Help: Can someone assist with load resistor calculation for vehicle LED lamp please?

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.

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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

Reply to
Lee Wilkinson
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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

Reply to
Tom Biasi

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.

Reply to
Michael Robinson

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

Reply to
Gyro

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

Reply to
Gyro

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?

Reply to
Michael Robinson

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).

Reply to
Dan Coby

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

Reply to
Gyro

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

Reply to
Gyro

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

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Reply to
BobW

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.

Reply to
John G

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.)
Reply to
Dan Coby

volts,

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.

Reply to
google

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

Reply to
Gyro

[snip]

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

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Reply to
BobW

14.4 volts,

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.

Reply to
Dan Coby

light'

components)

Ah, ok. Yes, you're right.

Tim.

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Reply to
Tim Woodall

Will do, and thanks again for your and everyone else's help.

Next step after that is to find something that will remove the white hazing that appears on the clear plastic around where 'super glue' (cyanoacrylate - spelling?) has been used to join the halves. As it's dried the clear plastic has gone milky - damn! If there isn't I need to find something that I can run into the joints to seal it together as I don't want to use a glue that will compress and ooze out as it looks nasty (especially on the inside where I can't trim it off)!

Lee

Reply to
Gyro

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