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LED driving questions.

Discussion in 'Electronic Basics' started by Boki, Apr 19, 2006.

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  1. Boki

    Boki Guest

    Dear All,

    My customer said:

    The LED circuit should draw <20uA under "battery good" condition.

    Does that mean my driving transistor should provide less than 20uA for
    LED when the LED in turn off state?


    Question2,
    I have to guarantee 10mA for turn on LED, but my supply voltage is only
    2.8V

    The cross voltage of LED is about 1.9V...

    I have only 0.9V for emitter ( pnp driving ) or collector ( npn driving
    )

    Even I use 330ohm resistor at collector with npn, the current seems
    still too small...

    Best regards,
    Boki.
     
  2. rob

    rob Guest

    An led requires a least 5mA to make it lite. so """"The LED circuit should
    draw <20uA under "battery good" condition"" is not possible. the following
    formula applies. R = (Vs - Vf)/If where Vs is the supply voltage, Vf is the
    forward current of the led. If is current required to lite the led. eg Vs =
    12V, Vs = 1.4 V, If = 5mA therefore R = 2120 ohms
     
  3. Naa.
    A Led is clearly visible with just 1mA.
    If switched on at 1mA, but 10uA average,
    this would mean a 100:1 deadtime.

    Further. The VCE of an NPN is as low as
    50mV saturated. Better than a resistor
    in terms of power is the use of PWM, a
    coil and a cap, in case you have a spare
    pin on your controller.

    Rene
     
  4. Deefoo

    Deefoo Guest

    To me that means that the LED driver should consume as little power as
    possible, so the LED has to be off in this condition and leakage currents
    should not exceed 20uA.



    You should not be using a bipolar transistor to drive the LED.

    --DF
     
  5. Boki

    Boki Guest

    Got you.


    but I saw a lot of design use bipolar transistor...

    do you mean use MOSFET to drive is better?

    Best regards,
    Boki.
     
  6. Deefoo

    Deefoo Guest

    Well, "should not" is a bit strong, but you are already in a "battery bad"
    condition when you have to activate the LED so you have to limit the power
    consumption as much as possible. If you can avoid a base current by using a
    FET, that's good. BTW there exist low current LEDs and personally I almost
    never need more than 2mA through a LED to get good visibility.

    --DF
     
  7. Fred Bloggs

    Fred Bloggs Guest

    This can only be talking about the LED driver circuit overhead current
    draw exclusive of the actual LED current. The overhead consists of the
    various bias and drive currents required to cause LED current to flow.
    I interpret it to mean that <20uA in either LED ON/OFF state.
    You mean "forward" voltage of the LED.
    For a 10mA LED current and a 20uA maximum driving circuit current, you
    will have to use a MOSFET with a maximum threshold Vgs less than your
    worst case (minimum) available gate drive voltage, or a very high beta
    transistor with gain > 10/0.02=500. A small MOSFET will just keep worst
    case leakage under 20uA. The simplest circuit is the LED in series with
    a current limiting resistor and MOSFET. At 2.8V you then have
    2.8V-1.9V=10mA*(R+RDS,ON)=10mA*R most of the time, making R=90 ohms or so.
     
  8. Chris

    Chris Guest

    Hi, Boki. There are dozens of battery monitor ICs that will do the
    job. Why not just go with a one IC solution?

    Many have MOSFET outputs that can sink 10mA with an Rds(on) of less
    than 20 ohms(if that's what the customer expects -- I'd go with a
    couple of mA strobed at 1Hz if all you want is a visible LED
    indication, and you want to conserve a low battery rather than drive it
    into the ground).

    If you need help in specifying the IC, possibly you could describe what
    kind of battery you're trying to monitor? Also you might want to tell
    a little more about your project. And since we're at it, how about a
    wish list of any other features you could use? This wheel has been
    reinvented so many times, manufacturers have added all kinds of bells
    and whistles to differentiate their products. You can easily take
    advantage of that, or just go with a bare-bones minimum if cost is your
    only consideration.

    Good luck
    Chris
     
  9. Fred Bloggs

    Fred Bloggs Guest

    If by this cryptic language you mean the LED is a LO battery indicator,
    then the requirements are more challenging than first impression. The
    20uA refers to the battery current draw by the *all* of the LO battery
    indicator circuit in the inactive state. These circuits will have some
    kind of threshold comparison for the LO battery condition and the
    challenge here is that most comparator circuits exhibit a large increase
    of bias current draw from the battery as the battery voltage slowly
    drifts close to that threshold. There are several varieties of
    specialized LO battery indicator ICs that overcome this drawback. Go to
    http://www.maxim-ic.com and check them out.
     
