How to measure current?

I have this BB700 V4U PR head that I am running off of NiMH AAs. It obviously is not in regulation, and that is the point because I don't want it to get hot. I was curious what it was running at, so I tested the current across the luxeon--it read 2.10A and 6.02V, that can't be right.
What am I doing wrong?

I think you are right. A Vf of U should pull 700ma between 6.87V and 7.35V

6.02 shoudl be under driven unless it's seriously overheated.

How many batteries are you using as imput to the BB700. You should feed it less than the Vf of the LED.

Did you put the DMM in series with the LED to read the current? That's the way to do it normally.

Daniel

It is on two AAs. I put the positive test lead on the positive side of the luxeon, and the negative on the negative. But I was thinking, that maybe the DMM was pulling as much as it could from the badboy--like when I tested a pila it read 9 amps, which is more than it would put out under normal circumstances.

You are measuring current incorrectly. In order to measure current, you must break the circuit, and put your meter in series with the circuit - so, to measure the current going through the luxeon, you would insert it inbetween it and the driver board on either the positive or negative lead.

So, 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.

sounds like you may have had a major conceptual error in measuring current... current is measured in SERIES and voltage in PARALLEL.. you described trying to measure both in parallel.. which will short out the lux when you attempt this maneuver.... did the lux blink out when you tried this?

If the lux is soldered in you can not measure current to it... you can only measure current from the battery since to measure DC current you must break the circuit and put the meter in series... you can estimate the current to the lux by measuring the battery current and voltage and multiplying and dividing by .85 to get power to the lux... you can then measure the voltage at the lux and divide the estimated power by that actual voltage.. it'll give you an estimated current.

If you tried to measure the current incorrectly like i think, you're lucky you didn't melt some component (or just lucky you have a tough driver designed to handle a short).. be careful doing current measurements... if you tried to do such a measurement on main house current something will melt for certain.

-awr

Man, that shows what 15 years of school is good for.

you didn't follow up on what you tried, or if you ever figured out the answer to your question... what's the point of not running in regulation? if you want to keep it cool... throttle back the output to your desired level with the resistor (sense) and have consistent brightness and runtimes.

What I really wanted was a BB500, but I am poor and the BB700 was very cheap--and I find that running it on 2 NiMHs will do the trick (for about 45 minutes). It is all potted, and I don't want to mess with it, because I have a bad track record for tinkering with electronics. I tried measuring it as you suggested, but there is not enough wire exposed to do so.

actually.. you said you measured the lux itself... so.. you just need to measure the battery current and voltage to determine power in... then measure the Lux voltage... to measure the bat voltage just use some wires outside the light...

the BB700 could be converted from what i understand.. just means changing the sense resistor... but sounds like a little late if you already potted it together.

The regulation calculation on the calculator page www.anlighten.com/shop/test.php will estimate the lowest battery voltage for a given combination of Vf, Battery voltage and battery capacity.

But, Almost any 5W combination the converter board will drop out of regulation around 4.5V or higher.

The required battery current at 3V is:

6.02V/3V * 1.3 ~= 2.61A.

Since the Converter stops at around 2A, it will not be able to deliver the power required to stay in regulation.

A general rule of thumb on the Badboy and a 5W is dropout occurs around 4-4.5V and you need to keep the battery higher than this to stay in regulation.

Another possible solution is to find a step up converter that will do 3-5A on the input side so that it will have sufficient margins to stay in regulation.

The only combinations that I'm aware of that work and work well are 2X123 and 4 AAs. 3AAs is starting to get real iffy.

I plugged in the following parameters to the calculator. 3V for Vin, 6.5V for Vf, desired LED current = 0.7 and battery capacity of 2 (2amp/HR).

The calculator came back with this:

"This configuration will not work with a Badboy since the input voltage of 3 is below the regulation drop out point of 3640 Volts
or, your input voltage is less than 1.6V"

Wayne

Thanks for chiming in dat2zip, I don't see you post very often outside of the Shoppe forum. That info would be very helpful if I didn't already know it all--I am all-too-familiar with the Shoppe calculator. I am not certain what you were responding to though, were you rebutting my statement that I could run the BB700 on 2AA? Because it is not in full regulation--not even close, but it does give a stable output. And it draws about 1.65amps from the batteries.

1.65A at 3.0V works out to be almost exactly 4W at the emitter using 80% efficiency... dividing 4W by 6.02v gives you the answer to your initial question... 660mA... which is darn close to the 700.. i wouldn't be surprised if that actually IS running in regulation.

Not 3V, more like 2.65V--under load 2.04V.

Yeah I was trying to guesstimate what kind of voltage drop there'd be under load... 2V with fresh cells? that's a little surprising... have you tried with NiMH batteries (oops... that's why the voltage discrepancy), they can handle 1.6A ok.. what kind of runtime do you get again? oh yeah 45min.. that sounds about right..

so.. 2.04 V x 1.65A = 3.36W multiply by 80% and you get about 2.11W... very under driven for a V.. too bad you didn't use a III.. much cheaper... so.. 2.11W divided by 6.02V is... 350mA... the answer to your initial quest... the current at your lux. that works out to 60% of max brightness based on lumiled's datasheet on the V... or 120lumens x 60% = 72... which is what a lux 3 is rated to output at about 2W also it seems... i guess that makes sense.. trade off is cost i suppose... if you are using a boost ckt though benefit of V is it works with a batt that won't with the III. .. FYI... a lux III has to be run at 600mA to output a similar amount of light, so the BB700 would do the trick as well.

-awr

At 350mA, your Lux V is very underdriven, and will run cooler (internally) and will last for a very long time.

I wonder whether the little heat it does generate is from the converter or the luxeon. Does anybody know whether I am going to overexert the badboy doing this?

Your badboy is probably running even less efficient than 85%, since the input current is so high.