# 240volt vs. 120volt

Does changing the voltage capability on my TS allow me to work the saw harder or does it simply help prevent overheating and burnout and a few breaker runs?
I have already blown out a capacitor on the motor, ugh. (120volt setting)
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By doubling the voltage, the current is halved. Therefore the I/R (voltage drop of the wire, which is a function of the current and resistance in wire) is cut in half. (more voltage to the saw). The saw starts better (the most current is on startup) runs better, cooler and should last longer. Just my \$0.02
Frank

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Why would the saw run cooler if there is more voltage to the saw?
To first order, won't the saw consume the same amount of power when using 120V or 240V (assuming of course the motor is wired correctly for the appropriate voltage)? That said, in reality I think that saw will consume a little more power (maybe 5-10% more for a typical installation) in the 240V configuration than it would in the 120V configuration.

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Yes.
240V
How do you figure that? It is not logical.
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wrote in message

using
Here's my rational for proposing that the 240V configuration gives the motor slightly more power.
First consider the 120V configuration: Lets suppose that the motor draws 16A from the 120V source. The motor windings will be connected in parallel, so for two windings in parallel, each winding will have 8A flowing through it and each will have close to 120V accross it (it is a little less than 120V because of the voltage drop in the wiring). If the wiring has a resistance of 0.3 ohms (not unreasonable for a run of 12 AWG wire), the voltage drop due to the wiring would be 0.3 ohm * 16 A = 4.8 V. So, the each motor winding would see 120V - 4.8 V = 115.2V. Note, at this point we can calculate the resistance due to the motor windings: 115.2 V / 8 A = 14.4 ohm. To double check the calculations: For a single winding of 14.4 ohm, the resistance of two winding in parallel will be 1/2 that of one (i.e. 7.2 ohm). The 120 V circuit "sees" a total resistance of 0.3 ohm + 7.2 ohm = 7.5 ohm; 120 V / 7.5 ohm = 16A (as we originally supposed). In terms of power, the wiring will use 16A * 4.8V = 76.8 Watts. The motor will use 16A * 115.2 V = 1843.2 Watts. Total power is 76.8 W + 1843.2 W = 1920 W, in agrrement with 16 A * 120 V = 1920 W for the total circuit.
Now consider the same motor configured for 240V operation: Now the two winding are in series with each other and the wiring, so doubling the voltage does not exactly halve the current (since the total circuit resistance changes). The total circuit resistance will be 0.3 ohm (assuming same run of wiring is used) + 28.8 ohm (two 14.4 ohm windings in series) = 29.1 ohm. The current will be 240 V / 29.1 ohm = 8.25 A. Now we can calculate the voltage drops: for the wiring we have 0.3 ohm * 8.25 A 2.48V (close to 1/2 the 4.8 V drop for the 120 V configuration); 14.4 ohm * 8.25 A = 118.8 V for EACH winding. The winding together in series give a drop of 2 * 118.8 V = 237.6 V; total circuit voltage drop of 2.475V + 237.6V = 240.075V (slightly off from 240V due to rounding of current to 8.25A). So, in the 240 V configuration, each winding sees a little more voltage AND a little more current, so therefore the motor gets a little more power. We can also calculate the power involved: the wiring uses 8.25A * 2.475 V 20.4 W and the motor will use 8.25 A * 237.6 V = 1960.2 W, for a combine total of 20.4 W + 1960.2 W = 1980.6 W.
To summarize: 120 V configuration: 1920 Watts total, 76.8 Watts for the wiring and 1843.2 Watts for the motor 240 V configuration: 1980.6 Watts total, 20.4 W for the wiring and 1960.2 Watts for the motor
Compared to the 120V configuration, when wired for 240 V the motor gets 1960.2 W - 1843.2 W = 117 W more power, or a (117/1843.2)*100 = 6.3% increase.
Granted, I've simplified the analysis by only considering the resistive loads (i.e. I've implicitly assumed that the reactance due to the motor's inductance has been balances out by the reactance of the capacitor). Also, I realize that motors are more complicated than I have described them; for example the amount of current they draw depends on the level of mechnical resistance they are working against (e.g. freely spinning saw blade versus cutting 2 inch thick maple). However, I hope I've shed some light on the an aspect of the 120V versus 240V debate in order to show that it is not as simple as "double the voltage, halve the current" response that is given so often.
Disclaimer: I am not an electrician, but I do have a background with some things electrical.
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1843.2
Interesting. I'm not sure about the "simplified" part though ;) Ed
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Because the current is less, and it's current that generates heat, not voltage.

To a first order approximation, yes.

