Induction Motor Speed Control

Different Speed control methods for Induction motor

• By changing rotor resistance (only possible in slip-ring induction motor (SRIM))

•By changing the applied voltage and frequency to the induction motor, while keeping the

ratio constant.

• By changing stator poles by reconnecting the stator coils

•By changing only the applied voltage to stator

•By recovering the slip power (only possible in SRIM)

Equivalent Circuit Modification

V1

I1

R1

jX1 jX2’

R2’

((1-s)/s)R2’

jXm

I2’

Im

Assuming stator drop is negligible (this is true except for

low speed operation when the stator resistance drop

becomes comparable to supply voltage) the equivalent

circuit of the induction motor can be modified to the

bottom figure from the top figure.

V1

R1

jX1 jX2’

R2’

((1-s)/s)R2’

jXm

I2’

Im

Output Torque

Ouput torque output neglecting rotational losses

= 𝑇𝑜𝑢𝑡 = 𝑃𝑜𝑢𝑡

𝜔=

𝑃𝑜𝑢𝑡

(1 − 𝑠)𝜔1=

𝑃𝑎𝑔

𝜔1= 𝐼2

′ 2𝑅2

′

𝑠𝜔1

From the equivalent circuit,

𝐼2′ =

𝑉1

𝑅1 +𝑅2

′

𝑠 + 𝑗 𝑋1 + 𝑋2

′

𝐼2′ =

|𝑉1|

𝑅1 +𝑅2

′

𝑠

2

+ 𝑋1 + 𝑋2′

2

Substituting the magnitude of current in torque expression,

𝑇𝑜𝑢𝑡 =|𝑉1|2

𝑅1 +𝑅2

′

𝑠

2

+ 𝑋1 + 𝑋2′

2

∙𝑅2

′

𝑠𝜔1

V1

R1

jX1 jX2’

R2’

((1-s)/s)R2’

jXm

I2’

Im

Maximum Torque 𝑇𝑜𝑢𝑡 =

|𝑉1|2

𝑅1 +𝑅2

′

𝑠

2

+ 𝑋1 + 𝑋2′

2

∙𝑅2

′

𝑠𝜔1

To obtain the maximum torque, the above expression of torque is differentiated with respect to

slip and equated to zero.

𝑇𝑚𝑎𝑥 =|𝑉1|2

𝑅1+ 𝑅12+ 𝑋1+𝑋2

′ 2

∙1

2𝜔1 ≈

|𝑉1|2

2𝜔1 𝑋1+𝑋2′

(neglecting 𝑅1)

Slip at which maximum torque occurs

𝑠𝑇𝑚𝑎𝑥 =𝑅2′

𝑋1+𝑋2′

This shows that while the maximum torque is independent of rotor resistance, the slip at which

the maximum torque occurs is dependent on rotor resistance.

Speed control by changing rotor resistance in SRIM

In case of a slip ring (wound rotor) induction motor

(SRIM) the speed can therefore be varied by changing the

rotor resistance. More on this will be discussed later.

s = 1 slip

T

Tmax

s = 0

TL

Increasing R2

Changing Tm profile with fixed

peak torque by changing R2

Decreasing speed with

increasing R2

Speed control by keeping constant flux (ratio of

applied voltage and frequency)

slip s = 0

s = 1

T

Tmax

Decreasing f

Changing Tm profile with fixed peak toruqe

with changing frequency and voltage but

with their ratio constant

Decreasing speed for decreasing voltage

and frequency but with their ratio constant

A different strategy is employed to achieve speed control for squirrel cage induction motor. It is

applicable for slip ring induction motor as well.

𝑇𝑚𝑎𝑥 ≈|𝑉1|2

2𝜔1 𝑋1 + 𝑋2′

𝜔1 = 2𝜋𝑓/(𝑝) where p is the pole pair number.

