DC motor with half wave rectified source

[Microfarad]

I am evaluating a shunt dc motor to determine if it will produce the min. torque required for my valves. It's ratings are .361 hp, 95 arm.volts, 4.7 arm.amps. It's being supplied by 104 Vac source (120V under min. volt conditions) that is half-wave rectified (through SCRs) by a Modutronic 10A controller. This gives me an average(DC) voltage about 35-40 V. Does it make any sense to supply a 95V motor with half its rated voltage? How will a dc motor react in this situation?

 

[RadarRay]

What your talking about is supplying the motor with single phase half wave. At 60Hz this is one voltage pulse every 8.3msec. With the size of motor you have, I doubt that there is much inductance in the armature, which means the voltage supplied will roughly be the same shape the current you get from the voltage pulse. Can you live with this much of a torque pulse? I you want to run the motor over half speed, you will not be able to do it, not enough voltage supply. Do you want to.

Are you going to have a current regulator to fire the SCR's? A Speed regulator to supply the current regulator with a reference?

With this type of supply, since it is hardly DC, you will have more heating than with a full wave supply. Can the motor take this?

So I guess the answer to your question will depend on the questions I asked. What is the application?

 

[jbartos]

Suggestion: The half-wave power supply for your motor would require very good ripple filtering. It also will produce a second harmonics that is nowadays forbidden by utilities since it causes motors and transformers to overheat and fail.
Therefore, it would be much better to have a full-wave rectification and reasonably good ripple filter. This would raise the voltage close to the motor terminal rated voltage. Consequently, the better motor operation will result.

 

[Microfarad]

This is a throttling valve for the aux-feedwater system that has been operating in the plant for 20 yrs. The field is full-wave rectified and always energized. The armature resistance is 5 ohms & inductance is 12 mH. The SCRs are in parallel with a 47 ohm resistor and a 0.1 F capacior. The Modutronic controller balances the position feedback signal from the operator against the milliampere signal from the process controller. Any difference causes the motor to run in the correct direction to position the valve under command of the process controller.

 

[ScottyUK]

I'm assuming that this motor drives through a worm and wheel, or some other form of self-locking drive, so the motor only drives when there is a demand?

The circuit sounds incredibly crude, it isn't fit to call it a servo as that would imply some finesse and precision.

The motor will produce somewhat less than 50% rated torque, and will get hot because of the huge AC component. A DC motor isn't normally designed to minimise eddy losses in the core, because a DC supply doesn't generate eddy currents in the iron in the way that AC does. As your controller is feeding the armature a mixture of AC and DC, the motor will heat up due to eddy losses, but not produce a power output commensurate with that heating effect. There will probably be significant torque ripple, although in this application you might not be too bothered.

The 47R / 100n network is a snubber designed to protect the SCR's from high dv/dt which would damage them.

I think you could make a useful modification by converting your circuit to a half-controlled bridge rectifier, which would use your existing SCR's and drive circuitry and add two power diodes of similar rating to your SCR's. This would give you a full-wave rectified supply which causes less harmonic pollution of your supply and give a far better DC supply to your motor. A win-win situation.

You need to do some more investigation into your firing circuit to determine what range of delay angle control it has, but the principle will work.

 

[jbartos]

Suggestion: The full wave rectification will need physically smaller ripple filters since the full wave rectification ripple is smaller.