Why would you think that?
-- Regards, Doug Miller (alphageek-at-milmac-dot-com)
Get a copy of my NEW AND IMPROVED TrollFilter for NewsProxy/Nfilter by sending email to autoresponder at filterinfo-at-milmac-dot-com You must use your REAL email address to get a response.
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Baffling. But I know people who swear that switching to 220 doubles a motor's power, too.
Charlie Self "The test and the use of man's education is that he finds pleasure in the exercise of his mind." Jacques Barzun
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The power is the same I (current) x E (voltage) 110V x 20 amps = 2200 watts 220V x 10 amps = 2200 watts. Period! There are other loss that come into play (power factor) but lets leave that stuff alone! Also, You are "using" the same current in the motor (at 110 Vs 220) but since the voltage drop doing to the motor increases (less voltage at the motor), it draws more current to make up for the power loss (2200 watts)! So it works harder and the power lines get hotter (that where the loss is going.)
Frank
Charlie Self wrote:

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leave that stuff

but since the

motor), it draws

harder and

This is almost correct, but your wording has me confused about what exactly you meant to say.
At 120V, when the current is approximately 2x, the voltage drop is greater in the feeder cable, so there's less voltage across the motor. That equates to less current per winding (they're in series here).
At 240V, since the current is roughly half, there's less voltage drop in the feeder, so more voltage is allowed across the motor, which will slightly increase the current per winding (which are now paralleled).
Of course, these variables all change with each incremental bit of developed power. So it's kinda moot.
And (to mention what others have alluded), it gets more detailed when you start considering the counter-EMF, mutual inductance, load rate of change, and power factor.
If you have the means to use 240V, just do it. It's advantageous.
Hope this helps. John Sellers
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John, series on 240, parallel on 120. When you have sag and current goes up, it's up in the feed as well as in the motor, so you actually do consume more power from the box, and a little more in the motor (the IR heating).
Wilson

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Whoops, got the motor connections backwards. And that was after I checked and swapped them - twice.
I agree about the increased power draw when the motor load increases; and so do the parasitic losses. But, that's why I said it's really a moot issue. There's no such thing as constant load with a woodworking machine. So, it doesn't make much sense to debate the issue of small current changes. Bottom line - 240V makes for easier use of the power, with less wasted energy.
Sorry for the confusion. John Sellers

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Oh, really?
Which generates more heat: 1) a 60-watt lightbulb (designed for 120V) running at 120v 2) a 60-watt lightbulb (designed for 240V) running at 240v
Note that #1 is drawing 1/2 amp, and #2 is drawing only 1/4 amp.
"Watts is watts", applies -- It doesn't matter how they're produced.
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Robert Bonomi wrote:

I agree that "Watts is watts" but wouldn't the temperature of the conductor(s) come into play since the resistance of a wire increases as the temperature rises?
-- Jack Novak Buffalo, NY - USA (Remove "SPAM" from email address to reply)
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"Could be... Could be", says he. <grin>
The light-bulb design already takes that temperature rise into consideration. The filament is designed to have the 'proper' resistance _at_operating_temp_.
Which is why incandescent bulbs _almost_ _always_ fail when they are turned on. The initial, or 'inrush' current is _many_, *MANY*, times higher than the 'operating current'. The 'cold' resistance of a 100 watt light-bulb is typically in the _low_ single digits.
As a point of engineering detail, the operating temperature of both bulbs will be fairly close to the same value. Incandescent bulbs of the same wattage are _amazingly_ close to each other in the 'color' of the light generated. which is _directly_ related to the temperature of the filament. A white light 'color temperature' difference of as little as 100 degrees C is easily detected by someone who is looking for it. Typical halogen white light color temperatures are around 4500 degrees. The 'white' on a color CRT is frequently in the mid 6000's. True daylight, if i recall correctly, is around 9500 degrees. Tungsten-filament incandescent, poor things, are down around 3000 degrees.
*IF* the filament in both bulbs is made of the same material, the 240v bulb has a filament that is longer and thinner than the 120V one.
Frequently, however, the higher voltage bulbs are made with a somewhat _different_ (higher resistance) composition of material for the filament. Allowing the filament construction to be thicker than the 120V 'counterpart'. This improves the mechanical stability, and the ability to withstand shock and/or vibration.
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snipped-for-privacy@host122.r-bonomi.com (Robert Bonomi) wrote:
All incandescent lamps use tungsten filaments. There are some trace elements added that tend to prolong life but that material is not altered for different voltages.
Halogen lampsd have one of the halogen gasses in the envelope. This creates a "halogen cycle" where the evaporated tungsten combines with a halogen gas molecule and then breaks down when the combined molecule strikes the filament again. Thus the lamp can operate at a higher temperature before failure.
Dick

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Robert Bonomi wrote:

I wasn't referring to the resistance of the filament but rather the conductors feeding a motor being operated at 110V verses 220V. The amperage being double when the equipment is operated at 110V would cause the conductors to generate more heat decreasing the overall efficiency. Granted, with a short run the difference would be minimal.
-- Jack Novak Buffalo, NY - USA (Remove "SPAM" from email address to reply)
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snipped-for-privacy@host122.r-bonomi.com (Robert Bonomi) wrote:
Keep in mind that the light bulb example is totally different than the saw. In the light bulb, the watts go into heating the filament. A 60 watt bulb is just that regardless of the voltage/current relationship to create the power.
In the case of the saw, most of the energy goes into cutting wood. Only a small amount goes into heating the motor. The motor heating goes as I^2*R. As the lower voltage requires a higher current, there is more heating in the lower voltage situation.
Dick

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This would be true if you didn't rewire the motor from a parallel connection to a series connection when going from 110 to 220. Current, therefore I^2R losses, should be the same. Most small motors, however don't change the connection to the start winding so it DOES carry twice the current during the brief period it's active.
rhg
Richard Cline wrote:

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snipped-for-privacy@host122.r-bonomi.com says...

Point taken; just the same, a dual-voltage motor *does* run cooler at 240V than it does at 120V.
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