Transfer Function of Dc Motor

TRANSFER FUNCTION OF ARMATURE CONTROLLED DC MOTOR Objective: To determine the transfer function of an armature controlled dc shunt motor. Reference:  “NAGARATH & GOPAL” -- Control System  “B.L.TERAJA” - Electrical Technology Apparatus Required:  Ammeters 

0-20A MC 0-5 A MC  0-2 A MC  0-50 MA (DMM) MI  Voltmeters 0-250 V MC  0-250 V MI  0-30 V MC  0-50 V MI  Rheostat 50 , 5 A  1000 , 1 A  50 , 5 A Theory: The transfer function is defined as the ratio of Laplace transform of the output variable to the Laplace transform of input variable. The DC motor converts electrical energy into mechanical energy. The electrical energy supplied at the armature terminals converted into controlled mechanical energy. (i)Armature controlled DC shunt motor In armature control, the field current is kept constant and the armature voltage is varied and hence the speed is varied. The field current If(t) is maintained constant by keeping the Vf(t) to a constant value Vf and the armature current Ia(t) is varied to change the torque Tm(t) of the load connected to the motor shaft. Thus the input variable of the motor is the armature voltage Va(t) and the output variable is the torque Tm(t). The speed of the DC motor is directly proportional to the armature voltage and inversely proportional to the flux in the armature. i.e. N  Va /  In the armature controlled DC motor, the desired speed is obtained by varying the armature voltage.

By Kirchoff „s law,

Transfer Function

Va = IaRa + La dIa(t)/dt + Eb(t) ……….. (1) Ra

Va

La

Ia

Taking Laplace transform of equation (1) , we have

If = constant

eb

Tm J,f

Va(s) = [Ra+sLa] Ia(s) + Eb(s) Ia(s) = [Va(s) - Eb(s)] / [Ra+sLa] ……… (2) Since the field current is kept constant , the torque developed is proportional to the armature current, i.e., The flux developed   If(t) , If is constant Hence torque developed T   Ia , T  Ia

T = Kt Ia ……. (3)

Taking laplace transform of equation (3), T (s) = Kt Ia (s) ……….(4) Where Kt =Torque constant, Kt = T /  Ia N-m / A …………. (5) The torque equation is given by, T(t) = Jd2/dt2 + f d/dt …………(6) Taking laplace transform of equation (6), T(s) = Js2(s) + f s(s). …………(7) Substituting equation (4) in (7), Kt Ia (s) = Js2(s) + f s(s) s(s) [Js+f] = Kt Ia (s) ………..(8) Substituting equation (2) in (8), s(s) [Js+f] = Kt [Va(s) - Eb(s)] / [Ra+sLa] ………..(9) Motor back emf is proportional to speed,i.e., Eb(t) = Kb d/dt = Kb(t) …………(10) Where Kb = back emf constant.

Taking laplace transform of equation (10), Eb(s) = Kb(s) …………………(11) Substituting equation (11) in (9), s(s) [Js+f] = Kt [Va(s) - Kb(s)] / [Ra+sLa] (s) [Js+f] = Kt [Va(s) - Kb(s)] / [Ra+sLa] , since d/dt = (s) & s(s)= (s)

(s) / Va(s) = Kt / { Kt Kb+ (Js+f) (Ra+sLa)} Where, Ra = Armature resistance ;  La = Armature Inductance ; Hendry Kt = Torque constant = T /  Ia ; N-m / A Kb = Back emf constant = Eb /  N ; Volts/(rad/sec) f = Frictional constant ; N-m/(rad/sec) J = moment of Inertia ; kg-m2 Block diagram: Using equations (2), (4), (7) & (11), the block diagram is drawn for armature controlled DC motor. Va(s) +

1/[Ra+sLa] -

Kt

1/[Js+f]

(s)

Eb(s)

Kb For finding the transfer function of the armature control DC motor, we have to find the values of J, f, Kt, Kb, Ra & La. PROCEDURE: (i).Load test to determine Kt & Kf 

Conduct the load test with two constants normal (rated value) & sub normal (say 80% of rated field current) values of field currents.



Give the connections as per circuit diagram.



Keep the field rheostat in minimum position & switch on 230 V supply.



Adjust the field rheostat to rated speed & consider the corresponding field current as rated field current.



For different loads note down Ia,Va & speed



Calculate the torque developed.



