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MODELING A DC MOTOR FOR SIMULATION WITH PSPICE 1. Equations for a separately excited dc motor The main equations for a dc motor whose speed is controlled by varying the armature voltage while the field current is kept constant are shown below. a a a a a a  E dt dI  L  I  R V  + + =  (1) m a  K  E  =  φ  (2) m m load elec  B dt d  J T T  + + =  (3) a e  I K T  φ =  (4) Equation (1) represents the armature circuit, where V a ,  I a ,  R a ,  L a  and  E a  are the average voltage applied to the armature, the average armature current, the armature resistance, the armature inductance and the induced emf in the armature ( speed voltage) respectively. The latter, as shown in (2), varies linearly with the motor speed ( m ) given in rad/s if one assumes that the flux ( φ ) is constant, what is quite reasonable since the field current is kept constant. Equation (3) presents the dynamics of the mechanical part of the motor and load. It shows that the electrical torque ( T elec ) is equal to the load torque ( T load ) plus the inertial and the friction torques. The constants  J  and  B are the moment of inertia (kgm 2 ) of the rotating masses and friction coefficient (Nms). As one can see in (4), the electrical torque is a function of the armature current. Example #1 - Steady-state characteristics of a dc drive Consider a separately excited dc motor supplied by a dc-dc step-down chopper. What is the chopper duty cycle so that one obtains a speed of 1000 rpm (104.72 rad/s)? The  paramet ers of the ch opper, motor and load are as follo ws: - Chopper: Supply voltage of 440 V and switching frequency of 500 Hz.
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DC Motor drive

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• MODELING A DC MOTOR FOR SIMULATION WITH PSPICE

1. Equations for a separately excited dc motor

The main equations for a dc motor whose speed is controlled by varying the

armature voltage while the field current is kept constant are shown below.

aa

aaaa EdtdI

LIRV ++= (1)

ma KE = (2)

mm

ae IKT = (4)

Equation (1) represents the armature circuit, where Va, Ia, Ra, La and Ea are the

average voltage applied to the armature, the average armature current, the armature

resistance, the armature inductance and the induced emf in the armature (speed voltage)

respectively. The latter, as shown in (2), varies linearly with the motor speed (m) given

in rad/s if one assumes that the flux () is constant, what is quite reasonable since the

field current is kept constant.

Equation (3) presents the dynamics of the mechanical part of the motor and load.

It shows that the electrical torque (Telec) is equal to the load torque (Tload) plus the inertial

and the friction torques. The constants J and B are the moment of inertia (kgm2) of the

rotating masses and friction coefficient (Nms). As one can see in (4), the electrical torque

is a function of the armature current.

Example #1 - Steady-state characteristics of a dc drive

Consider a separately excited dc motor supplied by a dc-dc step-down chopper. What is

the chopper duty cycle so that one obtains a speed of 1000 rpm (104.72 rad/s)? The

parameters of the chopper, motor and load are as follows:

- Chopper: Supply voltage of 440 V and switching frequency of 500 Hz.

• - Motor: PShaft = 26 kW, Van = 400 V, Ian = 75 A, Ra = 0.432 , La = 5.53 mH

and n = 1800 rpm (188.5 rad/s). The friction torque is proportional to the

speed.

- Load: The load torque is proportional to the speed, absorbing a mechanical

power equal to the rated power of the motor at rated speed.

Solution:

The duty cycle of the chopper is calculated as:

? , 440 , 7.104_7.104_

7.104 === aSupplySupply

am VVVV

VD

? ?, , 7.104_7.104_7.104_7.104_7.104_ ==+= aaaaaa EIERIV

? ,7.104_7.104_ == KKE ma

The factor K is constant and is calculated for the rated condition. From (1) - (2):

Nm/A 95.1188.5

432.0 x 75400 ==

=mn

aanan RIVK

V 22.2047.104_ =aE

The armature current at the desired speed is calculated from (3) (4).

? ?, , 7.104_7.104_

7.104_7.104_7.104_7.104_ ==

TTK

TI

Nm 6.76 Nm, 93.137 , 104.7_x_

n

mn

PTTT

The friction torque at rated speed is obtained from the power balance equation.

Nm 33.8 ,2x_

x_ =

==n

naanananfricn

mn

mfricnfric

PRIIVTTT

Nms 0.0442 Nm, 63.47.104_ ===

mn

fricnfric

TBT

AIa 66.4195.163.46.76

7.104_ =+

=

V 22.22222.2040.432 x 66.417.104_ =+=aV 0.50544022.222

7.104 ==mD

• 2. Creating a simplified model of a dc motor for PSpice simulations

Figure 1 shows a model of a dc motor. It was implemented with basic blocks of

PSpice according to the equations that represent the dynamics of the machine. The

average value of the voltage applied to the armature winding can be controlled with a

simple VPULSE voltage source. The armature winding is modeled with a resistance, an

inductance and a Voltage Controlled Voltage Source, block E of PSpice. This models the

effect of the motor speed on the induced emf in the armature or speed voltage (Ea). The

gain of block E is the K factor of the motor.