Also, 𝑋1 = 2𝜋𝑓 𝐿1 and 𝑋2′ = 2𝜋𝑓 𝐿2

′

Therefore, 𝑇𝑚𝑎𝑥 =𝑝|𝑉1|2

8𝜋2 𝐿1+𝐿2′ 𝑓2

=𝑝

𝑉1 𝑓

2

8𝜋2 𝐿1+𝐿2′

Thus if the ratio of supply voltage to frequency , i.e., 𝑉1 𝑓 maintained constant,

the peak flux of the motor (essentially the magnetizing current) and hence the maximum torque

can be held constant.

Speed control of Induction motor: Pole changing

By changing poles the synchronous speed and hence the motor operating speed can be changed.

Poles can be changed by changing coil connections of the stator. Normally, poles can be changed

in the ratio of 2:1. The speed can be changed only in discrete steps. Hence it is not very popular.

The figure above shows how to reconnect a 4 pole motor as a 2 pole one.

-

++

-

+

-

+

-

--

+

+

N

S

+

+

+

+

+

+

-

-

--

-

-

NS

NS

Speed control of Induction Motor: By varying stator supply voltage only

By using ac voltage controller as shown above (using back-to-back connected SCRs )

the three phase stator supply voltage can be smoothly controlled. Speed also

can be reversed by using the connections shown by dotted lines, with the other

back-to-back connected SCRs left ungated.

However, since the torque depends on the square of the voltage, the output torque

is reduced greatly when the voltage alone is reduced as can be easily understood from

the following expression of maximum torque

Speed control of Induction Motor: By varying rotor resistance in a SRIM

Example:

A 3-phase, 460V, 60 Hz, 6 pole SRIM drives a constant load torque of 100 N-m at a speed of

1140 rpm when the rotor terminals are short circuited. It is required to reduce the speed of the

motor to 1000 rpm by inserting resistances in the rotor circuit. Determine the value of the

external resistance if 𝑅2, the effective rotor winding resistance is 0.2Ω. The stator to rotor turns

ratio is 1:1.

Solution :

𝑁1 = 120𝑓

𝑃=

120∗60

6= 1200 rpm.

∴ slip at 1140 rpm = 𝑠1 =1200 − 1140

1200= 0.05

∴ slip at 1000 rpm = 𝑠2 =1200 − 1000

1200= 0.167

Y

R

B

Motor

Example on SRIM speed control by varying rotor resistance

Now

𝑇𝑜𝑢𝑡 =|𝑉1|2

𝑅1 +𝑅2

′

𝑠

2

+ 𝑋1 + 𝑋2′

2

∙𝑅2

′

𝑠𝜔1

We can keep the speed constant by keeping 𝑅2′

𝑠 constant.

∴𝑅2

′

𝑠1=

𝑅2′ + 𝑅𝑒𝑥𝑡

′

𝑠2

Since stator to rotor turns ratio is 1:1,

∴𝑅2

𝑠1=

𝑅2 + 𝑅𝑒𝑥𝑡

𝑠2

or 0.2

0.05=

0.2 + 𝑅𝑒𝑥𝑡

0.167

∴ 𝑅𝑒𝑥𝑡 =0.2 ∗ 0.167

0.05− 0.2 = 0.468Ω.

Electronic control of Rotor resistance in a SRIM

B

R

Y

Motor

RM

By varying the duty cycle ‘d’ of the MOSFET M the effective rotor resistance can

be changed by 0.5(1-d)R per phase.

Slip Power Recovery Scheme

C1

C2

Slip Power Recovery Scheme(2)

Instead of dissipating power in resistance (external resistance added in the rotor circuit), power

can be returned using a phase controlled rectifier.

Power delivered back to the source is 𝑉𝑑𝑐 𝐼𝑑𝑐

Effective resistance added to the rotor is given by 𝑅𝑒𝑓𝑓 = 𝑉𝑑𝑐

𝐼𝑑𝑐

Assuming that the rotor resistance𝑅2 is negligible compared 𝑅𝑒𝑓𝑓 , the entire slip power is fed

back to the source.

𝑠𝑃𝑎𝑔 = 𝑉𝑑𝑐 𝐼𝑑𝑐

For a three phase diode bridge rectifier, the average voltage is given by the expression

𝑉𝑑𝑐 =3 2

𝜋𝑠𝑎1𝑉1 = 1.35𝑠 𝑎1𝑉1, where 𝑎1 is the motor rotor to stator turns ratio and 𝑉1 is the

stator line-line voltage.

