Alternating Red/Blue LED Flasher Circuit

[Jamie Jackson]

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.

Thanks,
Jamie

 

[Eeyore]

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.

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

 

[Jamie Jackson]

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

 

[Jonathan Kirwan]

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).

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.

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.

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.

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

For other interested, let me post the ascii schematic you are
referring to:

: V+ V+
: | |
: | |
: | |
: Q1 e>| |<e Q2
: 2N3906 |----, ,----| 2N3906
: c/| | C1 | |\c
: | | 22u | |
: | | || | |
: +----, +----||---------+
: | | | || | |
: | | | | |
: \ | | C2 | \
: R3 / | | || 22u | / R4
: 220 \ '---||-----+ \ 220
: / | || | /
: | | | |
: | | | |
: | \ \ |
: --- / R1 / R2 ---
: \ / D1 \ 100k \100k \ / D2
: --- LED / / --- LED
: | | | |
: | | | |
: | | | |
: gnd gnd gnd gnd

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

 

[Eeyore]

I think that blue's produce some 3-4 times the lumens per watt, roughly
speaking.

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

Graham

 

[Jamie Jackson]

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.

 

[Jamie Jackson]

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.

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.

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

For other interested, let me post the ascii schematic you are
referring to:

: V+ V+
: | |
: | |
: | |
: Q1 e>| |<e Q2
: 2N3906 |----, ,----| 2N3906
: c/| | C1 | |\c
: | | 22u | |
: | | || | |
: +----, +----||---------+
: | | | || | |
: | | | | |
: \ | | C2 | \
: R3 / | | || 22u | / R4
: 220 \ '---||-----+ \ 220
: / | || | /
: | | | |
: | | | |
: | \ \ |
: --- / R1 / R2 ---
: \ / D1 \ 100k \100k \ / D2
: --- LED / / --- LED
: | | | |
: | | | |
: | | | |
: gnd gnd gnd gnd

Thanks!

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.

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.

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.

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

 

[Don Klipstein]

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

Graham

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:

http://www.misty.com/~don/ledrgb2w.html

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 ([email protected])

 

[[email protected]]

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:

http://www.misty.com/~don/ledrgb2w.html

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 ([email protected])

 

[[email protected]]

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
www.forrestmims.org
www.sunandsky.org

 

[Jamie Jackson]

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.

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. ;-)

 

[neon]

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?

 

[Rich Grise]

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.

Just FYI, "Pinewood Derby" is done indoors - the cars will fit in the
palm of your hand:
http://www.google.com/search?q="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!
Rich

 

[Rich Grise]

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.

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

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

Cheers!
Rich

 

[John Fields]

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.

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.

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

For other interested, let me post the ascii schematic you are
referring to:

: V+ V+
: | |
: | |
: | |
: Q1 e>| |<e Q2
: 2N3906 |----, ,----| 2N3906
: c/| | C1 | |\c
: | | 22u | |
: | | || | |
: +----, +----||---------+
: | | | || | |
: | | | | |
: \ | | C2 | \
: R3 / | | || 22u | / R4
: 220 \ '---||-----+ \ 220
: / | || | /
: | | | |
: | | | |
: | \ \ |
: --- / R1 / R2 ---
: \ / D1 \ 100k \100k \ / D2
: --- LED / / --- LED
: | | | |
: | | | |
: | | | |
: gnd gnd gnd gnd

Thanks!

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.

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.

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.

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.)

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

..+9V>----------+---------+------------+----------+
.. | | | |
.. +---+--|------+ | | |
.. | | | | | | |
.. [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] |
.. | | | |
..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.

 

[Jamie Jackson]

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.

Let me know if you have ideas, or need more information.
For other interested, let me post the ascii schematic you are
referring to:
: V+ V+
: | |
: | |
: | |
: Q1 e>| |<e Q2
: 2N3906 |----, ,----| 2N3906
: c/| | C1 | |\c
: | | 22u | |
: | | || | |
: +----, +----||---------+
: | | | || | |
: | | | | |
: \ | | C2 | \
: R3 / | | || 22u | / R4
: 220 \ '---||-----+ \ 220
: / | || | /
: | | | |
: | | | |
: | \ \ |
: --- / R1 / R2 ---
: \ / D1 \ 100k \100k \ / D2
: --- LED / / --- LED
: | | | |
: | | | |
: | | | |
: gnd gnd gnd gnd

Thanks!

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.

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.

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.

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.)---

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

.+9V>----------+---------+------------+----------+
. | | | |
. +---+--|------+ | | |
. | | | | | | |
. [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] |
. | | | |
.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.

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

 

[John Fields]

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

.+9V>----------+---------+------------+----------+
. | | | |
. +---+--|------+ | | |
. | | | | | | |
. [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] |
. | | | |
.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.

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!
---
Excellent!
---


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.)

---
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!