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Lowest draw "power on" indicator?

Discussion in 'Electronic Basics' started by Ken, Jan 10, 2005.

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  1. Ken

    Ken Guest

    What "power on" indicator consumes the least current?

    An LED draws 20 mA. In a flasher circuit (LM3909) this can be reduce
    to 0.4 mA, but at an expense of around $5 in components.

    Is there anything in between?

    I need an indicator for a circuit that itself only draws 20mA.


    Ken
    (to reply via email
    remove "zz" from address)
     
  2. John Fields

    John Fields Guest

    ---
    High efficiency LEDs are available with a rated current of 2mA which
    are clearly visible with less than 1mA going through them. High
    intensity LEDs are also available which should give you a nice output
    for much less than that.

    Also, if you want to go mechanical, power ON-OFF switches are avaiable
    with an indicator which changes color depending on whether it's ON or
    OFF. Zero power required for the indicator.
     
  3. Clarence_A

    Clarence_A Guest

    I often use LEDs at as little as 1mA for indicators. Just aren't
    as bright!
     
  4. Miles Harris

    Miles Harris Guest

    How is that brainwave going to save on power?
     
  5. Ken

    Ken Guest

    Won't the drop be 2 volts?

    Assuming I can make this work, how much power in milliwatts will the
    LED actually consume?


    Ken
    (to reply via email
    remove "zz" from address)
     
  6. There are LEDs that will require only .5mA to 2mA with decent appearing light.
    You could try those high-efficiency red types. A few discretes could achieve
    the blinking (two BJTs and three or four passives come to mind.)

    And in no way is the 3909 *worth* $5 -- cripes!

    Jon
     
  7. JeffM

    JeffM Guest

    if you want to go mechanical, power ON-OFF switches are avaiable
    The ones I've seen are push-on/push-off
    with 2 dayglow green (or yellow) bits that come together like eyelids.
     
  8. 20mA * 2V or 40mW.

    Probably, this series deal isn't a good idea. The use of a high-efficiency red
    led is probably much better. You can tweak it's current down to say .5mA and it
    will probably look okay. Another option would be to add the few discrete parts
    it would take to blink the LED to further reduce the average current. (I've got
    the schematic for a duplicate of the 3909, for example.)

    The simplest is probably the high-efficiency red led, though.

    Jon
     
  9. John Fields

    John Fields Guest

     
  10. John Fields

    John Fields Guest

     
  11. Jamie

    Jamie Guest

    you could try running the LED in series with the Power source if your
    circuit does draw that much,.
    you will get a slight drop in voltage how ever if that is not a problem.
     
  12. You usually do not need 20 ma to have a useful power on indicator.
    And the main claim to fame of the 3909 was that it could boost a low
    voltage high enough to light the LED, as well as flash it. You can
    use an LMC555 as a flasher, and put the LED in series with the
    discharge pin to give a flash often enough to indicate power on with a
    sub milliamp average drain and much less than $5 price tag. This
    assumes that you have more than about 4 volts to work with.
     
  13. Jamie

    Jamie Guest

    like i said, you will get a drop in voltage to your circuit, but since
    i don't know your circuit totally i am only guessing that you could
    recalucate for that problem.
     
  14. John Fields

    John Fields Guest

    ---
    Well, my boy, first consider:


    +V
    |
    +------+--E1
    | |
    | [RS]
    | |
    [RL] +--Eled
    | |
    | [LED]
    | |
    +------+
    |
    GND

    E1² (E1 - Eled)
    Pd = ----- + E1 * -------------
    RL RS


    Assuming that +V is 10V, that RL is the load, (which draws 20mA at
    10V) that Eled is 2V, and that the LED draws 20mA, if we plug in those
    numbers we'll have:


    10² (10V - 2V)
    Pd = ------ + 10V * ------------ = 0.4W
    500R 400R




    Then, if our circuit becomes:


    +V
    |
    +--------E1
    |
    [RL]
    |
    +--------Eled
    |
    [LED]
    |
    GND


    We'll have:


    E1 - Eled
    Pd = E1 * -------------
    RL


    Now, keeping everything as it was before and assuming that the
    circuitry represented by RL can work with 8V across it, if we plug in
    the numbers, this time we'll have:


    10V - 2V
    Pd = 10V * ---------- = 0.16W
    500R



    So, Einstein, since 0.16W < 0.4W, there's your answer.
     
  15. Steven Swift

    Steven Swift Guest

    For safety reason, I think mechanical indication is required on line powered
    equipment.
     
  16. Or a simple toggle labeled ON / OFF. (Optional extra: colour the
    labels.)
     
  17. Miles Harris

    Miles Harris Guest

    On Mon, 10 Jan 2005 17:44:20 -0600, John Fields

    [snip]
    Thanks, knob-head.
     
  18. Joel Kolstad

    Joel Kolstad Guest

    'Required?' There's tons of common household equipment that uses pushbutton
    switches that have no mecahnical indication of their powered/unpowered
    state. (In fact, I have a monitor where this is actually kinda annoying...
    the power LED changes color depending on the mode, and it can be hard to
    tell red's 'off' from orange's 'on, but in power save mode' at times!)

    The original poster might want to go a Google news groups search for the
    thread about ultra-low power LED flasher circuits some months back.
    Winfield Hill had plenty to contribute to it, which is a pretty good
    indication that the results were high quality and robust!
     
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