Slip Power Recovery Scheme(3)

For a three phase fully controlled SCR bridge converter, the average voltage is given by

𝐸𝑎𝑣 =3 2

𝜋𝑎2𝑉1 cos 𝛼 = 1.35𝑎2𝑉1 cos 𝛼 , where 𝑎2 is the transformer secondary to primary

turns ratio.

Since under steady state 𝑉𝑑𝑐 = 𝐸𝑎𝑣

∴ 𝑠𝑎1 = 𝑎2 cos 𝛼

∴ 𝑠 =𝑎2

𝑎1| cos 𝛼 |

Thus the speed of the motor can be controlled by changing the firing delay angle of the converter.

Also

𝑠𝑃𝑎𝑔 = 𝑉𝑑𝑐 𝐼𝑑𝑐

or 𝑠𝑇𝜔1 = 𝑉𝑑𝑐 𝐼𝑑𝑐 = 1.35𝑠𝑎1𝑉1 𝐼𝑑𝑐

or 𝑇 =1.35

𝜔1𝑎1𝑉1 𝐼𝑑𝑐

or 𝑇 ∝ 𝐼𝑑𝑐 .

Slip Power Recovery Scheme( Example) An 8 pole, 440V, three phase, 60 Hz, slip ring induction motor is being used for slip power

recovery scheme. The stator to rotor turns ratio is 1:0.75. The stator of the motor is connected

in delta, while the rotor is connected in star. The power is fed back to the utility supply

through a star-delta transformer after the inverter, with the star side (primary) connected to the

460V utility supply. The primary to secondary turns ratio for the transformer is 2:1. Calculate

a) the power sent back to the utility supply, b and b) the speed of the motor and c) the input

power drawn by the motor if the dc link current is 200A.𝛼 = 135°.

-

+

-

+

+

-

Motor

Three Phase Transformer

440

330√

3s

440*0.75*s

440/√3

220/

√3

Vdc

Eav

R

Y

B

Idc

Slip Power Recovery Scheme( Example)(2)

𝑉𝑑𝑐 = 1.35 ∗ 0.75 ∗ 440 ∗ 3 ∗ 𝑠 = 1.35 ∗ 330 3 ∗ 𝑠 = 771.63𝑠 V

𝐸𝑎𝑣 = 1.35 ∗220

3∗ cos 135° = 1.35 ∗ 89.8 = 121.25 V

∵ 𝑉𝑑𝑐 = 𝐸𝑎𝑣

∴ 𝑠 =121.25

771.63= 0.1571

∴ Power sent back to the source = 𝑉𝑑𝑐 𝐼𝑑𝑐 = 121.25 ∗ 200 W = 24.25 kW.

𝑇 =𝑉𝑑𝑐 𝐼𝑑𝑐

𝑠𝜔1=

24250

0.1571∗2𝜋∗900

60

= 1637.45 Nm.(∵ 𝑁1 =120𝑓

𝑃=

120∗60

8= 900 RPM)

Now Motor output power = 𝑉𝑑𝑐 𝐼𝑑𝑐 ∗(1−𝑠)

𝑠= 24250 ∗

1−0.1571

0.1571= 130.11 kW.

∴ Input power to the motor = Motor output power + Power sent back to the source

= 130.11 + 24.25 kW = 154.36 kW.

