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Leaving Cert engineering potentiometer help - quite urgent

  • 10-02-2022 11:04am
    #1
    Registered Users, Registered Users 2 Posts: 6,186 ✭✭✭


    Hi all,

    so my son is building a racing car as part of his LC and needs to vary the speed on the DC motor.

    It's a 9v Motor, 9v battery but needs a potentiometer to vary the speed ?

    Current setup is like this, but keeps burning out the potentiometer, so question is a) is this the correct setup (needs to be absolute basic) and also, what size/rating potentiometer should we get. Plan to buy from RS and want the cheapest possible.

    Thanks in advance



    Post edited by beer enigma on


Comments

  • Registered Users Posts: 270 ✭✭Irish_wolf


    Only a hobbyist myself when it comes to electronics but can you add an additional resistor (1k Ohm is a standard go to) in series with the potentiometer to reduce the current passing through it. It should reduce the overall current passing through the motor (reducing the top speed/power although not massively) but would still allow for the control of current and by extension the speed of the motor to adjusted using the potentiometer.



  • Registered Users, Registered Users 2 Posts: 6,238 ✭✭✭Cordell


    It will not work like that, you can't wire the potentiometer like that and a 1k resistor won't provide enough current to move the motor.

    Simple DIY circuits here: https://www.homemade-circuits.com/dc-motor-speed-controller-circuits/

    Maybe a 100 ohm 10W potentiometer might kind of work work - but wired as a variable resistor, i.e. remove the black wire from the battery to the potentiometer. But from what I see they are very expensive. https://ie.rs-online.com/web/p/potentiometers/8509413

    You can buy ready made circuits like this https://ie.rs-online.com/web/p/motor-controllers/4179728



  • Registered Users Posts: 95 ✭✭Tri Oidhcheannan Eile


    As Irish_wolf says, a 1k resister or a 1k pot.

    Also, since you're solely using the potentiometer as a variable resister, you don't need the -9v connection to pot lug 1 in your diagram above.

    As a VR, you only need wiper (pot lug 2) and either lug 1 or lug 3 connections. (IIRC, the different possible wirings will determine the direction of the pot... clockwise or anticlockwise for less or more resistance.)




  • Registered Users, Registered Users 2 Posts: 6,186 ✭✭✭beer enigma


    Thanks, i was wondering about the ready made one just hoping for something cheaper as the flipping model has cost me a fortune so far (so much for free schooling !). I suspect it may be the easiest option



  • Registered Users, Registered Users 2 Posts: 6,186 ✭✭✭beer enigma


    We had it wired as per the top diagram with a 1k pot, but no resister and the pot went up in smoke. I presume the resister will solve that issue ? - there's a comment above that the 1k resister will stop enough juice to the motor ?



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  • Registered Users, Registered Users 2 Posts: 6,238 ✭✭✭Cordell


    The potentiometer in that configuration can only be used to provide a control voltage for something else i.e. a power circuit that will actually power the motor - this is what happens in the schematics I've sent. In other applications it is used to control very low power signals like audio. It will never work to control a DC motor directly like that.

    A 1k resistor will limit the current to a max of 9mA which means no more than 81mW of power can be drawn from the battery, not nearly enough to spin the motor shaft, let alone drive the car :)



  • Registered Users, Registered Users 2 Posts: 6,077 ✭✭✭kirving


    1 1kohm pot will have 1kohm at the top end of it's resistance, but much less at the low end.

    At 1k ohm setting, it would need to dissipate 0.081W

    At 50ohm setting, it would need to dissipate 4.05W - probably more that it is capable of handling.

    https://ohmslawcalculator.com/ohms-law-calculator

    Check out TinkerCAD - it's free from Autodesk.

    Full disclosure - I'm no electronic engineer! What's happening below in your version (top) is that power is constantly flowing around the resistive track of the pot, and the motor is a tap off that. Does it actually run at all before burning out?

    With the lower version, power is flowing into the track of the pot, and then out the wiper to the motor


    If you keep the resistance setting high high enough, it should be fine. But you'd need to do this mechanically with a stopper, but you'd then always have some low voltage turning the motor at the lowest setting.

    Without getting too far into stuff I don't understand, you could use a transistor to cut off the power, if the pot goes below a certain level.

    I'd recommend doing this, and showing the TinkerCAD & wiring diagram and calculations in the paper report.


    You should add a cut off DP/DT switch to the car too, so it can reverse or be turned off. Plenty of diagrams online of these.



  • Registered Users, Registered Users 2 Posts: 6,238 ✭✭✭Cordell


    The clip shows a variation of the schematic below, without the protection diode and the limiting 470 ohm resistor that protects the potentiometer. This is the bare minimum, you absolutely positively need a transistor, it will not work with just a potentiometer.





  • Registered Users, Registered Users 2 Posts: 6,186 ✭✭✭beer enigma


    It runs fine for about 5 mins on and off but you can smell the burning and then eventually just fails - I have an on/off switch now connected.



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  • Registered Users, Registered Users 2 Posts: 6,238 ✭✭✭Cordell


    A purpose build pot that can handle the power will work but it looks to be much more expensive than a proper circuit. Those used so far - like the one in the OP image - are the kind that are not designed to deliver any power directly.



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