AMC7135 1400mA

Hi guys. Im building my first P7 light and have a couple of questions.
I am going to use 2 1400mA driver boards to get 2800mA.
Problem is these boards have a recommended input voltage range of 3.6V~4.5V and I would idealy like to run it all off a 12V NiCad 5Ah or 14.4V NiMh 4.2Ah battery I already have.

Does anyone know how I can regulate this higher voltage to 4.5volt with no or minimal loss of power?
I could always break the battery down and rewire the cells to get 4.8 volt but then I would have to series paralle them and I hear this isnt such a great idea. I could just use 4 cells all in series but then I will not have the capacity.
Keeping them as is seems a better option and means I dont have to build or buy another charger and I can still use them to drive haolgens or HIDs.

Any ideas?


http://www.dealextreme.com/details.dx/sku.1886

Yeah, just rewire them to 4.8V. If it's a "permanent" sort of pack, make sure you have the right pack charger.

The pack will be permenent so it will be hard to charge if I use more than 4 cells.
I read an artical that said Series Parallel of NiMH and Nicads causes one string of cells to get charged fully and the rest remain discharged.
I can set the cells up in a way where it will charge at 12V and run at 6V but rather not do this unless I have too.

Im now looking at the Li Ion option as they seem more tolerant to be wired in parallel.

So my next question is how does the AMC7135 driver work?
Does it boost voltage if needed?
If my Li Ion pack drops to 3.2 volt will the driver still put out 1400mA?
Id rather not serries the Li Ion pack if I dont need to as then I will be getting up to 8v on a fresh charge twice as many cells and no real extra capacity.

Im hoping to get at least 10,000mA

So who has run a P7 on a single 3.7v Li Ion battery? What did you find compared to 2 batteries in serries?

The AMC7135 does not boost voltage. It takes a higher battery voltage and reduces it to the Vf of the LED at whatever drive current you want, dumping the excess as heat. For example, a single AMC7135 (350mA) running an XR-E off three NiMHs will take a battery voltage of around 4.2V (when freshly charged) and reduce it to the 3.2V the LED needs to run at 350mA. If the battery voltage is less than the Vf of the LED, it'll go into direct drive. Continuing the example, when the three NiMHs drop below 3.2V, they'll be running the LED with declining (unregulated) output, as if the AMC7135 wasn't there (mostly).

It'd be ideal to run a P7 off a single Li-Ion, or multiple ones in parallel. Output declines over time, but it's not too bad. If you used multiple Li-Ions in series, you'd need a driver, and the AMC7135 would not work for that (battery voltage would be too high).

Excellent that is what I needed to know.
One other thing though. How much current can I expect to get out of 8 7135s driven off 10000mAh worth of Li Ion cells in parallel on a single P7.
Will I get 2800mA or will I need a higher voltage?
And how many lumen can I expect?

That depends on your P7's Vf at 2.8A. As soon as battery voltage falls below that Vf, the light will go into direct drive.

At 2.8A, you should get around 6-700 lumens (very rough estimate).

Well my P7 arrived today.
I hooked it up direct to a 18650 and lens and I was suprised at its output.
Much better than my 10w HID and as bright as my35w 14.4v halogen.
Li Ion was 3.5v no load and 3.1v 2 with a 2amp load.

I then tried 8 amc 7135s and got 2800mA @ 3.7v from 4 NiMh.
Didnt notice a whole lot extra light. Anyway I am charging up the Li Ions and will test again. My feeling at this stage is that the LI Ions may be the way to go.

Anyone know if removing the diodes off the 7135s will gain me anything?
Do these diodes drop .6v?
Are they are only there for reverse polarity protection?

Yep, those diodes are for reverse polarity protection. You can find out how much voltage one drops with a basic voltmeter.

Removing the diodes would drop the low-voltage cutout threshold so that it would remain regulated for slightly longer. Personally, I don't think I'd remove the diodes because I would probably fry the 7135s with reverse polarity as soon as I did.

If running a 4.8v pack I wouldnt but if running a 3.7v Li Ion pack or3.6 NiMh removing the diode MAY give you more power to the LED.

More specifically, it'll give you more power to the LED when the battery is low. It's not going to go any higher than what it's regulated to, but it'll stay in regulation for a little longer if the diode is removed (I would suppose). On the flip side, the diode would take some of the stress off the AMC7135s, as those have to dump excess voltage as heat, and they've been known to get pretty darn hot when run at close to their maximum Vin.

TigerhawkT3 said:

More specifically, it'll give you more power to the LED when the battery is low. It's not going to go any higher than what it's regulated to, but it'll stay in regulation for a little longer if the diode is removed (I would suppose). On the flip side, the diode would take some of the stress off the AMC7135s, as those have to dump excess voltage as heat, and they've been known to get pretty darn hot when run at close to their maximum Vin.

Click to expand...

Their thermal protection kicks in at 150˚C.

TigerhawkT3 said:

More specifically, it'll give you more power to the LED when the battery is low. It's not going to go any higher than what it's regulated to, but it'll stay in regulation for a little longer if the diode is removed (I would suppose). On the flip side, the diode would take some of the stress off the AMC7135s, as those have to dump excess voltage as heat, and they've been known to get pretty darn hot when run at close to their maximum Vin.

Click to expand...

Yeah I was more reffering to driving off a single ( or paralleled ) Li Ion where you may only get a high voltage for a short time. Once the voltage drops the diode (and the AMC 7135's) just get in the way.
I wouldnt recommend removing the diode on higher voltage batteries.