Transistor confusion

Discussion in 'General Electronics Chat' started by seanrs, Dec 17, 2012.

  1. seanrs

    Thread Starter New Member

    Dec 15, 2012
    3
    0
    Hello all, I'm new to this forum and to electronics and I've been experimenting with changing the speed of a small DC motor using PWM on an arduino board. I've also been going through the Forest Mims books and this circuit came up. I didn't have a buzzer so I replaced it with the dc motor that was on my desk. And to my surprise, turning the 10k pot changed the speed of the motor. Could someone please explain to me why this works or direct me to some place that would give me a clue.
    Thank you for any help that you may have to offer.

    Sean Problem.jpg
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,790
    The pot acts as a voltage divider and, in rough terms, whatever voltage appears at the wiper will be used by the transistor in such a way that the voltage at the emitter (the side connected to the motor) will be about 0.7V below that. Thus, as you change the pot you change the voltage on the motor. Change the voltage on the motor, you change the speed.
     
  3. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    Another thing the transistor does is supply a current gain, so a small current supplied by the pot is amplified to provide sufficient current to run the motor..
     
  4. seanrs

    Thread Starter New Member

    Dec 15, 2012
    3
    0
    thank you for your response, I did understand that it affected the voltage, but maybe I'm confused in the idea that there needed to be a minimum current for the motor to even turn, and I know that resistors affect current.
     
  5. seanrs

    Thread Starter New Member

    Dec 15, 2012
    3
    0
    ah ha, thanks tube guy, this is cray how much is involved with something so simple. I have a hard time getting my head wrapped around it all.
     
  6. #12

    Expert

    Nov 30, 2010
    16,283
    6,793
    Adding a second transistor allows more of the supply voltage to be applied to the motor. With this circuit, you don't lose that .7V that the first circuit uses up for its base drive.
     
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