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Old 07-03-2012, 12:23 PM   #21
OJgsxr
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Quote:
Originally Posted by anatram View Post
No, you need two resistors, one for each LED assembly, then you will need an additional 2 larger resistors to act as a load balancer.

If you hook the LEDS in Parallel, you will blow them out, becasue you are introducing too much current/amps to the LED, they will blow one by one, LEDS are not designed for a continous 12 volts, without stepping down.

When you do this they glow for about 1/2 a second and them pop and smoke.
You can put as many LEDs in parallel as you want, as long as the current through them is not greater than 20mA. You accomplish this by putting a current limiting resistor in series with each LED string.
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Old 07-03-2012, 01:48 PM   #22
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Quote:
Originally Posted by OJgsxr View Post
The forward breakdown voltage of an LED is 0.7 volts DC. The amperage needed for most LEDs at full brightness is 20mA. Given a 12 volts DC source you could a string of 10 LEDs then a 250 ohm resistor inline. This would set the current at 20mA. You could put as many of these in parallel as you wanted as long as your relay can handle the (number of parallel strings) * (20mA) amperage. If you want any more info let me know.
The LED's im using have 4 chips in them and require 2V and 80mA. Also that is the thing, I dont know how much amperage the relay can handle.

Quote:
Originally Posted by anatram
No, you need two resistors, one for each LED assembly, then you will need an additional 2 larger resistors to act as a load balancer.

If you hook the LEDS in Parallel, you will blow them out, becasue you are introducing too much current/amps to the LED, they will blow one by one, LEDS are not designed for a continous 12 volts, without stepping down.

When you do this they glow for about 1/2 a second and them pop and smoke.
So they will blow out even if each one has a resistor connected to it?

I used the link you gave me and got this:

Solution 1: 5 x 8 array uses 40 LEDs exactly

The wizard says: In solution 1:
•each 27 ohm resistor dissipates 172.8 mW
•the wizard thinks 1/2W resistors are needed for your application
•together, all resistors dissipate 1382.4 mW
•together, the diodes dissipate 6400 mW
•total power dissipated by the array is 7782.4 mW
•the array draws current of 640 mA from the source.

So would a load balancer be needed for that? Since the total amperage is going to be 6.4?


If I need a balancer, then can I do this:
Make 4 groups of 5 LEDs with a resistor in each series then conntect all their wires together and connect it to the load balancer and connect the main wire to the load balancer?
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Old 07-03-2012, 02:45 PM   #23
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[QUOTE=sik9;3302637]The LED's im using have 4 chips in them and require 2V and 80mA. Also that is the thing, I dont know how much amperage the relay can handle.
QUOTE]

So if you place them in series, the amperage required is not additive it will be 80mA through the loop.

If you have them in parallel the amperage is additive.

For those LEDs (I haven't seen them before) on a 12VDC source you can place up to 6 of them in series and the total power consumption will be 12VDC and 80mA for a total of 0.96 watts. If you place 2 strings of these in parallel it will pull 160mA at 1.92 watts, 3 will pull 240mA at 2.88 wattts and so forth. You may still want a current limiting resistor in series with each string but without seeing a data sheet I wouldn't know what value they want.
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Old 07-03-2012, 02:47 PM   #24
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Quote:
Originally Posted by OJgsxr View Post
Quote:
Originally Posted by sik9 View Post
The LED's im using have 4 chips in them and require 2V and 80mA. Also that is the thing, I dont know how much amperage the relay can handle.
So if you place them in series, the amperage required is not additive it will be 80mA through the loop.

If you have them in parallel the amperage is additive.

For those LEDs (I haven't seen them before) on a 12VDC source you can place up to 6 of them in series and the total power consumption will be 12VDC and 80mA for a total of 0.96 watts. If you place 2 strings of these in parallel it will pull 160mA at 1.92 watts, 3 will pull 240mA at 2.88 wattts and so forth. You may still want a current limiting resistor in series with each string but without seeing a data sheet I wouldn't know what value they want.
If the amperage in a series is not additive then why do I need a current limiting resistor?

Here are the ones I'm talking about! http://www.oznium.com/four-chip-led
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Old 07-03-2012, 03:06 PM   #25
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Quote:
Originally Posted by sik9 View Post
If the amperage in a series is not additive then why do I need a current limiting resistor?

