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driving a relay

Discussion in 'General Electronics Discussion' started by Kstarloomo, Nov 28, 2012.

  1. Kstarloomo

    Kstarloomo

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    Jun 28, 2012
    i wantd to drive two transistor using the circuits in d diagram but the one with a voltage divider has a base resistor which is an approximate value of the required calculated value while the one without the voltage divider is an exact value. i fill the first one is better but not sure. which one is better in design and more consistent, why?
     

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  2. BobK

    BobK

    7,645
    1,663
    Jan 5, 2010
    BJTs are current driven devices not voltage. You need to calculate how much current goes from the base to the emitter, based on how much current you need in the collector circuit.

    You don't need a voltage divider, you only need a base resistor that can be calculated as follows. Use your second diagram, but eliminate the 10K resistor to ground.

    Ic is the desired collector current, this is the current required by your relay coil
    Ib is the required base current. This is Ic / Hfe
    Vin is the voltage the base resistor will be connected to.
    Rb is the base resistor.
    0.6V is the typical voltage drop from the base to the emitter.

    From Ohm's law:

    (Vin - 0.6) = Ib Rb
    Rb = (Vin - 0.6) / Ib

    This would give you the base resistor needed if the Hfe was exactly right. You want to actually use about 1/2 that to double the current to make sure the transistor is saturated.

    Example:

    Ic = 100ma
    Hfe = 100
    Ib = Ic / Hfe = 1ma
    Vin = 5V

    Rb = (5 - 0.6) / 0.001 = 4400 Ohms

    Use 2.2K Ohms to make sure there is plenty of base current to saturate.

    Bob
     
  3. Kstarloomo

    Kstarloomo

    23
    0
    Jun 28, 2012
    Thanks. Hw about if the source is from an ic
     
  4. BobK

    BobK

    7,645
    1,663
    Jan 5, 2010
    Same calculation. Use the output high voltage of the IC where I used Vin, and make sure it can source the required amount of current.

    Bob
     
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