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6 : DC Motor Speed Control


A separately excited DC motor has a field system having a separate power supply. The rated field current here is 1.6 Amp with 75 Ohms resistance. Accordingly Vf is chosen.

The armature gets an excitation from the rectifier according to the firing angle chosen. As the firing angle decreases from 90 degrees towards zero, the voltage applied to the armature increases due to which the speed also increses. If the torque demanded is increased, then the armature current (la) drawn by the motor also increases. The inductance of the armature decided whether current can sustain itself f in a continuous manner or not even though firing angle may be large.

The basic equations of the DC motor are

Va = Eb + IaRa............................................(1)

Where Eb = back emf of the motor = K.If.w     (w being the speed in radians per sec, If being the field current and  K being a motor constant)

Va= Armature voltage
Ia = Armature current and Ra = armature resistance
T = K If Ia


Speed w = (Va/Kif)  - (TRa/K2If2).........................................(2)

From the above equation, it can be said that as the torque demand increases for a given Va, the speed ‘w’ decreases. The aim of this experiment is to check how the DC motor speed is changing as the firing angle of the converter changes , i.e., as Va changes.
The output of the converter is given by Va = (2*Vm *Cosα)/π

Where Vm = peak of the supply voltage and  α is the delay angle of the converter. Here the firing angle is chosen to be 80 degrees.

Cite this Simulator:

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