-

+

-

+

+

-

Motor

Three Phase Transformer

440

33

0√

3s

440*0.75*s

440/√3

220/

√3

Vdc

Eav

R

Y

B

Idc

Speed Control Scheme Using Slip Power Recovery

Transformer

Tachometer

Speed

Controller+

-+

-ωref

ω

Vd Vi

Id

SRIM

R

Y

B

aFiring

Angle

Controller

LimiterLimiter

Mechanical

load

Id,ref

Scalar v/f method of speed control of induction motor using a voltage source inverter(VSI) :

(i) Below Base Speed

V1

R1

jX1 jX2’

R2’

((1-s)/s)R2’

jXm

I2’

Im

𝑇𝑜𝑢𝑡 =|𝑉1|2

𝑅1+𝑅2′

𝑠

2

+ 𝑋1+𝑋2′

2

∙𝑅2′

𝑠𝜔1

Assuming slip to be small,

𝑅2′

𝑠≫ 𝑅1 &

𝑅2′

𝑠 ≫ 𝑋1 + 𝑋2

′

∴ 𝑇𝑜𝑢𝑡 ≈|𝑉1|2

𝑅2′

𝑠

2

∙𝑅2′

𝑠𝜔1=

|𝑉1|2

𝑅2′ ∙

𝑠

𝜔1

Multiplying and dividing by 𝜔1

𝑇𝑜𝑢𝑡 ≈ |𝑉1|

𝜔1

2

∙𝑠𝜔1

𝑅2′

Scalar v/f method of speed control of induction motor using a voltage source inverter(VSI) (2)

From the above equation it is clear that if the ratio of voltage to supply frequency is maintained

constant, then torque is proportional to slip speed 𝑠𝜔1. The approximation is valid as long as the

stator resistance 𝑅1 ≪ 𝑋1 + 𝑋2′ so that the maximum torque capability of the motor given by

𝑇𝑚𝑎𝑥 =|𝑉1|2

𝑅1+ 𝑅12+ 𝑋1+𝑋2

′ 2

∙1

2𝜔1 ≈

|𝑉1|2

2𝜔1 𝑋1+𝑋2′

=𝑝 𝑉1 𝑓 2

8𝜋2 𝐿1+𝐿2′ is not severely compromised.

For low frequency operation, 𝑉1 has to be boosted in order to accommodate for the increasing

influence of 𝑅1. The effect of not boosting 𝑉1 can lead to drastic reduction of peak torque as

shown by the figure below.

𝑇𝑜𝑢𝑡 ≈

|𝑉1|

𝜔1

2

∙𝑠𝜔1

𝑅2′

Scalar v/f control using VSI: Example (1) A three phase, induction motor drives a constant torque over a 4:1 speed range by frequency

control with the ratio of 𝑉1 𝑓 maintained constant. If the motor’s reactance to stator resistance

ratio is 5 at rated frequency of operation, calculate the effect on the peak torque of operation at

the lowest speed. What value of 𝑉1 𝑓 is required at the lowest speed if the peak torque is to

remain constant? Solution:

At rated frequency,

𝑇𝑚𝑎𝑥 =|𝑉1|2

𝑅1+ 𝑅12+ 𝑋1+𝑋2

′ 2

∙1

2𝜔1=

|𝑉1|2

𝑅1+𝑅1 1+ 𝑋1+𝑋2′ 𝑅1

2

∙1

2𝜔1=

|𝑉1|2

𝑅1+𝑅1 1+ 5 2 ∙

1

2𝜔1 Nm

or 𝑇𝑚𝑎𝑥 =1

1+ 1+ 5 2

|𝑉1|2

2𝑅1𝜔1= 0.082

|𝑉1|2

𝑅1𝜔1 Nm

At 1/4th

rated frequency, if the voltage is also reduced to one-fourth of its rated value then

maximum torque developed is

𝑇𝑚𝑎𝑥 = 𝑉1 4 2

𝑅1+ 𝑅12+ 𝑋1+𝑋2

′ 2

∙1

2 𝜔1

4

= 𝑉1 4 2

𝑅1+𝑅1 1+ 𝑋1+𝑋2′ 4𝑅1

2

∙2

𝜔1=

𝑉1 4 2

𝑅1+𝑅1 1+ 5/4 2 ∙

2

𝜔1 Nm

or 𝑇𝑚𝑎𝑥 =1/8

1+ 1+ 1.25 2

|𝑉1|2

𝑅1𝜔1= 0.0481

|𝑉1|2

𝑅1𝜔1 Nm

Scalar v/f control using VSI: Example (1) Contd...

