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Sci.Electronics.Basics -> Help: Can someone assist with load resistor calculation for vehicle LED lamp please?

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Author: Lee Wilkinson
Date: 13:44 04-06-08

Hi folks,

Don't have much electronics experience and have forgotten most of what I
learnt at school (well it was 25 years ago!) and am hoping I can find some
assistance in getting an LED lamp for my motorbike working correctly....

There is an aftermarket lamp made by Clear Alternatives that functions
correctly but has a very poorly designed LED board as far as vision is
concerned (very narrow beam and mis-aligned LEDs as well as poorly made lamp
body).

I've made my own LED board and have found lots of resources on the web to
tell me what resistors to use with the LEDs to make the lamp function
correctly with the 14.4V my motorbike supplies the lamp. My problem has
arisen due to the fact that my BMW motorbike has a CanBus wiring system that
has a bulb failure warning circuit. To get round this the aftermarket lamp
has a load resistor across each of the running and brake circuits - fooling
the onboard system into thinking there's a standard 5/21W tungsten bulb
fitted.

When connected, my lamp functions correctly with lower level running light
and full level brake light and looks great. I guessed on the 100 ohms for
the running light but it looks about the same as the 5W tungsten to my eyes.

If I take one of the aftermarket board load resistors and connect it across
the brake circuit, the lamp still functions correctly and it fools the
warning system. However when I connect a 2nd load resistor across the
running light circuit the LEDs all go out!! I presume that as the LEDs are
already dim on the running light circuit, that they get too little current
when the load resistor is insterted and so go out.

What I need to know is what load resistors to use, and if necessary what
resistors to change on the actual lamp board so they can all work together.

Even better, if someone were actually able to explain to me how it's worked
out, I can learn something for the future too!

Here are the diagrams of the aftermarket lamp that works and my new home
built one for reference.
http://www.postimage.org/image.php?v=aV2gM5w9
http://www.postimage.org/image.php?v=Pq295j6A

Can anyone help me with the above or point me at any reference material,
online forums / guides on this that can help?

You'll make a very frustrated man happy!!

Cheers,
Lee

Author: Tom Biasi
Date: 15:55 04-06-08

Author: Michael Robinson
Date: 16:46 04-06-08

Author: Gyro
Date: 16:51 04-06-08

On 04/06/2008 20:55, in article
e9OdncMvBq3WbtvVnZ2dnUVZ_h_inZ2d@giganews.com, "Tom Biasi"
<tombiasi***@optonline.net> wrote:

>
> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
> news:C46C9301.389%thepanda@btconnect.com...
>> Hi folks,
>>
>> Don't have much electronics experience and have forgotten most of what I
>> learnt at school (well it was 25 years ago!) and am hoping I can find some
>> assistance in getting an LED lamp for my motorbike working correctly....
>>
>> There is an aftermarket lamp made by Clear Alternatives that functions
>> correctly but has a very poorly designed LED board as far as vision is
>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
>> lamp
>> body).
>>
>> I've made my own LED board and have found lots of resources on the web to
>> tell me what resistors to use with the LEDs to make the lamp function
>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
>> that
>> has a bulb failure warning circuit. To get round this the aftermarket lamp
>> has a load resistor across each of the running and brake circuits -
>> fooling
>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>> fitted.
>>
>> When connected, my lamp functions correctly with lower level running light
>> and full level brake light and looks great. I guessed on the 100 ohms for
>> the running light but it looks about the same as the 5W tungsten to my
>> eyes.
>>
>> If I take one of the aftermarket board load resistors and connect it
>> across
>> the brake circuit, the lamp still functions correctly and it fools the
>> warning system. However when I connect a 2nd load resistor across the
>> running light circuit the LEDs all go out!! I presume that as the LEDs are
>> already dim on the running light circuit, that they get too little current
>> when the load resistor is insterted and so go out.
>>
>> What I need to know is what load resistors to use, and if necessary what
>> resistors to change on the actual lamp board so they can all work
>> together.
>>
>> Even better, if someone were actually able to explain to me how it's
>> worked
>> out, I can learn something for the future too!
>>
>> Here are the diagrams of the aftermarket lamp that works and my new home
>> built one for reference.
>> http://www.postimage.org/image.php?v=aV2gM5w9
>> http://www.postimage.org/image.php?v=Pq295j6A
>>
>> Can anyone help me with the above or point me at any reference material,
>> online forums / guides on this that can help?
>>
>> You'll make a very frustrated man happy!!
>>
>> Cheers,
>> Lee
>>
>>
>
> Hi Lee,
> Without knowing what you monitor circuit is looking for we can only make
> some assumptions based on the circuits that work and the problems you have
> with your circuit.
> It would probably be easier to start with a high value load that works for
> the LEDs and reduce it until it "fools" the monitor and the go a little
> lower.
>
> Tom
>
>
Hi Tom,
I can only presume it's looking for a 5w and 21w load on each circuit, the
same as supplied by a tungsten bulb. I have no idea how to calculate how
this equates to the 46 ohm load resistor that CA used on their board.

They used a low 43 ohm resistor on their running section and yet the LEDs
are too dim in mine and others opinion. If I were to drop on my board from
100 ohm down closer to the 43 used by CA, it may be too bright as a running
light without the load resistor but would that then be pulled back into line
when a load resistor was put across? - just a thought as a newbie!

What sort of value would you suggest as a high load across? I know these
need to be higher wattage resistors and can get hot but would a high value
resistor just force the current through the LEDs (an easier route so to
speak) or would it just dissipate the current as heat? Sorry if that sounds
silly but this really is new to me!

Cheers,
Lee

Author: Gyro
Date: 17:11 04-06-08

On 04/06/2008 21:46, in article g26uv8$8aa$1@aioe.org, "Michael Robinson"
<kellrobinson@yahoo.com> wrote:

>
> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
> news:C46C9301.389%thepanda@btconnect.com...
>> Hi folks,
>>
>> Don't have much electronics experience and have forgotten most of what I
>> learnt at school (well it was 25 years ago!) and am hoping I can find some
>> assistance in getting an LED lamp for my motorbike working correctly....
>>
>> There is an aftermarket lamp made by Clear Alternatives that functions
>> correctly but has a very poorly designed LED board as far as vision is
>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
>> lamp
>> body).
>>
>> I've made my own LED board and have found lots of resources on the web to
>> tell me what resistors to use with the LEDs to make the lamp function
>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
>> that
>> has a bulb failure warning circuit. To get round this the aftermarket lamp
>> has a load resistor across each of the running and brake circuits -
>> fooling
>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>> fitted.
>>
>> When connected, my lamp functions correctly with lower level running light
>> and full level brake light and looks great. I guessed on the 100 ohms for
>> the running light but it looks about the same as the 5W tungsten to my
>> eyes.
>>
>> If I take one of the aftermarket board load resistors and connect it
>> across
>> the brake circuit, the lamp still functions correctly and it fools the
>> warning system. However when I connect a 2nd load resistor across the
>> running light circuit the LEDs all go out!! I presume that as the LEDs are
>> already dim on the running light circuit, that they get too little current
>> when the load resistor is insterted and so go out.
>>
>> What I need to know is what load resistors to use, and if necessary what
>> resistors to change on the actual lamp board so they can all work
>> together.
>>
>> Even better, if someone were actually able to explain to me how it's
>> worked
>> out, I can learn something for the future too!
>>
>> Here are the diagrams of the aftermarket lamp that works and my new home
>> built one for reference.
>> http://www.postimage.org/image.php?v=aV2gM5w9
>> http://www.postimage.org/image.php?v=Pq295j6A
>>
>> Can anyone help me with the above or point me at any reference material,
>> online forums / guides on this that can help?
>>
>> You'll make a very frustrated man happy!!
>>
>> Cheers,
>> Lee
>
>
> Check your wiring.
>
> If you connect the "running light" load resistor to the wrong end of the
100
> ohm resistor, it will make the lights go out.
>
>
Hi Michael,

