[jguillen08]

Hello All,

I am wondering if any one has used a Variac to power a DC motor. My game plan is to place an isolation transformer between the variac and the rectifier(I read some where that is a good Idea). The variac would control the voltage out of the isolation transformer and the rectifier would convert this to DC voltage. I would then use a capcitor to smooth out the dc voltage. This sounds simple enough, but I am not sure this will work. I would appreciate any input from any one who has done this.

Thanks,
Joe

[CJL5585]

I have a variable voltage DC supply that I built years ago using the above system.

It will work IF:

The Variac and Transformer current rating is greater than the current drawn by the DC motor.

The rectifier used is a FULL WAVE BRIDGE and the capacitor voltage rating is 20% greater than the rated output voltage of the isolation transformer.

I would not recommend this setup for stepper motors, but it will enable you to have a variable speed DC motor setup.

Hope this helps.
Jerry

[JRoque]

Hello. It's also possible that by the time you're done buying the variac, isolation transformer, caps, diodes, etc, you might be over the price of a ready made DC power supply - unless of course, you already have all of these components. Check eBay for inexpensive DC power supplies that work great for your intended application.

JR

[gar]

060310-1140 EST USA

jguillen08:

You can run a permanent magnetic DC motor from any source that has an average DC voltage. This can be with or without filtering, half or full wave rectified AC, or some pulsed source. Obviously there are some assumptions to this statement.

A typical motor has a moderately long mechanical time constant due to its inertia. If the electrical input pulse rate to the motor is fast relative to the mechanical time constant, then you will get moderately smooth mechanical output speed. 60 Hz and most motors meet this criteria.

The advantage of full wave rectification and filtering is that you reduce heating within the motor.

If you use sufficient capacitance in a capacitor input filter to make the ripple small, then for a sine wave input your average output DC voltage is approximately sq-root of 2 (1.414) times the input Vrms.

Whether the ripple is high or low, and half wave or full wave rectification your capacitor voltage rating needs to be substantially above the AC peak voltage given by the above 1.414 calculation.

.

[jguillen08]

Thanks For all the input

[H500]

Since there is no feedback, the speed regulation would be terrible.... the motor speed would drop as soon as it is loaded.

[gar]

060210-2205 EST USA

H500:

A brush type DC motor with fixed field excitation ( a PM field falls in this category ) has an output speed of RPMout = K ( Vinput - ( Ia * Ra ) ) where K is a speed constant, Ia is armature current which is proportional to load torque, and Ra is the motor internal resistance.

This type of motor will probably be designed to have maybe a 5 to 10 % drop from no load to full load. If Vinput is designed with a low impedance, maybe 1/4 to 1/10 of the armature resistance, then it will not contribute much to speed variation.

Yes, if you apply feedback you can get much tighter speed control. The important question is how stable do you need the speed to be. The application will define this.

A very simple regulator is an SCR in a less than half wave rectifier and use the counter EMF of the motor as the tachometer. During the off half cycle the counter EMF is a farily accurate tachometer.

.

[H500]

I assume he is planning to use the variac for controlling the spindle speed on some sort of cnc machine. The loading change can be several hundred percent when the tool contacts the work.

If the speed was set slow, the motor will stall easily when loaded. In order to get the desired cutting speed, it would be necessary to set the no-load setting way up. This is very inconvenient for machining.

If gearing was not the primary mode of speed reduction, then I think some sort of feedback is a must.

[jguillen08]

Yes this will be for a spindle on a mill/drill conversion. The motor is 1-1/2 Hp. The name plate values are 3200 rpm at 110 vdc and 11.5 amps. I am having a hard time finding a dc motor controler that will match this and still run on 110 AC. There will be some gearing. I will have two options. One will be 1:1 and the other Will be 2:1.

[gar]

060311-1258 EST USA

jguillen08:

You need to define an acceptable speed variation that you will tolerate.

If you want to test this motor at full speed simply connect a bridge rectifier to the 120 line, and the motor. Load the motor and see if the speed variation is tolerable.

A full-wave rectified sine wave produces an average DC output voltage, Vdc = ( 0.636/0.707 ) * Vac RMS. Thus with 120 Vrms input the DC average output = 108 V.

The heating in the motor due to ripple means that you need to derate the motor to about 75% of the 1.5 HP at full speed. In reality the motor is current limited and not power limited for rating.

The change in speed vs load is a constant independent of speed. Thus, if X #-ft of torque produces 5% change at 3200 RPM, or 160 RPM, resulting in a loaded RPM of 3040, then at 500 RPM no load the X #-ft load will drop the RPM to 340.

Since you plan to use gearing you may not need a large speed range from the motor. Basically a PM DC motor is torque limited for continuous duty. But at lower speeds you need to add external cooling because the internal fan is less effective.

.

[gar]

060313-1334 EST USA

You may wonder where the constants 0.636 and 0.707 come from.

The average value of a half sine wave, that is from 0 deg to 180 deg or Pi radians, related to the peak of the sine wave is 0.636 . You can use calculus to determine the value or you can approximate it graphically. Integral calculus provides you a means of calculating area under a curve. If you evaluate the definite integral of a sine wave from 0 to Pi you find it is exactly 2. This has to be divided by Pi to get the average. Average = 2/3.1416 = 0.6366 .

RMS stands for root-mean-square. This means you take the square root of the mean value (average) of the instantaneous square of the sine wave. Again calculus is the exact means to obtain this value. The integral of sine squared u du = 1/2 u - 1/4 sin 2u + C . Evaluating this from 0 to Pi gives us Pi/2 . The mean is Pi/(2*Pi) = 1/2 . The square root of 1/2 = (sq-root of 2)/2 = 1.414/2 = 0.707 .

.