Thus the maximum torque developed has gone down at lowest speed by about 58.5%. In order to

compensate for the maximum torque the supply voltage has to be boosted. This can be estimated

from the following:

𝑇𝑚𝑎𝑥 = 𝑘 𝑉1 2

𝑅1+ 𝑅12+ 𝑋1+𝑋2

′ 2

∙1

2 𝜔1

4

= 𝑘 𝑉1 2

𝑅1+𝑅1 1+ 𝑋1+𝑋2′ 4𝑅1

2

∙2

𝜔1=

𝑘 𝑉1 2

𝑅1+𝑅1 1+ 5/4 2 ∙

2

𝜔1 Nm

or 𝑇𝑚𝑎𝑥 =2𝑘2

1+ 1+ 1.25 2

|𝑉1|2

𝑅1𝜔1= 0.769𝑘2 |𝑉1|2

𝑅1𝜔1= 0.082

|𝑉1|2

𝑅1𝜔1 .

𝑘 = 0.082

0.769 =0.326

Thus instead of 25%, 32.6% of the of rated voltage is to be applied if the maximum developed

torque has to remain constant at 1/4th

rated speed.

Scalar v/f control using VSI: Example (2) A 450 V, 50 Hz, 1450 rpm, 25 kW, Y connected, 3 phase induction motor delivers rated torque

at all speed. The motor equivalent circuit parameters at rated frequency are 𝑅1 = 0.1 Ω, 𝑅2′ =

0.17 Ω, 𝑋1 = 0.3 Ω, 𝑋2′ = 0.5 Ω, 𝑋𝑚 = 23.6 Ω. The machine flux is kept constant at rated value

under all operating conditions. Calculate the line current, power-factor, efficiency at 1/5th

the

rated speed.

1/5th

the rated speed =1450 5 = 290 rpm.

Since torque is constant at rated value under all speed, slip speed or slip frequency will have to

be the same at the rated value.

Now synchronous speed 𝑁1 =120𝑓

𝑃=

120∗50

𝑃> 1450 rpm.

∴ 𝑃 <120 ∗ 50

1450< 4.13

Since 4 is the closest and lesser even number to 4.13, 𝑃 = 4.

∴ 𝑁1 =120 ∗ 50

4= 1500 rpm.

Thus slip speed =𝑁1 − 𝑁 = 1500 − 1450 = 50 rpm.

Scalar v/f control using VSI: Example (2) Contd...

Thus synchronous speed at 1/5th

rated speed 𝑁1/5 = 290 + 50 = 340 rpm.

Therefore slip at 1/5th

rated speed =50

340= 0.1471.

Since the motor delivers rated torque at 1/5th

rated speed, therefore it delivers 1/5th

rated power at

this speed.

∴ 3 𝐼2′ 2 𝑅2

′ (1−𝑠)

𝑠= 3 𝐼2

′ 2 0.17(1−0.1471 )

(0.1471 )=

25000

5= 5000 W.

∴ 𝐼2′ =

5000 ∗ 0.1471

0.51 ∗ 0.8529= 41.12 A

Also at 1/5th

rated speed the new values of 𝑋1 = 0.3 340

1500 = 0.068 Ω , 𝑋2

′ = 0.5 340

1500 =

0.1133 Ω, 𝑋𝑚 = 23.6 340

1500 = 5.35 Ω.

Scalar v/f control using VSI: Example (2) End

Thus the equivalent circuit looks like the following:

V1

I1

R1=0.1 Ω

jX1=j0.068

ΩjX2

’=j0.1133 Ω

R2’=0.17 Ω

((1-s)/s)R2’ =0.99ΩjXm= j5.35 Ω

41.12<0

Im

Thus 𝐼𝑚 =41.12∠00(0.17+0.99+𝑗0.1133 )

𝑗5.35=8.96∠ − 84.42°A = 0.87 − 𝑗8.92 A.

