I have a mini motor that's rated at 3 volts. Does this mean that the 3 is the number of volts that will produce the desired amperage for the motor? Could I technically make it work by plugging it into a 10,000 volt circuit and add resistance until it's at the same amperage as it would be for 3 volts?

  • 3
    \$\begingroup\$ If the load torque increases, the motor current increases, and the motor voltage will decrease due to additional voltage drop across the resistor \$\endgroup\$
    – Chu
    Commented Aug 5, 2015 at 11:44
  • \$\begingroup\$ Remember that motor current draw can go from almost infinite (when stalled) to BELOW ZERO (when outside force is applied, the motor will become generator). No constant resistor can have same voltage drop at currents ranging from infinity to zero. \$\endgroup\$
    – Agent_L
    Commented Aug 5, 2015 at 12:41
  • \$\begingroup\$ @Agent_L: the stall current is finite and equal to the voltage applied divided by the winding resistance. Almost infinite is really stretching it. \$\endgroup\$
    – user42875
    Commented Aug 5, 2015 at 14:21
  • \$\begingroup\$ @user42875 yeah, it is stretching - but winding resistance is extremely small compared to regular operating conditions. \$\endgroup\$
    – Agent_L
    Commented Aug 5, 2015 at 15:12
  • 2
    \$\begingroup\$ If you put a resistor between your 10000V supply and your 3V motor, you're no longer feeding 10000V to the motor. \$\endgroup\$ Commented Aug 5, 2015 at 22:35

6 Answers 6


No, motors don't work that way.

Generally speaking, the voltage that you apply to a motor determines the speed at which it runs, but the current that it draws depends on the mechanical load (torque) that it is driving.

If you applied 10,000 V to a 3-V motor, even through a resistor, it would try to spin far too fast, probably destroying itself.

  • \$\begingroup\$ Depends on what it is driving. If the load is something like a fan or propeller which has increasing drag with speed then rpm (and motor voltage) will stabilize at the same point it would have done on fixed voltage at the same current. Without a load the motor will speed up until it reaches the speed and voltage that produces the same no-load current - probably way too high so the motor will over-speed and destroy itself. \$\endgroup\$ Commented Aug 5, 2015 at 20:51
  • \$\begingroup\$ But isn't it the number amps that the motor cares about? I thought that there wouldn't be a difference if I plugged it into a 3V circuit with 10 Ohms of resistance compared to a 10,000V circuit with ~33333V Ohms of resistance, since the number of amps would be the same. \$\endgroup\$ Commented Aug 6, 2015 at 5:39
  • \$\begingroup\$ @Justsomeguy: For motors, the amount of amps it draws depends on the amount of volts you supply to it.. motors don't work like other things. If there's not enough amps the motor will simply stall (stop spinning, though the stopping may be momentary so you may see it as stuttering instead of actually stopping). So what may happen is that the motor will try to spin too fast, stall, try to spin too fast, stall completely and stop and at that point you've basically connected your + and - terminals to a length of copper wire (the motor coil) and whatever is limiting the current will fry \$\endgroup\$
    – slebetman
    Commented Aug 6, 2015 at 5:47

Although in theory that works, it is not practical ... at all.

Let's say your load is 10 Ohms (Making this simple) and it is rated for 10V. This would yield a current of 1A. So let's say I have 10,000V available and want 1A to run through my load, I would add let's say a 10,000 Ohm resistor in series which yields approximately 1A.

Even though this work, now let's talk about power.

$$P = I \cdot V = I^2R$$

For the 10 Ohm resistor, we would dissipate \$P = 1^2 \cdot 10 = 10 \;watts\$. Where in the 10,000 Ohm resistor we would dissipate \$P = 1^2 \cdot 10,000 = 10,000 \;watts\$.

Of course this is not desirable because our power transfer is terrible. We would like maximum power delivered to our load. Therefore we often use a transformer to step down the voltage to an appropriate level.

