Resistor or driver for this?

sailor612

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Jan 23, 2009
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Ok...with DX taking so long to get the AMC7135 1400mA Regulated Circuit Board http://www.dealextreme.com/details.dx/sku.1886 off of back order...i'm contemplating going with the original idea that I was given.
Just to give yall a background, this is what i've got:
1 MC-E running off 3.6V

The original idea was just to use 4 1ohm 1W resistors like this:
mce.jpg


The next idea I had with the 1400mA driver was:
mce2.jpg



Since it is taking so long...and who knows when DX will actually have these in stock, I'm thinking of going back to the resistor idea.
Anyone got any pointers of which I should do? Will I lose noticable power with the resistors? I figured i'd get about a 10W light with the drivers...will that be the same with resistors? What is the difference between direct drive w/ resistors and using a driver?

OR...does anyone know anywhere else (online or in the US) that stocks those drivers??
Any help would be great!! This is my first mod and now i'm kind of let down with the fact that DX has taken so long (customers are actually complaining on their site!)
 

Greg G

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Have you thought about using a HipFlex or HipCC from TaskLED? Or a SharkBuck from the Sandwich Shoppe?
 

BillyNoMates

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With a 3.6V battery, you are not going to have sufficient head room for either the driver or the resistor.

For example, assume a single LED die in the MC-E running at 700mA. The required Vf for this will be (approx) 3.5V. With a 3.6V battery, you have 100mV headroom. Apply ohms law to find the current through the resistor and you get 100 mV / 1 Ohm = 100mA.... Now the same current must flow through the resistor and the LED die so what will happen in reality is that the circuit will settle at the current where the Vf of the LED and voltage across the resistor correspond to the same current. This will be some where about 300mA (Vf ~ 3.3V, Vresistor = 0.3V). The actual numbers will vary in practise, but this is a good start.

Problems come when we look at the battery voltage. The 3.6V is probably an average voltage - the real voltage will vary as the battery discharges (or as it is loaded more). Just considering the state of charge for now, a Li-Ion battery will be 4.2V when fully charged and 2.8V when fully discharged. The resistor solution will drive about 700mA at full charge in this case (Vf=3.5V, Vresistor = 0.7V). As the battery discharges, the current will drop until the voltage gets too low to produce any light (won't go much below 3.2V with the resistor). This is OK, but you won't get the full charge out of the battery with the resistor solution.

With the driver, it will limit the current to 700mA (per die) while the battery voltage is above the Vf of the LED at 700mA, below that level, you get what ever current the LED will pass given the battery voltage. This is why the drivers are a good solution to this problem.

I used to just use resistors, but I got fed up with the drawbacks and I always use drivers now. In the case of a single Li-Ion cell I use buck-boost converters (rather than linear regs) to keep the current constant over a larger voltage range (even when the battery voltage is below the Vf of the LED).

Alternatively, just use the resistors until the drivers turn up........
 
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kuksul08

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Apparently Kai Domain has shipped the amc7135 recently. I made an order and it's packing right now, I'll let you know if they actually ship.

I waited over 5 weeks for mine from DX and still never got them, but did get other things shipped separately.
 

sailor612

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I got to looking at those 1400mA drivers last night...they are only 3W 1400mA drivers.
Is that going to cause a problem? Since i'm going to run 2 dies in parralell at 1400mA (700mA per die) with a recommend Vf of 3.4V. That means at 1.4A and 3.4V (recommended) it is 4.76W. So a 3W 1400mA driver wouldn't be enough to run 2 parralell die at optimium levels, correct?

So, my other thought is this using 4 700mA drivers (replacing the 1 1400mA with 2 and having them wired to the seperate Vin/die) with 2 connected with a (-) jumper, and the other 2 connented together with a (-) jumper, and basing my MC-E Vout off of the original schematic. Would that work?
That would keep me in the 3W range of the drivers....

I drew up my schematic idea, but just can't figure out how to post it on here!
 
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Benson

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As already stated, you should have higher input voltage for the amc7135 regulators; something like 1V headroom works well, so they're real good for driving from 4xNiMH (4.8V) pack, not real good for a 1xLi-Ion (2.8-4.2V); they may regulate a bit on a full cell, but it quickly drops out of regulation, and will be dimmer than a direct-drive for most of the cell's life. DD may not work so well either, though, as the voltage drop when driving 2 dies will be less than with all 4 dies, so 2-die will be more than half brightness, and more chance of overdrive (depending on your cell and wiring resistance) on 2-die. It costs more, but if the battery choice is fixed, you're probably better off getting some sort of buck-boost driver, or running dies in series with boost drivers.

I got to looking at those 1400mA drivers last night...they are only 3W 1400mA drivers.
Is that going to cause a problem? Since i'm going to run 2 dies in parralell at 1400mA (700mA per die) with a recommend Vf of 3.4V. That means at 1.4A and 3.4V (recommended) it is 4.76W. So a 3W 1400mA driver wouldn't be enough to run 2 parralell die at optimium levels, correct?
If that's actually a power limit on the driver, it's for the power dissipated in the driver, not the power dissipated in the LED. (Actually, knowing DX, it quite likely means this is meant for 3W LEDs, albeit with some overdrive, and says nothing about the driver.) So if you drive from a 4.8V source, and and have Vf of 3.4V, the power in the driver is only 1.4V * 1.4A = 2.0 W; you're still good.

So, my other thought is this using 4 700mA drivers (replacing the 1 1400mA with 2 and having them wired to the seperate Vin/die) with 2 connected with a (-) jumper, and the other 2 connented together with a (-) jumper, and basing my MC-E Vout off of the original schematic. Would that work?
Not quite sure what you mean about the (-) jumpers, but assuming you mean the obvious 1 driver per die approach, yes, that would work, although you don't need it for power considerations. The only reason to do this is that it could be seen as a little more robust against any die mismatch. (This is only sane, though, if you think there's more variation between the dies in one MCE than between separate drivers not known to even be of the same lot...) But if you do go with this, you've already got the 1400mA drivers on order, so just modify them; cut the trace that crosses under D1, and you've got two 700mA drivers on one board, with the battery side already in parallel!

(Another couple thoughts: you have two switches, but your current circuit makes no distinction in output between switch A alone and switch B alone. For better brightness options, what about rigging the dies as 3 (2100mA) and 1 (700mA) instead of 2 and 2? Then you get 1, 3, or 4... Or even run all 4 in parallel, with two 1400mA drivers connected in parallel, and connect 3 and 5 AMC7135s, for 1050mA, 1750mA, and 2800mA? (Cut the outer ring trace to separate the battery side while keeping the emitter side paralleled.) The emitter will be more efficient when running all dies at half current than half of them at max current...)
 

HarryN

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I think you will be ok with the 1 ohm resistors with a target of 500 - 700ma per die junction. It takes a healthy cell to put out almost 3 amps and not sag, but it is ok.

I did more or less the same thing with my "Breeze" project, using a Lux V (2S 2P die), a pair of 2S CR2 Li Ion cells, and 2 ohms. The advantage I had was that the Vf bin of the Lux Vs I bought were quite low - likely lower than the MC-E you have, but it is hard to say.

Will it be perfect - no. Will it light up bright as heck - yes. :thumbsup:

While more power is always great to have, you probably cannot actually hold a small light for very long that is dumping 10 watts into your hand, and while things are warming up, LED efficiency drops off fast.

BTW - the resistors I used were special 1206 size (reasonably small SMT) but could handle the power due to their special ratings. Most SMT resistors cannot deal with much power for long.
 
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