I was connecting it the same way as CA connected theirs - earth to the feed
side of the 'running light' 100 ohm reduction resistor not the LED/Brake
light feed side.

If I connect it to the LED side next to the brake feed I'll effectively be
adding another load resistor in parallel to the brake load won't I, and
halve the load resistance? I'm not sure how it would be 'seen' on the
running light monitor?

Cheers,
Lee

Author: Michael Robinson
Date: 17:31 04-06-08

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46CC395.744%thepanda@btconnect.com...
>
>
>
> On 04/06/2008 21:46, in article g26uv8$8aa$1@aioe.org, "Michael Robinson"
> <kellrobinson@yahoo.com> wrote:
>
>>
>> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
>> news:C46C9301.389%thepanda@btconnect.com...
>>> Hi folks,
>>>
>>> Don't have much electronics experience and have forgotten most of what I
>>> learnt at school (well it was 25 years ago!) and am hoping I can find
>>> some
>>> assistance in getting an LED lamp for my motorbike working correctly....
>>>
>>> There is an aftermarket lamp made by Clear Alternatives that functions
>>> correctly but has a very poorly designed LED board as far as vision is
>>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
>>> lamp
>>> body).
>>>
>>> I've made my own LED board and have found lots of resources on the web
>>> to
>>> tell me what resistors to use with the LEDs to make the lamp function
>>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
>>> that
>>> has a bulb failure warning circuit. To get round this the aftermarket
>>> lamp
>>> has a load resistor across each of the running and brake circuits -
>>> fooling
>>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>>> fitted.
>>>
>>> When connected, my lamp functions correctly with lower level running
>>> light
>>> and full level brake light and looks great. I guessed on the 100 ohms
>>> for
>>> the running light but it looks about the same as the 5W tungsten to my
>>> eyes.
>>>
>>> If I take one of the aftermarket board load resistors and connect it
>>> across
>>> the brake circuit, the lamp still functions correctly and it fools the
>>> warning system. However when I connect a 2nd load resistor across the
>>> running light circuit the LEDs all go out!! I presume that as the LEDs
>>> are
>>> already dim on the running light circuit, that they get too little
>>> current
>>> when the load resistor is insterted and so go out.
>>>
>>> What I need to know is what load resistors to use, and if necessary what
>>> resistors to change on the actual lamp board so they can all work
>>> together.
>>>
>>> Even better, if someone were actually able to explain to me how it's
>>> worked
>>> out, I can learn something for the future too!
>>>
>>> Here are the diagrams of the aftermarket lamp that works and my new home
>>> built one for reference.
>>> http://www.postimage.org/image.php?v=aV2gM5w9
>>> http://www.postimage.org/image.php?v=Pq295j6A
>>>
>>> Can anyone help me with the above or point me at any reference material,
>>> online forums / guides on this that can help?
>>>
>>> You'll make a very frustrated man happy!!
>>>
>>> Cheers,
>>> Lee
>>
>>
>> Check your wiring.
>>
>> If you connect the "running light" load resistor to the wrong end of
the
>> 100
>> ohm resistor, it will make the lights go out.
>>
>>
> Hi Michael,
>
> I was connecting it the same way as CA connected theirs - earth to the
> feed
> side of the 'running light' 100 ohm reduction resistor not the LED/Brake
> light feed side.
>
> If I connect it to the LED side next to the brake feed I'll effectively be
> adding another load resistor in parallel to the brake load won't I, and
> halve the load resistance? I'm not sure how it would be 'seen' on the
> running light monitor?
>
> Cheers,
> Lee
>
>
Well, something doesn't jibe.

If I understand you correctly,

when the brake is not applied, and you don't have a load resistor installed
on the running light feed, the running light works;

but then, when you insert a load resistor from the "+14.4V Running Light"
node in your circuit to ground, the running light dies.

However, "+14.4 Running Light" is a low impedance power source, and
diverting some small fraction of an amp to ground from that power source
should have negligible effect; the voltage at that node should hardly change
at all with the addition of some small load. Either something's out of
kilter with your circuitry or else I misunderstood your post.

The points on your diagram labeled in red "+14.4V" signify a connection to
your bike's positive power source, do they not?

Author: Dan Coby
Date: 17:59 04-06-08

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...
> Hi folks,
>
> Don't have much electronics experience and have forgotten most of what I
> learnt at school (well it was 25 years ago!) and am hoping I can find some
> assistance in getting an LED lamp for my motorbike working correctly....
>
> There is an aftermarket lamp made by Clear Alternatives that functions
> correctly but has a very poorly designed LED board as far as vision is
> concerned (very narrow beam and mis-aligned LEDs as well as poorly made lamp
> body).
>
> I've made my own LED board and have found lots of resources on the web to
> tell me what resistors to use with the LEDs to make the lamp function
> correctly with the 14.4V my motorbike supplies the lamp. My problem has
> arisen due to the fact that my BMW motorbike has a CanBus wiring system that
> has a bulb failure warning circuit. To get round this the aftermarket lamp
> has a load resistor across each of the running and brake circuits - fooling
> the onboard system into thinking there's a standard 5/21W tungsten bulb
> fitted.
>
> When connected, my lamp functions correctly with lower level running light
> and full level brake light and looks great. I guessed on the 100 ohms for
> the running light but it looks about the same as the 5W tungsten to my eyes.
>
> If I take one of the aftermarket board load resistors and connect it across
> the brake circuit, the lamp still functions correctly and it fools the
> warning system. However when I connect a 2nd load resistor across the
> running light circuit the LEDs all go out!! I presume that as the LEDs are
> already dim on the running light circuit, that they get too little current
> when the load resistor is insterted and so go out.
>
> What I need to know is what load resistors to use, and if necessary what
> resistors to change on the actual lamp board so they can all work together.
>
> Even better, if someone were actually able to explain to me how it's worked
> out, I can learn something for the future too!
>
> Here are the diagrams of the aftermarket lamp that works and my new home
> built one for reference.
> http://www.postimage.org/image.php?v=aV2gM5w9
> http://www.postimage.org/image.php?v=Pq295j6A
>
> Can anyone help me with the above or point me at any reference material,
> online forums / guides on this that can help?
>
> You'll make a very frustrated man happy!!
>
> Cheers,
> Lee

The short form of the answer is that you need to reduce the 100 ohm resistor.
If I were you, I would try 43 ohms (i.e. the value from the original circuit.)

