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Running a 6V DC Motor on 0-30V variable supply?

Discussion in 'Sensors and Actuators' started by rbatty91, Nov 18, 2013.

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  1. rbatty91

    rbatty91

    3
    0
    Nov 18, 2013
    Hi Everyone,

    I'm fairly new to electronics and have a project I'm working on.

    I have a DC motor that I need to run from a few revs to 2000 rpm (unloaded).
    The motor reaches 2000 rpm at 6V and draws 30mA.

    Now, if I want to be able to use the full range of voltage of my variable supply from 0 - 30V to control the motor's speed could I simply place a resistor in series to drop the voltage applied to the motor?

    If so is it simply a case of applying ohms law to the circuit to calculate the required resistance value?

    Thanks for your help in advance.

    Richard.
     
  2. Boltar

    Boltar

    12
    0
    Nov 18, 2013
    If the motor draws 30mA at 6V then it has a load of 200 ohms. To reduce 30v to 6v you'd need a 4:1 ratio, so you'd have to put 800 ohms in series.with the supply.
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,505
    2,849
    Jan 21, 2010
    Ignoring for a moment that PWM would probably be more effective, varying the voltage between 0 and 6 volts will have reasonable effect.

    Varying the voltage from the power supply works because the power supply has a low output impedance. As you load the motor, the power supply can increase the current and the voltage across the motor stays fixed.

    If you put a resistor in series with the motor so that (say) the motor when unloaded has 6v across it and the resistor has 24v, then loading the motor has a perverse effect. Let's assume the motor wants 10% more current. That means 10% more current flows through the resistor, so it drops 10% more voltage. It now drops 26.4v. That only leaves 3.6v for the motor. The power to the motor has fallen! That pretty much means the motor is going to stall very easily.

    So a resistor in series is a bad thing and probably won't meet your needs if there is a variable load.

    Even with a fixed load, you would likely find that the motor didn't start until the voltage was turned up quite a bit, and then it would run at high speed. This problem would exist to some extent even with a 0 to 6v variable supply.
     
  4. Laplace

    Laplace

    1,252
    185
    Apr 4, 2010
    Shall we assume that the motor is a permanent magnet motor with commutator brushes rather than something exotic like a brushless DC motor? If that is the case then one thing to realize is that the motor is also a generator. Attach a voltmeter to the motor terminals and manually turn the shaft in the same direction that the power supply would make it turn, and notice that the generated voltage polarity is in opposition to the power supply. See attachment for a discussion of this generated voltage - the Back EMF (BEMF).

    The current through the motor is proportional to the shaft torque, and the torque is equal to the load plus internal friction. Since you are running the motor with no load, torque equals friction. I would expect that the internal friction is independent of motor speed so that the 30 mA no-load current is also the minimum current needed to start shaft rotation.

    So if you put an 800 ohm resistor in series with the 30 volt variable power supply, the voltage would need to be raised to 800 x 0.030 = 24 volts before the motor would start turning. But even a very small increase in the load torque would be enough to stall the motor at 24 volts. These are just generalizations that ignore the effect of air resistance on the turning rotor, so the starting friction might be less than at 2000 RPM. Without more detailed performance measurements on the motor it would be difficult to predict what might actually happen when adding an 800 ohm resistance, but it is generally considered bad practice to add resistance to a motor circuit.
     

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  5. rbatty91

    rbatty91

    3
    0
    Nov 18, 2013
    Thank you all very much.

    I have set up my circuit with 800 ohm resistor and it works just as you said it would; from 22v to 30v.

    I realise it is bad practice but it's just what my application requires.

    Thank you.
     
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