How to wire 9 xp-g for direct drive?

[olrac]

I am looking to run 9 xp-g off of 2 AW black C cells (8.4V) by direct drive and I always get confused by this, can anybody can help?

 

[steve6690]

I've given this some thought and I don't see a way to DD this number of led's, apart from wiring them up in a 9P configuration and using a whacking great resistor. Far from ideal though. What led drive current do you want to use ?

 

[Linger]

I'm not sure how much you'll gain by using that many xpg's.
Try 4s2p, this is a very safe configuration. You might be happier with the output at 3s2p, but alternately you could set 3s3p.
wire them up 9s. Note the output (compare to fixed light and take beam shots) and remove an xpg from the series. keep working your way down until you're happy with the set-up.

 

[Linger]

9P configuration using whacking great resistor

Sorry, I totally disagree. IF he's using 2 cells in series (8.4v) you'll be hitting each led with 8.4v if you wire the emitters in parrallel. This is far above the usual 3.5v. I have no idea what you're seeing that makes this seem like a good idea.

 

[steve6690]

Sorry, I totally disagree. IF he's using 2 cells in series (8.4v) you'll be hitting each led with 8.4v if you wire the emitters in parrallel. This is far above the usual 3.5v. I have no idea what you're seeing that makes this seem like a good idea.

I said it's not a great idea. You missed the part about the resistor to drop the voltage...

4s2p doesn't equate to 9 led's..

How do you propose that 9s is going to work ? Vf of around 30v, direct driven from 8.4v...

None of your suggestions are feasible

 

[Justin Case]

That's what the resistor is for. I see steve6690 beat me to it.

The 3S2P or 3S3P configuration is maybe possible, at least initially before voltage drop kills you. If you assume no voltage drop at all for your 8.4V power source, then 3S (assuming electrically identical LEDs) means 2.8V Vf for each XP-G. Looking at the datasheet, that corresponds to about 150mA drive current. That looks like about 40% relative luminous flux, or about 50 lumens per XP-G R4 bin. If you have 9 XP-Gs wired in 3S3P, then that could theoretically mean 450 lumens. However, if you suffer any voltage drop at all (which you undoubtedly will for a battery source for the current draw required to power 9 XP-Gs at 150mA per XP-G), e.g., to 7.6V or 7.4V, then your output could plummet. The problem is that the XP-G datasheet for If vs Vf stops at 2.75V. However, extrapolating to 2.5V Vf, I think you are going to get very little output. With 4S2P, you are looking at a Vf of 2.1V and I doubt that the XP-G will light up at all.

 

[steve6690]

That's what the resistor is for. I see steve6690 beat me to it.

The 3S2P or 3S3P configuration is maybe possible, at least initially before voltage drop kills you. If you assume no voltage drop at all for your 8.4V power source, then 3S (assuming electrically identical LEDs) means 2.8V Vf for each XP-G. Looking at the datasheet, that corresponds to about 150mA drive current. That looks like about 40% relative luminous flux, or about 50 lumens per XP-G R4 bin. If you have 9 XP-Gs wired in 3S3P, then that could theoretically mean 450 lumens. However, if you suffer any voltage drop at all (which you undoubtedly will for a battery source for the current draw required to power 9 XP-Gs at 150mA per XP-G), e.g., to 7.6V or 7.4V, then your output could plummet. The problem is that the XP-G datasheet for If vs Vf stops at 2.75V. However, extrapolating to 2.5V Vf, I think you are going to get very little output. With 4S2P, you are looking at a Vf of 2.1V and I doubt that the XP-G will light up at all.

Thanks Justin. I did the sums and came to exactly the same conclusion.

 

[HarryN]

Not identical, but I have run Lux V emitters (which are wired 2S2P) with 2S LiIon cells and a 2 ohm resistor. This is pretty close in concept.

Each LED string (2S) needs 2 ohms and you are really close. Of course - that pushes you to 4 strings of 2 (8 of them) or 10 of them in a 2S x 5P. I like to add a diode into each string to prevent accidentally reversing the battery and blowing the whole thing, so that takes a touch off of the Vf as well.

 

[olrac]

What do you think about running it 9P using two AW C cell in parallel instead. 3.7-4.2V (6600 mAh)

 

[steve6690]

The problem with wiring led's in parallel is that they all get exactly the same voltage. This means that unless the Vf of each led is exactly the same, they will draw different currents and hence have differing brightness levels. This could even mean that one or more of the led's is substantially overdriven. Something else to seriously consider is the total current. If you wanted to run each led at 1 amp, your total current draw would be 9 x 1A = 9 Amps (assuming matched Vf's). This would need thick gauge wire. If you were going to attempt this, I'd suggest choosing a drive current well within the led's capability - say 350ma - and see how you get on with that. You will need to consider voltage sag in your calculations too. I haven't looked to see if suitable resistors are available. The first thing I'd do is hook the led's up to a bench power supply and measure the Vf at a range of drive currents.
I think it'd be a nightmare though, if I'm honest, but you might find it an interesting experiment.

