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Alternating Red/Blue LED Flasher Circuit

Discussion in 'Electronic Basics' started by Jamie Jackson, Jan 21, 2007.

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  1. My son and I are working on a Pinewood Derby (Boy Scout) car, and he
    wants to put police lights on top. We're trying a dual LED flasher
    circuit from a Forest Mims mini-notebook ("Optoelectronics Circuits" --
    page 16, if you've got it).

    This circuit has two mirrored halves, and each half blinks its LED in
    an alternating fashion.

    We're simply trying to modify the circuit to replace a red LED with a
    blue one (resulting in alternating blue/red blinking). The problem is
    that the red LED becomes so much dimmer than the blue, and I can't
    figure out what to do to balance them out.

    Let me know if you have ideas, or need more information.

  2. Eeyore

    Eeyore Guest

    It's probably down to the high efficiency of the blue led making the rid one
    appear dim.

    Try a high efficiency / high output red led.

  3. Okay, I'll have to go get a couple, thanks.

  4. Yes, I have it. It's a multivibrator circuit on the bottom half of
    the page with two red LEDs specified and 220 ohm current limit
    resistors connected to them.
    I'm surprised that you are getting anything bright enough for outdoor
    use to start with. What battery supply are you using? What voltage?
    Higher currents will be present with higher voltages used.
    For other interested, let me post the ascii schematic you are
    referring to:
    The currents through the LEDs will be set by the difference in voltage
    across them divided by the 220 ohm resistors. With the same type of
    LEDs on both sides, this current should be roughly the same. However,
    you are using different types of LEDs and the voltage they require is
    different by at least a volt and probably more.

    Without going to look and having little experience to draw upon, I'd
    guess that a red LED is probably in the 670nm wavelength ballpark and
    the blue is probably 425nm, so if the red requires some 2V, your blue
    will need at least 670/425*2 or maybe 3.1V-3.2V. I think they are
    actually a little higher than that, maybe closer to 3.5V, but I
    haven't tested them so I don't know. This ratio of wavelengths
    completely discounts purely resistive aspects that also generate a
    voltage when current passes through, so don't take it as a very
    accurate way to think about it. But it gets you kind of close.

    Since the collector will be close to your supply voltage, and since
    the different kinds of diodes take differing voltages themselves with
    blue needing more voltage, I'd guess that the current through your
    blue LED is actually __less__ than what is flowing through your red
    LED, just now. Yet you observe that the red LED appears much dimmer
    by comparison. So why?

    Well, I don't know for sure. Some thoughts are this: I think that
    blue's produce some 3-4 times the lumens per watt, roughly speaking.
    Since watts is amps*volts, the red LED will be about 2*(Vbat-2)/220
    [which is (2*Vbat-4)/220] and the blue LED will be 3.3*(V-3.3)/220
    [which is (3.3*Vbat-10.89)/220]. Relatively speaking, the 220 ohm
    resistor cancels out and, applying the 3-4X factor, you have:

    3.3*Vbat - 11
    [3..4] * -------------
    2.0*Vbat - 4

    Or thereabouts. With Vbat=12V, this is about 4.3-5.7 times as bright.
    With Vbat=6V, this is 3.3-4.4 times as bright.

    In other words, it does actually appears to me that you __should__ see
    the blue LED as brighter. And you do. Okay, so no surprise.

    What to do?? Well, one recommendation might be to select a red LED
    with higher lumens per watt -- maybe Agilent's AlGaAsP? That would
    perhaps help even the field. Another idea might be to try and greatly
    increase the current through your red LED by reducing its 220 ohm
    resistor, but I suspect you will only be able to go 'so far' in that
    regard and you won't reasonly get the 4X current you want, that way,
    without other problems showing up. So you could reduce the current
    into the blue LED by increasing its 220 ohm resistor to say 1k or
    larger and see what happens. But my bet is on selecting something
    like Agilent's technology for more lumens/watt in the red LED.

    But what may also help is if you could drive your LEDs separately, set
    them up so that they are each invidually at the brightness you want to
    see them operating at, and then measure their voltage [across] and
    their current [through] and post the results of those measurements.
    Knowing what currents and voltages are required for your needs will
    provide needed input for a design.