Without more information about the monitor circuit, we can only guess about
the values for the green resistors. The 46 ohm value in the original circuit seems
to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5 watts).
Since this value works, you can continue to use it. If you want, you can try higher
values. Higher values will give you less power being wasted in these resistors.
However at some point with higher values the monitor circuit will decide that
the 'light bulb' is burned out.

You chose the 100 ohm resistor to give the correct brightness for the running
lights. However, I assume, that the second 46 ohm resistor was not connected
when you chose that value. Put the second 46 ohm resistor into your circuit and
try lower values for the current 100 ohm resistor until you get the brightness
desired.

The current combination of a 46 ohm resistor and a 100 ohm resistor creates
a 'voltage divider' circuit. I.e. (view in a fixed font):

_____
------| 100 |-----+--------o
| ----- |
+ ___
14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
- | 6 |
| ---
| |
|__________________|________o

The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4 volts,
R1 is 100 ohms and R2 is 46 ohms.

For your values, this gives 4.5 volts. However your LEDs need a minimum of 7.2
volts (4 x 1.8 volts) before then will produce light. So you need to change the
ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you probably
cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).

Author: Gyro
Date: 18:10 04-06-08

On 04/06/2008 22:31, in article g271k3$imu$1@aioe.org, "Michael Robinson"
<kellrobinson@yahoo.com> wrote:

>
> "Gyro" <thepanda@btconnect.com> wrote in message
> news:C46CC395.744%thepanda@btconnect.com...
>>
>>
>>
>> On 04/06/2008 21:46, in article g26uv8$8aa$1@aioe.org, "Michael
Robinson"
>> <kellrobinson@yahoo.com> wrote:
>>
>>>
>>> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
>>> news:C46C9301.389%thepanda@btconnect.com...
>>>> Hi folks,
>>>>
>>>> Don't have much electronics experience and have forgotten most of what
I
>>>> learnt at school (well it was 25 years ago!) and am hoping I can find
>>>> some
>>>> assistance in getting an LED lamp for my motorbike working
correctly....
>>>>
>>>> There is an aftermarket lamp made by Clear Alternatives that functions
>>>> correctly but has a very poorly designed LED board as far as vision is
>>>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
>>>> lamp
>>>> body).
>>>>
>>>> I've made my own LED board and have found lots of resources on the web
>>>> to
>>>> tell me what resistors to use with the LEDs to make the lamp function
>>>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>>>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
>>>> that
>>>> has a bulb failure warning circuit. To get round this the aftermarket
>>>> lamp
>>>> has a load resistor across each of the running and brake circuits -
>>>> fooling
>>>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>>>> fitted.
>>>>
>>>> When connected, my lamp functions correctly with lower level running
>>>> light
>>>> and full level brake light and looks great. I guessed on the 100 ohms
>>>> for
>>>> the running light but it looks about the same as the 5W tungsten to my
>>>> eyes.
>>>>
>>>> If I take one of the aftermarket board load resistors and connect it
>>>> across
>>>> the brake circuit, the lamp still functions correctly and it fools the
>>>> warning system. However when I connect a 2nd load resistor across the
>>>> running light circuit the LEDs all go out!! I presume that as the LEDs
>>>> are
>>>> already dim on the running light circuit, that they get too little
>>>> current
>>>> when the load resistor is insterted and so go out.
>>>>
>>>> What I need to know is what load resistors to use, and if necessary
what
>>>> resistors to change on the actual lamp board so they can all work
>>>> together.
>>>>
>>>> Even better, if someone were actually able to explain to me how it's
>>>> worked
>>>> out, I can learn something for the future too!
>>>>
>>>> Here are the diagrams of the aftermarket lamp that works and my new
home
>>>> built one for reference.
>>>> http://www.postimage.org/image.php?v=aV2gM5w9
>>>> http://www.postimage.org/image.php?v=Pq295j6A
>>>>
>>>> Can anyone help me with the above or point me at any reference
material,
>>>> online forums / guides on this that can help?
>>>>
>>>> You'll make a very frustrated man happy!!
>>>>
>>>> Cheers,
>>>> Lee
>>>
>>>
>>> Check your wiring.
>>>
>>> If you connect the "running light" load resistor to the wrong end
of the
>>> 100
>>> ohm resistor, it will make the lights go out.
>>>
>>>
>> Hi Michael,
>>
>> I was connecting it the same way as CA connected theirs - earth to the
>> feed
>> side of the 'running light' 100 ohm reduction resistor not the LED/Brake
>> light feed side.
>>
>> If I connect it to the LED side next to the brake feed I'll effectively be
>> adding another load resistor in parallel to the brake load won't I, and
>> halve the load resistance? I'm not sure how it would be 'seen' on the
>> running light monitor?
>>
>> Cheers,
>> Lee
>>
>>
> Well, something doesn't jibe.
>
> If I understand you correctly,
>
> when the brake is not applied, and you don't have a load resistor installed
> on the running light feed, the running light works;
>
> but then, when you insert a load resistor from the "+14.4V Running Light"
> node in your circuit to ground, the running light dies.
>
> However, "+14.4 Running Light" is a low impedance power source, and
> diverting some small fraction of an amp to ground from that power source
> should have negligible effect; the voltage at that node should hardly change
> at all with the addition of some small load. Either something's out of
> kilter with your circuitry or else I misunderstood your post.
>
> The points on your diagram labeled in red "+14.4V" signify a connection
to
> your bike's positive power source, do they not?
>
>
I'll connect all up again in the morning and double check, but yes I'm sure
you understand me & I've described the symptoms correctly. The circuits are
definitely as per my diagrams and I'm at a loss to work out why the CA
manufactured board works and mine doesn't - I know it's probably the
resistor values but don't know how to work them out hence posting.

For something being out of kilter, maybe it's my choice of the 100 ohm to
reduce the power for the running light. I don't know how to calculate what t
should be so took a guess by connecting a few different resistors until I
got something that 'looked' right. That's why I wondered whether changing it
for a lower value would be better.

Will check again as I say and come back.