  10. Why not just ask your customer what was meant?

    It is hard for me to imagine that you are supposed to keep the
    quiescent current down below 20uA, if the LED is going to draw at
    least 10mA when the battery is good. For one thing, 10mA is a lot to
    waste on an LED if a circuit is running off of some batteries (and I'm
    imagining a Lithium button, right now, which pushes the total circuit
    draw to a few mA at most, anyway.) For another, 20uA is such a small
    price to pay if you are already paying the price of 10mA for the LED
    that I can't easily see why that would be such a concern as to get
    spec'd like this.
    If this is a circuit which must somehow indicate battery good by
    lighting the LED, but can only draw 20uA to do that, you will have to
    accept pulsing the LED. A 10mA guaranteed pulse current with an
    average draw of 20uA implies a 500:1 period, in ideal conditions. And
    if this is the desire, then that still leaves out what should be done
    when the battery is determined to not be good. Would the circuit
    simply be OFF, in this case?

    If this is really a circuit driven off of a lithium button and that is
    the reason that the battery good LED circuit must draw <20uA during
    normal operation, then you will need to pulse and this will imply a
    large R and a modest C to start. Something basically like this:

    +bat
    |
    |
    \
    / 22k
    \
    | LED 10
    +---|>|--/\/\---------+-------------,
    | | |
    --- | |
    --- 47uF ,---------, ,--------,
    | | | | |
    | | voltage |->-| SCR |
    gnd ,--->| trigger | | switch |->--,
    | | | | | |
    | '---------' '--------' | 4.7nF
    | | | ---
    | gnd gnd ---
    | |
    '--------------------------------'
    positive feedback

    You can get this to run at under the 20uA and off of 2.8V, generating
    pulses, and use normal, cheap BJTs throughout (such as 2N3906 and
    2N3904) -- including the "SCR" which can be fabricated from the same.
    In fact, I set down a possible circuit on the above model that in
    LTSpice draws about 20uA, pulsing the LED at 2Hz or so with RC decay
    shaped pulses having half-power widths of 1ms or so, with peak pulses
    of 12mA. It uses (2) 2N3906 and (1) 2N3904 and a couple of 1N4148s.

    I'm baffled why you can't ask the customer for more details, though.
    That seems to be the place to go for clarification.

    Jon
     
  11. Sorry, that should be large C and modest R.

    Jon
     
  12. Rich Grise

    Rich Grise Guest

    As has been mentioned, 20 uA will not light an LED. This will work if
    the LED off means "battery good" and the LED on means "battery low".
    You must have missed the part where he says Vf is 1.9V and Vs is 2.8V. He
    also seemed to indicate that he needs 10 mA ILED, so R = (2.8 - 1.9) / .01.
    And, Vsat (of the driver transistor) can go as high as .3V at 15 mA for
    the PN2222, so at 10 mA it will be less and that's the max so it could be
    way less. So it should be R = ((Vs - Vf) - Vcesat)/If, or 60 ohms if your
    Vcesat is .3, or 80 if it's .1. So, use a 75 ohm resistor.

    Cheers!
    Rich
     
  13. Sjouke Burry

    Sjouke Burry Guest

    Google for a LED switching supply.Those can
    convert up from the powersupply, and sense the
    resulting current at ground connection.
    If a Led fails "shorted",the others just keep
    on working.
    They work by switching a coil current into
    the stack of LEDS,and adjust switching freq.
    to get the wanted current.
    Note that it works without series resistor,
    so you use your energy very efficiently.
     
  14. John Fields

    John Fields Guest

     
  15. Boki

    Boki Guest

    Do we have any more state ? : )

    Agree.

    Best regards,
    Boki.
     

  16. Typical indicator LEDs won't light in any way usefully if at all at 20uA,
    but the technology has been improving.

    I've got some white 5mm "26,000mcd" 20 degree viewing angle clear LEDs from
    here:

    http://www.lck-led.com/

    There is some noticeable part to part variation at extremely low currents,
    but for the most part they seem to light quite visibly in subdued lighting
    conditions all the way down to less than 1uA. They glow very nicely at
    20uA, and in fact I use them as indicators at that current level, even for
    quite bright office fluorescent standard lighting conditions. Higher
    current than a few tens of microamps isn't really desirable for indicator
    use since they quickly start to become dazzling to look at straight on.

    There is one drawback however... Typical transistors/MOSFETs and whatnot
    can often have fully off state leakage currents exceeding 1uA. This means
    you can't always guarantee the LEDs to be fully extinguished when switched
    with typical transistors/MOSFETs, unless you place a parallel resistance
    next to the LED to soak up the leakage without producing visible light.
     
  17. Hmm... Redoing the math, I take part of that back. I use them at 200uA as
    indicators in office standard lighting conditions. At 20uA they still glow
    very much visibly, but only if you look at them within the 20 degree viewing
    angle. As indicators I prefer to easily be able to establish if they are
    lit or not when viewed on or off angle.
     
  18. Or drive with a CMOS output. Not like you'll be needing a big
    transistor for 20uA!

    Best regards,
    Spehro Pefhany
     
  19. I know of LEDs that light damn well and bright at half a milliamp!
    Please mention upper limits and lower limits of voltage available to the
    LED or upper end of the LED's dropping resistor! Also color requirements
    - deep pure red can require a good couple milliamps but with lower voltage
    drop in the LED! Worst is where the color is required to be yellow or
    yellowish.

    - Don Klipstein ()
     
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