Help determining what potentiometer to use - Page 1
Forum section: electronics / beginners

author: Mighty Burger
author_rank: Regular Contributor
timestamp: March 04, 2016, 05:46:04 pm
content: Hello! I'm a very new starter and I've tried to design a very simple circuit.It's suppose to spin a low power moter, in this case a small fan, and have a little LED turn on when it's powered. There is a potentiometer to limit current and slow down the fan. I have questions though. If I want to have the fan speed range from full speed to around half speed, what should the potentiometer's max resistance be, and what will the voltage drop and current be in the worst case? And how did you come up with those answers, so I can figure it out myself in the future?And another small question: I put the diode there because motors are inductive and it would otherwise cause a voltage spike when it's switched off. Is this necessary?

author: Simon
author_rank: Unknown
timestamp: March 04, 2016, 05:51:09 pm
content: what sort of motor is it ? This is really not the way to do it. Controlling motors with resistors is tricky and a trial and error thing. The electronics world came up with PWM for a very good reason.

author: NilByMouth
author_rank: Regular Contributor
timestamp: March 04, 2016, 05:56:06 pm
content: I would start by looking at how other people have done it and work from there. Learn from other people's mistakes.

author: MK14
author_rank: Super Contributor
timestamp: March 04, 2016, 06:41:33 pm
content: By and large (with some rare exceptions, such as some massive Rheostats, used in Physics experiments and other stuff). Low cost, standard potentiometers (pots), have extremely low dissipation ratings and are NOT meant to pass any kind of serious current. E.g. 0.1 W max power dissipation rating.They tend to be a bit unreliable (especially after heavy use and/or many years since original manufacture). Passing any kind of current through them (even if allowed on the datasheet), could worsen their lifetime, still further.Usually only small (rule of thumb, think < 10 ma) currents are usually passed.You want to put in some type of transistor/fet or something, to act as an output driver. So that little current goes through the pot, and most of the current goes through the output device, into the motor.As mentioned above, PWM could be used. But that is getting even more complicated. Especially if you are not using a microcontroller yet.The diode (or maybe other stuff, such as more diodes/capacitors) may be needed. Depending on how much back emf, and interference suppression, you need at this stage. It is a complicated area to try and describe quickly. The motor datasheet (if available), and/or application guides and stuff, may be of help with it.Cheap motors tend to be very (electrically) noisy, compared to quality/expensive ones, usually. Your mileage may vary.One way of doing it ...Are you familiar with an emitter follower, type of transistor circuit. With the pot feeding the base of the transistor, an adjustable voltage (perhaps, between the supply rails, for the outer two pot terminals). A current limiting resistor, between the pot and the base. Then connect the emitter to one side of the motor (the other motor lead to ground). Collector to the +ve. (Find a suitable NPN transistor).555 timers can generate PWM, is another way. Because PWM is normally always fully on or off, little heat is generated in the output driver, which is good. Especially for big motors.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: March 04, 2016, 09:32:14 pm
content: As others have said, most common inexpensive potentiometers aren't able to handle much current. Even though your motor only draws 140 mA this may be too much and could burn the pot, making it intermittent or even fully inoperative.The standard way to control speed of brushed DC motors is by using Pulse Width Modulation. The PWM system can control the motor's speed from zero to full speed easily using very inexpensive components. A slightly more complicated circuit is required, with probably about 3 dollars worth of parts, but will be much more reliable and consistent than a simple resistive control.Here's an example 555-based PWM circuit that works fine with small motors.(In your original circuit the reverse-biased diode across the motor isn't really necessary because there isn't anything to protect from the inductive spike noise. In the PWM circuit using a mosfet driver, it is good practice to include that diode in order to protect the mosfet from these spikes. Also, the resistor value for the LED you have calculated will give you more than 20 mA through the LED. If the LED has a Vf of 2 volts, then your resistor is calculated by R=V/I where V = 12-2=10 and I = 0.02A, so R = 10/0.02 = 500 ohms. But with modern LEDs that is still going to be extremely bright. 10 mA is probably enough current for reasonable brightness so you could use a 1k resistor. )

author: uncle_bob
author_rank: Unknown
timestamp: March 05, 2016, 12:21:41 am
content: HiIf your motor pulls 140 ma at 12V while running, it likely pulls a bit more than that as it's starting up. How much more? That depends on the type of winding the motor has and just what limits it's speed / current draw. To really *know* how big a pot you need, the "stalled" current on the motor is what you need to know. On an AC motor this can be pretty crazy. On a DC motor, you can either measure it or guess.Let's guess that the motor pulls 250 ma stalled. There is no way to support that guess, it is just a guess. It is sure to be > 140 ma, it is unlikely to be more than 1.4A.Next, you need to know how much voltage the motor takes to get going with your load on it. That could be 10V. It also could be 3V. The easy thing to do is to measure it. Instead we'll guess again. Let's guess that it will get going at 3V. No way to support that, it's a guess.The two numbers would imply that at the lowest voltage (3V) the motor should pull the stalled current (250 ma). That's (like all the above) a guess. There are cases where this is not true.To drop 9 (=12-3) volts with 250 ma, you need a 36 ohm (from R= E/I) resistance. At that resistance you will have 81 (= E^2) / 36 (ohms) of power in the resistor. That's just a bit over 2W.Derating any resistor 2:1 is a good idea, that gets you to about 5W.Now off to shop for pots. Do you have a need for a million pieces? Hmmm ... not so much. You will have to settle for standard value parts rather than have one custom made to your spec. 50 ohms looks like the next standard value. It's about 1.5X your target, so the power rating will have to go up accordingly.Mouser shows:http://www.mouser.com/ProductDetail/Ohmite/RES50RE/?qs=sGAEpiMZZMtC25l1F4XBUxn1oJseK7HHlaNOy3a6vlE%3das the low cost leader at $40.87. At 12.5W it should do the job pretty well. For another $8 you can bump up to a part that has twice the power rating.Remember that this is all based on a bunch of guesses. The real numbers could be less current and a higher voltage. That would result in a bit less power. It might result in a 25 ohm pot doing the job.Bob