Vf refresher

Can someone please refresh my memory on how to measure the Vf of some of my Lux III's? I remember it involved using these .10 ohm 1% resistors I bought from Sandwich Shop, but I forgot how to measure them, what voltage to use, etc..

No need for resistors. While the LED is lit (running at rated current), simply measure the voltage across the two leads of the luxeon using your DMM set to the voltage scale.

If you don't have a nice power supply with a current meter, then you probably want to do this:

Connect the Lux III and .10 ohm resistor in series.
Connect (watch polarity of the Lux) to your power supply.
Connect a meter to your resistor.
Adjust voltage across resistor to be Vr=I-lux * R
-->example: Vr=0.300 * 0.1 = 30 mV (for 300 mAmp bias current)
Measure voltage across the Lux III for Vf (with your operating temperature and heatsink)

Be very careful not to connect the Lux III backwards--most LED's respond badly to more than a couple of volts of reverse bias.

If you have two DVMs, you can simply put one in place of the resistor and read the current directly.

The shunt resistor method is probably more accurate if you only have one DVM and you are using a voltage source (as opposed to a current source). At low currents, most meters have a relatively high internal resistance--and if you set up the current and then remove the meter--the actual current will probably change in your circuit--thereby changing your Vf measuring point.

Diodes are very non-linear devices. A small change in Vf (or even the diode's temperature) can change the current dramatically.

-Bill

That's what I needed - Thanks!

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BB said:
The shunt resistor method is probably more accurate if you only have one DVM and you are using a voltage source (as opposed to a current source). At low currents, most meters have a relatively high internal resistance--and if you set up the current and then remove the meter--the actual current will probably change in your circuit--thereby changing your Vf measuring point.



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While it's true that measuring the actual current at the time you measure Vf is the most accurate, the 'relatively high resistance' part is not. At least not on high current ranges (like we're using here).

The ten amp range on common DMMs is actually *much* smaller than the .1 ohm suggested above (the internal shunt is .01 ohms), there's no doubt more resistance in leads and other connections. I know the 'conventional wisdom' here at CPF is different, but the facts are the internal resistance is very low (it has to be, or power dissapation would cause serious problems).

Still, any flahlight junkie worth his salt should have several $3 meters about by now, why not leave one in circuit and use a second to measure Vf?

FWIW, Vf doesn't really change all that much for (small) changes in temperature. .002 Volt drop per degree C rise is a typical number. Given a modest rise (say a few dozen degrees), the change in Vf isn't all that large. Provided you're not living over the edge (ie 'Direct Drive') chances of thermal run away are fairly small. For a true current regulator there is no change in current (and therefore no risk), and even a fairly modest resistored system is fairly safe. It's the fixed voltage (or DD) guys that are at risk.

Doug Owen

OK, now that I know how to measure the Vf I have another question. How high over the Vf can I DD a LuxIII? I'm building a DD LuxIII into a Dorcy 4AA aluminum. I can either use a lower Vf with 3AA and 1 dummy, or a higher Vf with 4AA. I'm not sure what the differences would be as far as brightness and runtime. Can I DD 4AA without a resistor safely? What Vf should I use? (I have K and L right now) I need one of you to make up my mind!!!! Good discussion on Vf by the way. I do have 4 DMMs (2 cheapos, 1 pretty good, and 1 great fluke) Sometimes it would be great to know the exact Vf and in other cases I am just sorting unknown LEDs and just want to know where they stand relative to each other (exact Vf not as important) Nice to know I have options. Thanks again.

Please notice what I said about using a Voltage source vs. a Current source.

Voltage sources, by definition, have zero resistance. Current sources, by definition, have infinite resistance.

With large wires and no other sources of resistance (other than the meter and leads), with a voltage source (and no or high current limit set), you would be direct driving an LED. And minor changes in effective Vf can result in large changes in current. Thereby making accurate measurements impossible.

The problem is, without knowing the exact setup, I gave the method he (or she) was looking for--with appropriate warnings to help ensure accurate and repeatable readings.

Sincerely,
-Bill

At the risk of offending some of the good folks here abouts, I'm sorry you're on your own when it comes to Direct Drive.

I'm not even sure I think it's OK on coin cell keychain lights. For sure I consider it a highly questionable practice for 'normal' use, *as to the LED makers*.

Counting on ill defined internal LED resistances, uncontrolled cell resistance and often dicey contact and wiring resistances to regulate current and save the device goes against my training and experience.

Doug Owen

Normally, I fully agree with you. All of my other lights use some type of regulation, usually current (BB) since it gives the most consistent light output (which is the real reason we love these things - right?) In this case I'm just looking for a relatively cheap, small form factor light to use around the house. (plumbing, electrical repairs, etc..) I want it to have long runtime. I'm considering the Madmax plus, but I want more than 800ma output. I'd like to just direct drive it but not sure if I can do that with 4 AA batteries. (1.2v NiMh and/or 1.5v Alks)

Fine by me.

I was really speaking to 'should you' rather than 'can you', the decision to go DD (or any other route) is, of course, yours to make.

I just don't support the idea of DD, I consider it poor engineering. Yeah, I know there's a lot of guys doing it (with differing levels of success), then again there's a lot of guys dating married women......

