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AFAIK each of the world leading cordless power tools manufacturers produces several "product lines" of cordless power tools with different voltages. For example, Bosch currently produces tools with Li-Ion batteries with 10,7V, 14,4V, 18V and 36V output and the higher the voltage the more powerful a tool is.

Now those tools are powered by batteries that are assembled from cells with lower voltage (something like 3,7 volts for Li-Ion cells I guess) and cells are connected in sequence until the target voltage is reached.

They could instead connect cells in parallel. They would have the same voltage, but higher current and that would again yield higher power.

Why do they choose higher voltage over same voltage and higher current to get higher power in electric tools?

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    \$\begingroup\$ Voted to reopen. @Kortuk, you might not have closed this if Bosch tools weren't mentioned. It's a good question for EE IMO: why cells in series, and not parallel. \$\endgroup\$
    – stevenvh
    Jul 21, 2011 at 11:21
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    \$\begingroup\$ @Kortuk: Consumer electronics application is there as an example only. I can't see why it's not a pure electrical engineering question. \$\endgroup\$
    – sharptooth
    Jul 21, 2011 at 11:25
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    \$\begingroup\$ Shouldn't the process be "wait for enough close votes to close" rather than "wait for enough reopen votes" after forcing it closed? \$\endgroup\$
    – endolith
    Jul 21, 2011 at 16:20
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    \$\begingroup\$ @Kortuk closing something is not taking a vote. Taking a vote would be stating your personal opinion and casting (1) close vote yourself. To say that what you do is "not permanent" misses the point - there's a huge "potential barrier" between something that is open until enough people decide to close it, vs. something that is closed until enough people decide to re-open it. \$\endgroup\$
    – Chris Stratton
    Jul 21, 2011 at 20:40
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    \$\begingroup\$ The wording is wholly different but this is effectively why do you use series batteries over parallel. I will try to find a link later to share. Please feel free to contact me in chat if you would like to discuss this more. \$\endgroup\$
    – Kortuk
    Jul 21, 2011 at 22:18

4 Answers 4

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Matt already explained that using a higher voltage you'll have a lower current for the same power rating. This means thinner and less heavy wires, which means savings (copper is expensive). You may have to pay attention to better insulation, but that doesn't outweigh the advantage mentioned.

It's also much easier to place cells in series than parallel. When placed in parallel the voltages have to be exactly equal, otherwise you'll have high currents running from one cell to the other, causing big power losses and reducing the cell's life.

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  • \$\begingroup\$ +1 another reason, I would think, is the ease of building down converters from a higher voltage. \$\endgroup\$
    – kenny
    Jul 21, 2011 at 13:59
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    \$\begingroup\$ @kenny - compared to buck-boost converters, yes. Compared to boost converters, no. The latter are no more complicated than buck converters. And I think this is older than switchers, too. \$\endgroup\$
    – stevenvh
    Jul 21, 2011 at 14:03
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    \$\begingroup\$ Standard boost converters have one problem that buck, buck-boost, and SEPIC don't: they pass current to the load when the converter is not switching, preventing a simple shutdown function. In other news, all the bold words make it a tad harder to read. ;p \$\endgroup\$
    – Mike DeSimone
    Jul 21, 2011 at 19:00
  • \$\begingroup\$ @Mike - I always use bold words to highlight key words, and I never had any complaints. I may have overdone it here a little. But then, yeah, life can be tough! :-) \$\endgroup\$
    – stevenvh
    Jul 22, 2011 at 4:37
  • \$\begingroup\$ @stevenvh I'm not questioning your answer, because I know from my experience this is true. However, I can't seem to figure out how to demonstrate this mathematically. Could you provide an example? \$\endgroup\$
    – MGZero
    Jul 22, 2011 at 15:10
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TLDR: Higher voltage is needed for higher rpm.

I disagree with copper efficiency argument (above/below)

The only reason is back EMF of motors at high rpm. No matter how much current the batteries can supply, their current translates to torque for motor, but not velocity. At top velocity, theoretical lossless motor has back emf exactly equal to the voltage of supply and consumes current approaching zero while having zero torque.

Higher power tools perhaps have higher rpm, rotational velocity figures, and they need higher voltage.

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  • \$\begingroup\$ I like this answer. If you think about copper then less current = less magnetic force, you make up for this with more turns. I think the total mass of copper stays about the same. Even RPM can be addressed thru gearing, I think most tools gear down. \$\endgroup\$
    – russ_hensel
    Jul 21, 2011 at 15:00
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    \$\begingroup\$ I checked with Bosch Professional Tools site - RPM is not very different for different product lines. The reason is very high RPM is useless - driving screws is plain impossible at very high RPM and drilling produces lots of heat that damages both the material and the drill bit. \$\endgroup\$
    – sharptooth
    Jul 25, 2011 at 13:04
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For exactly the same reason the power companies transmit power around the country at many hundreds of thousands of volts instead of just doing it all at 110/230v.

A higher voltage means a lower current for the same amount of power.

A lower current means smaller components and thinner wires, thus making it cheaper and more efficient.

For example, take a DC motor.

For a 12V motor to develop the same amount of power as a 24V motor it would have to draw twice as much current. This would mean that the windings in the motor would have to be made up of thicker wire. This would increase both size and cost.

The electricity companies transfer power at a high voltage so the current is low so they can use small diameter cables. It's all the same principle.

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The higher is the voltage, the lower is the current, at same power. When you suck lot of current from a battery, you'll create a poorly changed area inside. That leads to a worse performance.

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  • \$\begingroup\$ I can't see why this is a problem. I have many cells in parallel now, so although I drain more from the pack I still drain the same from each cell. \$\endgroup\$
    – sharptooth
    Jul 21, 2011 at 10:30
  • \$\begingroup\$ @sharptooth - placing cells in parallel is not a good idea, see my answer \$\endgroup\$
    – stevenvh
    Jul 21, 2011 at 11:10

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