Repeat the same for 80% of excitation current.



Plot a graph between T(Y-axis) and Ia (X-axis) for both the cases.



From this graph, obtain Kt for any one field current.

[Note: The reason why we conduct the experiment for two field currents is to derive Kf from armature control graph. ] 

From the graph deduce two points (T1,If1) and (T2,If2) for the same armature current.



Plot the graph between T Vs If



From this graph, obtain Kf.

(ii).Retardation Test to find J & f The total losses can be divided into two parts,viz., constant losses and variable losses. The constant losses include frictional & inertia losses. Total losses in a circuit= VI – Ia2R - VIf Losses = V(Ia + If) - Ia2Ra - VIf, Energy = (Losses * t ) = ½ J (12 -22) N1 =1600 RPM ; N2 = 1400 RPM ;  = 2πN /60 J = (Losses * t *2 ) / (12 -22) N1 = N * e –t1/τm

N2 = N * e –t2/τm



Give the connections as per circuit diagram.



By closing the switch , make the measurement of V ,Ia and If.



Make the motor to run at a speed greater than 1600 RPM.



Open the switch suddenly.



Using stop watch, note down the time taken for the speed of the motor to fall down from 1600 RPM to 1400 RPM. (1 =1600 RPM,2 =1400RPM)



Plot a Graph showing relation between speed & time.



The moment of inertia , J is obtained from the relation between loss ,time and .



The friction constant , f is obtained using the exponential relation between speed , time and time constant.



Calculate no load input power using the values read by ammeter & voltmeter connected to armature circuitry. To find J

: [V(Ia + If) –Ia2Ra -VIf ] * t = (1/2) J(12 -22)

To find f : J / τm ;

τm = (t2 – t1) / (ln N1 – ln N2)

(iii). V-A method to obtain Ra, La 

Give the connections as per circuit diagram.



Measure Ra & Rf. [As the field winding resistance is of the order of 250300ohms and it can withstand a current of 1A,the circuit shown can be used for measurement of Rf Similarly , as the armature resistance is of order of 0.21 ohms and it can be measured using the circuit shown]



To measure Lf & La, give the connections as shown.



Apply an ac voltage & measure the field reactance Zf & armature reactance Za.



Calculate Xf =Square root of Zf2 –Rf2 and Lf = Square root of (Zf2 –Rf2 ) / 2f



Calculate Xa =Square root of Za2 –Ra2 and La = Square root of (Za2 –Ra2 ) / 2f,



where f= supply frequency=50 Hz

RESULT:

QUESTIONS: (COMMON TO ARMATURE / FIELD CONTROL) 1. Distinguish between DC motor & DC servomotor. 2. State the assumptions made while obtaining the transfer function of an armature controlled DC motor. 3. In field controlled DC servomotor the reversible operation is possible by reversing the field current – State TRUE or FALSE. 4. What are the characteristics of servomotors? 5. What is the field time constant of field controlled servomotor? 6. What is the motor gain constant of field controlled servomotor? 7. What is the use of transfer function? 8. What are the characteristics of feedback? 9. What is meant by reset time? 10. What is the purpose of retardation test?

(I). LOAD TEST TO DETERMINE KT & Kb Circuit Diagram

+

20A Fuse

220V D DC P Supply S T

3 point starter L F A

(0-20A) +A S1

1000Ω/1A +

S2

A1

V

F1 (0-300V) F2

M A2 Brake Drum

+ A

-

If (A)

Rated Value

80% of rated value

(0-2A) -

Ia (A)

Va (V)

Tabulation N (rpm)  =2N/60 (rad/sec)

Eb = Va-IaRa (V)

T=EbIa/ (N-m)

(II). RETARDATION TEST TO FIND J & F Circuit Diagram 20A Fuse 220V DC Supply

D P S T

(0-20A) + A

L F A

1000Ω/1A

S

F1 V

+

M - 0-300V

F2 + A -

(0-2A)

Tabulation Retardation Test For

Range of Speed (rpm)

J only

1500 to 225

Both J & f

1500 to 225

Time (Sec)

(III) .DETERMINATION OF RA Circuit Diagram 5A + fuse 30V DC Supply

D P S T

(0-5A) + A

500Ω/5A

+

A1

(0 –30)V V -

M A2

-

Tabulation S.No. Va (V)

Ia (A)

Calculation by least square method

Ra (Ohms)

