DIY Home Improvement Forum banner

Do you think a tool running at 240v works better than at 120v?

9K views 39 replies 16 participants last post by  micromind 
#1 ·
My brother and I had an interesting conversation, I purchased an old DeWalt radial arm saw wired for 240v and on the way home we got into discussing it can be wired/run at 120v as well. The motor says wired at 240v it draws 6.5 amps, if wired for 120v it draws 13 amps. The saw is wired for 240v from the factory.

On the long ride home came the discussion will the motor run better at 240v or 120v? Person A said 13 amps at 120v is the same as 6.5 amps at 240v they're the same and the motor will run the same. Person B said using 240v you have 2 opposite legs with push & pull and the motor runs better, why else does it come wired for 240v from the factory?

What's your thoughts or can you explain how different the motor will run wired either way? Thanks
 
#8 ·
It won't make a noticeable difference. It might not make any difference at all. Power out = efficiency x power in.

Correction; I-squared-R loss will be 4x greater at 120v, so the efficiency is better at 240v. This is why cross-country xmission lines run at very high voltages.

The reason that high voltage is used for transmission lines is because the higher the voltage the lower the amperage the smaller the wire needed.
 
#5 ·
I don't know with technical certainty, but I don't believe it matters. I think it just gives you a second wiring option for machines that use up a lot of juice and won't fit on a 120v 15 or 20A circuit. So instead of having to wire a 120v 30A circuit, you can do a 240v 15A.
 
#6 · (Edited)
His post said:

Requires 13 amps at 120 volts, so for him a 15 amp circuit would work 14 ga, 20 amp 12 ga, is better. It's about wire size. It's cheaper to run smaller gauge wires than heavier gauge. I will always use the 220 v if given the option. Some tools require it, usually 3 hp or greater and have the thermal safety switch/voltage drop shut down. BUT... 220v breakers cost more and take up more slots in the panel. So, maybe it's a toss up on the cost?:huh: bill
 
#7 ·
Motor wil perform the same both ways.

13 amps is a good size draw and will require larger supply wiring then the 240 to run properly.

If it is to be used in the field, the 120 option would make more sense but you would need to use at least 12 gauge cord and try to plug into 20 amp circuits to be safe.
 
#10 ·
Being a guy who owns a table saw that can be wired for 120 or 240, I can say that it definitely works better on 240. It seems to get up to speed faster, and it runs cooler.

Also, I would think that since the current drawn at 240 V is less, there would be less self-inductance in the windings, which may be why it is able to start faster. But that's just speculation on my part.

And by the way, 120 V "pushes and pulls" with a hot and neutral just like 240 V does with two hots. Current flows to AND FROM a neutral just like any hot wire. The idea that the neutral is some kind of dead, limp stand-around is one of the most wide spread misconception about AC.
 
#20 ·
Large commercial buildings use 347 volts for the flourescent lighting, for the simple reason that more fixtures can be run from a #14 conductor fed from a 15 amp breaker, than those powered by 120 volts. This means major savings in wire and conduit sizes.
Horse power has a direct relationship with wattage. So if you can achieve the same wattage on small gauge wire by using a higher voltage, less power is lost in the supply conductors and more is available to the load.
This would be of greater advantage on a long feeder run vs. a short run.
 
#21 ·
The heat loss in a motor (part of which is called the copper loss) represents a loss of energy available to perform useful work. The copper loss is due to the resistance of the windings, and is computed as IxIxR, where I is the current flow through the windings, and R is the resistance of the windings. At 120 volts, the current flow through the windings is twice as much as the current flow at 240 volts. The resistance of the windings is essentially independent of the voltage, at least at relatively low voltage like 120 volts. Therefore, the copper loss is 4 times as great at 120 volts as at 240 volts, representing less energy available to drive the tool, meaning lower performance and higher cost to run the tool when comparing 240 volts versus 120 volts.

The main reason small motors are built to run at 120 volts is because it costs more to insulate the motor windings at higher voltage, and 240 volts are not always available for small tools like a drill or a saw. The same tool would work better at higher voltage, however only in commercial establishments are you likely to encounter 480 volts and higher. The tool will run better using three phase power than single phase, for different reasons than the copper loss, but three phase power is rarely available in residential settings, hence you don't see three phase motors very often outside of commercial settings.
 
#23 ·
Y'all think about ohms law for a minute.
The nameplate says 6.5 amps at 240 volts and 13 amps at 120 volts. Did the copper resistance of the windings change by a factor of 4? :no:

Of course not. You're paralleling the windings at 120 volts and seriesing (however you spell that) them at 240 volts. Each winding sees the same voltage and current in either case. Therefore the losses in the motor itself are the same in either case. From an infinite buss, there is no difference.

In the real world, I prefer 240 volts. At startup, the inrush will be less at 240 than 120 causing less voltage drop in the circuit which in turn (as noted previously) gets you a quicker start with less flicker in the rest of the house.
Secondly, you get the same advantage when you hit that huge knothole that bogs the saw down. Obviously the advantages of this vary with the size and length of the circuit.

