help, Can I drive 2 Cree R2 with AMC1735 1400ma?

[ma_sha1]

Want to use 1 boost IC to drive 2 R2 in parallel, with 3V input
(2x E2 Enegizer AA li ion)

Will the AMC1735 1400 DX board work? The 1735 spec sheet says 2.75 v min input,

Anyone tried it?

Looks like 2 R2 in parallel will get 700MA each, if one led fail,
The remaining R2 should be fine with 1.4A.

thanks for the help

 

[spencer]

It probably won't work. Two e2 lithiums don't supply enough voltage. The AMC chip is a linear voltage regulator. That means it burns excess voltage off as heat and can't raise the voltage. You would be best off with probably 3Nimh or 1 li-ion.

 

[Benson]

Want to use 1 boost IC to drive 2 R2 in parallel, with 3V input
(2x E2 Enegizer AA li ion)

They're not Li-ion, just lithium (specifically, lithium/iron-sulfide). 2 Li-ion cells would be 7.4V.

Will the AMC1735 1400 DX board work? The 1735 spec sheet says 2.75 v min input,

Anyone tried it?

No, it won't work; the amc7135 is a linear regulator, not a boost convertor. You'd need 3 1.5V cells to make this work.

Looks like 2 R2 in parallel will get 700MA each, if one led fail,
The remaining R2 should be fine with 1.4A.

Unless you have reason to believe your LEDs are quite closely Vf-matched, I wouldn't do this (even if you were using a high enough voltage input) because any imbalance will give you, e.g., 800mA and 600mA instead, with a corresponding brightness difference. However, you can cut traces on a 1400 mA board to get two independent 700 mA channels, and use one for each LED.

(OTOH, this dual-channel arrangement will drop to half-brightness (1 LED at 700mA) if one LED fails -- if you actually want to overdrive the remaining LED to keep the brightness up, I suppose you could connect the output channels with back-to-back diodes...)

 

[cerberuss]

Yes vf matching might be a problem.. but I doubt there would be much visual difference in real life. If you arent too fussy with that, then you can use a variation on the circuit that StefanFS uses here.. It shouldn't be hard to tweak slightly for your purpose (supposing you use rechargeable li-ions). This seems like the cheapest most robust option.

http://www.candlepowerforums.com/vb/showthread.php?t=212835

 

[ma_sha1]

Thanks Guys,

The DX site says the 7135 IC can be used with 2AA or 2AAA, is this not true?
If this isn't boost, then there's no Boost IC that can deliver 1.4A when powered by 3V?

My current set up in the works is to use 2 IC to deliver 700MA to each R2,
but I am having hard time making 2 chips fit.

If I Run 2x R2 parallel out of 700Ma boost driver,
so each R3 get 350 mA, would this work? not worried about off balance.,
try to blend the bean into one spot anyways.
Can I expect 300LM? Still more light than 1 R2 at 700MA?

If one R2 fail, then all 700Ma goes to one R2, ~180 lumens left?

 

[Benson]

Thanks Guys,

The DX site says the 7135 IC can be used with 2AA or 2AAA, is this not true?

No, that's perfectly true; it will limit the current from 2AA or 2AAA to 350mA (per AMC7135, 1.4A for the whole board) by dropping the excess voltage. So if you had a red LED, where Vf is typically 1.8V IIRC, these would be fine.

If this isn't boost, then there's no Boost IC that can deliver 1.4A when powered by 3V?

Don't know of any right off, but I haven't really been looking. If you get boost drivers whose outputs may be connected in parallel, you can add them to get the required current, but most drivers can't.

My current set up in the works is to use 2 IC to deliver 700MA to each R2,
but I am having hard time making 2 chips fit.

If I Run 2x R2 parallel out of 700Ma boost driver,
so each R3 get 350 mA, would this work? not worried about off balance.,
try to blend the bean into one spot anyways.
Can I expect 300LM? Still more light than 1 R2 at 700MA?

Each LED would run at 350mA, which conveniently is the If value for binning. The datasheet doesn't show R2, but has the maximum flux for Q5 as 114 lm, so you could expect a minimum of 228 lm from matched R2s. Total luminous flux would fall off only slightly for small imbalances due to Vf mismatch, but in the extreme case (when one fails completely) would fall to about 85% of that, or 190 lm.

Could you connect them in series, and drive with one 700mA boost driver?

 

[ma_sha1]

Could you connect them in series, and drive with one 700mA boost driver?

Thanks a bunch for providing answers!

This is the boost driver I am using. 700MA. Works great driving 1 R2
from 1.5V Alk C.

Can this be used to manage 2 R2 in series?

 

[Benson]

Thanks a bunch for providing answers!

This is the boost driver I am using. 700MA. Works great driving 1 R2
from 1.5V Alk C.

Can this be used to manage 2 R2 in series?

Was there supposed to be a link in there, perchance?

 

[ma_sha1]

Sorry,

here is the link.

http://www.dealextreme.com/details.dx/sku.4735

Which way to get more lumens drive 2 R2 in series or parallel?

using 3V (2x1.5V) e2 lithium

 

[Benson]

Not good -- that driver won't do two LEDs in series.

The only thing I know to do (assuming you can't find another driver that would work) is to change your power source. If you're stuck with the 2xAA size, you could try direct-drive off 3x 2/3AA or using AMC7135s off 4x 1/2AA (or maybe 3x 2/3AA, but I don't think you'll get the current you're after.