What power rating do I need for my potentiometer?
October 29, 2005 1:54 PM   Subscribe

What kind of potentiometer do I need? Electronics and electrical math confuse the hell out of me, so I'm just going to lay it all out.

I want to vary the speed of a 6 volt motor from full forward to full reverse using a potentiometer. If the pot is turned all the way to the left, the motor runs full forward, as it moves to the right it slows to a stop (at the middle of the pot), then starts up again in reverse, getting up to full speed when the pot is fully to the right.

I've got it working close to the way I want using a dual 1K slide pot that's supposed to be used for stereo audio applications. I cut the resistance track down the middle, and have normal and reverse polarity connections on either side, with a connection to the motor being variably resisted. The problem with the pot I'm using now is (a) I have no idea if it's rated correctly to handle the voltage and I don't want it to break down or blow up and (b) the speed of the motor cuts way too fast, and the motor is dead for probably the middle 80% of the pot.

It seems to me the kind of pot I need is one that will do a linear taper of 6 volts from 100% to 0% at the halfway point. Maybe this is the same as a pot that handles 12 volts along the full length? What power rating should the pot I get have? Are there any other specifications I need to consider?
posted by cloeburner to Technology (7 answers total)
Best answer: Voltage isn't the issue, you need to know the current that the motor draws. Measure the current the motor draws when turning under load, multiply that by the voltage to get the watts, then see what wattage the pot is rated for. Chances are you'll need to use a wire-wound pot to get the required wattage rating, which are more expensive, and a little harder to find, so you may need to order it online.

Note also that using a pot at all is the wrong way to control the speed of a motor, especially if using batteries - it will waste a lot of power. Doing it nicely (such as through pulse-width-modification) take several extra components though, and it sounds like you just want it working, and aren't too concerned about that, but I thought I'd mention it :)

Then, yes, use a linear taper dual-pot where the motor is full on at the start point, and as slow as you want it (stopped?) at the halfway point, and cut the tracks like you did.
posted by -harlequin- at 2:06 PM on October 29, 2005

Best answer: Also, to find what the resistance value of the pot you're looking for is, without math, connect the motor to a pot, adjust the pot until it has slowed the motor to the desired slowest speed, then disconnect the motor without adjusting the pot and measure the resistance of the pot with a multimeter, then multiply that value by two (since you'll be cutting the tracks in half) and pick the pot with the nearest resistance value to that, with a suitable wattage rating as already described.
posted by -harlequin- at 2:12 PM on October 29, 2005

Response by poster: Awesome, thanks!
posted by cloeburner at 2:26 PM on October 29, 2005

What can you tell us about the motor you are trying to control? I infer it is a DC permanent magnet type, because you are reversing it by effectively changing the polarity of the armature current. But without current, voltage and rated power specs for the motor, I would just be guessing as to the size and ratings of any pot.

If the motor is rated for 6 VDC operation, it achieves its full rated RPM at that voltage, and generates enough back EMF internally to limit its armature current to safe levels (the mechanical components of the motor can safely dissipate the heat they create, the bearings will hold up, and the motor case won't exceed rated temperature, etc.) Unless a DC motor is specially designed for variable speed operation, the "dead" condition you are seeing may simply be that the motor isn't getting enough voltage to overcome cogging and hysteresis effects. For a reversible, variable speed application, you should select a motor designed for such operation.

Connecting a 12 VDC power supply across a potentiometer, and using the tap as a simple voltage divider, against a second fixed voltage divider network is possible, but the power efficiency of such simple passive control networks is poor. In theory, you'd dissipate at least as much power in the control network as you delivered to the motor at the maximum speed setting, but actually, you'd probably do a little worse than this, since there would always be at least a small series resistance to the motor. Therefore, if power efficiency is important, you may want to look at electronic motor controllers.
posted by paulsc at 2:40 PM on October 29, 2005

The reason your pot cuts out is because it is meant for audio applications. That means it will have either a log taper or audio taper (they're supposed to be different, either way it is not what you want). You want a linear taper pot.

Look at it like this:

A regular 10k potentiometer:
(0% twist) -> 0 Ohms
(25% twist) -> 2.5k Ohms
(50% twist) -> 5k Ohms
(75% twist) -> 7.5k Ohms
(100% twist) -> 10k Ohms

Your potentiometer [if it's 10k] (a guess, you do the math, I'm lazy):
(0% twist) -> 0 Ohms
(25% twist) -> 100 Ohms
(50% twist) -> 1k Ohms
(75% twist) -> 2k Ohms
(100% twist) -> 10k Ohms

For a real motor you are probably going to be looking at rheostats for power handling capability. Rheostats, as harlequin mentioned, are usually wirewound. They're also expensive! You're probably going to start at about $10 - $20 for a rheostat and could pay thousands for some insane industrial one that takes two people to lift. The really heavy duty ones may have a slider instead of a knob. Remember the watts you are dissipating are going out as head and design any enclosure accordingly. :^D
posted by shepd at 3:13 PM on October 29, 2005

Response by poster: Thanks for the answers--this problem appears to be more complex than I originally thought. Varying the speed and direction of the motor is just a bonus feature of the project itself, so if it can't be done without a new motor or expensive components, it's not a big deal.

The motor itself is for a portable cassette player. It says "6V DC" on top. Other than that, I don't know anything about it. Because it's a cassette motor, I'm assuming it's designed for variable speed/direction (fast-forward and rewind).

It's getting it's power from the circuitboard, which takes an AC plug, and not directly from a power source, if that makes a difference.
posted by cloeburner at 4:41 PM on October 29, 2005

Also, wire-wound pots are often called rheostats, for reasons that aren't important here. Not all rheostats are operating by a knob like a pot, but if it looks like a pot but is called a rheostat, don't assume it's not what you're looking for :)
posted by -harlequin- at 7:57 PM on October 29, 2005

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