Cheers, Lee

Author: Gyro
Date: 18:18 04-06-08

On 04/06/2008 22:59, in article
__-dnbPOTpbNjdrVnZ2dnUVZ_qqgnZ2d@earthlink.com, "Dan Coby"
<adcoby@earthlink.net> wrote:

> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
> news:C46C9301.389%thepanda@btconnect.com...
>> Hi folks,
>>
>> Don't have much electronics experience and have forgotten most of what I
>> learnt at school (well it was 25 years ago!) and am hoping I can find some
>> assistance in getting an LED lamp for my motorbike working correctly....
>>
>> There is an aftermarket lamp made by Clear Alternatives that functions
>> correctly but has a very poorly designed LED board as far as vision is
>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made lamp
>> body).
>>
>> I've made my own LED board and have found lots of resources on the web to
>> tell me what resistors to use with the LEDs to make the lamp function
>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>> arisen due to the fact that my BMW motorbike has a CanBus wiring system that
>> has a bulb failure warning circuit. To get round this the aftermarket lamp
>> has a load resistor across each of the running and brake circuits - fooling
>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>> fitted.
>>
>> When connected, my lamp functions correctly with lower level running light
>> and full level brake light and looks great. I guessed on the 100 ohms for
>> the running light but it looks about the same as the 5W tungsten to my eyes.
>>
>> If I take one of the aftermarket board load resistors and connect it across
>> the brake circuit, the lamp still functions correctly and it fools the
>> warning system. However when I connect a 2nd load resistor across the
>> running light circuit the LEDs all go out!! I presume that as the LEDs are
>> already dim on the running light circuit, that they get too little current
>> when the load resistor is insterted and so go out.
>>
>> What I need to know is what load resistors to use, and if necessary what
>> resistors to change on the actual lamp board so they can all work together.
>>
>> Even better, if someone were actually able to explain to me how it's worked
>> out, I can learn something for the future too!
>>
>> Here are the diagrams of the aftermarket lamp that works and my new home
>> built one for reference.
>> http://www.postimage.org/image.php?v=aV2gM5w9
>> http://www.postimage.org/image.php?v=Pq295j6A
>>
>> Can anyone help me with the above or point me at any reference material,
>> online forums / guides on this that can help?
>>
>> You'll make a very frustrated man happy!!
>>
>> Cheers,
>> Lee
>
> The short form of the answer is that you need to reduce the 100 ohm resistor.
> If I were you, I would try 43 ohms (i.e. the value from the original circuit.)
>
> Without more information about the monitor circuit, we can only guess about
> the values for the green resistors. The 46 ohm value in the original circuit
> seems
> to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5
> watts).
> Since this value works, you can continue to use it. If you want, you can try
> higher
> values. Higher values will give you less power being wasted in these
> resistors.
> However at some point with higher values the monitor circuit will decide that
> the 'light bulb' is burned out.
>
> You chose the 100 ohm resistor to give the correct brightness for the running
> lights. However, I assume, that the second 46 ohm resistor was not connected
> when you chose that value. Put the second 46 ohm resistor into your circuit
> and
> try lower values for the current 100 ohm resistor until you get the brightness
> desired.
>
> The current combination of a 46 ohm resistor and a 100 ohm resistor creates
> a 'voltage divider' circuit. I.e. (view in a fixed font):
>
> _____
> ------| 100 |-----+--------o
> | ----- |
> + ___
> 14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
> - | 6 |
> | ---
> | |
> |__________________|________o
>
> The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4
> volts,
> R1 is 100 ohms and R2 is 46 ohms.
>
> For your values, this gives 4.5 volts. However your LEDs need a minimum of 7.2
> volts (4 x 1.8 volts) before then will produce light. So you need to change
> the
> ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you
> probably
> cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).
>
>

Dan,

Thanks ever so much for the comprehensive reply- I'm going to check all
again in the morning anyway and will study what you've put here to make sure
I understand properly and make changes.

Cheers,
Lee

Author: BobW
Date: 20:45 04-06-08

"Lee Wilkinson" <thepanda@btconnect.com> wrote in message
news:C46C9301.389%thepanda@btconnect.com...
> Hi folks,
>
> Don't have much electronics experience and have forgotten most of what I
> learnt at school (well it was 25 years ago!) and am hoping I can find some
> assistance in getting an LED lamp for my motorbike working correctly....
>
> There is an aftermarket lamp made by Clear Alternatives that functions
> correctly but has a very poorly designed LED board as far as vision is
> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
> lamp
> body).
>
> I've made my own LED board and have found lots of resources on the web to
> tell me what resistors to use with the LEDs to make the lamp function
> correctly with the 14.4V my motorbike supplies the lamp. My problem has
> arisen due to the fact that my BMW motorbike has a CanBus wiring system
> that
> has a bulb failure warning circuit. To get round this the aftermarket lamp
> has a load resistor across each of the running and brake circuits -
> fooling
> the onboard system into thinking there's a standard 5/21W tungsten bulb
> fitted.
>
> When connected, my lamp functions correctly with lower level running light
> and full level brake light and looks great. I guessed on the 100 ohms for
> the running light but it looks about the same as the 5W tungsten to my
> eyes.
>
> If I take one of the aftermarket board load resistors and connect it
> across
> the brake circuit, the lamp still functions correctly and it fools the
> warning system. However when I connect a 2nd load resistor across the
> running light circuit the LEDs all go out!! I presume that as the LEDs are
> already dim on the running light circuit, that they get too little current
> when the load resistor is insterted and so go out.
>
> What I need to know is what load resistors to use, and if necessary what
> resistors to change on the actual lamp board so they can all work
> together.
>
> Even better, if someone were actually able to explain to me how it's
> worked
> out, I can learn something for the future too!
>
> Here are the diagrams of the aftermarket lamp that works and my new home
> built one for reference.
> http://www.postimage.org/image.php?v=aV2gM5w9
> http://www.postimage.org/image.php?v=Pq295j6A
>
> Can anyone help me with the above or point me at any reference material,
> online forums / guides on this that can help?
>
> You'll make a very frustrated man happy!!
>
> Cheers,
> Lee
>

Lee,

I should be able to figure this out since I'm an EE and I have a BMW RT. The
keyword is "should".

I sure love my RT. There's nothing on this planet like them. Great
acceleration, nimble in the twisties, smooth as silk at high speeds, great
gas mileage, comfortable, yada yada...

So, here's what I think is going on.

First, I'm making the assumption that the original circuit looks like this
for both the BRAKE and RUNNING circuits:

12V->SWITCH->LAMP->0V (aka GND), and there's also a high value
"sense"
resistor across the SWITCH.

Assuming that this is true, then when the BRAKE SWITCH is off and the LAMP
is good then the monitor circuit will see a "low voltage" across the LAMP.
If the LAMP is burned out (i.e. it is open) then the monitor circuit will
12V across the LAMP even though the SWITCH is off (thanks to the "sense"
resistor that's across the SWITCH).

If the above operation is correctly described then you'll have a problem
with your circuit since you're trying to use the same LED string for both
the BRAKE and RUNNING circuits. Specifically, a problem will occur in your
circuit if the RUNNING circuit is activated (RUNNING LIGHTS on) and then the
BRAKE CIRCUIT is tested for a blown lamp. In this condition, even though the
BRAKE SWITCH is off the monitor circuit will see something much greater than
a "low voltage" and will thus declare a blown BRAKE LAMP.

The solution would be to put a diode between the BRAKE load resistor and the
lower LED drive resistor. The anode would connect to the BRAKE load
resistor. This way, the BRAKE monitor circuit will see its "low voltage"
even though the RUNNING circuit is already turned on, and when the BRAKE
switch is on then the LED string will brighter.