∴ 𝐼1 = 𝐼𝑚 + 41.12 = 0.87 − 𝑗8.92 + 41.12 = 42.93∠−11.99° A

∴ 𝑉1 = 𝐼1 0.1 + 𝑗0.68 + 𝑗5.35 ∗ 𝐼𝑚 = 42.93∠−11.99° ∗ 0.121∠34.22° + 𝑗5.35 ∗ 8.96∠ −84.42° = 5.19∠22.23 + 47.94∠5.58° = 52.93∠7.19° V.

∴ Power − factor = cos 11.99° + 7.19° = 0.94 lag.

Efficiency = 𝑃𝑜𝑢𝑡

𝑃𝑖𝑛=

5000

3𝑉1𝐼1 cos 𝜃=

5000

3∗52.93∗42.93∗0.94= 0.78.

Induction Motor Speed up to Rated Speed Using Slip Speed Control

Pulse Width

Modulated

(PWM) Inverter

V1,ref

w1,ref

Function

generator

SCIMMechanical

load

+

+Slip Speed

Controller

-

+wref

ω ω

w1,ref

Limiter

Vd

Scalar v/f control using VSI: High Speed Region (From Rated Speed Up To 1.5-2 Times Rated Speed)

Beyond rated speed we cannot increase 𝑉1 since this is the rated value. As the mechanical

structure (bearings, rotor ) is usually sturdy increasing beyond rated speed is not a problem.

Strategy of speed control:

Above base speed the maximum allowable value of s is at its rated value (𝑠𝑟𝑎𝑡𝑒𝑑 ) and the supply

frequency is increased. Thus the maximum allowable steady state value of torque in this region is

𝑇𝑜𝑢𝑡 = |𝑉1|

𝜔1

2

∙𝑠𝑟𝑎𝑡𝑒 𝑑 𝜔1

𝑅2′ =

|𝑉1|2𝑠𝑟𝑎𝑡𝑒𝑑

𝑅2′ ∙

1

𝜔1=

|𝑉1|2𝑠𝑟𝑎𝑡𝑒𝑑

𝑅2′ ∙

1−𝑠𝑟𝑎𝑡𝑒𝑑

𝜔∝

1

𝜔.

Thus torque developed by the motor is inversely proportional to the speed. Therefore the motor

works in constant power region.

This region continues as long as the slip speed does not exceed the maximum allowable slip

speed 𝜔2 determined by the breakdown or maximum torque capability of the machine.

Scalar v/f control using VSI: Very High Speed Region (Beyond 1.5 or 2 times Rated Speed)

This region starts when slip speed reaches 𝜔2 = 𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1 (the maximum allowable slip speed).

In this speed zone the slip speed is limited to 𝜔2 , the maximum allowable slip speed or the

breakdown slip speed. Thus the maximum allowable steady state value of torque in this region

𝑇𝑜𝑢𝑡 = |𝑉1|

𝜔1

2

∙𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1

𝑅2′ =

|𝑉1|2𝜔2

𝑅2′ ∙

1

(𝜔1)2

With small value of 𝑠, the actual speed of the motor is very close to synchronous speed.Thus

torque is inversely proportional to the square of speed, while power is inversely proportional to

speed.

Summary of speed control philosophy in the three regions of operation

• Below rated (base) speed (Constant torque region)

1. Change slip speed to get required torque for any speed. 2. Obtain frequency as sum of speed and slip speed. 3. Determine supply voltage by adding stator resistance drop to the component that is proportional to frequency.

• Above rated (base) speed up to 1.5-2 times rated speed (Constant power region)

1. Keep supply voltage constant at rated value. 2. Limit slip to its rated value and adjust frequency such that slip speed does not exceed the breakdown or maximum allowable level of slip speed.

• Very high speed region beyond 1.5-2 times rated speed

1. Keep supply voltage constant at rated value. 2. Adjust frequency by limiting slip speed to its breakdown or maximum allowable level.

Scalar v/f control using VSI: Example (3) A 1740 rpm, 3 phase, 60 Hz, 4 pole, 460 V, 3 hp Y connected induction motor with 𝜔2 =

1.8𝜔1𝑠𝑟𝑎𝑡𝑒𝑑 is operated from an inverter and is expected to work at high/ very high speed

regions. Draw T vs. Speed, slip speed vs. Speed, slip vs. Speed, power vs. Speed indicating the

high speed and very high speed region.

