ = (Ti + Ts + Tc)

: Average delay in the converter (1.67 mS for a 6 pulse converter).Tc

db

1/Ti

ω 1/σ log ω

degree

ω log ω

Φm

Fig. 4 Asymptotic Bode Plot of Loopgain

The asymptotic bode plot of the overall loop gain is shown in Fig. 4. The proportional timeconstant is chosen to be a multiple of the equivalent natural time constant of the system.Tp σ

Tp = a2 σ

In such a case in order to obtain the best phase margin the proportional gain has to beKpchosen such that the gain cross over frequency is the geometric mean of the two characteristic

frequencies of the system. Phase margin is then given by1/Tp and 1/σ

ωc = 1Tp σ

= 1a σ = a

Tp

Φm = 180 + −180 + tan−1

ωc Tp − tan−1(ωc σ)

Φm = tan−1(a) − tan−1

1a

The above expression may be simplified to and put in the form of a table forconvenience.

Fig. 3 Current Controlled DC Machine

I* ω

ωfb

ω Ki KT Ks1+ s

KmJ s

Kp (1 + sTp)s Tp σ

Φm = tan−1

12

a − 1

a

a 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5

Φm 27.9 34.4 39.8 44.3 48 51.2 53.9 56.3 58.4 60.2 61.8 63.2 64.5

For the desired phase margin corresponding value for may be selected from the aboveaTable. The proportional time constant is then selected. A suitable value of is 2.Tp a

The magnitude of the loop gain is unity at the crossover frequency. From this theproportional gain is evaluated.Kp

G = 1 =KpKiKTKsKm

Tp

1+j ωcTp

(ωc)2 1+j ωcσ

1 = σ KpKiKTKsKm

1+a2

1+ 1

a2

Kp = 1σKiKTKsKm

1+a2

1+ 1

a2

Problem Set

1 Consider a permanent magnet dc servo motor with the following parameters.

;Trated = 10 Nm

;Nrated = 3700 rpm

;kT = 0.5 Nm/A

;kE = 53 V/krpm

;Ra = 0.37 Ω

;τe = 4.05 mS

;τm = 11.7 mS

;B = 0

Calculate the terminal voltage in steady state if the motor is required to developVta torque of 5 at a speed of 1500 rpm.Nm

(N. Mohan, P 11-1, pp 306)

2 Express the transfer function in the following form.G(s) = ω(s)Vt(s)

G(s) =

1kE

1+2sDωn + s2

ωn2

E6 201 Power Electronics

EE 211 IISc

Calculate and for the servo-motor parameters given in Problem 1. Plot theD ωnmagnitude and phase of by means of a Bode plot.G1(s)

(N. Mohan, P 11-2, pp 307)3 With the parameters given in Problem 1, calculate and plot the change in as aωm

function of time for a step change increase of 10 V in the terminal voltage of the uncontrolled unloaded motor.

(N. Mohan, P 11-3, pp 307)4 The servo-motor in Problem 1 is driven by a full bridge dc to dc converter

operating from a 200 V dc bus. Calculate the peak-to-peak ripple in the motorcurrent if a PWM bipolar voltage switching scheme is used. The motor isdeveloping a torque of 5 at a speed of 1500 rpm. The switching frequency is 20NmkHz.

(N. Mohan, P 11-4, pp 307)5 Repeat Problem 4 if a unipolar voltage switching scheme is used in the dc to dc

converter. (figure out the difference between unipolar & bipolar switching first).(N. Mohan, P 11-5, pp 307)

6 The speed of a separately excited dc motor is controlled by a single phase fullwave controlled rectifier. The field circuit is controlled by a full converter and thefield current is set to the maximum possible value. The ac supply voltage is 240 V,single phase, 50 Hz. The armature resistance is 0.5 ohm, the field resistance is 345ohm, and the emf constant is 0.71 V/A - rad/sec. Friction and no-load losses arenegligible. The armature and field currents are continuous and ripple free. Thedelay angle of the armature converter is 45° and the armature current of the motor is55 A. Determine

(a) torque developed by the motor, (b) speed of the motor in rpm, and (c) input power factor.

(M.H. Rashid, P 10-6, pp 336)7 A dc chopper controls the speed of a series motor. The armature resistance is 0.04

ohm. field resistance is 0.06 ohm, emf constant is 35 mV/A - rad/sec. The dc inputto the chopper is 600 V. If it is required to maintain a constant developed torque of547 Nm, plot the duty cycle of the chopper against the motor speed in the range of 0to 1200 rpm.

