MC-E 2S2P with a resistor?

[kosPap]

greetings all! I hope you are having a good time this season...

(and I hope it is the correct subforum I am posting)

So...

How about a MC-E connected 2S2P?
From what I can tell one goes for 1400mA output, 2xVf of the LED and a two cell Li-Ion or equivalent battery combination.

The big question is can it work with a resistor??

I used the Firestorm and LED Pro resistor calculators and from an input of 8.4 and Vin 3.6 (actually 7.2V for this connection) I should be using a resistor at 0.85 Ohms (correct?)
(though the calculator said 8.5 Ohms)

Which is big and my bigger problem is

How do I calculate the watt rating? Should I be using the W=I^2R? that would be W=1.4^2x0.85 Correct?
Than amkes 1.6666 watts that is not bad for the situation...

Now If I only knew that kinda of resistir dimensions? Can it fit inside a 17mm board flashligh pill?

Feel free to correct my math, please, Kostas

ETA: Seems that i will have to go with a higher Vf bin, or else that excess voltage will have to be spent on the resistor...

 

[2xTrinity]

I would use a separate resistor for EACH of the two strings, rather than one big resistor. In other words, I'd use two "1.7" ohm resistors, rather than a single 0.85 ohm resistor.

In this case, there will be a higher likelihood that both strings draw an equal amount of current.

You are correct on expected power dissipation. You could use two 1.6 Ohm (a standard value), 1 watt resistors. However, those tend to run fairly bulky.


Also, a few other things I would consider:

• Your battery will not ever be providing 8.4 amps under load, as it has some resistance of its own.
• You can't count on the Vf of LED until you actually test it. the MC-E's I've used have had Vf's as low as 3.3V @ 700mA (per die).
• Your battery loaded voltage will drop rapidly after the first few minutes, then stay steady for most of the runtime. I would plan your resistor around this "steady state" case, and tolerate some slight overdrive for the first minute or so of runtime.

So you can in all likelihood get away with a resistor value that is lower than the one suggested by this calendar. The resistance in the cell itself will do a lot of the "work" for you.

Another approach would be to use something like the 7135 drivers from Kaidomain -- these are low dropout linear regulators, and will in effect behave like "smart" resistors -- changing their value as the loaded voltage on your cell drops. You'll want a lower Vf bin if you can do this, as there needs to be some "overhead" of a few tenths of a volt for those drivers to stay in regulation.

 

[kosPap]

Thank you for your comments...
indeed the final value will be tested...

The application is for a 2 cell RCR123 cylidrical light. The 8.4 volt Value is the nominal.

How should the resistors be connected? Your value seems to suggest parallel. That means that each free end will power a 2 die cluster, doesn't it? Hmm Clever

 

[Mr Happy]

How about a MC-E connected 2S2P?
From what I can tell one goes for 1400mA output, 2xVf of the LED and a two cell Li-Ion or equivalent battery combination.

The application is for a 2 cell RCR123 cylidrical light. The 8.4 volt Value is the nominal.

I think you might want to consider a smaller load or a bigger battery.

Since the typical capacity of an RCR123 is about 650 mAh, the maximum recommended discharge rate of 2C would be 1300 mA, but lower would be preferable. It seems to me that you are stretching the limits of what is achievable with your proposed setup.

 

[kosPap]

hmm thanks....

You ahve been helpful before, so could you enxtend yourr generosity to verifying that my calculations are ok?
(I am talkig the resitor value and wattage...

 

[Mr Happy]

I have not ever done what you are trying to do, so I don't know how much I can help. However, here are some ideas to think about.

Your RCR123 cells will not have a fully charged voltage of 4.2 V for very long. Once you start drawing power from them they will start to discharge towards 3.6 V. So maybe it is better to assume something like 4.0 V for design calculations.

The cells also have internal resistance that will drop the voltage once current starts flowing. I don't know it will be exactly, but if we guess at an internal resistance value of 0.3 ohms, then at 1.4 A supplied to the load the voltage drop will be 0.42 V. This brings our operating cell voltage down to 3.6 V.

At this point, you are in the operating range of the Vf for the LED.

Therefore, you probably want to do an experiment with no extra resistance at all and measure the current to see what you get. Make sure the emitters have good heat sinking, and then if you do get higher currents for short periods you reduce the risk of damage.

 

[kosPap]

Thanks!

