Understanding how it fits together.

Discussion in 'General Electronics Chat' started by TheRoman, Sep 14, 2010.

  1. TheRoman

    Thread Starter New Member

    Jul 22, 2010
    Hi, my electronics knowledge is my weak spot, so I thought I would take a practical approach to getting the basics understand and also getting some help using forums. Basically I have a couple of model motors:
    Voltage Range: 1.5V to 4.5V nominal voltage 3V
    Nominal no Load Current: 0.35A
    Nominal full Load Current: 1.1A

    I want to drive them from a MCU of which its digital I/O pins:
    operate at 5 volts. Each pin can provide or receive a maximum of 40 mA and has an internal pull-up resistor (disconnected by default) of 20-50 kOhms.
    Now I was wondering if someone could explain to me how I use all this information, to know If the MCU I/O pins can drive the motors. I’m just a bit confused how I have to put all this information together.
    Any help would be appreciated.
    Thank you for your time in advance.
  2. SgtWookie


    Jul 17, 2007
    Well, your MCU can't drive the motors directly, but it can control a semiconductor device to control the current flow through the motors.

    You can use a logic-level N-channel enhancement mode power MOSFET. These are really quite easy to use with microcontrollers.

    Logic level MOSFETs are considered fully turned ON when their Vgs >=4.5v.
    Vgs = the voltage on the gate, measured using the source terminal as the reference point.

    A really handy MOSFET is an IRLD024. They come in a 4-pin DIP package, so they are really convenient for breadboarding and prototyping; and can sink up to 2.4A, with a Vdss of up to 60v.
    Vdss = voltage on the drain reference to the source terminal.
    MOSFETs are normally used as switches; when ON, they have a very low resistance between their source and drain terminals. They are easily damaged by static electricity, though.

    Another possibility is to use a Darlington transistor, such as a TIP122. You connect the base of the transistor to the I/O pin using a resistor (1k is a typical value), ground the emitter, and sink current from the motor via the collector. Darlington transistors have a Vce (voltage on the collector measured with respect to the emitter) from between 0.7v and 2.6v depending on the collector current, so their is a power loss penalty compared to MOSFETs.

    In addition to using either a transistor or MOSFET to sink current from the motor, you'll need to use a diode across the motor to provide a current path when the device is turned off; otherwise a high voltage may result that can destroy the semiconductor device.
    Last edited: Sep 14, 2010
    TheRoman likes this.
  3. TheRoman

    Thread Starter New Member

    Jul 22, 2010
    Hi SgtWookie, thank you for your help and time. Most appreciated.