newbie driver questions

[LittleBill]

im getting confused this driver
http://www.dealextreme.com/details.dx/sku.15880

can do .9 to 4.5v in put and has selectable current up to 1a

my question what the heck is the output voltage?

if using a 2xaa thats 2.4.

is the output then 2.4v@1a?

how exactly is this regulated i mean batterys have voltage drop, you would see this in a light with diminising light output.

can some give a quick rundown on this works?

as i would like to mod a 2d mag, but this driver crap has me confused

 

[sparkysko]

I wouldn't worry about the voltage. I'd expect any driver to provide at least 3.7 volts. Any excess voltage produced by the driver is just burned off as heat anyways, so it'll either light up or it won't (AFAIK).

 

[LittleBill]

I wouldn't worry about the voltage. I'd expect any driver to provide at least 3.7 volts. Any excess voltage produced by the driver is just burned off as heat anyways, so it'll either light up or it won't (AFAIK).

isn't that a issue though, i mean what if i want a driver for a 3w luxeon, 3.7 would burn it up


are pretty much all the current led's run at 3.7?

 

[sparkysko]

LED's are current operated devices, not voltage. Just like transistors, the juice won't start flowing through them until a minimum voltage is met. You get a voltage drop out the other end.

You could hook an LED to a 12 volt power supply with a resistor, you size the resistor for the current you want at that voltage and ignore the voltage.

Voltage doesn't burn up LEDs (Maybe VERY high voltage). Any excess voltage is just burned up as heat in the current limiting device. (A resistor would get VERY hot if you tried to power an LED off of 12 volt)

Ignore the voltage unless you're trying to size up a resistor, or if the light won't turn on.

LED's run from 3 - 4 volts more or less. The driver manufacturers know this. The brightness of the LED is determined by the amperage that goes through it.

 

[LittleBill]

LED's are current operated devices, not voltage. Just like transistors, the juice won't start flowing through them until a minimum voltage is met. You get a voltage drop out the other end.

You could hook an LED to a 12 volt power supply with a resistor, you size the resistor for the current you want at that voltage and ignore the voltage.

Voltage doesn't burn up LEDs (Maybe VERY high voltage). Any excess voltage is just burned up as heat in the current limiting device. (A resistor would get VERY hot if you tried to power an LED off of 12 volt)

Ignore the voltage unless you're trying to size up a resistor, or if the light won't turn on.

LED's run from 3 - 4 volts more or less. The driver manufacturers know this. The brightness of the LED is determined by the amperage that goes through it.

if voltage makes no difference then why does putting a 3.7 lipo aa in a fenix make a huge difference over a 1.2 nimh, if the current is regulated?

 

[Linger]

Hey,

A few things.

LED's have a number called 'voltage forward,' which is the voltage the emiter opperates on, for a given range of conditions. Most people ignore the conditions and just stick with a crude 'led's have vf of 3.7v'
The point of that is, an emiter has an optimal voltage it will be most efficient at.

Yes, a driver has an overhead, a small voltage drain (perhaps .2v) We can ignore this for most of this post, however.
That board is probably a buck / boost driver. It has an imput voltage range of .9v-4.5v. The driver is able to sense what the led's vf is, and works to keep v*out near the emiter's vf.
For v*batt (voltage from the battery) that is above the led's vf, the driver will 'buck' it down, so you can attach a lithium-ion cell at 4.2v and it won't fry the led.
The driver keeps bucking down voltage as the cell discharges. Eventually the v*batt (minus driver overhead) is less than the emiter's vf. Near this point the driver begins boosting voltage. For the rest of the discharge cycle, the driver will boost voltage up neared the led's vf. (This is an issue with rechargables as you may not notice dimming as cell nears end of its duty cycle, an excellent reason to know your lights and how long you have on it, and also to use protected cells so that the are not inadvertently over-discharged).

 

[Linger]

if voltage makes no difference then why does putting a 3.7 lipo aa in a fenix make a huge difference over a 1.2 nimh, if the current is regulated?

Voltage does make a difference. light is most efficient when voltage in = led's vf + driver overhead. Anything above or below that will have to be converted by the driver.

You might want to read up on an emiter's spec sheet to get a sense of an led's opperating parameters, as there are a few variables at play.

re: current is regulated - yes, it tries, but how efficient is it at that regulation? If a li cell is putting out 3.7v and the vf is 3.5, well that's pretty close. If a nihm is putting out 1.2v and the vf is 3.5, well, that driver is going to have to do a lot more work to get the voltage to the led up closer to 3.5v. Even if the battery is putting out 1000ma, if it's at 1.2v (and declining) the driver will struggle to boost that up. And you can see that the final current going to the emiter is no where close to the 3.7v li at 1000ma.