 

[Techjunkie]

Hi there, I'm reviving kosPap's thread again, as I find myself in a similar predicament. I've built several torches with MC-Es wired as 2s2p that direct drive from 2 li-ion batteries and have been so far, very successful. The problem that I now face is a combination of factors: 1) I've switched to IMR batteries which have very low internal resistance, showing almost no voltage sag under even very heavy loads. 2) The latest batch of emitters that I have received have very low Vf... despite excellent heatsinking, their tint shifts toward blue very easily compared to other MC-E emitter based torches that I have assembled in the exact same or very similar fashion.

Even a 3xMC-E torch, wired as 2s6p "gets blue" when the two IMR 18650s that power it are fully charged. I literally have to run them down to below 3.8V at rest (using a WF-500 Xenon torch) to be able to use them in the MC-E torches.

So I'm writing to find out how kosPap made out with the two 1.6ohm resistors in parallel and if there are any other opinions on how best to tackle the low Vf, issue when 7.2V 1.4A-4.2A drivers are not (commonly) available. Thanks.

 

[download]

I am using this setup with MC-E & 2x18650, 6x 7135 with multi-level, 2A draw from battery end.
~1A of each LED, very satisfied of this result.

You may see here more:
http://www.candlepowerforums.com/vb/showthread.php?t=201392

 

[Techjunkie]

I am using this setup with MC-E & 2x18650, 6x 7135 with multi-level, 2A draw from battery end.
~1A of each LED, very satisfied of this result.

You may see here more:
http://www.candlepowerforums.com/vb/showthread.php?t=201392

That setup might have worked nicely on one of my 2p2s configurations, if there was room inside the pill of the 35mm reflector (~17mm ID), but I'm sure there's not. The other 2p6s configurations wouldn't allow it for two reasons...

For one, my 2p2s configuration on each star is configured on the star by scratching two of the trace routes and soldering together two pairs of feet on each side. I'd have to scratch through the third trace route to break the series circuit and run new wires to insert the regulator between the parallel pairs. There's just not enough room around my stars for any more wires.

Secondly, I'd have to fit 3 regulator boards under the base in the pill area, and that just wouldn't fit either. Not to mention, heatsinking the regulators would be a challenge.

Here's what I've come up with that I'm actually pretty pleased with. I've engineered a removable resistance module that fits in the neck, under the base. I can use it when I want to run IMR batteries, and swap it for a non-resistive module when I want to use Li-ion or LiFePO4 batteries.

The resistors in the module were chosen based on the calculation of a reduction in voltage from 8.4v to 7.2v (2 3.6v in series) at a current of 4.2A (6 700mA in parallel). According to the calculation V/I=R, I needed .28 ohms resistance. I ran two .45 ohm 5 watt resistors in parallel. I know I'm pulling more than 10 watts (about 3x that) but I couldn't fit 10 watt resistors in the neck and Radio Shack didn't have any smaller than 1 ohm.

On two fully charged IMR 18650, the torch now pulls 4.22A from the batteries, right on spec.

Here's a pic of the results:

 

[VidPro]

just notes:
your pulling a total of ~15+ watts through the whole thing, but there shouldnt be 10watts at the resisters themselves. instead there is the current which is the same through the whole curcuit, and the voltage DROP across the resistance, that is the only wattage the resisters themselves have to be able to handle.
( i aint doin the math cause i will get it wrong, but you can)

so the 5watt are PERFECT, beings there isnt a lot of cooling for them anyways. so your right on spec again, according to my bad math.

 

[Techjunkie]

just notes:
your pulling a total of ~15+ watts through the whole thing, but there shouldnt be 10watts at the resisters themselves. instead there is the current which is the same through the whole curcuit, and the voltage DROP across the resistance, that is the only wattage the resisters themselves have to be able to handle.
( i aint doin the math cause i will get it wrong, but you can)

so the 5watt are PERFECT, beings there isnt a lot of cooling for them anyways. so your right on spec again, according to my bad math.

I just measured 4.5A at the tailcap using IMRs fresh off the charger, 4.25V at rest. 4.5A x ~7.2V = ~32.4 Watts. I assumed that because the current is split equally between the resistors in parallel that 16 Watts is running through them, but maybe that's not what's meant by the watt rating on the resistor. (I interpreted it as resistance within 5% of .45 ohms each at 5 watt current.)

Whatever the case, the amount of amps it's pulling is down considerably, and right on spec for 3 MC-E's running 2s6p at the appropriate voltage for a 2s configuration.

Now I can't wait until the 26500 IMRs are available. :)

As for my 2C sized 2p2s torches, I DX has a 25mm diameter 3A buck regulator that's meant for 1s4p. I measured - the 25mm will fit the neck (just not the pill) and the stars are easy enough to convert to 4p, so for a few bucks and a few weeks wait, the same issue with IMRs and those torches should be solved too.

