Testing Ampere draw with a multimeter

[wintermute]

When I measured current I took off the tailcap where the switch is located, and connected the leads to the body of the flashlight and the negative side of the battery.

 

[greenLED]

This thread taught me how to measure current. Must read. An example:
"...if you were to measure from the luxeon's positive lead, the connection would go:

driver board's Vout+, Meter + lead, meter - lead, luxeon + lead.

If you are attempting to measure current by putting the probes across the luxeon, you are shorting out the luxeon (and driver board) and getting an incorrect measurement. " (evan9612)

 

[member 6142]

Ya did it right, so it is valid what I said here
http://www.candlepowerforums.com/ubbthreads/showthreaded.php?Cat=&Number=819095&page=&view=&sb=5&o=&vc=1

 

[andrewwynn]

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.

-awr

 

[wintermute]

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.

Stock EL XM-3 Driver
Batteries 2.12v (under load) / 880mA = 1.866w
LuxIII 3.40v / 380mA = 1.292w
efficiency <font color="red"> 69% </font>

Single standard output MicroPuck 2009
Batteries 2.45v (under load) / 440mA = 1.078w
LuxIII 3.30v / 270mA = 0.891w
efficiency <font color="red"> 83% </font>

2 x standard output MicroPucks 2009 wired in parallel
Batteries 2.11v (under load) / 790mA = 1.646w
LuxIII 3.45v / 400mA = 1.380w
efficiency <font color="red"> 83% </font>

 

[wintermute]

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%.

[/ QUOTE ]

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).

 

[xenopus]

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.

Hope that helps!

 

[andrewwynn]

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.

-awr

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.

 

[member 6142]

One question to you mates: V under load must be misured in series, right?

 

[Doug Owen]

[ QUOTE ]
daberti said:
One question to you mates: V under load must be misured in series, right?

[/ QUOTE ]

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".

Doug Owen

 

[Doug Owen]

[ QUOTE ]
xenopus said:
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.


[/ QUOTE ]

Good point, although it makes your 'measured efficiency' higher than it really is (since you think it's drawing less current than it really is) typically. And most efficiency measurement is for 'bragging rights'......

Besides which, you really should have several $3 DMMs around so you can measure both. In fact, I've been known to rig up four meters when testing converters and drivers (although you only need 3 with linear drivers, as the current is the same).

Doug Owen

 

[sotto]

AWR:

How would you measure current to the LED head in a light like the Peak McKinley where the head is twisted to turn the light on?

Thanks.

 

[member 6142]

[ QUOTE ]
Doug Owen said:
[ QUOTE ]
daberti said:
One question to you mates: V under load must be misured in series, right?

[/ QUOTE ]

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".

Doug Owen

[/ QUOTE ]

Yes, I do know this is the rule, yet how to measure Voltage under load?

 

[andrewwynn]

sotto... two ways to measure that situation.. i use both doing the [http://peak.rouse.com]nanos[/url] which of course are twist-on.

1) non-destructive:

a) i take a piece of tin-foil aboug 1" square and clip the 'positive' lead to the ammeter to it.. set that on the bench..
b) clip the 'negative lead' from the ammeter to the negative wire on a battery clip holding the same arrangement that will be in the light (1 up.. 2up.. just has to match).
c) touch the positive battery lead to the 'anode patch' (typically solder bump).. on the battery head.

(if the light is coated.. you have to be more creative than the tinfoil patch.. you need to get the power through the threads.. you can wrap a long bare wire around the threads and twist it tight).

2) somewhat destructive..

a) drill a hole in the bottom of the light
b) put in a piece of pc board with a contact patch attached to a wire..
c) wire is put through the hole
d) assemble light.. measure current from wire coming out to some 'ground' on the light.

(this works really well with the peak AAA with the removable key-chain.. when you put it back on the light is actually 100% good as new.. the o-ring seal seals over the hole you make).

-awr

 

[Doug Owen]

[ QUOTE ]
daberti said:

Yes, I do know this is the rule, yet how to measure Voltage under load?

[/ QUOTE ]

I guess I don't understand you then.

Are you asking 'how do I physically make contact with all the places I need to?'?

When I have to make contact in lights like the Ultra and Arc I use a pair of contacts I made up some time back. They use 14 AWG 'house wire' with about 1/8 inch of the insulation stripped back. Masking tape is used to build up the diameter to center it in the tube (pruists would use duct tape, but I didn't have any handy....). In use, I clamp the flashlight nose down in the bench vice, drop the tool down the tube. clip a test lead to the threaded end and make the necessary connections.

I generally use one of my bench supplies so I can vary the voltage, but keep two and four cell holders handy. I also use a piece of wood with a groove in it to line up external batteries (sometimes with the help of some tape) sometimes.

I've also been known to use very thin insulated (wire wrap) wire. A piece is put up the tube with the end stripped and bent over before the cells are inserted. Done carefully, it gets trapped under the positive contact at the top of the tube. You need enough clearance to get the cells in, but it usually works.

Does that answer it?

Doug Owen

 

[member 6142]

@Doug Owen:
Yesterday I did measurements with series connection (DMM - to - of latest batt, DMM + to threads of torch, SF L6):
A= 0.27
V=6.17

Then I took batt. off the torch and measured them in parallel and the result was 8.07 .
Probably not a reliable way, but enough to get a rough estimate of efficiency.

NOTE: batteries are NOT brand new!

@andrewwynn:
Were you replying to me or to "sotto"?

 

[andrewwynn]

daberti: was answering sotto's question about 'how to basics' with a built light.. the system always says it's a reply to the previous post.

 

[member 6142]

Not that time -at least on my browser- thus my question /ubbthreads/images/graemlins/smile.gif

Thanks