Doug Owen

I have to agree 100% with Doug Owen. All it takes is a LM317 and a resistor to regulate the current. Power source would have to be at least 1.2 V above the Vf of the LED. There is a LDO version also but the LM317T can be found at Rat Shack. Alkalines could deliver over 2 A based on Energizer rated internal resistance. NiXx will deliver a whole lot more. And remember the NiXx voltage at full charge will likely be 1.45 V or higher. 1.2 V in the nominal voltage. Any voltage higher than Vf at the rated current is going to cause more current to flow. The real reason to find the Vf is not for voltage regulation but to find the minimum voltage would allow the full current to flow. When a DD's battery fall below this point the current, and brightness will drop and the color will shift. If you really what to do without the regulator at least use a resistor. An example: V[f] = 3.4 V, Battery is 4 AA cells and lets say we will only use Energizer alkalines. So the battery supplies us with 6 V and the LED drops 3.4 V. Now lets say we're only going to run 700 mA through the Luxeon. The resistor must drop the remaining voltage so 6 V - 3.4 V = 2.6 V. Ohm's law 0.7 A / 2.6 V = 3.713 ohms. Less the resistances of the battery (0.584 ohm) is 3.129 ohms. Nearest standard value 1% I think is 3.1 or 3.2. Wattage would be I^2 * R or 0.7 A ^ 2 * 3.1 ohm = 1.519 W. Normally you double so it dissipates the heat better so 3 W or larger. Please check the math. Notice with this you can at least be draining some power out while a DD solution (if you can call it that) will probably leave a large amount of power in the cells.

I am not the expert on CPF regulators (I do the bigger computer systems). But I would suggest asking in the Modified Flashlight and/or someplace like the Sandwich Shop.

With a "true" fixed voltage source and an LED (diode)--you could, in theory, adjust the voltage for the proper operating point on the I/V curve for the current you would want. In the real world, your voltage source is not perfect (battery with internal resistance), things change with respect to temperature and battery levels, etc. And even a "direct drive" still has a ballast resistor--it is just the parasitic resistor in the battery, wiring, and the LED itself. By upping the battery voltage and adding an external resistor, you will get a flatter output--but with more power lost as heat in the resistor.

The LM 317 make a great current source--but as GW says--you will be wasting much (if not 1/2 or more) of your total battery energy as heat through the LM 317.

If you can find a current mode (or power mode) regulator assembly that is close to your needs (right power and physical size)--you will be much happier. You will only loose a fraction of the energy (typically 10-25% depending on several factors) and you will end up with a light that has constant output up until there is no more energy in the batteries.

If you follow their instruction and keep wires short (and low resistance) the regulators should be pretty easy to work with.

Take care,
-Bill

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BB said:
..
The LM 317 make a great current source--but as GW says--you will be wasting much (if not 1/2 or more) of your total battery energy as heat through the LM 317.

If you can find a current mode (or power mode) regulator assembly that is close to your needs (right power and physical size)--you will be much happier. ..
-Bill

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Actually at 6 V in the effeciency is not that bad, about 60% with the Vf above. And improves as the battery voltage drops (I'm guessing around 75% just before dropping out of regulation). Compares well with a few of the switchers. Main issue is the drop out voltage because of the 1.2 V reference. Yes, a switcher with a higher effeciency is going to give a longer run time in regulation. But its going to cost more than $2 US.

George

Granted, it has been a very long time since I have used LM 317 regulators... But you would need to allow for the 1.25 volt band gap reference (when used in a current regulator mode) plus around 2.25 volts from input to output to properly bias the part for 700-800 mAmps... Or 3 volts total.

Fairchild LM 317 spec.

That plus the 3.4 Vf of the LED would require 6.4 volts of input battery voltage. I measured a couple unused Duracell AA's and are about 1.6 volts no load or 6.4 volts for four cells (again, no load). Basically, the LM 317 would have a difficult time driving the LED with 4 akaline cells at rated current and it would fall out of regulation very quickly as the cells were drained.

To reliably support 3.4 Vf in this case, you would need 6 alkaline cells (assuming a dead battery is 1.07 volts, excluding internal battery resistance). And, of course, the effeciency is even worst as the starting voltage would be somewhere in the 9V range (of which the LED is only seeing 3.4 volts of this energy) or somewhere south of 40% efficiency. The best efficiency will only be around 53% before the LM 317 starts to drop regulation.

A low dropout regulator will make the numbers somewhat better--Did I make any mistakes here?

-Bill

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BB said:
A low dropout regulator will make the numbers somewhat better--Did I make any mistakes here?


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No, Bill, once again yer spot on. We seem to forget about the headroom requirements of the 317 (although it is a neat part, 'ideal' in many cases).

Since it's fairly easy to get a LDO regulator to fly with .1 Volt of headroom you can get an honest 94% (3.4/3.6) from three NiMH or an average 71% (3.4/4.8) from four AA Alkalines (1.2 Volts each at 'half used'). The latter is hard to beat, the former 'impossible'.

IMO, given full options a proper LDO current regulator and 3 NiMH cells is 'the way to go' when driving 3.4 Volt Vf devices. When forced to use other batteries, the choice is not nearly so clear.

Also, IMO, DD is the 'solution of last resort'.

Doug Owen

You are right of course. Had a nagging feeling I was forgetting something in current mode. Just proves don't post when sleepy.

George