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If you attach a step-up circuit (like a boost converter) to a battery, and then run a brushed DC motor off of that higher voltage, will that motor have a higher mechanical power output than if you just hooked up the motor to the battery (without the circuit)?

In this post about how voltage and current relate to the torque and speed of a motor, the top answer states that:

For the same motor, ideally if you apply double the voltage you'll double the no-load speed, double the torque, and quadruple the power. This is assuming of course the DC motor doesn't burn up, reach a state which violates this simplistic ideal motor model, etc.

If this method would not work, is there any way to force a motor to pull more current from the battery, increasing power draw (and the energy in the battery would just run out faster) but increasing the mechanical power output of that motor?

I have heard of electric cars using boost converters to power higher voltage motors with lower voltage battery packs. If the answer to the above question is no, then why do these car manufacturers not just use motors that are the same voltage as their battery packs?

Finally, would stepping up the voltage of a battery to run a motor that would be rated for a higher voltage achieve more mechanical power? This was touched on in the post I linked to but not fully explained. Could this same increase in power be achieved with a different motor at the same lower voltage rating (the referenced user mentions stall current as another factor that affects motor electrical power draw and mechanical output)? In short, would stepping up voltage from a battery to run a higher voltage motor ever be useful to increase mechanical power output (ignoring efficiency or battery life)?

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Unfortunately, there is no such thing as ideal.

A motor has a maximum power that you can pump into it. The coil has a resistance and as such, you can only pump so much current through it or the thing will overheat and die.

A small DC motor rated at 12V is designed to withstand either being stalled, or running at some load torque, with the full 12V across the terminals indefinitely without overheating.

As such, doubling the voltage on a DC motor that is already running at full load will overheat the motor.

However, that does not mean you can not over-drive a motor. The trick is to limit the drive current to not exceed the maximum rated current. Notice, that maximum current value also sets the maximum sustainable torque you can achieve with the motor.

IF the 12V motor, when it settles at a speed at 12V, is not drawing all of the rated current, it is possible to increase the voltage, and speed, till that maximum rated current value is being delivered. However that takes some fancy electronics to do. Obviously, if the 12V motor is already drawing the rated current at full load at 12V, you can no longer increase the voltage.

Notice though, you should not just apply that higher voltage from stopped. Doing so will significantly increase the startup current of the motor potentially damaging it.

The drawback with doing that is increasing the voltage also increases arcing on the commutator, this significantly adds heat and reduces the life of the motor. It also voids any warranty you may have had.

As for commercial vehicles, I can not comment. However, they may be boosting the voltage because the motors themselves are rated at double the battery voltage.

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  • \$\begingroup\$ I am not necessarily talking about applying voltage that exceeds the motor's rating, but assuming overheating isn't a problem (say doubling the voltage applied across a 12v motor from 6v to 12v with a boost converter), would this voltage step-up increase power output? \$\endgroup\$ – Murey Tasroc Nov 28 '17 at 4:50
  • \$\begingroup\$ @MureyTasroc if it's a 12V motor and you have a 6V battery then yes, doubling the 6V gets you to the rated point of the motor and is the best thing to do. \$\endgroup\$ – Trevor_G Nov 28 '17 at 4:55
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For a brushed motor by itself, with a good power supply, if you double the voltage supplied to it, you'll roughly double the speed.

If the load is purely viscous, so a fan or propeller load, the torque required to turn the load rises as the speed squared, so the current drawn will rise by a factor of 4, and the power output by 8 times. For a pure friction load, the torque is constant, the current drawn will stay constant, and the power output will double. Real loads are not pure, but a mix of several types.

You can't force a motor to draw more current, other by increasing the load. Usually the load increases with increasing speed, which increases the current the motor draws.

Why voltage boost a battery pack? There are two ways to make (let's say) a fixed 7.4v. One is to have two 3.7v batteries in series. The other is to have the same 3.7v batteries in parallel, driving a boost converter. It's obvious that both schemes have exactly the same batteries, and so the same weight and stored power capacity. However, the boost converter weighs something, takes up space, costs money and wastes power.

So if you can get a suitable load matched to a suitable high voltage battery pack, doing without a boost converter is going to be better on those four counts.

However, if you want to control the motor with a variable voltage, then you are going to need some form of power electronics anyway. You could perhaps use a variable boost from low voltage instead of a variable buck from high voltage. But, a buck converter can control down to zero speed, whereas a boost converter cannot drop the output below the battery supply voltage, which makes buck the usual choice for motor speed control.

If you are selling RC models, and splash WITH POWER BOOSTER across the advertisement, I would expect some fraction of your buying public to say 'WOW' and your sales to increase. Just a thought.

There may be other less cynical reasons to choose a low voltage battery plus booster, which is availability of components. As a manufacturer, you will always be looking to reduce costs. If you suddenly get the opportunity to buy a really good deal on some batteries, but they're low voltage, you might find that it's worth using a booster to use them.

A reason intermediate between the above two might be to add a 'turbo' mode to an existing model quickly, where you need to keep the existing battery box mechanics unchanged.

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  • \$\begingroup\$ Thank you for the answer, I do know that you can wire batteries in series to increase resultant voltage but I'm specifically addressing step-up circuits. \$\endgroup\$ – Murey Tasroc Nov 28 '17 at 5:06
  • \$\begingroup\$ I am aware of your battery voltage limit. I find the contrast between series and parallel connections of like things, batteries, transformer windings etc, to be very useful in thinking about what changes and what stays the same when we try to do a like for like comparison. \$\endgroup\$ – Neil_UK Nov 28 '17 at 5:23
  • \$\begingroup\$ Right, but in your answer, which is purely what I am responding to, concludes that between just putting two batteries in series and using a boost converter, putting the batteries in series is better. I do appreciate you making an effort to answer my question, but if you were aware of my battery voltage limit, although you didn't have to be to know that I am not addressing series connected batteries in this question but step-up circuits, I don't know why you are suggesting I use two batteries in series. \$\endgroup\$ – Murey Tasroc Nov 28 '17 at 5:32
  • \$\begingroup\$ "if you double the voltage supplied to [the motor], you'll roughly double the speed... the torque required to turn the load rises as the speed squared, so the current drawn will rise by a factor of 4, and the power output by 8 times" -- so you claim that increasing the voltage will increase the power output. I think your suggestion of just increasing the prop/load to maximize current draw is the answer I was looking for regarding the competition. Nevertheless, if I cannot alter propeller pitch and length due to design restrictions, will stepping up the voltage also maximize current draw? \$\endgroup\$ – Murey Tasroc Nov 28 '17 at 5:43
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    \$\begingroup\$ You seem to have your thinking backwards. I am trying to get you thinking about physics, energy, efficiency, at the system level. I am not suggesting you put two batteries in series, I am saying contrast series and parallel to make it easier to think about what changes and what stays the same when voltage changes. For an autonomous hovering craft, system weight is at a premium. You can match the battery voltage and motor voltage without a booster if you can choose a suitable motor, but you may not be able to do this and are stuck with a specific motor kv. \$\endgroup\$ – Neil_UK Nov 28 '17 at 8:20

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