  5. Eeyore

    Eeyore Guest

    The other thing is the most of the blue leds around typically have narrower
    viewing angles so appear brighter because of that too.

  6. You're right, and I'm after an omni-directional effect, since the car
    isn't viewed from above. The first thing I did to the blues was scuff
    up the entire lens with fine sandpaper, and it made the whole bulb glow
    from all angles. Before I did that, it was hard to tell they were on
    from the side.
  7. This is an *indoor* thing, with little, 8" long cars (as opposed to the
    outdoor "soap box" racers).

    9V (6xAAs) in our Electonics Lab, but eventually moving to a single 9V
    in the car.
    I could swear that varying those 220 ohm resistors didn't have the
    dimming effect I thought it would. I could swear that it varied the
    blink rate. I could be wrong about that...

    I'm going to pick up some brighter red LEDs, to give them a shot.
    To get them visually even (@~6V):

    V Drop: 1.62 V
    I: 20.8 mA

    V Drop: 3.32 V
    I: 2.45 mA

    By the way, I just found a TLC555 IC in my stash of components, so I
    think that gives me another (more compact) way to skin the cat.
    However, I have only found NE555 dual flasher schematics, and they
    don't seem to work unmodified. (I don't know enough to modify it.)

  8. I want to mention another factor: Blue light sources have visibility
    and visual impact greater than would be indicated by their photometric

    The main reason is that blue light has more "color impact" (my words)
    than other color light of same photometric quantity.

    One example I can cite is a calculation that I did where a mixture of
    red, green, and blue LEDs added up to (providing adequate color mixing was
    achieved) a shade of white with color temperature of 5000 K.

    7% of the light by photometric measurements was from the blue LEDs.
    29% was from the red LEDs and 64% was from the green LEDs.

    This example (where % lumens from each color LED is not shown as of
    11:11 PM EST 1/21/07 but is easily enough dervivable from) is the left
    half of the 5000K row of the equal-current example set in:

    The right side of that same row uses a different, more pure blue LED
    (that appears to me to have run into disfavor from lower photometric
    figures, despite being more suitable than "regular blue LEDs" in RGB
    work). That specific example achieves 5000 K white with 5.1% of the
    photometric content being from the blue LEDs, 66.6% from the green LEDs
    and 28.3% from the red LEDs.

    I have done some experimentation and personal studying into "blue
    impact" translating to "noticeability" of blue LEDs.
    So far, it appears to me that "regular blue " LEDs (nominal wavelength
    470 nm) can easily appear 30-50% brighter than indicated by their
    photometric data, and that "royal blue" LEDs (although coming up dimmer
    than "regular blue" ones) can easily appear 45-90% brighter than indicated
    by their photometric data.

    Depending on ambient lighting and how adapted your eyes are to
    brightness/darkness and whether you are viewing the light with central
    vision or peripheral vision, scotopic vision ("night vision") may be
    coming into play. Its alternative, "photopic vision", has a spectral
    response well enough known and agreed upon at conventions to be basis of
    definition of photometric content in a given radiometric quantity of light
    as a function of its spectrum.
    Should scotopic vision be having a significant effect, then the blue
    LEDs can look extra-bright and the red ones a little dim. Keep in mind
    that there is controversy over significance of scotopic vision when
    ambient lighting level is high enough for photopic vision to be dominant.

    So, blue LEDs can appear extra-bright or extra-noticeable.

    One other factor - blue LEDs have a higher voltage drop than red ones,
    and for equal current the blue ones get more power.

    Another factor - if current is on the low side, most red LEDs (one
    exception being a "low current red" chemistry) lose efficiency and blue
    ones tend to not be so bad with efficiency loss at lower current - in
    fact, with moderate underpowering most blue LEDs gain efficiency!

    Another thing - choice of red LEDs. Ones with peak wavelength in the
    630's and dominant wavelength in the 620's of nm have a high tendency to
    have much more "luminous efficacy" (photometric efficiency) than ones with
    peak wavelength around 660 nm. And Radio Shack does not appear to me to
    have these really good red ones, despite the fact that they are now

    - Don Klipstein ()
  9. Guest

  10. Guest

    There's really not much I can add to the replies already provided.

