help with amc driver

[doctaq]

greetings cpf community

i have ordered a bunch of these drivers
http://www.dealextreme.com/details.dx/sku.1886
they run at 1400ma with 4 amc chips(sorry dont know the term) and i want to remove one chip so that it will run at 1050 ma, is there any more to it than clipping the connections around the chip? i am okay with wasting one.
or do i have to take the back panel off somehow and desolder it?
if i run these off of a 5v line lump power supply should i expect them to get very hot if i run 2 cree xr-e (3.7 vf) in parallel at 1400ma, 700ma each?
this is for a custom led lamp that will light up a tiny reef tank.
i have built a led fixture with a buck driver before and i also understand the danger of thermal runaway
TIA

 

[Benson]

greetings cpf community

i have ordered a bunch of these drivers
http://www.dealextreme.com/details.dx/sku.1886
they run at 1400ma with 4 amc chips(sorry dont know the term) and i want to remove one chip so that it will run at 1050 ma, is there any more to it than clipping the connections around the chip? i am okay with wasting one.
or do i have to take the back panel off somehow and desolder it?

There's no back panel. I would desolder it, but if you don't have the equipment or whatever, cutting the b+ trace to that chip should be enough.

if i run these off of a 5v line lump power supply should i expect them to get very hot if i run 2 cree xr-e (3.7 vf) in parallel at 1400ma, 700ma each?

Well, they'll be dissipating (5V-3.7V)*1.4A=1.8W. How hot they get depends on how you heatsink them.

But paralleling LEDs with that driver is just silly. Cut the output trace where it goes under one of the diodes, and you have two 700mA drivers on one board.

 

[TorchBoy]

i also understand the danger of thermal runaway

You won't get thermal runaway with a constant current driver - at least, not if you don't run the LEDs in parallel, which wouldn't be a good idea. I suppose it would be possible to burn the driver out as the Vf lowers but it would have to be pretty borderline to start with.

As for this particular driver, 3.7 V at 700 mA is an unusually high Vf for an XR-E, so it wouldn't surprise me if you were dissipating well over 2 W from the four AMC7135s. That's a lot of heat. What's a "line lump power supply" and why this driver?

Benson's suggestions are very good.

Benson, you've got an orphaned quote clipping there.

 

[TorchBoy]

Oh yeah - this thread should be in the Electronics forum, because it's about the drivers, not the LEDs.

 

[doctaq]

sorry i didnt know about the forum breaking down like that, i just saw leds and ran here,

will it be easy to tell where the lead from the first two drivers come out? i have a multimeter but will the current read if the driver is just powered up and not hooked up to an led?

is dissapating the power im talking about going to make it very hot? i would be willing to run a resistor if that would help, i was thinking about using one to deal with thermal runaway because i am running in parallel but i figure if i can draw current from two amcs on each chip then i wont have to bother with that, if anyone has any pictures of something like that that would be awesome, i cant quite understand a diagram and dont really understand resistors. if somehow i can reduce the current to the drivers to make them run cooler i would like to, otherwise i will put a tiny heatsink on them.

 

[DM51]

sorry i didnt know about the forum breaking down like that, i just saw leds and ran here

Welcome to CPF, doctaq.

Don't worry - you should soon get used to where different topics belong. I'll move your thread to the Electronics section for you.

 

[doctaq]

thanks mod

 

[TorchBoy]

How remiss of us... :welcome: doctaq.

Figuring out the tracks on the AMC7135 driver board should be possible with a bit of careful study of the board. All the same pins on the four individual driver chips are connected to each other, so you just have to figure out how they've done it, then work out the best way to cut a track to separate the chips into two pairs.

Alternatively, if you'd prefer to use resistors, you could place a small value resistor in series with each LED (for example, between the LED and battery positive), and then safely drive a couple of LEDs in parallel from the full 1400 mA. The resistors would mean slightly less voltage dropped across the AMC7135s.

Because the AMC7135 driver is in series with the LED, if a LED isn't connected the driver won't be drawing more than about 200 microamps, which is what it draws to operate. So that's all you'd read with a multimeter, depending on how and where you connected the meter.

You didn't answer my question about your power supply or why you picked this driver.

 

[doctaq]

sorry torch
a line lump is like those laptop power supplies that have a box in the middle of the cord, its 5v 5a, kinda hard to find. i chose this driver because it was cheap, the other options commonly available are buckpucks 20 bucks and meanwells 40 bucks, these guys are like 2 bucks, good for small arrays, as i am only trying to build a fixture with 4 or 5 leds. i thought that this PS was the best match with this driver, if i end up seperating the track and mount the led on each one will they each still draw the 3.3 or 3.5v they need?
i was thinking that a 2 ohm 5w resistor would drop the voltage enough to the driver that it wouldnt have to overcome such a large drop,
1 ohm would drop the voltage by .6v? thus getting closer to the optimal voltage?
i think this driver drops the current .5v and say 3.5 vf for the led so 4v +resistor value before that so only 4.4v is going in to the driver? does that make any sense?