Ra = [V1I1 +V2I2 +V3I3+V4I4 ] / (I12+I22+I32+I42)

(IV). DETERMINATION OF La Circuit Diagram 5A Fuse

(0 –5A) A

230V D 1Φ AC P Supply S T

-

A A1 E

(0-30V) V

M

A2

NL

c

Using LCD METER La = Model Graph: To find Kt Torque (N–m)

To find Kb Eb( V)

Rated 80% of rated Ia(Amps)

Model Calculation: 

Ra = ……..Ohms



Za = …….. Ohms

ω(rad/sec)



La = Square root of (Za2 –Ra2) / 2f =……. H



f = 50 Hz



From Graph,



Kt = Torque constant = T /  Ia = ………… N-m / A



Kb = Back emf constant = Eb /  N = ………. Volts/(rad/sec)



From Retardation test o Finding J = o Finding f = Transfer function of armature controlled motor  (s) / Va(s) = Kt / { Kt Kb+(Js+f)(Ra + sLa )  (s) / Va(s) =



Result: Thus the transfer function of dc shunt motor by armature-control method is determined to be

TRANSFER FUNCTION OF FIELD CONTROLLED DC MOTOR Aim: To determine the transfer function of a field controlled dc shunt motor Apparatus Required: 





Ammeters

0-20A MC 1 0-5 A MC 0-2 A MC 0-50 MA (DMM) MI Voltmeters 0-250 V MC 0-250 V MI 0-30 V MC 0-50 V MI Rheostat 50 , 5 A 1000 , 1 A 50 , 5 A

Theory: The transfer function is defined as the ratio of Laplace transform of the output variable to the Laplace transform of input variable. The DC motor converts electrical energy into mechanical energy. The electrical energy supplied at the armature terminals converted into controlled mechanical energy. (ii).Field Controlled DC Shunt Motor In field control method, the armature current I a(t) is maintained to a constant value Ia while the field voltage Vf(t) is varied to control the speed or torque of the motor. Thus the input of the motor is field voltage V f(t) and the output is the motor speed and the load displacement (t) or (t). Transfer Function The circuit for field controlled dc shunt motor is given in figure. Rf

Ia

Ef If T Let

J, f Rf = Resistance of field circuit ;

Lf = Inductance of field circuit ; Henry Ef = Excitation voltage ; Volts If = Excitation current

; Amperes

J = Moment of Inertia ;kg-m2 f = Co efficient of friction ;N-m / (rad/sec)  = Angular velocity = d/dt in rad/sec Applying Kirchoff‟s law, Ef(t) = RfIf(t) + Lf dIf(t)/dt ………..(1) Taking Laplace transform of equation (1) Ef(s) = RfIf(s) + Lf sIf(s) ,which implies If(s) = Ef(s) / [Rf+s Lf] …(2) The torque developed is proportional to the field current, since the armature current is constant. i.e.  If & T  Ia as Ia is constant, T   ,which implies T  If T =Kf If(t) ……(3) Where Kf = T /  If = Constant in N-m /amperes …..(4) Taking laplace transform of equation (3), T (s) =Kf If(s) ……(5) The torque equation is given by, T(t) = Jd2/dt2 + f d/dt = Jd/dt+f Taking laplace of the above equation, T(s) = Js(s) + f(s). ……(6) Substituting equation (5) in (6), Kf If(s) = Js(s) + f(s), which implies (s) = Kf If(s)/[Js+f]……….. (7) Substituting equation (2) in (7),

(s) / Ef(s) = Kf / {[Js+f] [Rf+s Lf]}

Block Diagram: Using equations (2), (5) & (6), the block diagram is drawn for field controlled dc motor.

Ef(s)

1 / [Rf+s Lf]

If(s)

Kf

T(s) 1 / [Js+f]

(s)

For finding the transfer function of field control method, we have to find the values of J, f, Kf, Rf & Lf. Kt & Kf can be found out using load test, V-A method cab be used to find Ra, Rf, La & Lf and retardation & Swinburne‟s test can be used to find J & f. Procedure: (i). Load test to determine Kt & Kf  Conduct the load test with two constants normal (rated value) & sub normal (say 80% of rated field current) values of field currents.  Give the connections as per circuit diagram.  Keep the field rheostat in minimum position & switch on 230 V supply.  Adjust the field rheostat to rated speed & consider the corresponding field current as rated field current.  For different loads note down Ia,Va & speed  Calculate the torque developed.  Repeat the same for 80% of excitation current.  Plot a graph between T(Y-axis) and Ia (X-axis) for both the cases.  From this graph, obtain Kt for any one field current. [Note: The reason why we conduct the experiment for two field currents is to derive Kf from armature control graph. ]  From the graph deduce two points (T1,If1) and (T2,If2) for the same armature current.  Plot the graph between T Vs If  From this graph, obtain Kf.

