# DC shunt wound motor speed increase by reducing flux

#### zazas321

Joined Nov 29, 2015
926
Hello guys. I have been looking at the DC shunt motor and I got a question. That is the Back emf formula formula E= (P.N.Z.Ø)/60A. Lets derive this formula to see the speed relationship between flux and we get N=E/kØ , k being constant that we cannot change k=PZ/60A. Okay so looking at the speed and flux relationship it is clear that the speed of the DC shunt motor is increasing because they are inversely proportional. But at the same time, Decreasing flux means decrease in Back emf because they are directly proportional to one another. If we look at the Voltage equation V=Eb-Ia*Ra . Va being total voltage, Eb - back emf and Ia and Ra armature current and resistance. So if back Emf is reduced , current must increase to keep ratio the same. So the increasing current should mean that the speed has gone down. So thats why it is confusing to understand. Why the Speed of the motor does not remain the same If the Flux is forcing motor to rotate faster but at the same time increased current is forcing it to rotate slower? Or am I confusing the Armature current with the Field current?

Last edited:

Joined Jul 18, 2013
23,917
The possible rpm is reached when the back EMF is equal (or almost) to the applied voltage, by reducing the field a higher rpm is required to produce the BEMF required to meet the value of the applied.
Max.

#### zazas321

Joined Nov 29, 2015
926
The possible rpm is reached when the back EMF is equal (or almost) to the applied voltage, by reducing the field a higher rpm is required to produce the BEMF required to meet the value of the applied.
Max.
Okay.. I have done some further research and found a solution. It really depends on what you assume to be constants.

Halving the field flux while holding armature voltage and load POWER constant will result in doubling the motor’s speed and halving its output torque - HOWEVER, the armature current will remain constant. Remember: Power in = Power out. Armature voltage is held constant, motor power is held constant, so V1A1 = V2A2. If V1 = V2 then A1 = A2. If speed doubles but power remains constant, torque must fall by half.

Thats what the article said. I did not assume that the current must be constant

#### zazas321

Joined Nov 29, 2015
926
But also I am confused how can you keep the load Power constant? How can you reduce flux keeping the Armature current same? Reducing flux by half should increase Armature current by the same factor to compensate the reduction in the flux.

#### LesJones

Joined Jan 8, 2017
3,571
Yes, assuming the motor is providing the same output torque. So as the speed increases and the torque is the same thoe output powere increases. (So the input power must also increase.)

Les.

#### zazas321

Joined Nov 29, 2015
926
But how do I know whether the outpur torque is gonna be the same or it is gonna change . And how do I know if the output power is gonna change or not? Do I have control on what constants to keep ? Lets take a real life example. If I am using a motor to drive conveyer belt lets say. It can be either light or very heave dependant how much weight is put on the conveyer. So essentially the weight is gonna be our load? If the load is not changing the power will remain the same . So reducing flux by half would increase the conveyer belt by 2. Am i understanding it right?

#### LesJones

Joined Jan 8, 2017
3,571
That is my understanding.

Les.