Of course motors are not resistors (Real and imaginary loads), but the demonstration works for this question. Motors are built with coils, behaving like an inductive load and can't be treated exactly like a resistor. Either way, it is best to deliver maximum power to your load or else you will be wasting valuable energy in your 10,000 Ohm resistor.

Hope that helps, please comment if you have further questions.

  • 1
    \$\begingroup\$ This is actually a rather poor answer. Rather than talking about "maximum power", you should be talking about efficiency. The problem is not that the load is not getting enough power (Pout is too low), but rather that you're wasting a lot of input power in the process (Pin is too high). And besides, motors and resistors are very different, but not in the way you describe -- it has a lot to do with back-EMF and very little to do with inductance. \$\endgroup\$
    – Dave Tweed
    Commented Aug 5, 2015 at 14:01
  • \$\begingroup\$ I am aware that we are interested in efficiency, that is why I mentioned a lot of energy would be wasted in the 10k resistor. I mention the inductance property purely for a basic answer, and just indicating that the motor cannot be treated like a resistor. \$\endgroup\$
    – Josh Jobin
    Commented Aug 5, 2015 at 14:34
  • 2
    \$\begingroup\$ Please don't assume I downvoted your answer, I would not do such a thing. Especially when I do not disagree with your answer. \$\endgroup\$
    – Josh Jobin
    Commented Aug 5, 2015 at 18:40
  • 1
    \$\begingroup\$ OK, I apologize and withdraw the comment. \$\endgroup\$
    – Dave Tweed
    Commented Aug 5, 2015 at 21:28

Short answer: not in practice. Long answer:

Effects of voltage and current

Voltage and current ratings generally protect against independent failure mechanisms, in most cases respectively insulation breakdown (especially for semiconductors), and damage due to excessive Joule heating.

Understanding motors

Some components are more flexible than others around their voltage and current rating, up to what is called "absolute maximum" ratings. DC motors are of that kind. Increase the voltage and the power of the motor will increase (meaning, either more speed for a given torque, or more torque for a given speed) - that is, assuming the motor can draw the current it needs. This means that the current is only limited by limiting the voltage across the motor (this is also true for other devices, it can be seen on current limiting power supplies).

You could run a 3V motor at probably 12V with a giant heatsink to make sure the heat gets away faster than it heats up the wires (otherwise they'll melt); however you cannot run such a motor at 10,000V because 1) the insulation will break, increasing drastically the current drawn by the motor 2) whether the insulation broke or not, the current will be so high the wires will melt. Note that depending on the shaft alignment and the bearings, too high a voltage may generate too much vibration and destroy the motor before the heat melts the coil.

Limiting via power resistor

Now, you suggest adding a device to drop the voltage from 10000V to 3V. That would make the motor happy: as long as it gets 3V across it, it will draw current up to its stall current and hopefully it has been designed to cope with that current. However, you are dropping 9997V. If any current is drawn through the device that drops that voltage, this is power that has to be dissipated somehow. 9997*I, even if I is as low as 50mA is 500W... To put that into perspective, 5W aluminium clad resistors are about as big as a finger. Note that power resistors will only drop a fixed voltage at a given current: change the mechanical load on the motor and the voltage across the motor will change, this will lead to unstability... Assuming a 200k resistor for a design current of 50mA, +/-15uA will have a +/-3V effect on the motor! Not to mention that 500W are wasted for 0.15W used.

Limiting via buck converter

You would need a converter with a better efficiency, like buck DC DC converters. Theoretically, they do not drop the voltage themselves, but switch between full voltage and no voltage to achieve the required voltage in average (in this case, the motor inductance filters the current, which filters the voltage via the motor resistance). However, this means switching ON 0.03% of the time - the motor inductance will give you how long full voltage can be applied for before the current is too high, and that will not be long. Therefore 0.03% of that time will be very short, chances are no converters will be able to. The vibrations of the motor due to the high current ripple would probably ruin the motor and its load anyway.


