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bi-color LEDs with mixed common lead

Discussion in 'Electronic Components' started by Carrie, Sep 18, 2006.

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  1. Carrie

    Carrie Guest

    Starting on a new PIC project using bi-color LEDs. First thought was to use
    a hex inverter to change colors between RED and GREEN. However it
    occured to me that if there were a bi-color LED (RED/GREEN) that had a mixed
    common lead it would save me 4 74ALS04 hex inverters, the board
    real estate and the board size. I can find common anode and common cathode
    LEDs but, so far, no mixed common. By mixed common I mean that the
    anode of one LED is internally connected to the cathode of the other LED.

    Anyone heard of such an animal??

    Thanks !
  2. I've not heard of them, but if you can make a rail of half your supply
    voltage (must source and sink, so op-amp driven is best), you could use
    two-lead bicolour LED's, and a single resistor for each. High output would
    send current through it one way, low would reverse it. You save a resistor,
    each LED needs only two connections, and although it might be overkill for
    a handful of LED's it might be ideal for lots of them. The main problem I
    can see is the need for a very low value resistor and a very well regulated
    supply if it's 5V, because half of that doesn't leave much headroom.
  3. Carrie

    Carrie Guest

    Hadn't thought of that. I am trying to save having to use 4 hex inverter
    chips. Each chip
    has 6 inverters so I am using a total of 24 bi-color LEDs. If I understand
    your suggestion
    correctly I will still need 24 op-amps????

    Thanks for your help!
  4. I hope not. :) Assuming 24*20 mA, you'll need maybe a quad op-amp with all
    four stages in parallel, or better, with 6 LED's per stage, to handle that
    much current.

    I don't know the PIC chips, but if their outputs can source and sink 20 mA,
    this will work if you choose a quad op-amp that can dissipate 1 watt. If
    you can get by with 10 mA or less per LED, a dual op-amp should be enough.
    All the op-amp is doing is providing a ground rail so a single logic output
    can supply the LED in either state.
  5. They are common, but becoming rarer.
    Historically, the first 'bi-colour' LEDs, used this design. Latter designs
    with three leads, are commonly called 'tri-colour' LEDs (though of course
    the 'bi-colour' units can also generate the third colour by being fed from
    AC). You can turn a tri-colour design into a bi-colour unit by just
    joining the outer pins on the package, and hence these are generally
    getting to be more common. Look at:
    What you have at the moment, is normally called a 'tri-colour' design.

    Best Wishes
  6. Wait a minute.

    An old style bipolar LED has the two led's arranged so
    the voltage one way lights up one LED, the opposite polarity
    lights up the other (both the left and right sides are connected

    But the three lead LEDs have them arranged so (with the common lead
    being either the left or the right side, I forget which):

    If you parallel the two other ends, all you get is two different
    color LEDs that light up at the same time. There is no way of
    changing color.

  7. But if you connect to one, the other or both you get red, green or yellow,
  8. Carrie

    Carrie Guest

    I am confused as well. I looked at the link provided and it showed a 2 lead
    LED. Description says it is a RED/GREEN LED. I will assume then that you
    are guaranteed a color will light up
    no matter what as long as one is positive and the other is negative. This
    will give you either a RED lite or a GREEN lite. Nice but once in curcuit
    there is no way to change the color. Expensive
    way to avoid having to figure out which is the anode or the cathode of as
    single color LED.

    No, what I need is a LED that will lite up RED OR GREEN, under a PIC
    control. It will never have to go off. It will either be RED or GREEN.

    Thanks for your help.
  9. Either connect them back to back and drive the control lead above and below
    zero or connect the two non-common leads to two output pins on the chip and
    get 3 colors (which is better).
  10. Now I'm confused.

    What I read in the post I replied to was the suggestion that old
    style two lead bipolar LEDs could be created by taking three lead
    bicolor LEDs and connecting the two uncommon leads together.
    The point of my post was that it wouldn't create a two lead
    bipolar because the three lead package starts with a common cathode
    or anode.

