T&L Publications

What component(s) are used to change the speed of a regular DC 12-24V cordless drill, as the trigger is pressed in?

Thank you,

Boris.

Hi Boris,


Most likely a PWM circuit that drives a high
current MOSFET. I've measured over 20 amps
in one of my drills once, under load.

Are you thinking of building one?


Take care
Al

I am building a motor controller capable of handling instantenous current of up to 120amps, ~ 40amp continuous (Dewalt Drill Motor). Since the mechanical part of the project is already complete I wanted to test it using some simple way of controlling motor speeds.

They definitely are PWM. Not only can you hear the drill "sing" as you pull the trigger, but it tears up your AM and SW radio reception as well if you're too close to the radio. PWM is the only efficient way to do variable speed on a DC, battery-operated tool since conserving charge is a primary goal.

Don't forget that if you're considering a more "remote" control (vs. right inside the tool) of speed, low voltage, high-power systems mean high current and that translates into fierce IR drops with poor connections, poor switches and undersized wire.

Dean

Hello again,


Dean:
Yeah, he he, mine sounds like a bird doing
different chirps...kinda cool really

Boris:
There are quite a few PWM controller chips
out there. You'll also need some high current
MOSFETS or perhaps parallel a few.
You'll want a controller chip that is made to
interface directly with MOSFETS or else you'll
want to use a driver chip that is made specially
for MOSFETS as the drive Z is low enough to
control MOSFETS fast enough for the kind of
currents you're talking here...without the right
drive you'll end up with hot hot MOSFETS or burn
out from the rise/fall times.
If you can do reverse with a switch that will
be easiest, rather than an H bridge...drill
control circuits use a switch.

I think i would measure armature current and
use that for feedback rather than voltage,
but maybe voltage would be easier to measure.


Take care,
Al

I have attached a light disk to a PCB drill.When I switch it ON, it moves instaneously. However if I attached a heavier disk, it does not move when I switch it ON. However with a slight push it moves nicely.
Question: Does the maximum speed of the lighter disk equals the heavier disk for the same voltage. Let's say the PCB drill runs on 5V, would the lighter/heavier disk influences the maximum speed of the disks at 5V?

Hi there,

The rotational inertia of the heavier disk is
greater than that of the lighter disk. The
top speed shouldnt be affected unless there is
more 'drag' with the heavier disk or more with
the lighter disk. The starting may be affected
because it takes more force to start the heavier
disk. Once it's started however it takes more
force to stop it too.

Take care,
Al

<small>[ February 19, 2006, 10:19 AM: Message edited by: MrAl ]</small>

The voltage should be the same, however the current draw might have a slight increase at the same speed, assuming the motor can keep the same speed for both disks. Its more likely that the current will go up slightly with the heavier disk and the speed will reduce slightly.

I assume "drag" here is the aerodynamic drag. I assumed that the current would be the same. As I = V/R. If the voltage and the resistance is the same, how can current change?

I would propose that the resistance in an AC drill motor should be more properly considered as impedance, the combination of the resistance of the wire plus the reactance of the wire coiled up as an inductor. I'm not sure this has anything to do with the topic, but it's obvious when my ShopVac suction tube clogs up, that the motor inside speeds up to compenate due to the increased load.

The increase in Load mass changes the current needed to drive it.

Same as a empty pick truck requiring so many horse power to go up the hill is different from a loaded pick up going up the same hill.

A larger mass requires more torque, thus more amps.

The voltage remains fairly constant because it comes from a constant current battery source and it wont change or change much, unless the mass changes drastically into a over draw situation.

For a small change in mass, the voltage will remain fairly constant, while the current will increase proportionally trying to find its adjusted RPM due to mass and its equilibrium based omn motor design.

The reason your shop vac speeds up is because the load decreases when it blocked up.

The cavitation in the turbine means the vanes stop working properly, and thus the load decreases, and the thus motor with the same power factor minus the load, spins faster.

If you place a vacuum gage on the hose near the motor turbines you will notice the vacuum drops drastically when the cavitation occurs.

Same goes for boats when the props spins too fast, you lose power to the water.

At the risk of being accused of being a complete, undeucated moron, I would venture to say that the comparison to a truck going up a hill is wrong. Once the heavier disk is up to speed, it shouldn't take extra energy to keep it there, assuming aerodynamic drag is the same as the lighter disk. But I do not have a degree in mechanical or electrical engineering, chemistry, or explosives, and I have only worked with the FBI in the field, not in the lab. I have noticed that on rc planes the addition of a heavy prop hub affected acceleration very slightly but not top rpm.


regards

Tom

The reason your shop vac speeds up is because the load decreases when it blocked up

Hmmm I though the load (on the motor)would *increase* when blocked because the vanes in the chamber were meeting resistance in pulling in air.

No, cavitation occurs, decreasing the load to almost zero, but not quite.

If you spin a prop in water too fast, time will not allow the surrounding pressure and water to fill in the holes where the water has been pushed out, there fore you spin the prop in a "hallow" [cavity] space containing no water and all air, and thus the prop fails to provide thrust.

Same goes for air.

The surrounding 14.7 pounds of air does not get enough time to replace the air thrown out from the vanes because 14.7 pounds is the limit to replace this force [there in no more], and the vane speed exceeds this limit, thus the vanes spin, empty of air and the chance to be replaced efficiently by the surrounding new air is at only 14.7 psi, which doesnt cut it.

Therefore it cavitates.[cavity]

UTE
From my mechanical engineering days...I can assure you that a greater load, according to the rules of Horse power, does require a greater force to push a "heavier truck" up the same hill.

Momentum doesn’t come into play because it takes more horse power to accumulate a greater momentum under a greater load.

If you wish to gain a momentum of say equal to 30mph with double the load, you will expend more HP doing so, and then that power is stored in the form of kinetic energy, [speed time mass] but it took more to get there than the light load did.

And it will expire twice as fast with the light load as with the heavy load.

If you climb a hill from stop and empty, and then again with a full load, you will use different amounts of horse power because horse power in it self states this fact.

X ....foot pounds per minute of lift..

If you double X, you need Double the horse power to do this.

If you are climbing a hill from start, like the lighter spinning wheel, and the HP of your motor is ten.

With a small disk you will accelerate up to normal speed which happent to be 10MPH and you continue up the hill using only 9 of your ten HP and only one minute to get to the top. Your power and your load are matched well.


However, same situation, same hill same truck, with a load of bricks in the back.

As you try to start up the hill the brick load is too heavy for your 10hp, so your crawl up to full speed in twice the time using your full 10hp, not just the 9.
You never reach full speed because the load is slightly above your limit, so you make it over the top at 7mph instead of ten, you used all ten HP doing so, and you got here in two minutes instead of one at a reduce speed using more HP.

<small>[ February 19, 2006, 10:03 PM: Message edited by: Chris Smith ]</small>