Using a 0.1 ohm resistor to measure current sometimes
works out better than the meter itself. The 0.1
ohm resistor is connected in series with the battery
and the voltage across the resistor is measured with
a digital voltmeter. The voltage measure reflects the
current, and is ten times higher than measured.
For example, if you measure 0.1 volts then the
actual current is 10 times that, or 1 amp.
If you still want to use the meter, connect it in
series with the battery.
The simple basics of measuring current is that you must interrupt the circuit and put the meter in series with the power source.. i.e. typically with a flashlight like a maglight that has a tail cap.. you can turn the light 'on'.. take off the tail cap.. set the meter to 'amps' and put the leads in the right holes.. at that point touch the 'negative' lead to the bottom of the battery and the 'positive' lead to the shell of the flashlight.. the meter acts as the tail cap and completes the current loop.. if you hook up the leads backwards most meters don't care they just display negative values.
The higher end meters have an alarm if you have the leads in the 'amp' connections and the meter set to anything but amps.. easy way to melt things is to hook a meter into a circuit thinking it's set to volts but it's still in the 'amps' mode.. to a circuit, an ammeter looks like a 'short'.. a voltmeter looks like an 'open'.
It will become 'second nature' if you do it enough... just remember.. amps you have to disconnect.. volts you don't.
You need to touch one meter lead to the minus side
of the battery inside the light while applying a little
pressure to keep the batteries in good electrical contact,
and the other meter lead to the INSIDE of the light where
the threads are. Note the outside is covered with a non
conducting paint so you cant touch the meter lead there.
Have you used the meter for other current measurements yet
to prove it works ok?
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I guess that Resistance is something different if between - of batteries and threads than with tailcap. Am I right?
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Resistance is never checked with a power source attached. Resistance is the measurement of how well something blocks electricity. The DMM has a small battery inside that it uses to check the resistance at a known voltage.
To practice, Take the lamp from your mag lite. Set the DMM to the 200 ohm scale, and touch one lead to the silver case and the other to the metal blob at the tip of the lamp. Your meter will then tell you the resistance of the bulb.
You're right and I spoke not so clearly: I meant to say that the spring in the tailcap is more conductive than the threads, but I had a second though....as both are used when closing my torch. uuuhh feel like a moron [img]/ubbthreads/images/graemlins/icon23.gif[/img]
How many batteries have you got?
Lets make an example:
assumed that ya're running at 6V
W=V*A (note: I'm simplifying, it should be something like V*A*0.8 instead)
W=6*0.980 thus ....?(Waiting for yor calculation [img]/ubbthreads/images/graemlins/smile.gif[/img] )
In any case given that LuxIII should run at 3W in any case you're overdriving it, as you're running at 5W levels (if I recall well 0.980 is the current of SF L5 which runs @6V).
if you measured 3.06V and 1000mA.. that is 3.06W.. but that's coming out of the battery... more than likely you are getting just about 85% efficiency with the mini pucks.. so that is 2.6W at the lux.. bright but could be brighter.. you'd have to interrupt the connection to the LED to get a more accurate measurement at the LED.. if you have a 'J' binned emitter i have a formula that works pretty well to estimate power and current only from knowing voltage on the emitter.
to find out the efficiency of the driver(s) you need to measure the current and voltage (under load)... both at the battery and at the emitter (the tricky one.. usually means cutting or unsoldering a wire).. multiply current x volts at the battery and at the emitter.. you will get watts for each.. now divide the watts at the emitter by the watts at the battery and multiply that by 100 and you'll have the % efficiency. Expect something between 70 and 90.. if it's way off you are probably measuring something wrong.
After further testing, I realized an error in my testing strategy and therefore calculations.
It seems that to accurately measure how much power the driver is pulling, I needed to test the voltage of the 2 x AA batteries under load. (thanks andrewwynn)
On the same circuit, under load, the 2 x AA batteries are putting out 2.05v and 1010mA on fresh batteries. Total power consumption would be 2.0705 watts. Assuming that the micropucks are 83% efficient at this voltage (which I found was the case when testing other MicroPucks) then the power consumption by the LED would be a mere 1.7185 watts. So just a little over half of what the LED is rated for.
Here's the kicker, the voltage at the LuxIII bin TWOJ was only 3.00v...which would make the current approx 580mA.
I did some testing with another unknown bin LuxIII (the stock LuxIII from the XM-3) and various other drivers and this is what I got. I measured the current directly, in the circuit between the batteries and the driver, then between the driver and the LuxIII. Then measured the voltage directly. So the only numbers that were estimated in these tests was the efficiency of the drivers.
So the stock XM-3 driver is very inefficient. More info on this driver can be found here. Koala's findings were:
[ QUOTE ] koala said:
...mine's at 380ma due to the lower vf(3.08v) of my Luxeon. My efficiency measurements at ~72%.
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And just to clarify, the reason I am getting so much more output with the driver in my XM-3 currently, I am using 2 x High-Output MicroPucks 2009-H (and the different LuxIII bin obviously).
Remember that there can be an error in calculations using your ammeter with low voltages, where the voltage drop across the shunt resistor in the ammeter can throw off your efficiency calculations.
What you might want to do is measure the voltage at the point of interest, then use a lab. power supply so generate the same voltage at the measurement point whilte measuring the current draw, and you'll notice that the lab power supply is reading slightly higher at the source. Your calculation of power from the current and voltage (measured at the point of interest) will be more accurate.
I think your micropuck is not as efficient at the voltage level.. I was only getting like 75% max at volts near 2.. maybe close to 90 with 3.0... here's the thing.. when i developed the voltage to current curve (formula) for the Tbin luxes i use.. it comes out to that if you have 3.0V on your emitter.. my math says it has only 122mA and .36W... that would be extremely low efficiency so i'm wondering if you took your voltage from like the 'neg' of the battery and the anode of the LED vs direct on both sides of the LED..
1.7w into a Jbin should be more like 3.41V and 498mA...
The formula i use is this: (Vf-2.6)^2*0.76 = I.. then i calculate power my multiplying the two. (I x Vf).
It never lets me down.. when i measure 450mA on my nanos.. it always is within a couple hundredths of 3.37V.. if i measure 3.37V it comes out to 1.52W..
i was going to develop a formula for K bin emitters as well but i only use J.
In any event.. that formula is a life saver dealing with emitters that are soldered in to a circuit.
oh.. PS.. if you are trying to do this with one meter... it won't be accurate... i hook up one meter for amps and one for voltage so i can get power.. (one day i'm building a 'power' meter that reads directly watts),,, 3 leads.. hook up the 'common, and the current loop.. internally the 'goesout' of the current loop is also hooked to the + voltage.. it reads both current and voltage and displays the product or power.
[ QUOTE ] daberti said:
One question to you mates: V under load must be misured in series, right?
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No. Voltage is always measured in parallel.
You read voltage (a pressure) *across* a part, you read current (a flow rate) *through* a part. The current meter must 'count' each electron, so they all have to flow through it. Voltage (what some call 'tension', as in 'high tension lines') can be sensed with very minor disturbance.
Them's the rules, "Voltage across, current through".