Alternating Red/Blue LED Flasher Circuit

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.

Graham

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

Jamie

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.

Jon

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

Graham

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.

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.

Thanks!

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):

Red: V Drop: 1.62 V I: 20.8 mA

Blue: 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.)

Thanks, Jamie

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

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 common.

- Don Klipstein ( snipped-for-privacy@misty.com)

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

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 backup.

Thanks, Jamie

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

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

.+9V>----------+---------+------------+----------+
.              |         |            |          |      
.       +---+--|------+  |            |          |     
.       |   |  |      |  |            |          |
.     [1M]]--+
.       |     1| 7555                 |A         |
.       |+     |                    [LED1]       |  
.     [1µF     |                      |         [R2]
.       |      |                     [R1]        |
.       |      |                      |          |
.GND>---+------+----------------------+----------+

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.

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

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! Rich

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

Cheers! Rich

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t=2E

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.)

Jamie

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Excellent!