LED Driver inside canister, wires to lighthead?

aleks

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Dec 10, 2007
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Just an idea i got to save space within the lighthead, is to put the driver inside the canister and then run the led output wires through the wire to the lighthead? doable? good/bad idea?
 

josb

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Aug 25, 2008
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Hi Aleks,
I don’t know if it’s just a matter of saving space but I think I would always prefer to have the control of light intensity (dimming / brightening) on the front end e.g. on my light head instead of having to reach back to the canister. For that reason I would say it’s a bad idea.

Technically I don’t think it’s impossible to extend the wires from the driver to the LED unit, I’m curious what the electronic technicians here on CPF have to tell.
 
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aleks

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Dec 10, 2007
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Well, since my switch is located on the canister (due to not risk swiching off the light during wreck penetrations) it´s not an issue for me...the question is more if it is doable in regards ro resistance in the wires, max cable lenght and so on.
 

SmokedCPU

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Jan 9, 2010
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Hi

I did this in a video lamp setup, i used a H6CC from georges to power two SSt90 from a 4cell 10A battery pack, the two leds were in serie so driver took care of wire loss.

there were no dimming, only on-off by magnetic toggle, inside of diver glove.

make yourself an aluminium plate to dissipate heat on a bigger area

i did not mesure effiency tought


good luck
 

350xfire

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Aug 14, 2008
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Yeah, I think oyu'll do OK with single mode. It's multi-mode I would be afraid of.
 

wquiles

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At a very high level, I would always prefer to put the driver as close to the LED, with the shortest wires as possible, rather than having the LED far away in the canister. Why? Power/efficiency due to the current involved.

Of course, I am assuming that you have a true, constant current LED driver, not a voltage regulator. If you have an LED driver:

Power to the LED = Power to the LED driver - efficiency losses at the driver (a good driver should be able to be 85% plus over a "wide" range of input voltage and loads).

But the equation above assumes short wires everywhere. So to minimize power losses and voltage drops, you should always keep the wires the shortest where the current is highest.

In the case of a canister providing a voltage to the LED driver, the voltage reaching the driver is almost always higher than the voltage needed at the LED (again, for the more common case of a buck regulator - if you have a boost regulator, it would be the opposite). So given that (except for the power loss at the regulator) the power is conserved, power into the regulator is roughly the same as power out to the LED, if the voltage at the input of the regulator is higher than at the output, the current at the input of the regulator "has to be" lower than the current at the output of the LED regulator going to the LED. Since power loss in wires is:

Ploss = I^2 * R (where I is the current in Amps and R is the wire's resistance in Ohms - and remember the resistance is times 2 since you have two individual wires)

For any given current, the way to minimize power loss in the wires is to keep the wires as short as possible. Since for a buck regulator the current is higher on the LED side, the most efficient solution is therefore to have the LED driver be as close to the LED as possible. This power lost/dissipated in the wires might not be a huge amount, but everything else being the same, this is why I would always put the LED driver next to the LED in the head, and not in the canister.
 
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350xfire

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Aug 14, 2008
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At a very high level, I would always prefer to put the driver as close to the LED, with the shortest wires as possible, rather than having the LED far away in the canister. Why? Power/efficiency due to the current involved.

Of course, I am assuming that you have a true, constant current LED driver, not a voltage regulator. If you have an LED driver:

Power to the LED = Power to the LED driver - efficiency losses at the driver (a good driver should be able to be 85% plus over a "wide" range of input voltage and loads).

But the equation above assumes short wires everywhere. So to minimize power losses and voltage drops, you should always keep the wires the shortest where the current is highest.

In the case of a canister providing a voltage to the LED driver, the voltage reaching the driver is almost always higher than the voltage needed at the LED (again, for the more common case of a buck regulator - if you have a boost regulator, it would be the opposite). So given that (except for the power loss at the regulator) the power is conserved, power into the regulator is roughly the same as power out to the LED, if the voltage at the input of the regulator is higher than at the output, the current at the input of the regulator "has to be" lower than the current at the output of the LED regulator going to the LED. Since power loss in wires is:

Ploss = I^2 * R (where I is the current in Amps and R is the wire's resistance in Ohms - and remember the resistance is times 2 since you have two individual wires)

For any given current, the way to minimize power loss in the wires is to keep the wires as short as possible. Since for a buck regulator the current is higher on the LED side, the most efficient solution is therefore to have the LED driver be as close to the LED as possible. This power lost/dissipated in the wires might not be a huge amount, but everything else being the same, this is why I would always put the LED driver next to the LED in the head, and not in the canister.

What he said... But one must also consider the design of the light. Sometimes, you just have to do what you have to do! lol
 

aleks

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Dec 10, 2007
Messages
13
At a very high level, I would always prefer to put the driver as close to the LED, with the shortest wires as possible, rather than having the LED far away in the canister. Why? Power/efficiency due to the current involved.

Of course, I am assuming that you have a true, constant current LED driver, not a voltage regulator. If you have an LED driver:

Power to the LED = Power to the LED driver - efficiency losses at the driver (a good driver should be able to be 85% plus over a "wide" range of input voltage and loads).

But the equation above assumes short wires everywhere. So to minimize power losses and voltage drops, you should always keep the wires the shortest where the current is highest.

In the case of a canister providing a voltage to the LED driver, the voltage reaching the driver is almost always higher than the voltage needed at the LED (again, for the more common case of a buck regulator - if you have a boost regulator, it would be the opposite). So given that (except for the power loss at the regulator) the power is conserved, power into the regulator is roughly the same as power out to the LED, if the voltage at the input of the regulator is higher than at the output, the current at the input of the regulator "has to be" lower than the current at the output of the LED regulator going to the LED. Since power loss in wires is:

Ploss = I^2 * R (where I is the current in Amps and R is the wire's resistance in Ohms - and remember the resistance is times 2 since you have two individual wires)

For any given current, the way to minimize power loss in the wires is to keep the wires as short as possible. Since for a buck regulator the current is higher on the LED side, the most efficient solution is therefore to have the LED driver be as close to the LED as possible. This power lost/dissipated in the wires might not be a huge amount, but everything else being the same, this is why I would always put the LED driver next to the LED in the head, and not in the canister.


that´s one hell of an explanation! :D thanks!
 
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