How much voltage can a LUXIII take?

Timson

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I don't know the Vf rating of my LuxIII.
I was wondering whether I could risk direct driving it from 4xD NiCd cells! As far as I am aware Lux III's are specd' at approx 3.9V.
I suspect it would probably be too much for it.
If so can anyone suggest how much resistance would be needed to make it OK.
TIA,
Tim
 

Doug Owen

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Wrong question, it should be 'how much *current* can it take'. Vf is a function of current. The LED is a current mode device.

Unless you are using a proper current regulator, you're going to need to know the Vf rating of your part.

Let's say it is 3.9 Volts and your battery really is 4.8 Volts there's a difference of .9 Volts. If you want to hold to 750 mA you're looking for .9/.75 ohms, just over one. I'd suggest starting with say 2 ohms and seeing what you get in the way of current. You can fine tune it by adding more resistance in parallel to lower the effective value (raising the current) in small steps. For instance 20 ohms will add about 10% more to the total. 10 ohms would add 20%.

Doug Owen
 

Illuminated

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Hello Tim, and welcome to CPF -

Adding 10 or 20 ohms of resistance in parallel with the 2 ohm resistor will *subtract* from (not add to) the total resistance.

Formula for resistors connected in parallel:

1/(1/R1 + 1/R2 + 1/R3 + ...etc.)= R total

so R total = 1/(.5 + .05) = 1.82 ohms for the 2 ohm and 20 ohm,

and...R total = 1/(.5 +.1) = 1.67 ohms for the 2 ohm and 10 ohm

Power dissipated in the resistor = Voltage across the resistor times current through the resistor, so roughly speaking, 1 amp through the resistor times .9 volts across it is .9 watts. A 1 watt resistor would get pretty warm, so I'd use a 2-watt for a single resistor, or 1 watters if using 2 in parallel, for example.

Hope this helps - John
 

Doug Owen

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[ QUOTE ]
Illuminated said:
Adding 10 or 20 ohms of resistance in parallel with the 2 ohm resistor will *subtract* from (not add to) the total resistance.


[/ QUOTE ]

Yup, John, that's what I said.....right?

'lower the effective value and raise the current'?

Doug Owen
 

Illuminated

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My bad Doug - I read it wrong. You did say add 10 or 20 ohms in parallel to add 10 or 20% to the total *current*.

Sorry /ubbthreads/images/graemlins/blush.gif

I'll go back to my corner now...
 

Doug Owen

Flashlight Enthusiast
Joined
Jan 30, 2003
Messages
1,992
[ QUOTE ]
Illuminated said:
My bad Doug - I read it wrong. You did say add 10 or 20 ohms in parallel to add 10 or 20% to the total *current*.

Sorry /ubbthreads/images/graemlins/blush.gif

I'll go back to my corner now...

[/ QUOTE ]

Not a bit of it, good to know we're checkin' up on each other. Enough problems communicating without stupid mistakes.....

Thanks for keeping me honest.

Doug Owen
 
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