Using variac to run corded brushed drill

[primegaps]

I saw someone on youtube plug in their corded black and decker drill into a variac, to demonstrate what the variac does. He showed that you could get any speed out of the drill from 0 to maxium, and then a bit more by engaging the overwindings.

Ive been wanting to make a crappy belt sander with variable speed for some time, will i be able to do this using a variac to control the input voltage to the drill?
I know many cordless drills motors actually are taking in PWM DC, will running the drill with the trigger squeezed all the way permanently, but controlling the speed with lower AC voltages damage the drill?

[xavier60]

Most, if not all mains powered motorized tools contain universal motors.
As far as I am aware, universal motors don't mind at all about being run on variable AC voltage whether from a  variac transformer or from  crudely chopped AC from a phase control fired triac.
My dremel rotary tool switches a diode in series with the motor for half speed. This is too crude as it has poor load regulation. So I leave it on the full speed setting and control the speed with a variac transformer.

[Berni]

These universal brushed motors can run on pretty much anything including DC voltage from batteries.

So using a variac or a lamp dimmer are fine for reducing the speed of them. The only problem with this is that you also reduce the toque of them considerably since you are effectively limiting how much power they can draw. But the only way to get around this is by having a actual regulator circuit that senses the speed of the motor and dials the power to it up and down to keep it constant. Not so simple anymore, but hey a belt sander is not terribly critical to run at a constant stable speed.

[Ian.M]

When running a universal motor power tool on reduced voltage, *NEVER* let it 'bog down' for more than a second as that can burn out the motor.  Reason: a universal motor draws more current when loaded, in an attempt to maintain the speed where its back-EMF is equal to the supply voltage (minus I*R drop), but the volume of air moved by the tool's internal cooling fan drops with the shaft speed.  You therefore end up in the situation where its drawing more current, so I²*R losses (Joule heating) are much worse, but cooling efficiency is *WAY* down.

Using a drill at reduced voltage to power a belt sander is just asking for trouble . . . .

[Gregg]

For a cheap variable speed motor suitable for a belt sander, look for broken treadmill.  Many treadmills are thrown away because they need repair other than the motor or they are taking up too much room.  The better ones use a permanent magnet DC motor and a feedback PWM speed control and they can run for extended time periods at less than full speed.  The motor shaft sizes tend not to be standard diameters which could be a problem. 

[Berni]

The original question was if reduced voltage AC is somehow bad for the motor instead of PWM control that is usually used in hand drills.

But yes running drills very slowly is a bad idea but i would assume you wouldn't be running a belt sander at such a low speed. As long as you keep it running at more than 20% of its full speed and don't put any very heavy loads on it when at low speed it should survive just fine. Hand drills are designed to run at lower speeds and i often use them with the power setting way down. For example i don't need full speed when drilling a large hole in metal and if the drill bit catches i woulnd't want the drill to twist my wrist.

That being said not all drills are as well built.  The drills i got are old overengineered ones that still work great to this day so they never got replaced with newer ones. The cheapest drill off amazon might be engineered to barely not overheat during casual normal use.

[Whales]

When running a universal motor power tool on reduced voltage, *NEVER* let it 'bog down' for more than a second as that can burn out the motor.  Reason: a universal motor draws more current when loaded, in an attempt to maintain the speed where its back-EMF is equal to the supply voltage (minus I*R drop), but the volume of air moved by the tool's internal cooling fan drops with the shaft speed.  You therefore end up in the situation where its drawing more current, so I²*R losses (Joule heating) are much worse, but cooling efficiency is *WAY* down.

Using a drill at reduced voltage to power a belt sander is just asking for trouble . . . .

This seems completely at odds with my understanding of brushed motors.

(1) When the motor is stalled there is no airflow whatsoever.  The dremel will last longer stalled at a lower voltage than at a higher one.

(2) Stall power = V^2 / R and Airflow ~= Rot speed * constant   ("Fan Law 1" from here).  Even when you factor in the lower V across the R due to the back EMF, doesn't this still lead to power (approx square) dropping faster than airflow (approx linear) when V is reduced?

[Ian.M]

@Berni,

It depends on the speed controller.   Advances in modern semiconductors make it viable to sense load current by tapping a small fraction of the MOSFET drain current and a variable speed trigger switch may have advanced features like closed loop speed control (by back-EMF sensing during the PWM off time) and stall protection, tripping on a sudden drop in speed and increase in current, or by sustained overload, reset by releasing the trigger.