As for line losses while it's running, the more current, the more line losses....but at this level it's fairly negligible.

As to the transmission line losses, all the above arguments are true (unless I missed a weird one). For a given size conductor, doubling the voltage halves the current and causes line losses to fall by a factor of four (not 2). Sure, you can put up bigger wire, but bigger wire means stronger towers, more towers (closer spacing due to weight), etc. It is all cost driven.
 
#24 ·
Y'all think about ohms law for a minute.
The nameplate says 6.5 amps at 240 volts and 13 amps at 120 volts. Did the copper resistance of the windings change by a factor of 4? :no:

While this is true, there must be some other factor involved, because I KNOW my saw works better at 240 than at 120. I'm thinking it must be some self inductance. Or it may just be the voltage sag on the line is less at start up or under heavy load, thereby allowing enough current in to compensate for the load.

Either way, I have done the test to see, and it seems pretty clear to me that my saw works better at 240.
 
#26 ·
Well, the answer is that the tool will run more efficiently at a higher voltage - and dual voltage motors will perform better at higher voltages.

However, you'd be hard pushed on a lot of tools to notice any real difference in real-world performance.

Here in the UK, a lot of professionals use an inverter to switch down the voltage from 240v to 110v and run their power tools at 110v. The benefit is the additional safety from running electrical systems at 110v rather than the higher voltage.
 
#27 ·
While the resistance of the motor is changed by a factor of 4
R=E/I 120/13= 9.23 ohms 240/6.5= 36.92
The copper loss or IIR is the same
IIR 13 X13 X 9.23=1559.87 6.5 X 6.5 X 36.92 = 1559.87
The wattage of the motor is the same so the same amount of heat is created and the same amount of work is produced.

14 ga wire has a resistance of 2.525 per 1000 feet and 12 ga wire has a resistance of 1.588 per 1000 feet the difference in resistance will equalize the cable loss for the 2 voltages. So there is virtually no difference in operation. A 1/2 HP motor produces 1/2 Hp on either 120 v or 240 volt.
 
#32 ·
All I can add is this: My Delta tablesaw motor is placarded to read: 1 1/2 hp at 110V and 2 hp at 220V. It runs better than twice as good on 220, spins up way faster, doesnt even come close to bogging down like it does on 110V. I never bothered to ask it why it does this I just really appreciate it :yes::yes::whistling2:
 
#29 ·
Not to beat a dead horse, but there appears to be some misunderstanding about losses in a motor. The effective resistance of a motor may be computed as V/I = R, where V is the RMS voltage, and I is the current. However, that is NOT the resistance of the windings, it is the total effective resistance of the motor including windings, stator, eddy currents, and back emf of the motor. You do not compute IxIxR losses based on the total current, since most of the current actually performs work (i.e. driving the tool).

The copper losses are computed based on the resistance of the windings themselves, which are always much less than the total effective resistance of the motor. The resistance of the windings is due to the resistance of the copper wire in the windings, which can be minimized through a variety of techniques, such as using larger diameter wire, using hollow wire for AC motors (check out skin effect for an explanation of why hollow wire is useful), and proper insulation of the wire. However, for a GIVEN MOTOR, the resistance of the windings (NOT THE RESISTANCE OF THE MOTOR) is essentially fixed, and is independent of the voltage applied.

So, the IxIxR copper losses in the winding are only a function of the current flow I. If you double the current, as you do if you run at 120 volts versus 240 volts, you multiply the winding losses by 4 times, and you decrease the power output of the tool. The confusion seems to be in using the effective resistance of the motor INSTEAD of the winding resistance to compute losses in the winding.

The total available power to the motor is VxI, so the total theoretical power available to the tool is the same whether you run at 120 volts or 240 volts. However, the tool only utilizes a percentage of the available power to produce work. Typically, the efficiency of the tool is somewhere around 80 to 85 percent. You measure efficiency by taking the actual shaft power produced by the tool (measure with dynamometer) and dividing by the power available at the input wires (VxI). Tool losses include motor efficiency loss (copper losses, eddy current) and mechanical losses in the tool (friction).

If you do the analysis carefully, you will find that the electrical losses are greater at 120 volts than at 240 volts, hence the tool will have more power at the shaft at 240 volts than at 120 volts. The exact difference depends on a variety of factors, ONE OF WHICH is the resistance of the windings.
 
