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I'm planning on using a 500 W DC motor powered by 2 x 12 V batteries in series. But I want to reduce the power even more by adding 2 Ohm resistance so the power will be 288 W instead.

However, adding a small 2 Ohm resistor will ignite it instantaneously and burn down my house in the process (Thanks to Murphy).

So I was wondering:

  • Is there a resistor which can handle those 288 W?
  • Is there a potentiometer which can aswell?
  • If no, what would be the right way to go about this? Buy a ~300 W motor instead?

Background

The motor will power a bicycle and I'm having troubles finding a suitable DC motor for this purpose. The only one I found was this 500 W DC scooter motor.

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4 Answers 4

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Is there a potentiometer which can aswell?

That's called a rheostat (basically a power potentiometer for the purpose you described) -- but I'd recommend against it, since they're pricey.

If no, what would be the right way to go about this? Buy a ~300 W motor instead?

Easy -- use pulse-width modulation by using a MOSFET (and freewheeling diode) to apply 24V to the motor some fraction D of the time. Vary D and you will vary voltage to the motor. (Just watch the motor current.)

The MOSFET should be rated at least 40V and a low-enough on-resistance that you don't have to worry about overheating at your maximum current. Here's a BUK9509-40B which is rated at 40V 75A 9milliohm max at 5V Vgs.

(For some more info about the basics of driving MOSFETs from microcontrollers, read this blog entry of mine.)

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  • \$\begingroup\$ Thanks for a good explanation and additional resources. Using PWM to control the motor seems spot on for this project! Is there anything I should look out for when using PWM with a DC-motor? Wouldn't the input to the motor be AC now? \$\endgroup\$
    – rzetterberg
    Feb 12, 2012 at 19:06
  • \$\begingroup\$ Input to the motor has a DC component and an AC component at the PWM frequency + harmonics. The normal PWM frequency is >= 20kHz so you don't hear it; you'll also need to keep the PWM frequency high enough so the ripple current (depends on motor inductance) is relatively small. Otherwise very little difference driving motors from PWM source vs. a linear amplifier. \$\endgroup\$
    – Jason S
    Feb 12, 2012 at 19:15
  • \$\begingroup\$ Ok, thank you! :) \$\endgroup\$
    – rzetterberg
    Feb 12, 2012 at 19:19
  • \$\begingroup\$ I'm learning so much from this! But looking at the Digi-Key info on that MOSFET link, it says max power is 157W. Was this just an example and the asker will need to find a more suitable part or am I misreading/misunderstanding something else? \$\endgroup\$
    – Erik Noren
    Feb 13, 2012 at 20:01
  • \$\begingroup\$ "max power" in a transistor refers to power dissipation, not power transfer. (it's also spec'd at particular circumstances e.g. 25 C case temperature, so may not be valid in your particular case) Calculate the power dissipation and the junction temperature rise above case temperature (delta T = Rjc * Pdiss, Rjc = thermal resistance in datasheet) to check if you're going to be ok. I picked that part as a relatively inexpensive part with what looked like plenty of room to carry the current you need. But it was a quick seat-of-the-pants selection. \$\endgroup\$
    – Jason S
    Feb 13, 2012 at 23:50
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I'm amazed nobody has yet pointed out the main reason for not using a resistor or rheostat. If you do that, you will waste half the power (288W) heating the resistor, which will halve the distance you can travel on your bike. Hence, the way to go is to with pulse modulation as explained by Jason. Of course, the mosfet circuit will be a lot more complex than a simple resistor, but it may not cost any more.

As for using batteries in parallel, that is not a good idea either, as no 2 batteries are ever identical. Their voltage and internal resistance will differ, so one battery will take the brunt of the load.

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  • \$\begingroup\$ Actually it's probably slightly better if they're in parallel than in series. Series operation tends to turn charge imbalance into damage: the battery/cell that is more full than the rest is overcharged during charging, and the battery/cell that is more empty than the rest is overdischarged during discharging. \$\endgroup\$
    – Jason S
    Feb 13, 2012 at 0:37
  • \$\begingroup\$ And in parallel you can always use diodes to isolate batteries... \$\endgroup\$
    – clabacchio
    Feb 29, 2012 at 12:25
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There is a company called arcol (www.arcolresistors.com) that produces resistors that can dissipate 700W.

You could also use a MOSFET with PWM, to act as a constant current sink.

There are other companies that produce them as well. I would suggest using a component supplier and searching for you requirements.

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  • \$\begingroup\$ Thanks for the tip of the company, I will certainly check them out! :) \$\endgroup\$
    – rzetterberg
    Feb 12, 2012 at 19:07
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Perhaps I am missing something, what about using the batteries in parallel?

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  • \$\begingroup\$ I thought about that at first, but then I could not produce the 288 W, it would only produce 144 W at maximum. Or am I missing something? I'm really new to electronics, so bear with me :P I wanted to being able to use the full 576 W, but wanted to start out a bit smaller \$\endgroup\$
    – rzetterberg
    Feb 12, 2012 at 21:12

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