It may be necessary to add a diode from the RUNNING circuit, too, in the
case that you've depressed one of the brakes as it's doing its diagnostics.
You'll have to play with this scenario, but (assuming this is all correct) I
would add the addtional diode just in case.

I hope this makes sense and I hope it solves your problem. After all, we're
like BMW motobrothers.

Bob

--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

Author: John G
Date: 23:10 04-06-08

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46CD344.756%thepanda@btconnect.com...
>
>
>
> On 04/06/2008 22:59, in article
> __-dnbPOTpbNjdrVnZ2dnUVZ_qqgnZ2d@earthlink.com, "Dan Coby"
> <adcoby@earthlink.net> wrote:
>
>> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
>> news:C46C9301.389%thepanda@btconnect.com...
>>> Hi folks,
>>>
>>> Don't have much electronics experience and have forgotten most of what I
>>> learnt at school (well it was 25 years ago!) and am hoping I can find
>>> some
>>> assistance in getting an LED lamp for my motorbike working correctly....
>>>
>>> There is an aftermarket lamp made by Clear Alternatives that functions
>>> correctly but has a very poorly designed LED board as far as vision is
>>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
>>> lamp
>>> body).
>>>
>>> I've made my own LED board and have found lots of resources on the web
>>> to
>>> tell me what resistors to use with the LEDs to make the lamp function
>>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
>>> that
>>> has a bulb failure warning circuit. To get round this the aftermarket
>>> lamp
>>> has a load resistor across each of the running and brake circuits -
>>> fooling
>>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>>> fitted.
>>>
>>> When connected, my lamp functions correctly with lower level running
>>> light
>>> and full level brake light and looks great. I guessed on the 100 ohms
>>> for
>>> the running light but it looks about the same as the 5W tungsten to my
>>> eyes.
>>>
>>> If I take one of the aftermarket board load resistors and connect it
>>> across
>>> the brake circuit, the lamp still functions correctly and it fools the
>>> warning system. However when I connect a 2nd load resistor across the
>>> running light circuit the LEDs all go out!! I presume that as the LEDs
>>> are
>>> already dim on the running light circuit, that they get too little
>>> current
>>> when the load resistor is insterted and so go out.
>>>
>>> What I need to know is what load resistors to use, and if necessary what
>>> resistors to change on the actual lamp board so they can all work
>>> together.
>>>
>>> Even better, if someone were actually able to explain to me how it's
>>> worked
>>> out, I can learn something for the future too!
>>>
>>> Here are the diagrams of the aftermarket lamp that works and my new home
>>> built one for reference.
>>> http://www.postimage.org/image.php?v=aV2gM5w9
>>> http://www.postimage.org/image.php?v=Pq295j6A
>>>
>>> Can anyone help me with the above or point me at any reference material,
>>> online forums / guides on this that can help?
>>>
>>> You'll make a very frustrated man happy!!
>>>
>>> Cheers,
>>> Lee
>>
>> The short form of the answer is that you need to reduce the 100 ohm
>> resistor.
>> If I were you, I would try 43 ohms (i.e. the value from the original
>> circuit.)
>>
>> Without more information about the monitor circuit, we can only guess
>> about
>> the values for the green resistors. The 46 ohm value in the original
>> circuit
>> seems
>> to have been chosen to give about a 5 watt load at 14.4 volts (actually
>> 4.5
>> watts).
>> Since this value works, you can continue to use it. If you want, you can
>> try
>> higher
>> values. Higher values will give you less power being wasted in these
>> resistors.
>> However at some point with higher values the monitor circuit will decide
>> that
>> the 'light bulb' is burned out.
>>
>> You chose the 100 ohm resistor to give the correct brightness for the
>> running
>> lights. However, I assume, that the second 46 ohm resistor was not
>> connected
>> when you chose that value. Put the second 46 ohm resistor into your
>> circuit
>> and
>> try lower values for the current 100 ohm resistor until you get the
>> brightness
>> desired.
>>
>> The current combination of a 46 ohm resistor and a 100 ohm resistor
>> creates
>> a 'voltage divider' circuit. I.e. (view in a fixed font):
>>
>> _____
>> ------| 100 |-----+--------o
>> | ----- |
>> + ___
>> 14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5
>> volts
>> - | 6 |
>> | ---
>> | |
>> |__________________|________o
>>
>> The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is
>> 14.4
>> volts,
>> R1 is 100 ohms and R2 is 46 ohms.
>>
>> For your values, this gives 4.5 volts. However your LEDs need a minimum
>> of 7.2
>> volts (4 x 1.8 volts) before then will produce light. So you need to
>> change
>> the
>> ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you
>> probably
>> cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).
>>
>>
>
> Dan,
>
> Thanks ever so much for the comprehensive reply- I'm going to check all
> again in the morning anyway and will study what you've put here to make
> sure
> I understand properly and make changes.
>
> Cheers,
> Lee
>
I agree completly with Dan Coby.

Surely the circuits should be the SAME except for a variation in the values
of the LED dropping resistors if your new LEDS are different to the original
circuit.

Why did you change anything beyond the geometery?
Did you change the TYPE of lead ie. for a brighter model?

The 2 loading resistor should still be 46 ohms as that is what works for the
check circuit which we can only guess about.
If the running lights go out then the 100 ohm resistor is the culprit and
should be lower.

John G.