    Regarding the higher voltage requirement of blue LEDs, all that's
    necessary is to power the flasher at, say, 6 or 9 volts. This will
    permit both the blue and red LEDs to operate, and the current limiting
    resistors can then be adjusted as described in other replies.

    If you can't find suitable LEDs at Radio Shack, try DigiKey or JameCo.

    Forrest M. Mims III
  11. Okay, thanks. I'll try modifying those current limiting resistors again
    tonight, but will pick up some bright red LEDs on the way home as a


    P.S. My son will get a kick out of your reply. You and J. K. Rowling
    are his favorite authors at the moment. ;-)
  12. neon


    Oct 21, 2006
    Not all LEDS are created equal. therefore when you buy leds you must buy LEDs as milicandle of brightness from a 300mm to 19,000mm the brightness is controlloed by current as opposed to voltage either way you never could match a 300 mm to a 19,000 mm. having say that buy a close match of mm to begin with then change the current trough each LED to closely match the visual output to satisfaction. got that?
  13. Rich Grise

    Rich Grise Guest

    Just FYI, "Pinewood Derby" is done indoors - the cars will fit in the
    palm of your hand:"Pinewood+Derby"

    As for the OP's Q, I concur with Eeyore - look for a more efficient LED;
    I might add, try different dropping resistors. :)

    Good Luck!
  14. Rich Grise

    Rich Grise Guest

    Don Lancaster and Winfield (Win) Hill hang out around here too. :)

  15. John Fields

    John Fields Guest

    This will work for a TLC555: (View in Courier)

    .. | | | |
    .. +---+--|------+ | | |
    .. | | | | | | |
    .. [1M]<-+ | | | | |A
    .. | |8 | | | [LED2]
    .. | 6+---+---+3 | | E |
    .. +--|TH OUT|--+--|--[10K]---B 2N3906 |
    .. | 2|___ _|4 | C |
    .. +-O|TR R|O----+ | |
    .. | +---+---+ +--[CR1>]--+
    .. | 1| 7555 |A |
    .. |+ | [LED1] |
    .. [1µF | | [R2]
    .. | | [R1] |
    .. | | | |

    I've shown a 1 megohm pot hooked up as a rheostat for the timing
    resistor but you can substitute a fixed resistor if you don't want
    to be able to adjust the flash rate.

    Using a 360 kohm resistor will give you about a half-second of red
    and a half-second of blue at a 50% duty cycle

    Because of the lower voltage drop of the red LED, LED2 should be the
    blue LED, and CR1 can be any common diode (1N4148, 1N40XX...)

    Select R1 and R2 to give you the brightness you need, and if you use
    a high-efficiency 2mA red LED like an HLMP-4700, R1 should be around

    Vcc - Vled 9V - 1.8V
    R = ------------ = ----------- = 3600 ohms
    Iled 0.002A

    The power the resistor will dissipate will be:

    P = Iled (Vcc - Vled) = 0.002A * 7.2V = 0.014 watts,

    so a standard 5% 1/4 watt carbon film resistor would be fine.
  16. Thanks so much for that diagram. My son thinks you're "really nice" for
    having created it. (I do too.)

    I used it successfully, and the car has police lights!

    I also bought some higher-efficiency reds, and I was able to balance
    them really well with the blues by adjusting the limiting resistors.

    One small note: I think your equations for R and P are slightly off. I
    think the current should be 0.02A... well, for my red (20mA) LEDs,
    anyway. (Let me know if I'm wrong about this.)

  17. John Fields

    John Fields Guest

    for a standard 20mA LED, you're right, it should be 0.02A.

    However, for the high-efficiency red HLMP-4700 I used as an example
    it should be 0.002A, since that's what it's rated for.

    In the end though, in order to balance the LEDs the current's going
    to have to be what it has to be as long as you don't overdrive
    either or both of the LEDs.

    I'm glad you got it working! :)
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