dynamo triple XPG , front and back.

capnahab

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Jul 11, 2010
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Maybe i am expecting too much but I have a red triple xpg on an indus star (ledsupply) and a white one the same. I thought I would try and use both of them for a front and rear set up using Martins circuit 3. The red led lights well and is replacing one of the diodes as instructed. The front led tho will not light . I have checked its working , and if I replace it in the reds position in the circuit its fine.
Am I expecting too much for my SON dynamo to power both of them.
I am a novice at this so patience appreciated.
 

Bobblehat

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Jan 7, 2010
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87
Maybe i am expecting too much ........... .

I take it your red and white triples are wired in series on the star? If they are, then your red LEDs that you put in as substitute for the standard diode do not have the same forward voltage as Martin's paralleled red LEDs ...... it is at least 3 or 4 times higher .... probably 9 volts in total. This is what is likely to be affecting your white LEDs.

Even 1 red xp-g as a rear light will be very bright (too bright?) and 3 xp-g's for the front will be great at speed but will not produce much light at lower speed. If the low speed performance is acceptable why not try Martin's circuit 7 where you have 1 red and 3 white LEDs all in series? Use Martin's values 4 LEDs in the table ...... they'll be close enough.
 

A10K

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Dec 20, 2011
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The voltage within the circuit is controlled by the Vf of the LEDs. This means that your maximum voltage is the first place the dynamo can sink all 500ma it can provide*, and in your case its the red LEDs** . I've had the same issue before; as soon as the potential difference of the dynamo gets to about 2V, the red LEDS light up and sink 500ma current. There dynamo never gets to the voltage that would light the white LEDs (the max Vf of most red LEDs is at the barest threshold of where white LEDs start to glow). What I had to do was to put a 2-Ohm resistor in series with the red LEDs to bring their effective voltage up to that of the white LEDs.

The size resistor you will need will be determined by the difference in voltage between the white and red LED groups at the maximum drive current (500ma). You will need a multimeter and power supply for this. Basically, (Vwhite-Vred) = Current * Resistance. If you divide (Vw-Vr) by .5 (aka multiply by two), you will get the proper resistance with which to balance the reds and the whites. The other option is to get (order from Digi-Key) a 10 Ohm, 5 Watt (minimum) potentiometer and use it as a variable resistor to determine the right value by trial and error.

*(Pardon my terrible use of terminology. I am not an electrical engineer and have only the barest theoretical knowledge)
**(which are XP-Es, there's no such thing as a red XP-G. This is important when discussing max current. An XP-G has a maximum current of 1.5A. A colored XP-E is half of that)
 

capnahab

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Joined
Jul 11, 2010
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9
Thanks Guys,
so , with a 2.2 OHM resistor I can get both lights to work . The red still predominates at this low speed. Doing the calculations (thanks A10K)
red volatge 2.2 , white 3.1 should use a 1.8 ohm resistor , - 2.2 was all I could get at short notice.
I think it still needs some tweaking but thanks to your help it is getting there.

 
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