Author: Dan Coby
Date: 03:51 05-06-08

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46CD344.756%thepanda@btconnect.com...
>
>
>
> On 04/06/2008 22:59, in article
> __-dnbPOTpbNjdrVnZ2dnUVZ_qqgnZ2d@earthlink.com, "Dan Coby"
> <adcoby@earthlink.net> wrote:
>
>> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
>> news:C46C9301.389%thepanda@btconnect.com...
>>> Hi folks,
>>>
>>> Don't have much electronics experience and have forgotten most of what I
>>> learnt at school (well it was 25 years ago!) and am hoping I can find some
>>> assistance in getting an LED lamp for my motorbike working correctly....
>>>
>>> There is an aftermarket lamp made by Clear Alternatives that functions
>>> correctly but has a very poorly designed LED board as far as vision is
>>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
lamp
>>> body).
>>>
>>> I've made my own LED board and have found lots of resources on the web to
>>> tell me what resistors to use with the LEDs to make the lamp function
>>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
that
>>> has a bulb failure warning circuit. To get round this the aftermarket lamp
>>> has a load resistor across each of the running and brake circuits - fooling
>>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>>> fitted.
>>>
>>> When connected, my lamp functions correctly with lower level running light
>>> and full level brake light and looks great. I guessed on the 100 ohms for
>>> the running light but it looks about the same as the 5W tungsten to my
eyes.
>>>
>>> If I take one of the aftermarket board load resistors and connect it across
>>> the brake circuit, the lamp still functions correctly and it fools the
>>> warning system. However when I connect a 2nd load resistor across the
>>> running light circuit the LEDs all go out!! I presume that as the LEDs are
>>> already dim on the running light circuit, that they get too little current
>>> when the load resistor is insterted and so go out.
>>>
>>> What I need to know is what load resistors to use, and if necessary what
>>> resistors to change on the actual lamp board so they can all work together.
>>>
>>> Even better, if someone were actually able to explain to me how it's worked
>>> out, I can learn something for the future too!
>>>
>>> Here are the diagrams of the aftermarket lamp that works and my new home
>>> built one for reference.
>>> http://www.postimage.org/image.php?v=aV2gM5w9
>>> http://www.postimage.org/image.php?v=Pq295j6A
>>>
>>> Can anyone help me with the above or point me at any reference material,
>>> online forums / guides on this that can help?
>>>
>>> You'll make a very frustrated man happy!!
>>>
>>> Cheers,
>>> Lee
>>
>> The short form of the answer is that you need to reduce the 100 ohm resistor.
>> If I were you, I would try 43 ohms (i.e. the value from the original circuit.)
>>
>> Without more information about the monitor circuit, we can only guess about
>> the values for the green resistors. The 46 ohm value in the original circuit
>> seems
>> to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5
>> watts).
>> Since this value works, you can continue to use it. If you want, you can try
>> higher
>> values. Higher values will give you less power being wasted in these
>> resistors.
>> However at some point with higher values the monitor circuit will decide that
>> the 'light bulb' is burned out.
>>
>> You chose the 100 ohm resistor to give the correct brightness for the running
>> lights. However, I assume, that the second 46 ohm resistor was not connected
>> when you chose that value. Put the second 46 ohm resistor into your circuit
>> and
>> try lower values for the current 100 ohm resistor until you get the brightness
>> desired.
>>
>> The current combination of a 46 ohm resistor and a 100 ohm resistor creates
>> a 'voltage divider' circuit. I.e. (view in a fixed font):
>>
>> _____
>> ------| 100 |-----+--------o
>> | ----- |
>> + ___
>> 14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
>> - | 6 |
>> | ---
>> | |
>> |__________________|________o
>>
>> The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4
>> volts,
>> R1 is 100 ohms and R2 is 46 ohms.
>>
>> For your values, this gives 4.5 volts. However your LEDs need a minimum of 7.2
>> volts (4 x 1.8 volts) before then will produce light. So you need to change
>> the
>> ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you
>> probably
>> cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).
>>
>>
>
> Dan,
>
> Thanks ever so much for the comprehensive reply- I'm going to check all
> again in the morning anyway and will study what you've put here to make sure
> I understand properly and make changes.
>
> Cheers,
> Lee

A few more comments:

I suggested earlier that you replace your 100 ohm resistor with a 43 ohm resistor
(the same as the original circuit). However that value may still be a little large. With
a 43 ohm resistor, the output of the voltage divider is 14.4 * 46 / (43 + 46) = 7.44
volts.
That voltage is still probably too low to get the desired brightness. (I did not do the
math earlier, I simply assumed that the original circuit had reasonable values and
that the LEDs were similar.) My guess is that something in the 33 to 36 ohm range will
be better.

The 46 ohm resistors will dissipate about 4.5 watts. The other resistor will also
dissipate
about that much heat. Obviously they need to be sized to handle that power level. I
suggest
at least a 10 watt power rating. Even so they will get warm. Larger resistors with more
surface area (or a heat sink) will run cooler.

The brightness of the LEDs in the 'running light' mode will be pretty sensitive to the
battery voltage. (The original circuit had the same problem.) You could add a diode
into the connection between the 'brake light' 46 ohm resistor and your 100 ohm resistor.
This would reduce the voltage sensitivity by eliminating the voltage divider. Thus you
could use the 100 ohm value that you already have. (The voltage divider is eliminated
since the diode is reversed biased when the brake light switch is not active.) The
voltage drop across this diode will reduce the drive signal to the LEDs by about 0.7
volts.
You might want to reduce the 130 and 270 ohm resistors by 5% to compensate.

Both the original circuit and your version have groups of LEDs arranged as two strings
of four LEDS in parallel sharing a single current limiting resistor. This is not really
a good
idea. (Its only advantage is minimizing the resistor count.) LEDs are very non-linear in
their current/voltage characteristics. It is likely that the groups of 4 LEDs will not
share
the current equally between the two strings. A better design would have a separate
270 ohm resistor for each string of 4 LEDs. (Note: Having the LEDs in strings of at
least 4 is good since it helps reduce wasted power.)

Author: google@woodall.me.uk
Date: 05:23 05-06-08

On Jun 4, 10:59 pm, "Dan Coby" <adc...@earthlink.net> wrote:
>
> The current combination of a 46 ohm resistor and a 100 ohm resistor creates
> a 'voltage divider' circuit. I.e. (view in a fixed font):
>
> _____
> ------| 100 |-----+--------o
> | ----- |
> + ___
> 14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
> - | 6 |
> | ---
> | |
> |__________________|________o
>
> The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4
volts,
> R1 is 100 ohms and R2 is 46 ohms.
>

Unless it's been wired up differently to the way it's been drawn then
this isn't the circuit the OP is using. He's using:

-----
------------------+--| 100 |------o
| | -----
+ ___
14.4 volts | 4 |
- | 6 |
| ---
| |
|__________________|________o

The only way I can see this failing is if there's a high resistance
somewhere (bad connection? Bad earth?). Occasionally you see a similar
problem in car rear light clusters where the indicator flashing causes
the brake light to flash in anti-phase.

Tim.

Author: Gyro
Date: 10:01 05-06-08

On 05/06/2008 08:51, in article
6JudnUt4Z5V7B9rVnZ2dnUVZ_qPinZ2d@earthlink.com, "Dan Coby"
<adcoby@earthlink.net> wrote:

> "Gyro" <thepanda@btconnect.com> wrote in message
> news:C46CD344.756%thepanda@btconnect.com...
>>
>>
>>
>> On 04/06/2008 22:59, in article
>> __-dnbPOTpbNjdrVnZ2dnUVZ_qqgnZ2d@earthlink.com, "Dan Coby"
>> <adcoby@earthlink.net> wrote:
>>
>>> "Lee Wilkinson" <thepanda@btconnect.com> wrote in message
>>> news:C46C9301.389%thepanda@btconnect.com...
>>>> Hi folks,
>>>>
>>>> Don't have much electronics experience and have forgotten most of what
I
>>>> learnt at school (well it was 25 years ago!) and am hoping I can find
some
>>>> assistance in getting an LED lamp for my motorbike working
correctly....
>>>>
>>>> There is an aftermarket lamp made by Clear Alternatives that functions
>>>> correctly but has a very poorly designed LED board as far as vision is
>>>> concerned (very narrow beam and mis-aligned LEDs as well as poorly made
>>>> lamp
>>>> body).
>>>>
>>>> I've made my own LED board and have found lots of resources on the web
to
>>>> tell me what resistors to use with the LEDs to make the lamp function
>>>> correctly with the 14.4V my motorbike supplies the lamp. My problem has
>>>> arisen due to the fact that my BMW motorbike has a CanBus wiring system
>>>> that
>>>> has a bulb failure warning circuit. To get round this the aftermarket
lamp
>>>> has a load resistor across each of the running and brake circuits -
fooling
>>>> the onboard system into thinking there's a standard 5/21W tungsten bulb
>>>> fitted.
>>>>
>>>> When connected, my lamp functions correctly with lower level running
light
>>>> and full level brake light and looks great. I guessed on the 100 ohms
for
>>>> the running light but it looks about the same as the 5W tungsten to my
>>>> eyes.
>>>>
>>>> If I take one of the aftermarket board load resistors and connect it
across
>>>> the brake circuit, the lamp still functions correctly and it fools the
>>>> warning system. However when I connect a 2nd load resistor across the
>>>> running light circuit the LEDs all go out!! I presume that as the LEDs
are
>>>> already dim on the running light circuit, that they get too little
current
>>>> when the load resistor is insterted and so go out.
>>>>
>>>> What I need to know is what load resistors to use, and if necessary
what
>>>> resistors to change on the actual lamp board so they can all work
together.
>>>>
>>>> Even better, if someone were actually able to explain to me how it's
worked
>>>> out, I can learn something for the future too!
>>>>
>>>> Here are the diagrams of the aftermarket lamp that works and my new
home
>>>> built one for reference.
>>>> http://www.postimage.org/image.php?v=aV2gM5w9
>>>> http://www.postimage.org/image.php?v=Pq295j6A
>>>>
>>>> Can anyone help me with the above or point me at any reference
material,
>>>> online forums / guides on this that can help?
>>>>
>>>> You'll make a very frustrated man happy!!
>>>>
>>>> Cheers,
>>>> Lee
>>>
>>> The short form of the answer is that you need to reduce the 100 ohm
>>> resistor.
>>> If I were you, I would try 43 ohms (i.e. the value from the original
>>> circuit.)
>>>
>>> Without more information about the monitor circuit, we can only guess about
>>> the values for the green resistors. The 46 ohm value in the original
circuit
>>> seems
>>> to have been chosen to give about a 5 watt load at 14.4 volts (actually 4.5
>>> watts).
>>> Since this value works, you can continue to use it. If you want, you can
>>> try
>>> higher
>>> values. Higher values will give you less power being wasted in these
>>> resistors.
>>> However at some point with higher values the monitor circuit will decide
>>> that
>>> the 'light bulb' is burned out.
>>>
>>> You chose the 100 ohm resistor to give the correct brightness for the
>>> running
>>> lights. However, I assume, that the second 46 ohm resistor was not
>>> connected
>>> when you chose that value. Put the second 46 ohm resistor into your circuit
>>> and
>>> try lower values for the current 100 ohm resistor until you get the
>>> brightness
>>> desired.
>>>
>>> The current combination of a 46 ohm resistor and a 100 ohm resistor creates
>>> a 'voltage divider' circuit. I.e. (view in a fixed font):
>>>
>>> _____
>>> ------| 100 |-----+--------o
>>> | ----- |
>>> + ___
>>> 14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
>>> - | 6 |
>>> | ---
>>> | |
>>> |__________________|________o
>>>
>>> The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is
>>> 14.4
>>> volts,
>>> R1 is 100 ohms and R2 is 46 ohms.
>>>
>>> For your values, this gives 4.5 volts. However your LEDs need a minimum of
>>> 7.2
>>> volts (4 x 1.8 volts) before then will produce light. So you need to change
>>> the
>>> ratio of R2 / (R1 + R2) to give you a higher output voltage. Since you
>>> probably
>>> cannot increase R2 (46 ohms), you will need to decrease R1 (100 ohms).
>>>
>>>
>>
>> Dan,
>>
>> Thanks ever so much for the comprehensive reply- I'm going to check all
>> again in the morning anyway and will study what you've put here to make sure
>> I understand properly and make changes.
>>
>> Cheers,
>> Lee
>
> A few more comments:
>
> I suggested earlier that you replace your 100 ohm resistor with a 43 ohm
> resistor
> (the same as the original circuit). However that value may still be a little
> large. With
> a 43 ohm resistor, the output of the voltage divider is 14.4 * 46 / (43 + 46)
> = 7.44 volts.
> That voltage is still probably too low to get the desired brightness. (I did
> not do the
> math earlier, I simply assumed that the original circuit had reasonable values
> and
> that the LEDs were similar.) My guess is that something in the 33 to 36 ohm
> range will
> be better.
>
> The 46 ohm resistors will dissipate about 4.5 watts. The other resistor will
> also dissipate
> about that much heat. Obviously they need to be sized to handle that power
> level. I suggest
> at least a 10 watt power rating. Even so they will get warm. Larger resistors
> with more
> surface area (or a heat sink) will run cooler.
>
> The brightness of the LEDs in the 'running light' mode will be pretty
> sensitive to the
> battery voltage. (The original circuit had the same problem.) You could add a
> diode
> into the connection between the 'brake light' 46 ohm resistor and your 100 ohm
> resistor.
> This would reduce the voltage sensitivity by eliminating the voltage divider.
> Thus you
> could use the 100 ohm value that you already have. (The voltage divider is
> eliminated
> since the diode is reversed biased when the brake light switch is not
> active.) The
> voltage drop across this diode will reduce the drive signal to the LEDs by
> about 0.7 volts.
> You might want to reduce the 130 and 270 ohm resistors by 5% to compensate.
>
> Both the original circuit and your version have groups of LEDs arranged as two
> strings
> of four LEDS in parallel sharing a single current limiting resistor. This is
> not really a good
> idea. (Its only advantage is minimizing the resistor count.) LEDs are very
> non-linear in
> their current/voltage characteristics. It is likely that the groups of 4 LEDs
> will not share
> the current equally between the two strings. A better design would have a
> separate
> 270 ohm resistor for each string of 4 LEDs. (Note: Having the LEDs in strings
> of at
> least 4 is good since it helps reduce wasted power.)
>
>

My sincere thanks to everyone that replied. This has helped me out a lot and
I now have it sorted as per Dan's recommendations - a 33ohm 10W resistor
rather than the 100 has everything functioning correctly.

I now understand how to work out stuff a little more too.

Will look to adding a diode as per Dan & Bob for 'belt and braces'.