    What confuses me in reading it now is what the purpose of that was.
    The original poster wants an easy way to switch an led between
    two colors. A two lead bipolar requires both sides to be driven,
    one by a buffer (or directly if the original output is strong
    enough) and the other side through an inverter. The three
    lead LEDs would just need that same inverter, just arranged

  11. (Michael Black) wrote in
    Carrie was trying to avoid inverters! That was the point of the original
  12. But it really seams that you want something that does away with
    an inverter.

    Whether the LED has two leads or three leads, you need the inverter (unless
    you go with the scheme of resistors that someone suggested).

    With the two lead LED, you drive one side directly (or via a buffer
    if more current is needed) and the other side via an inverter. When
    the side directly connected to the buffer is high, that pin will be
    high while the other pin will be low (because of the inverter). Switch
    the buffer output to low, and that pin of the LED will be low while
    the other side is high (because of the inverter). You are changing
    the direction of the voltage flow, and hence can switch the LED
    between red and green.

    With the three lead LED, you connect the common anode to the positive
    supply, and then connect one of the non-common leads to your buffer.
    Then also connect an inverter to that buffer, and connect the output
    of the inverter to the other non-common lead.

    When that buffer is low, it will switch on the first LED (by
    grounding it), while the inverter output will be high and thus
    the second LED can't turn on. Drop the buffer output to high,
    and the first LED will have both sides at the same potential,
    and hence not be on, while the second LED will see a low, and thus
    turn on.

    (It would be a lot easier with a diagram, but I can't be bothered
    trying to draw it here).

  13. I already explained a way. I since looked up PIC chip outputs, they can
    source AND sink, up to 50 mA either way. All you need to do is supply the
    PIC with 5 volts, and make a 2.5 volt rail to use as LED common connection.
    Feed each two-lead bipolar LED through a resistor selected for a 2.5V
    supply. High level gives one colour, low level gives the other.

    Take care to work out the power dissipation fpr the op-amp you'll need to
    make the 2.5V rail, use a quad op-amp with all 4 stages controlled as
    voltage followers by one pair of series resistors of equal value. If you
    don't need much LED current, you might get by with an 8-pin dual op-amp IC.
  14. (Michael Black) wrote in
    That 'someone' was me. There was a tad more to it than resistors though.
    I think Carrie's first post left no room for doubt that she knew how the
    inverters worked for this. What she wants is to avoid them, they take too
    much space on the board.
  15. I guess you could tie each of two pins to pull up resistors and pull down
    one or the other pin - but it seems perverse.

  16. Byron A Jeff

    Byron A Jeff Guest

    20 mA. Not 50 mA.

  17. (Byron A Jeff) wrote in
    True, it seems. I had the misfortune to first bump into this:

    "PIC pins will sink (drive to ground) or source (output to a ground-
    connected load) 50mA." Near the bottom of that page...

    20 mA is enough though, assuming it can sink as well as source.
  18. Byron A Jeff

    Byron A Jeff Guest

    The page is incorrect.
    The PIC can sink and source @ 20mA.

  19. (Byron A Jeff) wrote in
    Ok, while we're in the mood for heavy pedantry, I'll annouce formally and
    disticntly, I KNOW IT WAS INCORRECT. Happy? :)
  20. I'm not sure if have it figured out by now, but here is a simple way to
    figure it.

    Consider the led's red or green has a forward drop of about 2 volt, and its
    equivalent resistance at 20 ma will be 100 ohms. Now take 2) 100 ohm
    resistors to make a voltage divider ( + 5 volt to resistor a with resistor b
    in series then connecting to ground). now with the back to back led
    connected to the center of your divider being 2.5 volt and the other end of
    the led to the Pic output, A high or a low will give roughly 2.5 volt
    difference on the led. You may or may not even need a resistor to drop the
    extra .5 volt but if you do, you end up with 3 resistors and the led instead
    of the extra opamp's. I would still consider at least a transistor buffer on
    the Pic outputs to drive the leds.
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