Who has the formula for deciding resistance?
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PEU
Flashlight Enthusiast
you can ask a guy called Ohm /ubbthreads/images/graemlins/crackup.gif
r=v/a
r=7.5 ohm
Pablo
r=v/a
r=7.5 ohm
Pablo
Icebreak
Flashlight Enthusiast
I hadn't noticed that the sticky for led_pro.exe was gone until I read your post.
Programmed by: (jtice)
Concept by: (jtice), (rothrandir)
Algorithm by: (dat2zip)
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Introduction and discussion on CPF
Programmed by: (jtice)
Concept by: (jtice), (rothrandir)
Algorithm by: (dat2zip)
Logo by: (logicnerd411)
Download from jtice's site
Introduction and discussion on CPF
gadget_lover
Flashaholic
The formula if you want to do it by hand;
First, let me point out two problems. 1) Lumileds only publishes a range for the Vf. 2) This range is only valid for the design current. In the case of the LUXIII that current is 700ma.
Given:
B= battery voltage (6 v)
e= voltage to be consumed by the resistor (to be calculated).
Vf = voltage for the luxeon (from the limileds Bin charts).
i = current
r = resistance
e=i*r (Ohms law)
Solve for 700ma:
Assume a Vf of a K bin is 3.5 to 3.75, on average 3.625
e = B-Vf = (6 - 3.625) = 2.375
(e= i*r) = (e/i = r) = (2.375 / .7 = r ) = 3.392 ohms
You can guess at the approximate Vf for 800ma by adding 1/10 of a volt for every 100 ma over the target. So the 3.625 becomes 3.725.
Caveat #2; The Vf for your particular Luxeon may be as low as 3.5 or as high as 3.75. Plug those figures into the formula and you find the resistance can vary from a low of 3.124 to a high of 3.571.
Thoe only way to find the Vf for your particular LED (that I know of) is to hook it up to a variable power supply with a voltmeter and ammeter attached. It just occured to me that you could also find out by measuring the amperage when using a fresh 6 volt battery and a 3.3 ohm resistor. If you read more than 700ma, you have a lux with a Vf less than 3.625 volts. /ubbthreads/images/graemlins/smile.gif If you see less than 600ma, you have a VF that's higher. If it's exactly 700....
Daniel
First, let me point out two problems. 1) Lumileds only publishes a range for the Vf. 2) This range is only valid for the design current. In the case of the LUXIII that current is 700ma.
Given:
B= battery voltage (6 v)
e= voltage to be consumed by the resistor (to be calculated).
Vf = voltage for the luxeon (from the limileds Bin charts).
i = current
r = resistance
e=i*r (Ohms law)
Solve for 700ma:
Assume a Vf of a K bin is 3.5 to 3.75, on average 3.625
e = B-Vf = (6 - 3.625) = 2.375
(e= i*r) = (e/i = r) = (2.375 / .7 = r ) = 3.392 ohms
You can guess at the approximate Vf for 800ma by adding 1/10 of a volt for every 100 ma over the target. So the 3.625 becomes 3.725.
Caveat #2; The Vf for your particular Luxeon may be as low as 3.5 or as high as 3.75. Plug those figures into the formula and you find the resistance can vary from a low of 3.124 to a high of 3.571.
Thoe only way to find the Vf for your particular LED (that I know of) is to hook it up to a variable power supply with a voltmeter and ammeter attached. It just occured to me that you could also find out by measuring the amperage when using a fresh 6 volt battery and a 3.3 ohm resistor. If you read more than 700ma, you have a lux with a Vf less than 3.625 volts. /ubbthreads/images/graemlins/smile.gif If you see less than 600ma, you have a VF that's higher. If it's exactly 700....
Daniel