Cheers, Lee

Author: BobW
Date: 11:50 05-06-08

"Gyro" <thepanda@btconnect.com> wrote in message
news:C46DB058.775%thepanda@btconnect.com...
>
> On 05/06/2008 08:51, in article
> 6JudnUt4Z5V7B9rVnZ2dnUVZ_qPinZ2d@earthlink.com, "Dan Coby"
> <adcoby@earthlink.net> wrote:
>

[snip]

>>
>> The brightness of the LEDs in the 'running light' mode will be pretty
>> sensitive to the
>> battery voltage. (The original circuit had the same problem.) You could
>> add a
>> diode
>> into the connection between the 'brake light' 46 ohm resistor and your
>> 100 ohm
>> resistor.
>> This would reduce the voltage sensitivity by eliminating the voltage
>> divider.
>> Thus you
>> could use the 100 ohm value that you already have. (The voltage divider
>> is
>> eliminated
>> since the diode is reversed biased when the brake light switch is not
>> active.) The
>> voltage drop across this diode will reduce the drive signal to the LEDs
>> by
>> about 0.7 volts.
>> You might want to reduce the 130 and 270 ohm resistors by 5% to
>> compensate.
>>
>> Both the original circuit and your version have groups of LEDs arranged
>> as two
>> strings
>> of four LEDS in parallel sharing a single current limiting resistor. This
>> is
>> not really a good
>> idea. (Its only advantage is minimizing the resistor count.) LEDs are
>> very
>> non-linear in
>> their current/voltage characteristics. It is likely that the groups of 4
>> LEDs
>> will not share
>> the current equally between the two strings. A better design would have a
>> separate
>> 270 ohm resistor for each string of 4 LEDs. (Note: Having the LEDs in
>> strings
>> of at
>> least 4 is good since it helps reduce wasted power.)
>>
>>
>
> My sincere thanks to everyone that replied. This has helped me out a lot
> and
> I now have it sorted as per Dan's recommendations - a 33ohm 10W resistor
> rather than the 100 has everything functioning correctly.
>
> I now understand how to work out stuff a little more too.
>
> Will look to adding a diode as per Dan & Bob for 'belt and braces'.
>
> Cheers, Lee
>

Lee,

If what Dan and I suspect is truly going on, if you add the diodes then the
only values that would matter are the two fool-the-monitor-circuit load
resistors. These load resistors should only need to be some maximum value
(i.e. minimum sense current). After that, all the other brightness-setting
resistors will have no effect on the proper operation.

I would be concernced that without doing a lot of characterization with the
current circuit, you may get into trouble (e.g. lights not coming on) as
your battery voltage varies and the monitor circuit characteristics vary.
Again - if it were my bike I would spend the extra effort to see if the
diodes allow you to vary resistor values without concern for the operational
stability of your lights.

Bob
--
== NOTE: I automatically delete all Google Group posts due to uncontrolled
SPAM ==

Author: Dan Coby
Date: 13:27 05-06-08

<google@woodall.me.uk> wrote in message
news:c0827f1c-dd15-409c-9772-ee73d7f6b7c6@x35g2000hsb.googlegroups.com...
> On Jun 4, 10:59 pm, "Dan Coby" <adc...@earthlink.net> wrote:
>>
>> The current combination of a 46 ohm resistor and a 100 ohm resistor creates
>> a 'voltage divider' circuit. I.e. (view in a fixed font):
>>
>> _____
>> ------| 100 |-----+--------o
>> | ----- |
>> + ___
>> 14.4 volts | 4 | Vo = 14.4 * (46 / (100 + 46)) = 4.5 volts
>> - | 6 |
>> | ---
>> | |
>> |__________________|________o
>>
>> The voltage out of this circuit (Vo) is Vin * R2 / (R1 + R2) where Vin is 14.4
volts,
>> R1 is 100 ohms and R2 is 46 ohms.
>>
>
> Unless it's been wired up differently to the way it's been drawn then
> this isn't the circuit the OP is using. He's using:
>
> -----
> ------------------+--| 100 |------o
> | | -----
> + ___
> 14.4 volts | 4 |
> - | 6 |
> | ---
> | |
> |__________________|________o
>
>
> The only way I can see this failing is if there's a high resistance
> somewhere (bad connection? Bad earth?). Occasionally you see a similar
> problem in car rear light clusters where the indicator flashing causes
> the brake light to flash in anti-phase.
>
> Tim.

I did not include the 46 ohm resistor that is across the 'running light' switch
since it has no effect upon the LED brightness (unless as you noted there is a
bad connection.) The 46 ohm resistor that I did show is the one across the
'brake light' switch. When the 'running light' switch is on and the 'brake light'
switch is off then the circuit is:

_____
------------+----| 100 |-----+--------o
| | ----- |
+ ___ ___
14.4 volts | 4 | | 4 |
- | 6 | | 6 |
| --- ---
| | |
|____________|________________|________o

The 46 ohm resistor on the left does not affect (at least with ideal components)
the LED brightness.

Author: Tim Woodall
Date: 17:15 05-06-08

On Thu, 5 Jun 2008 10:27:55 -0700,
Dan Coby <adcoby@earthlink.net> wrote:
>
>
> I did not include the 46 ohm resistor that is across the 'running light' switch
> since it has no effect upon the LED brightness (unless as you noted there is a
> bad connection.) The 46 ohm resistor that I did show is the one across the
> 'brake light' switch. When the 'running light' switch is on and the 'brake light'
> switch is off then the circuit is:
>
> _____
> ------------+----| 100 |-----+--------o
> | | ----- |
> + ___ ___
> 14.4 volts | 4 | | 4 |
> - | 6 | | 6 |
> | --- ---
> | | |
> |____________|________________|________o
>
> The 46 ohm resistor on the left does not affect (at least with ideal components)
> the LED brightness.
>
Ah, ok. Yes, you're right.

Tim.

--
God said, "div D = rho, div B = 0, curl E = - @B/@t, curl H = J + @D/@t,"
and there was light.

http://tjw.hn.org/ http://www.locofungus.btinternet.co.uk/

Author: Gyro
Date: 13:41 06-06-08

On 05/06/2008 16:50, in article
8fydndWJ_MLbltXVnZ2dnUVZ_gidnZ2d@giganews.com, "BobW"
<nimby_NEEDSPAM@roadrunner.com> wrote:

>
> "Gyro" <thepanda@btconnect.com> wrote in message
> news:C46DB058.775%thepanda@btconnect.com...
>>
>> On 05/06/2008 08:51, in article
>> 6JudnUt4Z5V7B9rVnZ2dnUVZ_qPinZ2d@earthlink.com, "Dan Coby"
>> <adcoby@earthlink.net> wrote:
>>
>
> [snip]
>
>>>
>>> The brightness of the LEDs in the 'running light' mode will be pretty
>>> sensitive to the
>>> battery voltage. (The original circuit had the same problem.) You could
>>> add a
>>> diode
>>> into the connection between the 'brake light' 46 ohm resistor and your
>>> 100 ohm
>>> resistor.
>>> This would reduce the voltage sensitivity by eliminating the voltage
>>> divider.
>>> Thus you
>>> could use the 100 ohm value that you already have. (The voltage divider
>>> is
>>> eliminated
>>> since the diode is reversed biased when the brake light switch is not
>>> active.) The
>>> voltage drop across this diode will reduce the drive signal to the LEDs
>>> by
>>> about 0.7 volts.
>>> You might want to reduce the 130 and 270 ohm resistors by 5% to
>>> compensate.
>>>
>>> Both the original circuit and your version have groups of LEDs arranged
>>> as two
>>> strings
>>> of four LEDS in parallel sharing a single current limiting resistor. This
>>> is
>>> not r