# Using a Dimmer for Power Tools?

I want to make something for my 120V AC Black&Decker Jigsaw, But it will cover the trigger.

My question is:

Are there any problems with the fact that the jigsaw is an "inductive load"? If it is, Can I improve it with a capacitor?

My jigsaw uses 4.5Amps @120V, How many Watts is it? (IDK the formula for AC Watts...)

Is there any more information that I missed?

Thanks,

Yonatan

My question is:

**Is okay to use a 120V/110V AC Dimmer (PWM?) to control the speed instead?**I can keep the jigsaw's trigger at "full power" with a zip-tie, But the control the speed control will come through the dimmer...Are there any problems with the fact that the jigsaw is an "inductive load"? If it is, Can I improve it with a capacitor?

My jigsaw uses 4.5Amps @120V, How many Watts is it? (IDK the formula for AC Watts...)

Is there any more information that I missed?

Thanks,

Yonatan

active| newest | oldestIf you lower the voltage, the motor will try to draw more amps and you

could easily fry it or trip your breaker. It's why you can damage

powertools or melt the powercords when you use low-grade extension cords

for powertools (because of voltage sag). Not to say it can't be done, it just has to be done right and I don't know that the dimmer you have in mind would work properly. There are universal options that exist; i.e. specialized induction load dimmers. Some even have pedal accessories. You may be able to get by with a ceiling fan dimmer.

The usual formula for power is P=V*I, where P is power (in watts) V is voltage (in volts) and I is current (in amperes).

Just multiplying 120 V by 4.5 A gives 540 W, but if those numbers came from a label or a plate, then I would not take them too seriously.

Manufacturers tend to exaggerate these numbers. 300 W is probably closer.

If you read the numbers from a power measuring tool, like one of those Kill-a-Watt(r) brand gizmos, then I would consider those numbers more believable.

BTW, there's a slightly fancier version of the same formula, specifically for AC power, that looks like P=V*I*cos(theta), where the cos(theta), also called "power factor", is a measure of the phase difference (i.e. difference in timing) between the current and voltage waveforms.

Anyway, regarding the question of whether or not it is "okay" to use a lamp dimmer with your jigsaw,

I suggest you try it.I've used a lamp dimmer with just a few different AC loads, namely a soldering iron, a shop vac, and an electric mixer (a kitchen tool).

So I would NOT expect a jigsaw to instantly explode the moment it is plugged into a lamp dimmer. Nor would I expect this to instantly kill the lamp dimmer.

Essentially what a lamp dimmer does, is take the existing AC voltage waveform, which is a sinusoid, and replace it with a different waveform that is still AC, still alternating in the sense it has moments both of positive and negative voltage, but the waveform that comes out of the dimmer has lower RMS voltage. So now with the twist of a knob you can now vary the RMS voltage of your mains power, whereas before it was fixed at (approximately) 120 VAC. If you want a picture of what the modified waveform looks like, I found one here,

https://en.wikipedia.org/wiki/File:Dimmer_60_volts...

from:

https://en.wikipedia.org/wiki/Dimmer

RMS stands for root mean square. It is a kind of mathematical average used for AC signals, and some other things.

Anyway, if you want some vague advice on how to proceed, I suggest the lamp dimmer's highest settings will be the "safest", since these give a waveform with a shape closest to the original sinusoidal waveform the motor was designed to work with.

Also turning the dimmer output so low the motor stalls, is probably bad, but that's really not telling you anything new, if you knew its bad to stall an electric motor under other conditions.

Finally, if anything unsavory is happening, it will probably manifest itself as excessive heat. So, you know, check every so often to make sure there's not abnormal amounts of heat coming off from the motor.

Thank you, So for short, You're saying:

Try it, And use common sense...And about the "120V * 4.5A gives 540W", Isn't that the calculation for DC Voltage? I remember that my dad told me that for AC Voltage there is a different calculation formula

There is a lot of crap with AC voltage your mains are 120 volts but if you put a meter to it, you can get anything from 95 to 140 volts. If you put an oscilloscope to the mains you can get 340 to 400 volts peak to peak.

Everything is just a ballpark as long as you keep it on the field it works.

RMS 120 volts and DC ohms law keeps the ball between the foul lines.

Power is rate of energy flow. Like other flow rates ( e.g. speed in meters per second, salary in dollars per year) it is a quantity divided by some amount of time.

If a toaster is on for 60 seconds, and in that time it uses 60 KJ of electricity, the rate at which it is turning electricity into heat is P = (60 000 J)/(60 s) = 1000 J/s = 1000 W. Or rather this is a measure of

averagepower dissipation, over thetime intervalwhen the toaster was turned on.For defining a rate in general, what number should you pick for the time interval? A minute? A microsecond? An hour? A year?

Well, one way to answer that question, is to define your rate as time derivative.

https://en.wikipedia.org/wiki/Differential_calculu...

Which is what you get for a time interval that is arbitrarily small.

That's how

instantaneous electrical poweris defined, with the formula:P(t) = V(t)*I(t)

where the "t" in parenthesis is pronounced "of t", "of time", and the meaning is, each of those is an instantaneous quantity, right at a particular moment, e.g. t = 24.00 microseconds, after midnight, Tuesday morning... or whenever.

If V(t)=V_DC and I(t)=I_DC are DC signals, constant with time, then it doesn't matter what time it is, and you can throw away the "of t"s.

P_DC = V_DC*I_DC

For AC voltage and current, V(t) and I(t) are changing with time, but in a very predictable way. There is a special kind of average, called root mean square (RMS) whose definition you can look up (It is the square root of the mean value of the square of a signal over some time interval.)

https://en.wikipedia.org/wiki/Root_mean_square#Ave...

and it turns out, for RMS values, the formula for electrical power has

almostthe same form, as it did for instantaneous power, and DC power. This formula isP_RMS = V_RMS*I_RMS*cos(theta)

There is just one more factor, cos(theta), which was given the name "power factor". For the case when power factor, cos(theta) = 1, as is the case for a purely resistive load.

P_RMS = V_RMS*I_RMS

the formula has exactly the same form as the formulas for DC power and instantaneous power. For loads that have power factor approximately equal to 1, the formula is

approximatelycorrect.Variac, maybe. Standard triac-based dimmer, no. Just isn't nice to the equipment or dimmer.

+1 Everything Jack said...

Plus consider if the jig motor runs on brushes they will be worn out prematurely as you slow down the saw, because the phase control high RMS currents are not PWM..