Block Diagram and Transfer Function of DC Motor PDF

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Block Diagram and Transfer Function of DC Motor Armature Controlled DC Motor

Consider the armature controlled dc motor and assume that the demagnetizing effect of armature reaction is neglected, magnetic circuit is assumed linear and field voltage is constant i.e. if =constant Let

Ra= Arm ture resistance La= Armature self inductance caused by armature flux ia=armature current if =field current

E=Induced emf ∈armature V = Applied voltage

T =Torque developed by the motor θ= Angular displacement of the motor shaft J =Equivalent moment of inertiaof motor shaft ∧load referred ¿ the motor

B=Equivalent coefficient of friction of motor∧load referred ¿ the motor Apply KVL in armature circuit v =R a ia + L

d ia +E dt

Since, field current if an emf is induced

(i) is constant, the flux

∅

will be constant when armature is rotating,

E ∝∅ w E=k b w

E=

k b dθ dt

(ii)

where w=angular velocity k b =back emf constant Now, the torque T delivered by the motor will be the product of armature current and flux T ∝ ∅i a T =k ia

(iii)

where k =motor torque constant The dynamic equation with moment of inertia and coefficient of friction will be T=

J d 2 θ Bdθ + dt d t2

Take the Laplace transform V ( s )−E ( s )=I a ( s ) (Ra +S La ) E ( s )=k b Sθ(s) T ( s) =k I a (s) T ( s) =( S J + SB ) θ(s) 2

T ( s) =( sJ +B ) Sθ(s) The block diagram for each equation

(iv)

Combine all four block diagrams

Block diagram of armature controlled dc motor The transfer function is determined by block reduction

θ (s ) k = V (s) ( Ra + s La) ( Js + B ) s+ k k b s Equation (v) can be written as

(v)

θ ( s) = V ( s)

where

k

(

Ra 1+ La Ra

) ( )

s La sJ +k k b s sB 1+ B Ra

=τ a , time constant of armature circuit

J =τ , mechanical time constant B m θ (s ) k = V (s) s R a B ( 1+s τ a )( 1+s τ n ) +k k b s From the block diagram the effect of back emf is represented by the feedback signal proportional to the speed of the motor. Field Control DC Motor

1. A constant current ia is fed to the armature 2. Flux is proportional to the field current ∅∝ i f

∅=k f I f

(i)

3. Apply KVL to the field circuit Lf d I f (ii) dt 4. Torque developed by the motor is proportional to the flux and armature current T ∝ ∅ Ia V f =Rf I f +

From (i)

T =k I k f I a I f T =k k f I f

(iii)

where k =k I I a Dynamic equation of torque in terms of J & B T=

J d 2 θ Bdθ + dt d t2

(iv)

Laplace transformation of (i), (ii), & (iv) V f ( s) =R f I f ( s ) +S Lf I f (s) ¿ I f (s ) [R f +s Lf ] I f=

V f (s) R f +s L f

T ( s) =k k f I f (s)

(vi)

T ( s) =θ(s)[ s2 J + sB]

(vii)

Sub. (v) in (vi) T ( s) =k . k f .

V f (s) R f + s Lf

k kf θ(s) = V f (s) S ( SJ +B ) (R f +S Lf ) (viii) can also be written as θ(s) = V f (s)

k kf

[ ]

R f Bs 1+

(v)

sL sJ [1+ f ] B Rf

k kf θ(s) = V f (s) R f Bs [ 1+ s τ m][1+τ f ] J where τ m = =mechanical time constant B

(viii)

τm=

Lf =time constant for field circuit Rf

τ f=

Lf =time constant for field circuit Rf

Block diagram of field controlled dc motor...