The *OTHER* issue is that using a drill to power a belt sander will need some serious engineering to avoid the side load on the drill's output shaft FUBARing its bearing and taking out its planetary reduction gear. You'd need a flexible coupling between the drill and the drive roller, which would need bearings either side of it.

IMHO its going to be a lot easier to start with an appropriate gearmotor that has a gearbox rated for the side loads involved with driving pullies or sprockets.

@Whales,
(1) is true. If the voltage is low enough it may even survive.

(2) Your equation is the limiting case of totally stalled.    See your case (1)
However assuming the motor is still turning at a significant fraction of the desired speed, the current is likely to be similar to the full voltage case, as there will be a tendency for the operator to increase the voltage  to maintain the desired speed when loaded, so the motor is already running hotter than usual, because the current is about the same, but the cooling has reduced at least linearly with the speed reduction.

[Gyro]

Just dipping back to the OP for a minute...

If using a variac, it is vitally important to have a fuse on the output. Just because it is a variable voltage transformer doesn't mean that it can supply the same power at all settings, it is current restricted by the wire diameter and wiper to the same current rating as the input. Theoretically (for say, a 500W variac) it is capable of outputting 50A at 10V without blowing the input fuse... just for a few milliseconds, before bottom of the winding evaporates!

Forgetting this limitation has caused the death of many a good variac.

[primegaps]

Oh okay, cool i will do that. I have finished repairing and restoring all the variacs i salvaged from the dump to their former glory, and they all work fine right now. When fusing the output, should I use the same amperage as the input fuse? I have four 10 Amp variacs as well as four 7.5 amp variacs, and i have replaced all the fuses with fast blowing 10 and 7.5 amp fuses respectively.
Is fast blow the wrong choice? This is what the guy at the electronics store told me. I know upon start up there can be huge inrush current, so should I be using slow blow fuses for the input? Or will i be okay as long as i start up at low voltage and turn it up, to avoid the initial starting current?
What about the output? should i go for fast or slow blow?

Is it alright to use an inline fuse of this style on the output? And Just to be sure, I can just put this output fuse inline with one of the output prong receptacles right?

[oldway]

An universal motor works better with rectified current than in alternating current.
The motor has fewer losses and has better speed regulation.

Just add a rectifier bridge of 4 diodes 1N5406 or 1N5408.

But in DC, it is always necessary to start the motor from 0V otherwise the starting current will be excessive and may damage the motor (arc to the collector) and the diode bridge.

[Gyro]

Oh okay, cool i will do that. I have finished repairing and restoring all the variacs i salvaged from the dump to their former glory, and they all work fine right now. When fusing the output, should I use the same amperage as the input fuse? I have four 10 Amp variacs as well as four 7.5 amp variacs, and i have replaced all the fuses with fast blowing 10 and 7.5 amp fuses respectively.
Is fast blow the wrong choice? This is what the guy at the electronics store told me. I know upon start up there can be huge inrush current, so should I be using slow blow fuses for the input? Or will i be okay as long as i start up at low voltage and turn it up, to avoid the initial starting current?
What about the output? should i go for fast or slow blow?

Is it alright to use an inline fuse of this style on the output? And Just to be sure, I can just put this output fuse inline with one of the output prong receptacles right?

It's very tempting to use those fuseholders because they are so easy to fit and have decent current rating. Unfortunately they are intended for automotive use and the fuses are only rated to 32V AC/DC. I'm not sure what you have used for the input fuses but ideally you would use a panel mount fuseholded with an ordinary cylindrical ceramic mains rated fuse (normally 1"x1/4"). Alternatively, you may be able to find a mains rated in-line fuseholder, for example: http://cpc.farnell.com/bulgin/fx285/fuse-holder-in-line-1-x1-4/dp/FF02319  hopefully there is something similar available locally.

With regard to fuse rating... Yes the output fuse should be rated at the same current as the input fuse. As you say, inrush current on variacs can be high, so it wouldn't be unusual to use an anti-surge one for the input. I would want a fast blow fuse on the output though, for maximum protection of the wiper contact and winding. Hopefully you can move them across. In use, I would expect the output fuse to blow first on overload, to protect the variac (the input fuse should only normally blow if the variac goes faulty). That means that you probably want the output fuse to be the most accessible.

Yes, you want the fuse inline with the live (wiper) to output receptacle lead.

I hope this helps.

[Raj]

Could use,an inductor paired up with a bulb dimmer, with a tryac upgrade