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"Powered with a suction of 18 volts"

Isn't this just a nonsensical "bigger is better" measure?

The thing you really care about as a customer is the motor's torque or power or RPM or something like that.

Is there any direct relationship between supply voltage and one of these measures of motor performance or battery life? Doesn't seem like it to me, since every motor is different, could have a different number of windings, different number of coils, etc.

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closed as not constructive by Kellenjb, tyblu, clabacchio, Kortuk Jun 20 '12 at 22:52

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ It's probably more to give some idea of what replacement batteries are required. Volts by themselves are not much of a performance indicator, except that lower voltage causes higher current at the same power level. High currents cause various losses and therefore inefficiencies. \$\endgroup\$ – Olin Lathrop Jun 17 '12 at 17:10
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    \$\begingroup\$ @Olin - Replacement could be the reason sometimes, but a typical case here are dustbusters, whose batteries aren't user-replaceable. So there mus be other, obscure, reasons too. \$\endgroup\$ – stevenvh Jun 17 '12 at 17:41
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    \$\begingroup\$ I think it has to do with the fact that it is easier for a 12V battery to deliver n watts of power than for a 4V8 battery. Losses in regulators/switches/cables. (and of course the reality marketing people wants us to live in). \$\endgroup\$ – jippie Jun 17 '12 at 17:44
  • \$\begingroup\$ This seems like a consumer electronics/marketing question. No one here will have any first-hand knowledge about why this is done, it can't be researched or tested.... I think this is not a good question for this site. However, it was asked by a high-rep user, and it has 7 upvotes. What's going on? \$\endgroup\$ – Kevin Vermeer Jun 18 '12 at 13:55
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    \$\begingroup\$ "Is there a relationship between supply voltage and motor performance?" is not a "consumer/marketing question". \$\endgroup\$ – endolith Jun 22 '12 at 13:47
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For motors Power is proportional to torque times the rotational speed. So for a given rotational speed and torque the device produces a given amount of power.

To increase the amount of power two options exist. Generate the same amount of torque at a higher speed, or increase torque at a given speed.

For a cordless drill, the speed is normally variable and depends on the application. For instance high speed for steel, lower speed for masonry, and lower speed again for wide hole "auger" bits in timber.

Ok so to increase the power of a cordless drill you will not change the speed, as the drill needs to deliver power at a variety of speeds.

Two other factors to consider, in a DC motor, voltage is proportional to speed and current proportional to torque.

But all the designers are doing is increases the pack voltage. for a given coil resistance in the DC motor, increasing the voltage across the coil also increases the current, thus the torque delivered.

So increasing the voltage is a way that designers can increase the torque, thus power the end users can use!. So more volts the better! upto a point as more volts means more cells, and more cells means more weight, more weight means more user fatigue. So these tend to balance out, at the moment anywhere from 14.4 V DC to 18 V DC for a typical cordless drill.

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    \$\begingroup\$ This is a good answer, but "for a given coil resistance ... increasing the voltage ... also increases ... the torque delivered". So you really can't compare one manufacturer to another based on voltage specs, because the coil resistance could be different? \$\endgroup\$ – endolith Jun 24 '12 at 15:13
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It is nonsensical, and it doesn't say anything about the tool's power. You would think they use voltage because it has more impressive numbers, but I've seen devices (dustbusters) which mentioned in large digits "2.4 V", so that doesn't look impressive at all. The only other reason I can think of is that people may be more familiar with the word "volt" than with the word "watt" (not implying that they would know what either means).

edit
I think a number of answers is outside the question. The question as it was asked is "Why are they specified in volts?", not why they use high voltages. That has been covered in the past in at least one question (which I can't find for the moment). What this question is about IMO is this:

enter image description here

and that's nonsensical! It tells you nothing about the dustbuster's capabilities. Mine says proudly "2.4 V", and I can't believe this one has 9 times the suction power of mine. It would be capable of creating black holes if it were. Mine was cheap, and IMO Black&Decker released it to have a reference for their other dustbusters. A 3.6 V is better than a 2.4 V, so we can ask a higher price for it. Those marketing guys are no idiots. Ask Jane Doe which one is the most powerful and she'll say the one with the highest voltage. Wanna bet?

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  • \$\begingroup\$ If being more familiar with the word, why not use 'break horse power'? ;o) \$\endgroup\$ – jippie Jun 17 '12 at 17:41
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    \$\begingroup\$ But light bulbs use watts, so people must be fairly used to that unit. \$\endgroup\$ – Rocketmagnet Jun 17 '12 at 18:55
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    \$\begingroup\$ It is by no means nonsensical. For reasons such as most of the parasitic power losses being current-proportional, high-power versions of battery powered devices (be they R/C aircraft, drill/drivers, or portable computers) tend to use a larger number of moderate capacity cells, rather than a smaller number of extra large capacity cells. \$\endgroup\$ – Chris Stratton Jun 17 '12 at 19:04
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    \$\begingroup\$ @Chris - (Is the downvote yours?) I agree with what you say, but IMO that's not what's being asked. See the edit to my answer. \$\endgroup\$ – stevenvh Jun 18 '12 at 12:42
  • \$\begingroup\$ @ChrisStratton: For a motor to operate usefully at a higher voltage, it will generally need to have either stronger magnets, or more turns on its coils, than would be needed at lower voltage. Adding more turns would increase parasitic loss per unit current, but using better-quality magnets would not. Further, even adding turns is something manufacturers would generally not do unless it was necessary to achieve a desired level of product performance. \$\endgroup\$ – supercat Jun 18 '12 at 17:23
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Increasing battery voltage in portable appliances is partially driven by practicality and partially by marketing, but in the last decade or so marketing has definitely been the major factor.