(M.H. Rashid, P 10-12, pp 336)8 A dc series motor is powered by a dc chopper from a 600 V dc source. The

armature resistance is 0.03 ohm, field resistance is 0.05 ohm, emf constant is 15.27mV/A - rad/sec. The average armature current is 450 A without any appreciableripple. The duty ratio of the chopper is 75%. Determine

(a) input power from the source, (b) equivalent input resistance of the chopper, (c) motor speed in rpm, and (d) developed torque of the motor.

(M.H. Rashid, P 10-13, pp 336)9 A dc series motor is braked with a chopper to return the power into a dc bus of 600

V. The armature resistance is 0.03 ohm, field resistance is 0.05 ohm, emf constantis 12 mV/A - rad/sec. The average current through the motor is maintained constantat 350 A. The armature current is continuous and ripple free. The duty ratio of thechopper is 50%. Determine

(a) average voltage across chopper, (b) power regenerated to supply, (c) equivalent load resistance of the motor acting as a generator,

(d) minimum possible braking speed in rpm, (e) maximum possible braking speed in rpm, (f) motor speed in rpm.

(M.H. Rashid, P 10-14, pp 337)10 A dc series motor is controlled by a two phase chopper. The average armature

current is 250 A with negligible ripple. Each of the chopper is operating with 50%duty ratio. The individual choppers are appropriately shifted in phase. Eachchopper is operated at a frequency of 250 Hz. Sketch the input current drawn fromthe source. Evaluate the average input current and the fundamental component of theinput current. At the input a simple LC filter (L = 0.35 mH, C = 5600 microfarad) isused. Evaluate the rms fundamental component harmonic current injected into thesource.

(M.H. Rashid, P 10-18, pp 337)11 A 40 hp, 230 V, 3500 rpm, separately excited dc motor is controlled by a linear

converter of gain 200. The moment of inertia of the load is 0.156 Nm/ rad/sec,friction is negligible, armature resistance is 0.045 ohm, armature inductance is 730mH, The emf constant is 0.502 V/A - rad/sec. The field current is maintainedconstant at 1.25 A.

(a) Obtain the open-loop transfer functions ( ) and ( )ω(s)/Vr(s) ω(s)/TL(s)for the motor. (b) Calculate the steady state speed of the motor if the input voltage isVrset at 1 V, and the motor is loaded to 50% of its rated torque.

12 The following are the parameters of a permanent magnet dc motor.

; ; ;Trated = 12 Nm Nrated = 3000 RPM KT = 0.6 Nm/A ; ; ;RA = 0.575 Ω LA = 3 mH B = 0.003 Nm/(rad/ sec)

;J = 0.001 Nm/(rad/ sec2 )Evaluate the rated current and rated voltage of the motor.Evaluate the electrical time constant of the motor.Evaluate the mechanical time constant of the motor.Rated output power of the motor.Electrical and mechanical loss in the machine under rated operatingcondition.Efficiency of the motor under rated operating condition.Sketch the speed ( ) Vs torque ( ) characteristicsω rad/ sec Tgenerated Nwm

(only in the first quadrant) of the motor for the two conditions of at 200VAand 100 V.

13 The above motor is driven by a chopper with an input dc voltage of 250 Volts. Theoperating duty ratio of the chopper is 0.6, while the generated torque of the motor is9 . Evaluate the speed, load torque, output power, input power and efficiencyNwmof the motor. Assume the chopper switches to be ideal.

14 Evaluate the three different speed transfer functions for the motor, and the threedifferent current transfer functions for the motor. Present your results in anormalised form (dc gain, normalised zeroes, and normalised poles).

^ω^d

^iA^d

E6 201 Power Electronics

EE 213 IISc

^ω^

vg

^iA^

vg

^ω^tL

^iA^tL

15 Figure 15 shows a dc series motor driven through a chopper from a voltage source. The mathematical model of the motor is given byvg

vin = Ld iadt

+ ea

tg = J d ωd t

+ B ω + tL

ea = k1 ia ω

tg = k2 ia2

a. For a lossless motor show that .k1 = k2b. Write down the differential equations governing the system during i) ii) Ton Toffc. Write down the averaged equivalent system differential equations andstate the assumptions in arriving at the same. d. Express the system equations in the form

where is the system state vector , •x = f

x, ω, vg, d, tL x [ia ω]T

d = TonTon+Toff

e. Consider small perturbations in the inputs

; ;vg = Vg +^

vg d = D +^d tL = TL +

^tL

and the corresponding perturbations in the motor states

;ia = Ia +^ia ω = Ω + ^ω

Under the above perturbations, derive the steady state and small signal linear ac modelsin the form

