estimating current

[knifebright]

Hey all
I need a little help with measuring current. I've done a bunch of reading last night and I believe andrew w had some very good ways of estimating the current over the emmitter. However i need some more help. What i can measure is
current from battery end
forward voltage of lux
voltage of battery
voltage under sag

Now I know that power=voltage*current. I can use that formula to figure out wich ever variable is missing. So if i can get current from the battery end, know the voltages and can factor the power, where do i go from here to guestimate amperage at the emmitter end. The point of all this is i'm trying to figure out if and for how long my nexgen 750 is running in regulation, or just in safe mode.
thanks
jimmy

 

[HarryN]

The LEDs usually are only drawing around 1 amp. Some DMMs will allow you to just put the leads in line with the LED and measure current directly (often up to 10 amps).

I think the other way is to put a known, precision resistor in line, and watch the voltage drop acrossed it. I R = V drop.

 

[Leeoniya]

the point being that to get any sort of accuracy you need to break the circuit and put something in line with the output of the regulator and the LED. anything else would be approximations because the LED resistance probably changes with higher junction temperatures. this makes it impossible to associate any measured output voltage with a current. also, this limits the voltage-current correlation to only the first few moments when the junction temp doesnt change much and neither does R. you have V=IR, if R changes with temp, and V is changed by the regulator to keep I constant there are too many unknown variables.

without putting anything in-line you would need to record both the voltage output of the regulator, as well as the resistance of the LED at the same time. if you have a way of measuring both simultaniously, then you can measure current pretty accurately by dividing one into the other. it's usually just easier to do it in-line with one multimeter rather than in parallel with two.