Here are the ones I'm talking about! http://www.oznium.com/four-chip-led
Voltage = Amperage * Resistance or V=IR. The way LEDs work is they drop voltage across them (in your LED case 2V) but they do not limit current in anyway. So if you have enough voltage to overcome the forward breakdown voltage of 2V then the rest is just sinked to ground causing the amperage to be massive and burning out the LED quickly. You have to limit the amperage flowing through the LEDs by putting in a resistor to limit the current. In the case you have with those diodes the calculation needed is this:

(Source Voltage - (number of diodes * voltage drop across diodes)) / (current rating of diodes) = Current limiting resistor value

(12 - 5*2) / (0.08) = 25 ohm current limiting resistor. The wattage rating of the current limiting resistor must be at least 0.16 watts, there are tons of 1/4W or quarter watt resistors available.

The reason I did not put 6 diodes in is because you may not have enough forward voltage to illuminate them so left a 2V overhead. You can put as many of these in parallel as you want and the current consumption will be (number of strings) * (0.08).

If you want to balance the lights so they don't blink at a strange rate then you will have to put a resistor in parallel with all of your strings of diodes.

I know this is getting mildly technical but trying to let you know exactly how it works.

Last edited by OJgsxr; 07-03-2012 at 03:09 PM.
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Old 07-03-2012, 10:08 PM   #26
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(12 - 5*2) / (0.08) = 25 ohm current limiting resistor. The wattage rating of the current limiting resistor must be at least 0.16 watts, there are tons of 1/4W or quarter watt resistors available.
This is the resistor that would go in the 5 LED series? So it would be LED/LED/LED/LED/LED/.25W resistor?

Quote:
If you want to balance the lights so they don't blink at a strange rate then you will have to put a resistor in parallel with all of your strings of diodes.
This is the "load balancer" that will be connected to the main wire coming from the front of the bike? The 10 Watt 10 ohm resistors?
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Old 07-05-2012, 07:04 AM   #27
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Originally Posted by sik9 View Post
This is the resistor that would go in the 5 LED series? So it would be LED/LED/LED/LED/LED/.25W resistor?



This is the "load balancer" that will be connected to the main wire coming from the front of the bike? The 10 Watt 10 ohm resistors?
Yes and Yes. You would place the 10ohm resistor in parallel with the
LED/LED/LED/LED/LED/(25ohm 0.25W resistor) string.
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Old 07-05-2012, 10:52 AM   #28
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To me, using load resistors defeats the main advantage of using LEDs, which is their efficiency. It's a lot easier to get an electronic flasher that is a drop-in fit for your old flasher module.

http://www.superbrightleds.com/cgi-b...2Fflashers.htm

You can install a bike-full of LEDs as aux turn signals and won't be loading down the regulator.
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Old 07-05-2012, 11:23 AM   #29
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One of the main advantages of LEDs is their efficiency, but why would you care on a motorcycle. In a home, yes they can bring down your bill but on a motorcycle saving that 10W is insignificant. 746W is a HP so it's 0.01HP. The main advantages on motorcycles is life of the LEDs and customize-ability.
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Old 07-05-2012, 12:50 PM   #30
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Most bikes' regulator has limited capacity. Add a couple of driving lights and heated gear and you may get close to maxing it out. Plus some bikes have alternator stators that are sensitive to load and may wear out faster if constantly loaded.
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Old 07-05-2012, 01:06 PM   #31
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Most bikes' regulator has limited capacity. Add a couple of driving lights and heated gear and you may get close to maxing it out. Plus some bikes have alternator stators that are sensitive to load and may wear out faster if constantly loaded.
Correct. Only the standard lights and turn signals (that were originally incandescent) need a load balancing resistor so they flash at the right rate and the ECU doesn't think they are burnt out. All accessory LED strings can be run off the output of the accessory relay and will have a much lower power draw.
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Old 07-05-2012, 04:54 PM   #32
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im sorry....but why do you think you need a resistor? LEDs are 12volt and run off 12volt....the only thing the resistor is going to do is fix the "fast flash" rate of the blinkers....if thats what your trying to do just replace the flasher module with a digital flasher module....prolly $6 at auto zone....

all that chinese just hurt my head.....althought props for the info!
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Old 07-06-2012, 09:47 AM   #33
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im sorry....but why do you think you need a resistor? LEDs are 12volt and run off 12volt....the only thing the resistor is going to do is fix the "fast flash" rate of the blinkers....if thats what your trying to do just replace the flasher module with a digital flasher module....prolly $6 at auto zone....

all that chinese just hurt my head.....althought props for the info!
LEDs are not 12 volt, a typical PN junction is 0.7VDC.

I agree there are easier and cheaper ways to do it, I was just offering a way for someone to do it themselves. Sometimes the satisfaction of doing it yourself is greater than the cost savings, for me anyway.
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