; X = g (Vg, D, TL) X = [Ia Ω]T

Fig. 15 Chopper Driven Series Excited Motor

Vg

Lai aTof f

Ton

Le

V a

ig

J, Bω

T g

TL

^•x = A

^x + b

^vg + f

^d + h

^tL

f. The motor has a rated current of A and rated speed of rpm. The10 750machine constants are Volt sec/ Amp rad ; Nwm/Amp2.k1 = 0.1 k2 = 0.1The friction coefficient is NwM sec/rad. The machine whenB = 0.006driven from a chopper supplied with dc power at V is running at 150 750rpm delivering half the rated torque to the load.

i) Evaluate the steady state current drawn by the motor.

ii) Evaluate the steady state duty ratio of the chopper.

g. From the small signal model evaluate the following transfer functions

i) ii)^ω (s)^

tL (s) ^vg = 0,

^d =0

^ω (s)^d (s) ^

vg = 0,^tL =0

h. For mH and NwM sec2/rad, express the aboveL = 10 J = 0.003transfer function in the normalised pole zero form.

16 The series motor shown in Fig. 16 is controlled by a chopper ( switching atS and_S

high frequency with on and and off) and is running at 750 rpm, deliveringS2 S1 S3a load of 50% rated torque. The motor details are

A ; rpm ; SI unit ;Irated = 10 Nrated = 750 k = 0.1

; ;B = 0.006 NwM Sec/rad J = 0.003 NwM Sec2/rad

; ; ; TL = 50% Vg = 150 V L = 10 mH

a Evaluate the steady state current. Under this condition at t = 0, the chopperswitches are both turned off and dynamic braking is applied ( bothS and

_S

off and off and both on).S2 S1 and S3b Write down the dynamic equations of the system. c Write down the initial conditions of the system dynamic equations.d Write a program to numerically solve the system equations.e With the help of the program, evaluate the time taken for the motor speedto drop to 150 rpm for

i) ii) Rb = 0.1Ω Rb = 3.0Ω

f Evaluate the peak current reached in each case.g Evaluate the energy dissipated in the resistor in each case (till the timeN reaches 150 rpm).

E6 201 Power Electronics

EE 215 IISc

Fig. 5 Dynamic Braking of Series Excited Motor

LeVg

S

S

S1S2

S3

Rb J, B, Tg, TL

ωea

ia

h Establish the energy balance from start to end in either case (i.e. initialstored energy, final stored energy, and the energy dissipated or delivered inbetween).

17 Engineer Lam just graduated from Indian Institute of Science with an M.E. inCAPSAD (how glad?). He did not want to pursue further education because he wasconned by his teacher into practising what he has studied for ages. He got a job in aGovernment Defence Research Laboratory.

The place of posting is Hyderabad. He thought that the government wanted agenius like him to be kept secure far away from all the borders of the country.

The time is 1995 when the gods have been exhibiting their thirst if nothunger.

The first assignment for Engineer Lam was to design a position controlsystem. The purpose was supposed to be, to move a glass of milk and position it atthe mouth of Lord Ganesh. Engineer Lam being smart knew that the real purpose ofthe position control system is more sinister.

Engineer Lam did some research (i.e. took out his class notes and copied thesystem equations) and wrote down the following equations under constantexcitation.

Electrical Circuit.va = Ladiad t

+ Raia + eb

Mechanical system with no counteracting loadTg = Jd2θd t2

+ Bdθd t

Torque relationshipTg = KT ia

Back emf relationshipeb = Ked θd t

He took these equations to his boss. His boss is an Ex-Service Veteran - CaptainSham. The captain was pleased with his new recruit. From an old book which wasmeant only for the eyes of the captain he read out the following parameters forEngineer Lam to note down.

; ; ; ;Ra = 2 La = 0.01 J = 1 B = 5 ;Ke = 0.5 KT = 20

Engineer Lam was confused. His book said that and were normally equal.KT KeFinally he understood why they were not equal.

(1) Can you explain a possible reason why they are not equal.Then Engineer Lam reasoned that the field current may be kept constant so that lifeis simple. He knew that for such a linear set of relationship, Transfer function is ahandy way of presenting the dynamics of the machine.

(2) Evaluate the open loop transfer function .θ (s)va(s)

What Engineer Lam next did was a stroke of sheer genius. He neglected thearmature inductance because it was very small. Unlike what you obtained in (2)Lathe result he got was of the form

θ (s)va(s) = α

s (s +β)(3) Evaluate and .α β

Engineer Lam was happy that everything was moving great. He looked at theopenloop transfer function and realised that it will be impossible to control theposition with openloop control.

(4) Do you agree with Engineer Lam?