A "powerful" battery powered appliance (drills probably being the most common but not the most power intensive) may have a power rating of 100's of Watts.

Take 100 Watts as an example:
At 100 Watts 12V ~= 8A, 16V ~= 6A, 24V ~= 4A, 36V ~= 3A.
Losses in wiring and connections is mainly due to heat loss = I^R.
For the same resistance losses for 12/16/24/36 volts would be in the proportions
64/36/16/9 so a 36V system may notionally have 9/64 ~= 14% of the losses of a 12V system.
So in practice, as current goes down with increasing voltage, you get less losses with the same resistance or can tolerate somewhat more resistance and still be well ahead.

In a 12V 8A system a one ohm circuit resistance will dissipate I^@R = 8^2 x 1 = 64 Watts - so as that's 64% of total power it would be intolerable. Something more like 0.1 Ohm = 6.4% would be better. It is exceedingly easy to add 0.1 Ohms in wiring and connections, so a 100W 12V system gets annoyingly difficult to build. Even an 18V system with 2/3 the current = 4/9 = 44% of the losses is usefully better.

HOWEVER more voltage requires more battery cells and the room required for interconnects, extra loss in connections and the loss of effective available volume due to square-cubed law effects* means that above a certain Voltage the extra losses start to offset the gains. Marketing doesn't care and the engineers and marketers will have had a behind the scenes stoush to arrive at the final result.

A factor that makes higher voltages easier is the use of LiIon cells. These have a nominal voltage of say 3.6V/cell which is about 3 times that of NiCd or NimH so a 10 cell NimH battery will be 12V nominal but a 10 cell LiIon of the same size will be 36V nominal.

Top grade/quality/cost power tools such as De Walt (Black & Decker in disguise) use LiFePO4 (Lithium Ferro Phosphate) cells in some products with a 3.2V nominal voltage per cell. 10 would give 32 V nominal and this will be "almost sensible" in some applications.
An aside: I understand that De Walt use the industry leading A123 LiFePO4 cells. A123 cells are generally "hard to buy" on the retail market and I have heard of electric vehicle makers buying large numbers of De Walt battery packs to get the cells.


Square-cubed law:

Effects caused by changes in the ratio of area to volume as scale changes.
Volumes are proportional to edge^3.
Surface areas are proportional to egde^2.
so the ratio of volume to edge is proportional to edge^3/edge^2 = edge - which means that volume per surface area increases as objects get bigger.

Secondary effects of this are eg it's harder to cool big things by surface radiation.
Conversely, it's harder to keep small things warm when it's cold.
For a given surface thickness big things have less content per volume.
The latter effect affects batteries.
if a battery can be built with ABOUT the same wall thickness across a range of sizes then big batteries will have more active content per volume than small ones.

One only example.
Two cubes with 1mm thick walls and edges of 1cm and 4 cm.
Wall volumes = 6 x edge^3 x 1mm
Cube total volume = edge ^2
Inner cube inside walls volume ~~= (edge- 2 x wall_thickness)^3

1 cm cube inner/outer volume = (10-2)^3/10^3 = 512/1000 mm^2 = 51%
4cm cube inner/outer = (40-2)^3/40^3 = 54872/64000 = 85%. !!!
The 4 x larger edge cube is 85/51 = 1.59 x more effective a user of available volume than the small one.
Conclusion: High voltage battery packs that use NimH or NiCd may be a bad idea for this reason alone. There are others.

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  • \$\begingroup\$ I wonder if drill makers originally went from 12V to 18V simply because they had to keep mechanical compatibility with the part of the battery pack that goes inside the handle, so they added more of the same size cells to the part of the battery pack that's outside the hand grip. Relatively simple change, component volume went up & per-unit cost down, and a drill could mechanically accept both packs. Then marketing figured it was the simplest way to advertise "a difference worth paying for" without starting a war over which mechanical specs (torque etc) matter the most. \$\endgroup\$ – Matt B. Jun 29 '12 at 19:57
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Horsepower would be a smaller number, and the marketing department would prefer to use a bigger number than the competition, but not a big wattage number, since big wattages do not market as "Green" enough to some market segments. For equivalent technologies and battery pack cost, higher voltages might be more efficient or incur less wiring and conversion losses in producing usable tool power.

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The numbers applied to the exterior of power tools are for "bigger is better" purposes and likely also to differentiate various technologies the manufacturer uses.

In other words, purely for marketing purposes.

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That measure is indeed nonsensical. There're two reasons why it is used

  1. it allows for marketing comparison of tools of the same manufacturer with different voltages (see this related question) - like you pay this amount of money for this "crappier" 10,8 volts model or you can pay extra and get that "much more powerful and better" 18 volts model

  2. most tools have swappable batteries and only tools with the same voltage can use batteries of that same voltage - so if you have a 12 volts driver and you want to buy a saw of the same manufacturer then if you buy a 12 volts model you now have a total of 4 batteries (and 2 chargers!) that can be used with both tools and that's good because a separate battery or a separate charger costs a fortune

Other than that voltage is not a useful measure. You don't care of the voltage, you care of mechanical characteristics.

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