voltage/amperage.

[gar]

090305-1314 EST

Rick:

Conservation of energy is the controlling factor. You can not treat the synchronous motor with ohm's law and view the motor as an invariant resistor.

I have no problem creating a load that is constant torque at a constant speed. Given that my load is a constant torque and that a synchronous motor runs at a synchronous speed defined by the supply frequency, then as I lower the input voltage the component of current that is in phase with the excitation voltage must go up to maintain the same power input as is being consumed on the output.

This is also true of an induction motor but harder to describe an example to prove the point.

Here is another example with a different product. Consider a switching power supply with automatic regulation of voltage accomplished with the switching control providing 5 V at 20 A. That is 100 W output. Assuming no losses the input is 0.833 A at 120 V. At 240 V input the current is 0.417 A. This is an easy experiment to perform. Find one of the new supplies that is rated from 95 to 250 V input and anything between. Put a fixed load on the supply. Use a Variac to adjust the input voltage from 95 to 240 V and monitor the input current.

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[Besoeker]

Whoa, you've got your cause and effect a little backward from reality there. You have drawn the assumption that the motor output is constant and it's speed doesn't change. What then is the mechanism that causes the current to increase? .

Gar referred to synchronous motors.
For a synchronous motor the speed doesn't change. Hence synchronous.
So, for a given load, if the voltage drops the active component of current must increase.
You can, of course, play around with the excitation to change power factor to change total current.

 

[Rick Christopherson]

Given that my load is a constant torque and that a synchronous motor runs at a synchronous speed defined by the supply frequency, then as I lower the input voltage the component of current that is in phase with the excitation voltage must go up to maintain the same power input as is being consumed on the output.

Lowering the voltage on the stator windings is comparable to changing the rotor field, and the result is that the phase angle between the rotor and stator magnetic fields changes, which results in a change in the powerfactor. (I found a fairly good picture showing this from EngineersEdge.)

The motor still follows Ohm's Law, but the impedance has changed, and is a function of the Torque Angle, which in turn is a function of the torque load. The point being, is that something else has changed, and it is not simply a case where current and voltage follow inverse curves. The impedance is not fixed, but is a variable.