# Amperage reduction

Discussion in 'General Electronics Discussion' started by stevow, Nov 9, 2011.

1. ### stevow

3
0
Nov 9, 2011
Hi - just trying to figure something out.

I am fairly new to the electronics field and have undergone a project so please bear with me!

In a nutshell, I have two motors (from an electric golf cart) and a 12v 80amp heavy duty battery.

The cart did have a motor controller that has been completely busted and is beyond repair. What I want to do is reduce the amperage coming from the battery to 30amp so I can replace the motor controller with a more generic controller.

I understand I would need a current resistor, can anybody assist in what type/spec?

many thanks!

Last edited: Nov 9, 2011
2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
As long as the motors use less than 30A in total you need to do nothing.

The battery may be capable of 80A (I'd guess it's actually 80Ah -- which is a totally different thing) but it will only supply what is demanded.

3. ### stevow

3
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Nov 9, 2011
Thanks Steve.

If I decided not to go with the controller, and wire a potentiometer and fuse, how would I calculate the best resistor for the pot(presuming it would fry if it was just any pot switch?).

By the way, you was correct in saying 80ah!

Thanks again

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
A potentiometer will not work.

Well, it *might* work, but not very well.

Here's how you calculate the power requirement for the pot.

First you calculate the maximum current the motor will require. If it's a 360W motor running from 12V, then it requires 360/12 = 30A. (you really should provide some sort of fudge factor to this to allow for higher current when stalled, so let's double it -- 60A)

Then you need to decide what value the maximum resistance should be.
This is a little tricky because adding series resistance to a DC motor does not actually slow it down, it simply reduces the torque. And it does this in a disproportionate manner, so it's really hard to start without a jerk.

The motor's resistance is a fraction of an ohm, so the series resistance needs to be small. Let's pull a figure out of the air and say 5 ohms.

Now for the power requirement. The rating of a pot is the rating for the entire element. So, if it is rated for 50W, it can dissipate 50W, but only if the 50W is spread out over the entire resistance element. If you turn it down to half way, it can only dissipate 25W.

What you have to do is to calculate the maximum current the device will have to pass and use this to determine the power rating.

In this case we have calculated 60A and 5 ohms. So the power rating is 60 * 60 * 5 watts, or 1800 Watts.

If you can manage to find a 1800W 5 ohm pot, and you can lift it, you'll probably find your wallet much lighter than it would have been if you had bought a proper controller that would actually work in controlling the speed of your motor.

5. ### davelectronic

1,087
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Dec 13, 2010
Amusing post there Steve, i had to read on, hope the OP can figure it out to work it for his project / intended use.
Dave.
PS, 1800 watt 5 ohm potentiometer, i will keep an eye out for one.

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
It's not really intended to be amusing. (Oh, it might be if you know the punch line).

The method of calculating the rated power requirement for a pot is something that a lot of people don't really understand very well.

There is a reason (actually several) that these controllers exist, and that motors are not controlled by simple variable resistors. I thought it was worth exploring just one of those reasons.

The OP needs to know that the post is not meant to make fun of him or his idea. His misconception is very widely held and it's rare that we adequately explain why.

7. ### davelectronic

1,087
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Dec 13, 2010
Yes i understand the potentiometer value is what i found amusing, and your right a correct explanation on how to arrive at the correct value component for a given voltage and power rating.
Dave.

8. ### stevow

3
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Nov 9, 2011
oh man, this is going to be fun!

Thanks for the reply, and the details. I am fairly new to this side of electronics, so all a good learning curve for me.

Thanks again