The modeling of the armature winding is straightforward. The input voltage is the

sum of the voltage drops across the series impedance and speed voltage of the armature.

For the modeling of the mechanical part of the motor, one can represent the electrical

torque, by a Current Controlled Current Source, block F of PSpice. Thus, the other

torques should also be represented by currents that flow in parallel branches. The

magnitude of the voltage across the three branches is numerically equal to the speed of

the dc motor. Under this assumption, the friction torque can be emulated by a resistor

with a value that is equal to the inverse of the friction coefficient (B). The same idea

applies to the representation of the load torque when it is a function of the speed of the

motor. The inertial torque can be represented by a branch with a capacitor, since its value

depends on the variation of the speed of the motor. The value of the capacitor should be

equal to the moment of inertia of the rotating masses (J) in kgm2.

Fig. 1 Model of a dc motor using PSpice.

• Figure 2 shows the transient response of the armature current and motor speed

during start-up with rated armature voltage and after the armature voltage is reduced to

222.22 V, that according to Example #1, should yield a motor speed of 104.72 rad/s. One

can see the large positive armature current during start-up and the negative armature

current that flows in the circuit when the armature voltage is decreased. In such a case, a

regenerative braking takes place and the machine works as a generator, converting kinetic

energy from the rotating masses

= 2

21 JEk into electrical energy that is sent to the

converter - dc bus that is feeding the machine.

Time

0s 0.2s 0.4s 0.6s 0.8s 1.0s 1.2s 1.4s 1.5sI(La)

-1.0KA

0A

1.0KA

SEL>>

(741.438m,75.403)

(783.475m,-261.746)

(32.988m,758.211)

100V

200V

(1.4914,104.745)

(736.288m,188.431)

Fig. 2 Transient response for the system considering J = 1 kgm2.

A motor drive usually employs a main control loop, for regulating the speed of

the motor, and a secondary control loop that is intended to limit the current during start-

parallel types as shown in Figs. 3 and 4. In the cascade case, one designs first the inner

(current) loop and then the outer (speed) loop, with a lower crossover frequency. Note

that the reference for the current is the output of the speed controller and is limited to a

safe value by a limiter. In the parallel configuration, the two control loops are designed

• independently. A commutation system, with OR logic based on diodes, selects the output

signal of the controller with lower magnitude to be the PWM modulating signal.

Fig. 4 Parallel control loops.

3. Transfer functions of a drive system for a dc motor

The following transfer functions can be derived for a dc motor.

++

+

+

=

1 111

)()(

mamaa

a sssJL

KsVs

(5)

++

+

+

+

=

1 111

1

)()(

mamaa

m

a

a

sssL

s

sVsI

(6)

• where: ( )21

and ,

K

JRBJ

RL a

mma

aa === .

Considering that the dynamics of the motor and load are much slower than the

dynamics of a step-down dc-dc converter, the latter and its PWM modulator can be

modeled by a simple gain ST

Supply

cont

achop V

VsV

sVsG )()()( == .

)(sVcont is the control or modulating voltage of the PWM and is the peak

value of the carrier waveform, usually a ramp or sawtooth, which presents a minimum

value equal to zero.

STV

In order to measure the motor speed, a dc tachometer with a rated input to output

relationship of 188.5 rad/s to 10 V is used. It can be represented by a gain

A current transducer is also required in the drive. Considering that the

time constant of the armature (

( .053.0)( = sH )

a) is much bigger than the switching period of the dc-dc

converter (T), no signal filters are required and the current transducer can be modeled by

a simple gain ( ).1.0)( =sHi

• 4 Design of the controller for the Current Loop

The uncompensated loop transfer function for the current loop, including the

PWM modulator, the converter, the motor and the current sensor is given by

i

mamaa

m

ST

Supply

cont

meaa KsssL

s

V

VsV

sI

++

+

+

+

=

1

_

111

1

)()(

(7)

The frequency response of the loop transfer function considering J = 1 kgm2 is

20 log TFPIAn.

freqn0.1 1 10 100 1 103 1 104

40

20

0

20

40

arg TFPIAn rtd.

freqn0.1 1 10 100 1 103 1 104

100

50

0

50

100

Fig. 5 - Bode Diagram of the loop transfer function for current (MATHCAD).

