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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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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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: 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. Square-cubed law: Effects caused by changes in the ratio of area to volume as scale changes. Secondary effects of this are eg it's harder to cool big things by surface radiation. One only example. 1 cm cube inner/outer volume = (10-2)^3/10^3 = 512/1000 mm^2 = 51% |
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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
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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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
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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