Solution:

Synchronous speed 𝑁1 =120𝑓

𝑃=

120∗60

4= 1800 rpm.

∴ 𝑠𝑟𝑎𝑡𝑒𝑑 =1800−1740

1800= 0.033

𝜔1,𝑟𝑎𝑡𝑒𝑑 =2𝜋𝑁1

60=

2𝜋∗1800

60= 188.5 𝑟𝑎𝑑/𝑠

Therefore, rated slip speed = 𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1,𝑟𝑎𝑡𝑒𝑑 = 6.22 𝑟𝑎𝑑/𝑠

𝑇𝑟𝑎𝑡𝑒𝑑 =𝑃𝑜𝑢𝑡 ,𝑟𝑎𝑡𝑒𝑑

𝜔=

3×746

2𝜋×1740/60=

2238

182.21= 12.28 𝑁𝑚

Transition to very high speed region takes place when 𝜔2 = 1.8𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1,𝑟𝑎𝑡𝑒𝑑 . Therefore,

𝜔2 = 1.8 ∗ 6.22 = 11.2 𝑟𝑎𝑑/𝑠. Corresponding synchronous speed 𝜔1 =𝜔2

𝑠𝑟𝑎𝑡𝑒𝑑= 339.4 𝑟𝑎𝑑/𝑠.

Motor speed corresponding to this synchronous speed in rpm is given by

𝑁 =𝜔1 1−𝑠𝑟𝑎𝑡𝑒𝑑

2𝜋

60

= 3134 rpm . Torque at this speed is 𝑇 =𝑃𝑜𝑢𝑡 ,𝑟𝑎𝑡𝑒𝑑

𝜔=

2238

3134∗ 2𝜋

60

=2238

328.19=

6.82 Nm.

Scalar v/f control using VSI: Example (3) Contd...

The following figure shows the variation of the different parameters in the different speed zones

in details.

Torque vs. speed

12.3

1740 3134

6.8

Speed (rpm)

T (Nm)

Slip speed vs. speed

slip speed

(rad/s)

1740 3134 Speed (rpm)

6.2

11.2

Power vs. speed

P (W)

1740 3134 Speed (rpm)

2238

Slip vs. speed

1

1740 3134 Speed (rpm)

0.033

Slip

Scalar v/f control using Current Source Inverter (CSI)

R1

jX1 jX2’

R2’

((1-s)/s)R2’

jXm

I2’

ImI1

Recall that, neglecting rotational losses, 𝑇𝑜𝑢𝑡 = 𝑃𝑜𝑢𝑡

𝜔=

𝑃𝑜𝑢𝑡

(1−𝑠)𝜔1=

𝑃𝑎𝑔

𝜔1= 𝐼2

′ 2 𝑅2′

𝑠𝜔1

Now 𝐼2′ =

𝑗𝑋𝑚

𝑅1 +𝑅2

′

𝑠+ 𝑗(𝑋1 + 𝑋2

′ + 𝑋𝑚)

𝐼1

∴ 𝑇𝑜𝑢𝑡 = 𝐼2′ 2 𝑅2

′

𝑠𝜔1=

𝑥𝑚2

𝑅1+𝑅2′

𝑠

2

+(𝑋1+𝑋2′ +𝑋𝑚 )2

𝑅2′

𝑠𝜔1 𝐼1

2 ≈1

𝑅2′ .

𝑠𝜔1𝐿𝑚2

1+(𝐿1+𝐿2′ +𝐿𝑚 )2 𝑠𝜔 1 2

𝑅2′ 2

𝐼1 2; assuming 𝑅1 ≪

𝑅2′

𝑠.

Thus torque can be controlled by varying slip speed 𝑠𝜔1 as determined from the slip speed controller. Then

the following relationship between 𝐼1 and 𝐼𝑚 can be used to determine 𝐼1 for a given slip speed.