Read here:
http://www.dealextreme.com/forums/Default.dx/sku.15880~threadid.204845

efficiency / output ratings with various cell configurations.
Also note, those numbers change as the light is used and voltage drops

Here's an example of efficiency over the discharge curve. From the following item as reported by Hilarion :
http://www.dealextreme.com/details.dx/sku.10084

Uin Iin Uout Iout Pin Pout Eff
3,2 0,42 3,25 0,36 1,34 1,17 87,05
3,02 0,45 3,25 0,36 1,36 1,17 86,09
2,82 0,49 3,24 0,36 1,38 1,17 84,41
2,6 0,54 3,25 0,36 1,4 1,17 83,33
2,41 0,59 3,25 0,36 1,42 1,17 82,28
2,2 0,65 3,25 0,36 1,43 1,17 81,82
2 0,74 3,25 0,36 1,48 1,17 79,05
1,8 0,84 3,24 0,36 1,51 1,17 77,14
1,6 1 3,25 0,36 1,6 1,17 73,13
1,41 1,23 3,25 0,36 1,73 1,17 67,46
1,2 1,86 3,25 0,36 2,23 1,17 52,42
1,1 1,84 3,21 0,32 2,02 1,03 50,75
1 1,83 3,17 0,27 1,83 0,86 46,77

 

[LittleBill]

really appreciate it guys making a lot more sense

i guess the driver can push only so much depending on the voltage. i guess i didn't take in the driver efficency,

but whats a fenix driver efficiency rating? i mean to see double light difference really speaks volumes to how crappy the driver efficiency is if it loses that much power

 

[sparkysko]

Boost convertors charge up a coil, and when a transistor senses that the coil is charged up, it flips the switch, and dumps all the power out of the coil. The magnetic field around the coil makes the electrons speed up and it outputs a higher voltage than when it started. Unfortunately it has to constantly 'pulse' this. Then this power goes into a 7135 or similar chip which is set to a maximum current that it'll let pass. This is your current regulator.

If you try multiple types of NIMhs of different quality, some have lower internal resistance than others. If you had a super Nimh that had really low internal resistance, I bet the brightness would be the same as with a lithium cell on the same light. I've noticed a 50% difference in power sent to the emitter with different nimhs. The limiting factor really seems to be just how fast the battery is capable of shoving electrons into that coil.

As far as the LED's voltage, you don't have to worry about that, just pick what current you want.

If you want an idea on efficiency, then that's a different story.

 

[LittleBill]

Voltage does make a difference. light is most efficient when voltage in = led's vf + driver overhead. Anything above or below that will have to be converted by the driver.

You might want to read up on an emiter's spec sheet to get a sense of an led's opperating parameters, as there are a few variables at play.

re: current is regulated - yes, it tries, but how efficient is it at that regulation? If a li cell is putting out 3.7v and the vf is 3.5, well that's pretty close. If a nihm is putting out 1.2v and the vf is 3.5, well, that driver is going to have to do a lot more work to get the voltage to the led up closer to 3.5v. Even if the battery is putting out 1000ma, if it's at 1.2v (and declining) the driver will struggle to boost that up. And you can see that the final current going to the emiter is no where close to the 3.7v li at 1000ma.

Read here:
http://www.dealextreme.com/forums/Default.dx/sku.15880~threadid.204845

efficiency / output ratings with various cell configurations.
Also note, those numbers change as the light is used and voltage drops

Here's an example of efficiency over the discharge curve. From the following item as reported by Hilarion :
http://www.dealextreme.com/details.dx/sku.10084

Uin Iin Uout Iout Pin Pout Eff
3,2 0,42 3,25 0,36 1,34 1,17 87,05
3,02 0,45 3,25 0,36 1,36 1,17 86,09
2,82 0,49 3,24 0,36 1,38 1,17 84,41
2,6 0,54 3,25 0,36 1,4 1,17 83,33
2,41 0,59 3,25 0,36 1,42 1,17 82,28
2,2 0,65 3,25 0,36 1,43 1,17 81,82
2 0,74 3,25 0,36 1,48 1,17 79,05
1,8 0,84 3,24 0,36 1,51 1,17 77,14
1,6 1 3,25 0,36 1,6 1,17 73,13
1,41 1,23 3,25 0,36 1,73 1,17 67,46
1,2 1,86 3,25 0,36 2,23 1,17 52,42
1,1 1,84 3,21 0,32 2,02 1,03 50,75
1 1,83 3,17 0,27 1,83 0,86 46,77

i guess one last thing.


so is it better to stick around the 3.7v and just parallel battery's or is it better to go higher and let it buck down? where is it more efficient?

i assume near 3.7 since the driver doesn't really have to get involved except for current limiting?

 

[sparkysko]

Well, you don't know if your VF is 3.0, 4.0 or what, it's all random since they don't bin for VF.

The driver is always involved. I imagine at the minimum there are loses involved with the switching transistor and whatever losses from the current limiting.

The current limiter is just a fancy variable resistor, so I think the most efficient would just be direct drive with a 1 or 2 ohm resistor.