 

[VidPro]

geesh as if i could do the math before now you really got me confused :-)

i will take this slow so i dont fudge it up.

the MC-E is capable of mabey 4.5 amp MAX total drive , when all 4 dies are in Parellel and driven with a single 3.6v li-ion for example.
that is ~16 Watts MAX, before everything goes to heck, and probably should be driven more like 9-12W max.

|--@--|
|--@--|--------------------{======]-------
|--@--| < one parellel 4 die led . ^ one cells
|--@--|



if you SERIES parallel the LED die
which is what were doing here in this thread (if i understood it)

|--@--||--@--| <2 series Die
|--@--||--@--|---{======]--{======]------
2 parellel ^ die . . . . .. ^two cells

is still the same 16 watts , 2.25 amps through the 2 parellel dies, and 7.4v about through the 2 series led die.
2.25A x the 7.4v = ~16


now you put in some resistance

|--@--||--@--| <2 series die
|--@--||--@--|--{~~}-----{======]--{======]------
2 parellel ^ die . . . .^resistance .. ^two cells

IF the current though everything is ~2A , then the current though the resister is 2Amps.
IF the one resister there in my picture is dropping only ~1Volt, then it is handling 1V x 2A or 2Watts

each of the parellel led die is handling half the total amperage through the curcuit, each of the series led die is splitting up the voltage (about) between them


IF you spread the resitance between 2 resisters, then each resister is only handling 1/2 that total resistance wattage or about 1W .

|--@--||--@--|--{~~}----|
|--@--||--@--|--{~~}----|{======]--{======]------
2 parellel ^ led . . .^resisters . .. ^two battery

then when you dropped the voltage at the resitance, you lowered the total flow of current throughout the WHOLE curcuit, so your resistance is a good choice for the job, nice and overrated like it should be, and the total power through the whole curcuit is dropped because of the resitance added, and everything is peachy keno.
(About)

back to the resister, if it is twice the above example (like you have) , and your using 2 resisters, then you still are at a safe ~2Watts through each resister, which was the only thing i wanted to point out, that the wattage of the resister is it's heat handling of it's wattage , not the total amperage through it, or total voltage through it. so that is the voltage DROP through the resister 1v, times the 4+ amps your pushing through the curcuit, or 4 watts total split between 2 resisters. 2watts per resister.

the power handling specs of the resister, are only for what the resister has to do, the Voltage drop at the curcuits amperage = the resisters wattage dissipation needs.

i was just trying to explain the watts of heat dissapated by the resister, and how it is not the watts used in the entire curcuit.

 

[2xTrinity]

I just measured 4.5A at the tailcap using IMRs fresh off the charger, 4.25V at rest. 4.5A x ~7.2V = ~32.4 Watts.

This is wattage dissipated by your entire light, not in the resistor.

You can find power dissipated in a resistor (not the whole circuit) with the following formula:

Power = Current^2 * Resistance

P = 4.5A * 4.5A * 0.225 Ohm = 4.6 Watts


If you are dissipating 32.4W total, that means your LED is consuming 27.8 watts.

Two five-watt resistors is still good though -- you always want to have resistors rated for about double the wattage they will be absorbing most of the time. This "headroom" will help avoid excessively high temperatres.

 

[VidPro]

yes that is what i was trying to say :-) what trinity just said.

but now i see, i think he is running 3x 2p2s not just 2p2s

should i delete the other post :-) what a mess.

 

[Techjunkie]

VidPro and 2xTrinity, thank you both for the explanations. I get it now. It's good to know what the wattage rating on the resistors refers to.

I'm glad to know I (accidentally) got it right too - no worry of the resistors burning out or generating additional heat.

And yes, I am running 3 x 2s2p MC-Es in the one light.

Now that I know the actual meaning of the wattage rating on the resistors, I might be able to fit adequate resistors into the smaller torches that I did not think had enough room. It's good to know that if I'm not satisfied with the regulated output of the drivers I ordered, or if the drivers I ordered don't fit, I can revisit the resistor route on those as well.

As for the calculations, the IMR batteries (which total 8.5V at rest) sag very, very little under load, so in my particular application, the total wattage consumed from the batteries is closer to 37w than 32W on a fresh charge, but even still, that still leaves plenty of head room at the resistors.

This worked out so well, I'm sorry it's too late to cancel my orders for LiFePO4 batteries and charger. I'm glad I at least made that resistance module, well, modular. Although, in all likely hood, those extra batteries will probably end up just insipring another project. I'm already thinking 4x18650 in a 2x18mm bored 2D Mag with 4 or 5 parallel MC-Es on all series stars. Must resist...