To maintain flux at rated value at all speed up to base speed

𝐼𝑚 ≈

𝑅2′

𝑠

2

+ 𝑋1+𝑋2′

2

𝑅2′

𝑠

2

+ 𝑋1+𝑋2′ +𝑋𝑚

2

𝐼1 =

𝑅2′

𝑠𝜔 1

2

+ 𝐿1+𝐿2′

2

𝑅2′

𝑠𝜔 1

2

+ 𝐿1+𝐿2′ +𝐿𝑚

2

𝐼1 = constant ; assuming 𝑅1 ≪𝑅2

′

𝑠

Thus 𝐼1 =

𝑅2′

𝑠𝜔 1

2

+ 𝐿1+𝐿2′ +𝐿𝑚

2

𝑅2′

𝑠𝜔 1

2

+ 𝐿1+𝐿2′

2

𝐼𝑚

Scalar v/f control using Current Source Inverter (CSI) Contd...

From previous slide:

𝑇𝑜𝑢𝑡 =𝑥𝑚

2

𝑅1 +𝑅2

′

𝑠 2

+ (𝑋1 + 𝑋2′ + 𝑋𝑚)2

𝑅2′

𝑠𝜔1

𝐼1 2

To obtain the maximum torque, the above expression is differentiated with respect to slip and

equated to zero. This leads to

𝑇𝑚𝑎𝑥 =|𝐼1|2

𝑅1+ 𝑅12+ 𝑋1+𝑋2

′ +𝑋𝑚 2

∙1

2𝜔1 ≈

𝐼1 2

2𝜔1 𝑋1+𝑋2′ +𝑋𝑚

Slip at which maximum torque occurs

𝑠𝑇𝑚𝑎𝑥 =𝑅2

′

𝑅12 + 𝑋1 + 𝑋2

′ + 𝑋𝑚 2

≈𝑅2

′

𝜔1 𝐿1 + 𝐿2′ + 𝐿𝑚

Scalar v/f control using Current Source Inverter (CSI) End

Because of the presence of the extra term 𝑋𝑚 in the denominator of the 𝑠𝑇𝑚𝑎𝑥 expression the

𝑠𝑇𝑚𝑎𝑥 value is much lower compared to being driven by a VSI (see figure below). Consequently,

the region of stable operation between the s = 0 and s = 𝑠𝑇𝑚𝑎𝑥 becomes very steep. Thus the

current source inverter driven induction motors must be operated with the speed loop closed.

slip s = 0

s = 1

T

Tmax

VSI

TL

Tm1 Tm2 Tm3

slip s = 0

s = 1

T

Tmax CSI

TL

Tm1 Tm2 Tm3

Compare VSI and CSI sTmax :Example (4) A 450 V, 50 Hz, 1450 rpm, 25 kW, Y connected, 3 phase induction motor delivers rated torque

at all speed. The motor equivalent circuit parameters at rated frequency are 𝑅1 = 0.1 Ω, 𝑅2, =

0.17 Ω, 𝑋1 = 0.3 Ω, 𝑋2′ = 0.5 Ω, 𝑋𝑚 = 23.6 Ω. Compare the slip speed at which 𝑇𝑚𝑎𝑥 can be

obtained for i) a VSI and ii) a CSI.

Solution:

i) For a VSI slip speed at 𝑇𝑚𝑎𝑥 is given by

𝜔1𝑠𝑇𝑚𝑎𝑥 ≈𝑅2′

𝐿1+𝐿2′

=0.17∗2𝜋∗50

0.3+0.8 = 485.5 rad/s . For 50 Hz operation this would mean a

𝑠𝑇𝑚𝑎𝑥 =485.52

2𝜋∗50= 0.1545.

ii) For a CSI slip speed at 𝑇𝑚𝑎𝑥 is given by

𝜔1𝑠𝑇𝑚𝑎𝑥 ≈𝑅2′

𝐿1+𝐿2′ +𝐿𝑚

=0.17∗2𝜋∗50

0.3+0.8+23.6 = 2.16 rad/s . For 50 Hz operation this would mean a

𝑠𝑇𝑚𝑎𝑥 =2.16

2𝜋∗50= 0.0069.

Current Regulated Induction Motor Drive