 

[TorchBoy]

a line lump is like those laptop power supplies that have a box in the middle of the cord, its 5v 5a, kinda hard to find.

Ah, I'm with you now. Is it switchmode (thus having a constant 5 V output) or is it really heavy, with a transformer inside (non-constant voltage output)?

1 ohm would drop the voltage by .6v? thus getting closer to the optimal voltage?
i think this driver drops the current .5v and say 3.5 vf for the led so 4v +resistor value before that so only 4.4v is going in to the driver? does that make any sense?

Sadly no. The driver is a linear regulator so it draws the same current it feeds to the LED, which is 1.4 A. Ohm's Law says a 1 ohm resistor will drop 1.4 V at that current. A driver wouldn't drop the current by 0.5 volts. The AMC7135 has a 0.12 V overhead (the amount the input voltage needs to be above the Vf of the LED).

 

[doctaq]

are there any markings on this that will easily tell me where everything is? and after i cut the track do i make the connection on a metal pin that comes out of one of the amc

 

[doctaq]

Ah, I'm with you now. Is it switchmode (thus having a constant 5 V output) or is it really heavy, with a transformer inside (non-constant voltage output)?


Sadly no. The driver is a linear regulator so it draws the same current it feeds to the LED, which is 1.4 A. Ohm's Law says a 1 ohm resistor will drop 1.4 V at that current. A driver wouldn't drop the current by 0.5 volts. The AMC7135 has a 0.12 V overhead (the amount the input voltage needs to be above the Vf of the LED).

ah, i meant volts, not current, so would it help if i added the 1 ohm resistor? dropping the drivers Vin by the 1.4V so 3.6 which would still be enough to overcome the overhead? i thought i read somewhere that the overhead was higher thats why. a .5 ohm resistor could drop it only .7v and definateley allow me to power the leds .

do all the amc run off of the same power source on the back if i cut the track?
i dont quite know about the power supply being switching or transformer, it looks and feels very similar to a laptop power supply? what kind are those?

 

[TorchBoy]

You know, if you're running the LEDs from a 5 V power supply you could just use a resistor. Sites like http://ledcalc.com/ will help you work out what value resistor you need for each LED.

 

[doctaq]

You know, if you're running the LEDs from a 5 V power supply you could just use a resistor. Sites like http://ledcalc.com/ will help you work out what value resistor you need for each LED.

i already have the drivers on the way but does adding the resistor make any sense?

 

[TorchBoy]

It'll help keep the AMC7135s a little cooler by burning off some of the excess voltage itself instead of the AMC7135s doing it, and they'll help balance current across LEDs if you run LED pairs in parallel from a single driver instead of cutting the track on the board. You'll generate the same total amount of heat as if you just used resistor; it'll be about 70% efficient either way. Since you've already got the drivers on the way, you may be able to buy/use slightly less expensive resistors if they'll just be used to drop a little voltage rather than all the work at ~1.5 V each.

If you're really worried about heat, you could use inexpensive buck drivers (prices start from $6.97 for 4) and get ~90% efficiency, so roughly a third of the heat as using linear regulators and/or resistors.

 

[doctaq]

i like the idea of cutting the track but dont quite know how its going to be done, maybe ill have a better idea when they get here, which feels like its taking forever, the anticipation is killing me.
i guess ill post the question again when i get it, with pictures or something haha, thanks torch i appreciate it, truth be told, im not even that worried about a single led overdrawing that much since ive read anecdotal evidence that a single xr-e running at 1400ma doesnt immediateley explode.

 

[doctaq]

alright, so i think i would like to run a resistor before the driver to drop some of the voltage, the driver seems to me like it has to overcome about 1.5v,
and if i get a .5 ohm resistor it would drop it .7v at 1400ma? is this true and is it a good idea? how many watts does the resistor have to be?
i found a .75ohm resistor that can handle 5w for 3 for a dollar, which would drop the voltage about 1v? 4v going into the driver should be perfect right?

theoretically what would happen if the driver did not get enough voltage to overcome its overhead and the vf of the led at the given current?

 

[TorchBoy]

P = I V.
Power = current x voltage.
Watts = amps x volts.

0.7 V x 1.4 A = 0.98 W

So a 1 W resistor would probably do, but that's getting very close to its rated power handling.

 

[doctaq]

thanks again torch

would 4v going into this do what i need it to? so should i spring for the .75 ohm?
how about the driver and led not getting enough voltage question, just hypothetical

 

[TorchBoy]

Have you got the LEDs on hand? Can you feed them 700 mA (one by one) and measure their forward voltage?