resistor question

[big vin]

I am assembling a uv light using the NCSU033A by nichia, and have a few questions about the resistor to use.

The led is a Rank Ua6PL, wich means a Vf of 3.4~3.6V at 500mA

when i direct drive it, using a single 3.96V 18650 it consumes about 0.82A.
when i use a 1 ohm resistor i get 0.27A.
This is measured at the led itself, using a 30$ MM at the 10A position.
As i understand it the resistance of the MM affects the amount of current passing through the led, what would likely be the actual current through the leds with this readings?

i would like to get around the neigbourhood of 0.5A when using with a 18650 going for the long haul, but it shouldn't peak at more then 0.8A with fresh battery's to prevent burning out.

What would be a nice value resistor to choose?



Thanx, Vincent

 

[Mr Happy]

At a very rough estimate, a meter on the 10 A scale will drop about 0.5 V when passing 10 A. This depends also on the quality of the test leads and lack of oxidation on the banana plugs and sockets. In general you might assume the resistance of the meter and test leads is between 0.05 ohms and 0.1 ohms. Therefore you could add 0.1 ohms to the final resistor value you choose to compensate for the inclusion of the meter in the circuit.

 

[big vin]

okay, thats clear.

do you have any suggestions on the resistor value?

 

[Mr Happy]

do you have any suggestions on the resistor value?

Somewhere between 0 and 1 ohm?

There are too many unknowns in this kind of setup to calculate a value with any accuracy. Your best bet is to do what you are doing: take a freshly charged battery of the kind you will use, take a selection of resistor values, and measure the current with your meter until you find a resistor value you like.

Calculations have a place, but final answers are always arrived at on the test bench.

 

[qwertyydude]

Resistors are cheap since it'll only be dropping a miniscule amount of voltage you'll only need maybe a 1/2 watt resistor. You can buy those real cheap at an electronics store.

 

[big vin]

yes, i already guessed that that would be the method to go. The only thing is that i need 30 pieces, so i have to buy 30 pieces in 2 or 3 values, making it a bit expensive. (or i buy 1 of each value and need to drive back to the store (would cost even more). Bjut hey, if thats whats needed...

think i will get some 0.3 ohm, 0.5 ohm and 0.7 ohm, if those are existing values.

 

[HKJ]

i would like to get around the neigbourhood of 0.5A when using with a 18650 going for the long haul, but it shouldn't peak at more then 0.8A with fresh battery's to prevent burning out.

Measure the voltage across the resistor and use ohms law to calculate the current: current=voltage/resistance
Then you will not have any problem with voltage drop in the meter.