For a crossover frequency of 46 Hz (289 rad/s) one calculates the gain as 8.4 dB

and the phase as 74.88 . In order to achieve a phase margin of 45 with a type 2

controller, the k factor and the frequencies of the zero and pole are calculated from:

oo 88.2990 =+= TFPIApmboost (8)

• 73.12

45tan =

+= boostk o (9)

z kk

(10)

The dc gain of the controller (Kci) can be calculated as the opposite of the gain of

the partially compensated (Kci = 1) loop transfer function at the crossover frequency. In

MATHCAD it was calculated as 35.92 dB, yielding Kci = 62.5. The Bode plot of the

controller and compensated loop transfer function are shown in Fig. 6.

Fig. 6 - Bode Diagram of the current controller and compensated loop transfer function.

The simulation with PSpice to verify the dynamic performance of the controller

was carried out with the schematics shown in Fig. 7, that resulted in the waveforms

shown in Fig. 8. There one sees that in the start-up the armature current is limited to 150

A and is kept in this value for some time. The motor speed increases slowly due to the

limited current and electrical torque. The duty cycle increases slowly and become equal

to 1 at t = 0.95 s. From this point on, the motor speed, and the speed voltage (Ea), reach a

• certain level, decreasing the current. Nothing can be done to keep it at 150 A, since the

duty cycle has already reached its maximum level. This positive error makes the output

voltage of the controller (Vc) increase more and more. At 1.2 s, when the reference

current is decreased, the output voltage of the controller starts do decrease, but since its

value is too high, it is not able to reduce the duty cycle of the main switch

instantaneously. One sees that the duty cycle remains equal to 1 up to 2 s. This could be

avoided by using an anti-windup integrator whose output would not increase indefinitely.

Fig. 7 - Schematics diagram of the closed loop current control of the dc motor.

• Fig. 8 Waveforms for the dc motor operating with closed loop current control: (a)

Output voltage of the controller; (b) Duty cycle of the main switch; (c) Motor speed and

(d) Reference voltage for the current and feedback current signal (Ki = 0.1).

5 Design of the controller for the Speed Loop

The uncompensated loop transfer function for the speed loop, including the PWM

modulator, the converter, the motor and the speed sensor (tachometer) is given by

++

+

+

= K

sssJL

KV

VsVs

mamaa

ST

Supply

cont

mea

1 111

)(

)(

(11)

The frequency response of the loop transfer function considering J = 1 kgm2 is

• 20 log TFPW n.

freqn0.1 1 10 100 1 103 1 104

150

100

50

0

50

arg TFPW n rtd.

freqn0.1 1 10 100 1 103 1 104

200

150

100

50

0

Fig. 9 - Bode Diagram of the speed loop transfer function (MATHCAD).

For a crossover frequency of 5.4 Hz (33.9 rad/s) one calculates the gain as -10.34

dB and the phase as 96 . In order to achieve a phase margin of 45 with a type 2

controller, the boost angle, the k factor and the frequencies of the zero and pole are

calculated using (8-10) as 51, 2.82, 12 rad/s and 95.7 rad/s respectively,

The dc gain of the controller (Kc) can be calculated as the opposite of the gain of

the partially compensated (Kc = 1) loop transfer function at the crossover frequency.

In MATHCAD it was calculated as 36.05 dB, yielding Kc = 66. The Bode plot of the

controller and compensated loop transfer function are shown in Fig. 10.

• Fig. 10 - Bode Diagram of the speed controller and compensated loop transfer function.

The simulation with PSpice to verify the dynamic performance of the controller

was carried out with the schematics shown in Fig. 11, that resulted in the waveforms

shown in Fig. 12. There one sees that the start-up is done with duty cycle equal to 1, so

the armature current and motor speed evolve as if there was no converter. During the

interval the reference speed was higher than the motor speed, the output of the controller

kept rising, reaching high values. This way, when the actual speed exceeds the reference

and the duty cycle should be reduced, it is not because the output of the controller that is

decreasing, still implies a duty cycle equal to 1. Again an anti-windup integrator should

be used. After this transient, the motor speed finally reaches the steady state value with

zero error. At t = 1.2 s the speed reference is increased with a step change and one can

see that the transient to this small signal variation is good, with a small overshoot due to

the 45 phase margin. It could be reduced with a controller designed for a bigger phase

margin. Observe that the output voltage of the controller and duty cycle do not saturate.

• Fig. 11 - Schematics diagram of the closed loop speed control of the dc motor.

Fig. 12 Waveforms for the dc motor operating with closed loop speed control: (a)

Armature current; (b) Output of the speed controller (c) Duty cycle of the main switch

and (d) Reference voltage for the speed and feedback speed signal (K = 0.053).

• 6 Speed and current loops in parallel

Example #1 - Steady-state characteristics of a dc drive