Reduced voltage effect on energizing a coil

[mull982]

I understand that during motor starting, a reduction in voltage leads to a reduction in current becasue of the fact that the motor windings are seen as a fixed resistance during starting.

Is this the same case when energizing a coil on say a starter or whatever. Will a reduction in voltage lead to a reduction in current when starting, or energizing a coil, or will the reduced voltage lead to increased current when energizing a coil? Is a coil also seen as a fixed resistance during energizing or inrush?

 

[K2500]

What type of motor, what type of starter. That might help better explain things.

 

[Jraef]

You are leaving out a big important part of the picture: load.

In a motor, reduction of voltage reduces current in direct proportion, but also reduces the TORQUE by the square of the voltage reduction.

In a coil, voltage reduction does reduce the coil current, but also reduces the magnetic holding / closing strength of the coil, also by the square of the voltage reduction. So what happens is, at 70% voltage (typical "drop out" specifications), the coil strength drops to less than 1/2 of design spec. (.70 x .70 = 49%) and the coil can no longer overcome the magnetic and spring forces that are attempting to force the contacts apart, so the starter opens up. Then because the motor load goes away, the voltage drop does too, and without a 3 wire control circuit the coil gets full voltage again and closes, the motor comes on-line and causes the voltage to drop again, which again drops out the starter, repeating ad nauseum until the coil fries or the contacts weld, or both.

 

[mull982]

You are leaving out a big important part of the picture: load.

In a motor, reduction of voltage reduces current in direct proportion, but also reduces the TORQUE by the square of the voltage reduction.

In a coil, voltage reduction does reduce the coil current, but also reduces the magnetic holding / closing strength of the coil, also by the square of the voltage reduction. So what happens is, at 70% voltage (typical "drop out" specifications), the coil strength drops to less than 1/2 of design spec. (.70 x .70 = 49%) and the coil can no longer overcome the magnetic and spring forces that are attempting to force the contacts apart, so the starter opens up. Then because the motor load goes away, the voltage drop does too, and without a 3 wire control circuit the coil gets full voltage again and closes, the motor comes on-line and causes the voltage to drop again, which again drops out the starter, repeating ad nauseum until the coil fries or the contacts weld, or both.

If I understand what you are describing, it involves what happens to a starter coil with a voltage drop under normal operation. It sounds like you are describing what happens to a coil when it drops to below 70% of its rated voltage. I was not aware that the coil strenth droped as a square of the voltage.

I was more curious what happens during the period of when a coil is first energized or during the "inrush" period. I'm not necessarily talking about a motor starter, but rather any coil in general, weather its a coil on a relay, transformer coil, etc... I wanted to find out what happend to the current during the brief coil energization/inrush period if the voltage was reduced say 10%.

Say I have a relay coil rated for 120V and before I send power to the coil, I measure my voltage source to be 108V (10%) If I then send this power to energize the relay coil, what effects on current should I see. Should I see more or less current then I would see if I has 120V during the coil inrush? I'm assuming that during normal operation of a relay coil that less voltage corrosponds to more current.

 

[mull982]

As I was looking at some motor datasheets here at our plant I got even more confused about this topic but with a different application.

I was always under the impression that when a motor is running (after starting) weather loaded or unloaded a reduction in voltage at the motor terminals results in an increase in current being drawn by the motor.

However on the motor datasheet for a 4kv 6500hp 865FLA motor, it showed test data for the motor unloaded operating at different voltages. When looking at the values it appeared that with no load on this motor the amp readings were 206A. During the test however as they reduced the voltage the current reduced on the line in direct proportion with the voltage.

This leads me to the question of weather or not the motor voltage vs current relationship is different for when the motor is running with no load vs with having a load.

 

[wirenut1980]

Any current drawn during no load is considered losses of the motor including mechanical losses and armature and stator impedances. And since the motor is not trying to maintain torque under no load, I believe the "load" acts the way you describe in the test, such that as V and I act linearly. Or maybe a better way would be to say the motor does not have to work hard to maintain torque under no load.