Engineer Lam realised the need for a closed loop controller for position control. Alittle more research (!) indicated that a simple proportional compensator will besufficient to get zero steady state error in position.

(5) Can you substantiate the above claim?Engineer Lam sketched a simple proportional controller with a gain of as shownKin Fig. 1. He remembered that for second order systems where proportionalcontroller is used, root locus is a good technique to select with a knowledge ofKopenloop poles and the desired closed loop poles.

(6) Sketch the root locus for the closed loop poles as a function of Kvarying from 0 to .∞(7) Select such that the closed loop response is a second order responseKwith a damping factor of 0.7. Evaluate the natural frequency of responseunder this condition.

Engineer Lam was by now ecstatic. He decided to impress the captain by throwingin a simulation also. He had this nice package TUTSIM available (pirated from IISclab).

(8) Write a simple program simulating the motor, and the closed loopcontroller (the simplified model obtained after the stroke of genius from Er.Lam).(9) Verify from the simulation the closed loop response for unit step input in

, the natural frequency of response, and the overshoot correspondingTheta∗to the designed damping factor of 0.7. The official Nos. may be obtainedfrom any control system book.

Er. Lam completed all this and showed his results. Captain was thoroughly pleased.He asked Er. Lam to build the system.

18 Engineer Lam visited the place where the position control system was going to beinstalled (remember the Ganesh temple where the system was supposed to beinstalled ?). He checked with the user that the position commands will always bewithin 2 units ( is limited to be within 2 units). Er. Lam ran his± Theta∗ ±position control simulation program to find out the maximum voltage to be appliedto the motor and the maximum current drawn by the motor.

From your program evaluate the maximum current and maximum voltage tobe supplied to the motor if the position command input is restricted asabove.

Er. Lam then selected a power amplifier accordingly, which is capable of meetingthe above requirements. He wanted the position feedback signal to the controller tobe in the range of . He selected a suitable potentiometer for the purpose as±10Vshown in Fig. 1.

What is the value of the feedback gain ? What will be the value of theβamplifier gain so that the response is the same as designed in the previousKproblem?

Er. Lam got hold of an amplifier to serve as the Block K. He checked its workingindependently that it gives the required gain, and capable of delivering the required

E6 201 Power Electronics

EE 217 IISc

K MotorΘ*

Θ

va Θ

Fig. 1 DC Motor with Proportional Controller

current. The amplifier also has a knob (remember this knob) at the front which canbe adjusted to get the required gain? He connected the motor to the amplifier, andthe position feedback potentiometer to the motor shaft. He built a small summingcircuit as shown below in Fig. 2. He used a linear power supply giving ( ) to±15Vprovide the power required for the summing amplifier. Next he built a small stepsignal injection circuit as shown below in Fig. 3.

Indicate how the above may be used to inject a unit step input to positioncommand.

Er. Lam tested his design by injecting a step input of 2 V to the position command.

Just to be sure he also tested his design with the actual gain one half of the designedvalue. He ran his simulation program again to check the peak current and voltage

requirements from the amplifier if the gain were doubled. Then ensuring that hisamplifier can supply the required voltage and current, he tested the system withdouble the designed gain.

From your simulation program give the response for the above two cases.What are the peak currents, voltages and settling time in each of thesecases?

Being satisfied with the design, Er. Lam arranged for a demonstration for the benefitof his boss. The demonstration went off perfectly. Er. Lam showed the designedresponse, response with half the gain, and with double the gain. The captain wasquite impressed. He looked at everything. He wanted to throw the switches S1 andS2 in different sequences and see for himself the position control system. Then his

−θ∗

+θ∗

10k

10k

10k

To Amplifier+15V

-15V

-

+

Fig. 2

K Motor

β

θ∗

θ

θva

Fig. 1

−10V ≤ θ∗ ≤ + 10V

-

+To −θ∗

S1S2

5k

10k

+15V

-15V

+15V

-15V

Fig. 3

eyes landed on the gain setting knob of the amplifier. He wanted to know its role inthe overall plan. Er. Lam said that by simply cranking up the gain, it is possible tomake the system respond faster. The captain gleefully cranked up the gain.

It all started with a small whimper from the amplifier. The whimper grewinto a growl. Then someone noticed an imperceptible shiver in the motor shaft. Theshivering grew, the shaft started shaking violently, fumes started to come out of themotor, and mercifully the fuse in the amplifier blew before more damage was done.

Explain what happened. Take into account the neglected inductance of themotor, and evaluate the gain threshold which the Captain exceeded.Add the effect of the motor inductance in your simulation program andcreate an 'action replay' of what happened in the system.

E6 201 Power Electronics

EE 219 IISc