#30 ·
So, the IxIxR copper losses in the winding are only a function of the current flow I. If you double the current, as you do if you run at 120 volts versus 240 volts, you multiply the winding losses by 4 times, and you decrease the power output of the tool. The confusion seems to be in using the effective resistance of the motor INSTEAD of the winding resistance to compute losses in the winding.
I disagree....not in your statement about current vs. losses, but in the fact that the current through the windings is not changing.
There are two windings in the motor. Each is designed for a specific current range and voltage, in this case 120 volts.
When you change the taps at the junction box, you are simply either connecting these windings in series (i.e., 240 volt input, but 120 volts per winding as described in Kerchoffs Law), or in parallel for 120 volts(each winding still sees 120 volts and the same current per winding.) Sure, the line current feeding the motor changes, but the windings of the motor see the same voltage and the same current PER WINDING in either configuration. If that stays the same, the losses stay the same.

The only practical thing you are impacting is how the circuit feeding the motor reacts to additional voltage drop, and how the motor reacts to that voltage drop. With that said, the 240 volt hookup minimizes the voltage drop through lower line current. Ergo (it's not often I get to use "ergo") the 240 volt hookup is better, IMHO. Marginally, perhaps, but still better.

If going 240 volts means re-wiring the entire circuit, go 120. It's probably not going to make or break your saw either way unless you have an unusually long circuit.

Hook it up 240 and enjoy your saw.
Hook it up 120 and enjoy your saw.
 
#31 ·
Mea culpa

After considerable review, I have concluded that WFO is exactly correct, and I am completely incorrect about current loss in a dual voltage tool. As WFO explains, virtually all dual voltage motors are designed with two coils, which can either be connected in series or in parallel. As such, WFO is exactly correct, in that the current in each coil is identical regardless of whether the tool is operating at 120 volt (low voltage) or 240 volts (high voltage). My mistake was in assuming that tools were built with a single winding that could be operated at dual voltages, which is apparently not the case for conventional 120/240 tools.

I am curious, however, if there are any tools built to run on variable voltage with a single winding. It would appear that this would offer the opportunity to control the speed of the tool, i.e. with a variable speed router or drill. If that is a technique for controlling speed, it would seem as though those type of tools would have lower copper losses at high voltage (full speed) than at low voltage (low speed). Anyone have any ideas on this?
 
#39 ·
....WFO is exactly correct.....
To me, this proves that Daniel has one of the most brilliant minds on this forum.:thumbsup:
However, I have shown this to my wife and she is skeptical. Perhaps she remembers the time I took on a bottle of tequila, and bottle of syrup of ipacac, and a toilet bowl lid and went 0 for 3. :icon_cry:


You (micromind) must have one humongous Variac.:eek:
...now she wants to meet micromind....... :huh:
 
#34 ·
As stated above, a dual-voltage motor is simply two motors (so to speak) contained in the same frame. Lower voltage = higher current, higher voltage = lower current. Watts and HP is the same at either voltage.

Suppose we had a 2 HP dual-voltage motor that was driving a fan or centrifugal pump. East to start, and no chance of overload. During operation, I doubt if you could measure any difference between 115 volts and 230 volts. Input watts, shaft RPM, efficiency, winding temperature, etc. would all be very close.

If the same motor was used to power a table saw, it'd be a different story. A table saw, by its nature, is frequently overloaded. Sometimes to the point of stalling.

A typical induction motor has a breakdown torque rating of about 250% of full-load torque. Breakdown torque is the maximum the motor can produce before a rapid reduction in speed. Anyone who has ever ripped 2Xs on a table saw knows right where this occurs!

This 250% of full-load torque is accompanied by about 300% of full-load current. This is where the difference is noticeable. Our 2 HP motor, that is now producing around 5 HP, is drawing about 35 amps on 230 volts, or about 70 amps at 115. Reduced shaft torque (also known as HP) is more a function of percentage of voltage reduction that anything else. Therefore, at the lower voltage we're causing a greater number of volts lost (because of the higher current), and an even greater percentage of loss (because of basic math). This holds true on starting as well, not just overload.

Unless the wire is grossly oversized and a short distance from the source, the difference will be noticeable. Longer lengths of normal sized wire will have an even more pronounced effect on starting and overload performance.

If I have some spare time, I'll rig up a torque meter to a dual-voltage motor, and power it with a variac (to maintain an exact voltage at the motor terminals), and read torque vs. current at constant voltage and varying loads. Should be interesting!

Rob
 
#36 ·
Sorry, I tend to get a bit technical at times. Maybe it's because I thoroughly enjoy winning arguments with electrical engineers!

I suspect the test will reveal about the same performance at either voltage (because it's laboratory conditions), but in real life you'll find better performance, cooler operating temperatures, and longer life at the higher voltage.

Rob
 
#40 ·
OOPS! Maybe I should explain!

A variac is a variable voltage transformer. There's a dial, usually on top, that you can set to a voltage and it'll maintain it. Provided the line voltage is stable. It's about the only way to vary an AC voltage.

I have two. Both are 240 in, max 260 out. The small one is 2 KVA, single phase. The larger one is 9 KVA, 3 phase.

Rob

P.S. Which one ya thing she'd like to see first?!??
 
This is an older thread, you may not receive a response, and could be reviving an old thread. Please consider creating a new thread.
Top