Basic voltage drop question

[xtree]

OK,

Im trying to go by the saying that the only stupid question is the one not being asked here...so please bear with me here.

I want to use ultrabright leds to make a flashlight. I was trying my hand with just one led and I wonder at the results:
Power supply: 3 AA batteries. Measured voltage was 4.6 volts
My led works at 3.6 volts and a current of 20 mA.
I have used a LED calculator and it said I need a 56 ohm resistor.
I ve measured and put the resistor up, hooked everything up and measured the current across the LED and I got to 3.0 volts.
Problem is that math/physics dont make sense here: resistance=voltage drop/current. I got a voltage drop of 1.6 (4.6-3.0) by using a resistor of 56 ohms which leads me to the conclusion that the led is getting 28 mA instead of the 20 mA (which was planned). I cant measure mAs with my digital multimeter so I cannot check that for sure.
What am I missing??
/ubbthreads/images/graemlins/mad.gif

 

[hawkeye]

Have you considered the internal resistance of the batteries, the resistance of the meter, and the resistance of the 6 connections? You're only missing 24 ohms between actual and theoretical, I think.

 

[Lynx_Arc]

there are many factors involved. First unless you measure the resistor it could be as much as 10% off, next the actual Vf of the LED can vary some also from 3.6 to 3.2-4.0 I have one white LED that was in a light I thought went bad but the Vf just was too high to work with the others. I am unsure why your DMM cannot measure milliamps, most of them can unless you blew the fuse inside of it somehow.

Also 28ma on a white LED isn't that bad, it is overdriving it but it should last a few thousand hours at that rate I figure AND when the batteries start sagging and dying it will continue to put more light out at lower voltages.

 

[Doug Owen]

Measure the *voltage* across the resistor and trust Ohm's Law. Current (in the LED, as everywhere in the series circuit) is that measured voltage (in volts) divided by 56 ohms (answer in amps).

For the numbers you give, 28 mA.

For sure you didn't "the current across the LED and I got to 3.0 volts.".....

Doug Owen

 

[xtree]

Thank you for your answers. Ill check that fuse.
As for [ QUOTE ]
Doug Owen said:
For sure you didn't "the current across the LED and I got to 3.0 volts.".....Doug Owen

[/ QUOTE ]
I just put the DMM to volts and then the + and minus directly to the LED terminals. Isnt that the current the LED gets?

 

[MikeLip]

That's the voltage drop across the diode, which is independent of current. Voltage will remain roughly the same across a diode in conduction for a wide range of currents. Better to measure the voltage across the resistor; then V/R will reliably give you current, within the tolerance of the resistor and your voltmeter.

 

[xtree]

OK, I have replaced the fuse and I am now able to measure mAs too (thanks for the idea).
Here are the results, and what I am going to use:
Power supply: 4.5 V (3xAAA)
3 Ultrabright 3.6 V (manufacturer spec.)leds connected in parallel
1 68 Ohm resistor (placed it between the battery and the plus terminals of the LEDs)

I tested this and it works. I was able to measure 58 mA across the resistor. I assume that is split by 3 so I should get just under 20 mA for each of the led.
Does that make sense math wise?
I am asking because its not anywhere close to what the online led calculators tell me (which is like 18 Ohms for the resistor).

 

[MikeLip]

Trust ohms law! /ubbthreads/images/graemlins/grin.gif If you get 58mA by measuring voltage drop across your resistor, you should be OK.