 

[Linger]

That's what the resistor is for.

-minor follow-up: I didn't know use of resistors qualified as direct drive. As such, was trying to think of viable set-ups. Thus 3s2P, yes which isn't 9 xpg's, is atleast a config that doens't . Same as 9s, while 9 won't work, I did instruct that the series xpg's be reduced down until a feasible set-up is discovered. In short, with my initial assumption that direct drive meant no resistors as well as no boards, I was hinting that 9xpg's off 2s C batts was a challenged format.

 

[jasonsmaglites]

are you sure 2 lithium ion c cells aren't just 4.2 right off the charger, cause i thought they were 3.6 or 3.7 like the 18650s. if they are 3.6 or so, i'll tell you in theory how you could dd them, although ive never done it per say. you could wire 2 in series, 4 times, with each group being run parallel to the next. this would give you twice the voltage of the led as the drive vf, or about 7 volts(depending on vf bin of course) and you would slightly overdrive it, just dont run them hot off the charger. let the batteries rest for a day. this would get you 8x xp-g dd.

you could do 9 if you wired them 3 in series, 3 times, with each group of 3 being run parallel to each other. assuming a standard vf of 3.5 again, you'd have a drive vf of 10.5 and you would be underdriving the xp-gs but this would give you good runtime and probably be right in the heart of their efficency (probably more like in sscp4 peak eff. of 350ma but ok). im sure it would still be plenty bright but this is one way you could pull that off. just a theory, good luck.

edit: i only skimmed this thread before posting. similar suggestions have been made. good luck nonetheless

 

[TorchBoy]

-minor follow-up: I didn't know use of resistors qualified as direct drive.

It doesn't, because the resistor is a current limiting device, so the LED(s) would not be running directly from the battery. I'm going to go out on a limb and claim 3s3p is the only safe genuinely direct driven option that would actually make some light. :nana:

 

[olrac]

Thanks to everyone for your contributions. This LED and battery config. is not a DD candidate.

 

[jasonsmaglites]

sure it is, and in my humble opinion, a resistor still qualifies as direct drive. maybe a regulated 7135 doesnt, but a good old fashion heat wire does.

 

[Justin Case]

-minor follow-up: I didn't know use of resistors qualified as direct drive. As such, was trying to think of viable set-ups. Thus 3s2P, yes which isn't 9 xpg's, is atleast a config that doens't . Same as 9s, while 9 won't work, I did instruct that the series xpg's be reduced down until a feasible set-up is discovered. In short, with my initial assumption that direct drive meant no resistors as well as no boards, I was hinting that 9xpg's off 2s C batts was a challenged format.

Not sure why you decided to quote me instead of steve6690, but since you did....

Whether or not use of a dropping resistor meets the definition of DD is not relevant. The point is that you seemed to believe that steve6690 was suggesting that sending 8.4V into a 3.5V LED load was ok. He said nothing of the sort. Here is what you wrote about steve6690's post:

Sorry, I totally disagree. IF he's using 2 cells in series (8.4v) you'll be hitting each led with 8.4v if you wire the emitters in parrallel. This is far above the usual 3.5v. I have no idea what you're seeing that makes this seem like a good idea.

But this is what steve6690 wrote:

I've given this some thought and I don't see a way to DD this number of led's, apart from wiring them up in a 9P configuration and using a whacking great resistor. Far from ideal though. What led drive current do you want to use ?

The problem with your suggestion of a 4S2P is that it is so safe that it probably won't even light up. For 3S2P or 3S3P, of course you don't need any resistor since the load voltage is already high enough so that the LEDs won't be overdriven. The problem is that you are running at the bottom of the LED's If-Vf curve and small changes in Vbatt can result in a big drop in output. There is safe, and there is safe and practical.

Regarding the definition of DD, If I had a junky old Mag that had high enough contact resistances such that I got 2.8A at the tail for an SSC P7 build, is that direct drive or not? Now suppose I have a brand new, clean Mag with low contact resistances and get 3.6A at the tail. If I add a dropping resistor, somehow this is different?

 

[Linger]

1)noted
2)I'm not concerned what definitions of dd are, i was just stating my assumptions, which I will henceforth endeavor to do in the original post