(ii).Retardation Test to find J & f The total losses can be divided into two parts,viz., constant losses and variable losses. The constant losses include frictional & inertia losses. Total losses in a circuit= VI – Ia2R - VIf Losses = V(Ia + If) - Ia2Ra - VIf, Energy = (Losses * t ) = ½ J (12 -22) N1 =1600 RPM ; N2 = 1400 RPM ;  = 2πN /60 J = (Losses * t *2 ) / (12 -22) N1 = N * e –t1/τm

N2 = N * e –t2/τm

 Give the connections as per circuit diagram.  By closing the switch , make the measurement of V ,Ia and If.  Make the motor to run at a speed greater than 1600 RPM.  Open the switch suddenly.  Using stop watch, note down the time taken for the speed of the motor to fall down from 1600 RPM to 1400 RPM. (1 =1600 RPM,2 =1400RPM)  Plot a Graph showing relation between speed & time.  The moment of inertia , J is obtained from the relation between loss ,time and .  The friction constant , f is obtained using the exponential relation between speed , time and time constant.  Calculate no load input power using the values read by ammeter & voltmeter connected to armature circuitry. To find J

: [V(Ia + If) –Ia2Ra -VIf ] * t = (1/2) J(12 -22)

To find f : J / τm ; τm = (t2 – t1) / (ln N1 – ln N2) (iii).V-A method to obtain Ra, Rf, La & Lf  Give the connections as per circuit diagram.  Measure Ra & Rf. [As the field winding resistance is of the order of 250300ohms and it can withstand a current of 1A,the circuit shown can be used for measurement of Rf Similarly , as the armature resistance is of order of 0.21 ohms and it can be measured using the circuit shown]  To measure Lf & La, give the connections as shown.  Apply an ac voltage & measure the field reactance Zf & armature reactance Za.

 Calculate Xf =Square root of Zf2 –Rf2 and Lf = Square root of (Zf2 –Rf2 ) / 2f  Where f= supply frequency=50 Hz (I). Load Test to Determine KT & KF Circuit Diagram 20A Fuse

+ 220V D DC P Supply S T

(0-20A) A+

L F A

S1

1000Ω/1A + V

- M (0-300V)

F1 F2

Brake Drum

+ A

-

(0-2A) -

From load test Ia (Amps)

If (Amps)

Va(Volts)

(II). Retardation Test to find J & F Circuit Diagram 20A Fuse 220V DC Supply

S2

D P S T

L F A

1000Ω/1

+

(0-20A) A

S

F1 + V

+ A -

M

F2 (0-2A)

Tabulation Retardation Test For

Range of Speed (rpm)

J only

1500 to 225

Time (Sec)

Both J & f

1500 to 225

(III). DETERMINATION OF Rf Circuit Diagram

1000Ω/1A 2A

(0-2)A + A

+ Fuse

+ 220V DC Supply

D P S T

F1

V

(0-300V)

Rf F2

-

Tabulation S.No. Vf (V)

If (A)

Rf(Ohms)

Using Least square method Rf = (IV) . Determination of Lf Circuit Diagram 1A P 230V 1Φ AC 50Hz

(0-100)mA A

D P S T

A

F1 (0-300)V 1/[Rf + s Lf]

Zf

E C

F2

N By using LCR meter Lf = Model graph: To find Kf Torque (N-m)

If (amps)

Model Calculation:  Rf = ……..Ohms  Zf = …….. Ohms  Lf = Square root of (Zf2 –Rf2 ) / 2f = ……. H  f = 50 Hz  From Graph,  Kf = T /  If = …………… N-m /amperes\  From Retardation test o Finding J = o Finding f =  Transfer function of field controlled motor (s) / Ef(s) = Kf / {[Js+f] [Rf+s Lf]} (s) / Ef(s) = Result: Thus the transfer function of dc shunt motor by field-control method is determined to be