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7809 replacement

Discussion in 'Electronic Components' started by Marcellus Pereira, Nov 17, 2003.

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  1. Dear ALL,

    I'm using a 7809 ( voltage regulator ) with 18~20V In and 40mA ( 1A,
    50 ms peaks ).

    But it is very hot... I can't lower the input voltage. Any ideas, like
    a pin-compatible more robust device?

    Thanks a lot,
  2. Michael

    Michael Guest

    First, scope it to satisfy yourself that the regulator isn't
    oscillating. You *are* asking that poor little device to dissipate a
    heck of a lot of power. Help it out; introduce it to a good heatsink.
  3. Tim Wescott

    Tim Wescott Guest

    I assume that you mean a 78M09, and that your average current is 40mA; a
    TO-3 package should be able to dissipate 440mW without a heatsink.

    In a linear regulator such as the 7809 the difference between the input and
    output voltages is burned off as heat. So no matter what regulator you
    choose you're going to dissipate the same amount of heat. If you use the
    same package as your current 7809 it'll get just as hot. The only things
    that you can do to control the situation are to control the amount of power
    that you're dissipating with a pass resistor or transistor, or to control
    the surface area that you're dissipating from with a heatsink.
  4. What you can do is to couple an PNP power-transistor in parallell with your
    7809 like this:

    | |
    | E |
    | | T1 -> PNP
    | |<
    | .------------||
    | | B |\
    | | |
    | | ____ K |
    | ___ | | | |
    Vin R |____| Vout
    created by Andy´s ASCII-Circuit v1.24.140803 Beta

    When the voltage over R exceeds ca 0.6V, T1 starts conducting, so the most
    of the current goes through T1 instead of the 7809 ic.
  5. Thank you all, guys!

    I'm trying a heatsink. If not works, I will use the transistor.

  6. Thank you all, guys!
    Just one more thought, If the transistor is a very powerful one, it's gain
    may be little (less than 20 in some cases), wich means that your 7809 may
    still conduct more current than it can endure. What you have to do then is
    to use a second transistor. There's two ways to that (as far as I know).
    Here is how:


    o E o E
    | |
    | |
    | |
    | |
    |< Power .---------o
    .----| transistor | |
    | |\ | |
    | | B |< |
    B |< | o------o-| |
    o------| | |\ |
    |\ | Small | |/
    A smaller | | PNP '-------| Powerful
    transistor | | |> NPN
    '---o--' |
    | o
    | |
    | |
    o K |

    This one use a combination
    of an PNP and NPN
    transistor. Much the same
    as the Darlington.
    created by Andy´s ASCII-Circuit v1.24.140803 Beta
  7. Jim Adney

    Jim Adney Guest

    If you're sure about this being the max load current, then add a 120
    Ohm, 1 Watt resistor in series with the input to the 7809. This will
    allow about 1/2 of the power to be dissipated in the resistor and the
    other half in the 7809.

  8. Guest

    The trouble with that is his 1 amp peaks. The 120 ohm resistor
    wouldn't allow that. However, a 5 ohm 10 watt resistor in series
    with the input would be a good choice, and might solve the
    problem. It would move 5 watts of heat away from the 7809
    during the 1 amp peaks. 440 mW in a TO-3 should not get
    hot, so the heat is either coming from the 1 amp peaks,
    or he's using the smaller package.

    The best solution, if there is room, is a TO-3 7809
    on a heat sink.
  9. Jim Adney

    Jim Adney Guest

    You're right, I overlooked the 1A peaks. In that case, just change the
    resistor to 100 Ohms and add a 2000 uF capacitor to ground between the
    resistor and the 7809. I'm not sure if the 2000 uF is big enough to
    give you sufficient "ride-thru" but you can increase it if it's not

    I admit that this is not the most elegant solution, but given what he
    has installed there already I think it's the easiest and simplest.

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