Whether a power resistor or a switching converter is considered, both will have to have 10,000V insulation, which means your product will be huge to respect the creepage distances and insulators thicknesses.

Unless you mean 10,000VAC from the supply grid to 3VDC? In that case, you will have to use a step down transformer and then a rectifier, but I genuinely hope that you know what you are doing.


Depends on what your goal is. If you want relatively constant speed with varying loads, then a constant voltage source (or even slightly negative series resistance) is best. The motor will be happy unless you load the shaft too heavily, provided you give it no more than the rated voltage. Too heavy a load and it will overheat.

If you want a constant torque with varying loads, then a constant current source is best. The motor will be happy unless you load the shaft too lightly, provided you give it no more than the rated run current. Too light a load and the RPM will wind up and it could damage itself (the motor voltage will exceed the rated voltage under those conditions). A large series resistor behaves more like a constant current source. A >10K source resistance source is easily simulated electronically without the very high voltage and insane resistor.

In reference to your original question, in neither case are you actually exceeding the rated voltage at the motor terminals (the only thing the motor cares about).

  • \$\begingroup\$ What is negative series resistance? \$\endgroup\$ Commented Aug 5, 2015 at 12:20
  • \$\begingroup\$ @ScottSeidman Output voltage goes up slightly as current draw increases, so it behaves like an ideal voltage source in series with negative value resistor. Not a physical negative resistor, of course. The total resistance in the loop should be positive for stability. \$\endgroup\$ Commented Aug 5, 2015 at 12:46
  • \$\begingroup\$ @SpehroPefhany: Motors behave interestingly when the loop resistance goes negative. >:*3 If the net loop resistance is exactly zero, applying a certain amount of torque will cause the motor to produce equal and opposite torque, thus remaining where it is. If it's slightly positive, the motor will produce opposing torque that's not quite sufficient to prevent it from moving. If it's negative, the motor will produce opposing torque which makes it turn in the direction opposite the applied torque, which can be fun as long as the motor is always free to move. \$\endgroup\$
    – supercat
    Commented Aug 5, 2015 at 19:44

Motors can be damaged by exceeding any of the following maxima:

  1. Mechanical force on components thereof (may cause instant failure)

  2. Amount of magnetic flux (may cause instant failure)

  3. Rotational speed, vibration, etc. (may greatly accelerate wear, or in some cases generate forces resulting in instant failure)

  4. Amount of heat generation in the windings (intermittent operation may allow motor to tolerate more power (heat/second) than sustained)

  5. Amount of heat generated in the bearings (intermittent operation may allow motor to tolerate power (heat/second) than sustained).

It may be possible for a "3-volt" motor to be operated safely at higher voltages under some conditions if those conditions reduce the stress factor which was limited operation to 3 volts and do not raise any other stress factor enough to cause trouble. What's important is not keeping the motor voltage below the rating, but ensuring that all of the other stress factors stay within their corresponding limits.

  • \$\begingroup\$ Insulation breakdown is another possibility \$\endgroup\$
    – teambob
    Commented Aug 6, 2015 at 0:37

A motor, electrically speaking, is somewhat like an inductor (ignoring any motor loading effects). Inductors resist any change in current so when you apply a voltage to the inductor (read motor), the current will initially be zero but rise exponentially (see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indtra.html). This results in the full 10,000 volts (assuming you could apply this instantaneously) being applied across the motor windings for a short period of time regardless of what resistor value you use. The high voltage would likely breakdown the winding insulation and result in short circuits across the motor windings.

Practically speaking, a circuit using 10,000 V and a resistor to drive a 3 V motor would be hugely inefficient. The motor current will depend on how much the motor is being loaded. In other words if you selected a resistor that would give 3 V across the motor at no motor load, then added some mechanical load, the voltage across the motor would drop because its trying to draw more current. This would slow the motor down and it would try to draw more current reducing the voltage further and eventually stalling the motor at much less than its rated torque.


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