Resistor vs buck puck Battery Brain Question

[Sifu John]

Hi everyone,
I had an interesting thought this morning.
If I understand correctly, the Buck puck is the reason that the LED's continually draw a small amount of power, yes? So, If a set-up was made using the resistor instead of the buck puck when off there would not be a continual draw?
Does anyone know the approximate comparison to how long the Li-Ion batteries will last in a saber with sound using the resistors vs using the buck puck (due to the loss of efficiency in the resistor)?
Thanks,
Sifu John

[Greywolf]

Well, there is a small misconception about what a buck puck does...

Basically it is a small circuit which modifies its output voltage to get its target amperage (700mA, 1000mA, etc). In doing so it is more efficient than a regular resistor which just turns 'excessive' voltage into heat^TM.
Also a buck puck can handle varying input voltages and provide a constant output while (to put it simply) resistors drop input voltage by a 'fixed' level.

Now here's the catch:
Full batteries start with a higher voltage and drop voltage while draining; that means that
- a resistor (with its 'fixed' setting) slightly overdrives the LED at the beginning, hits its 'sweet spot' afterwards and then underdrives the LED when the battery dies.
- a buck puck is constantly busy working but provides its target amperage (= constant LED brightness) all the way till the (battery's) end.

That said I would guess that a resistor solution might provide more run time, but only with severely reduced brightness during the last third
-> Does that answer your question or raise even more ?

[Racona Nova]

In a sound saber it's not only the LED that has a constant drain - the sound board has it, too.

If I'm correct, due to excessive heat development in the Resistor setup it would be logical that you have a reduced runtime of your Li-Ions because a certain amount of the voltage and mAh provided by the batteries aren't used for the LED but transformed into heat. So a BuckPuck setup, which has a decreased heat development and thus increased efficiency, would last longer than a Resistor setup because the energy that would otherwise become heat can be used for the LED.

[Sifu John]

Thanks Greywolf !
I appreciate the help 

Well, there is a small misconception about what a buck puck does...

Basically it is a small circuit which modifies its output voltage to get its target amperage (700mA, 1000mA, etc). In doing so it is more efficient than a regular resistor which just turns 'excessive' voltage into heat^TM.
Also a buck puck can handle varying input voltages and provide a constant output while (to put it simply) resistors drop input voltage by a 'fixed' level.

Now here's the catch:
Full batteries start with a higher voltage and drop voltage while draining; that means that
- a resistor (with its 'fixed' setting) slightly overdrives the LED at the beginning, hits its 'sweet spot' afterwards and then underdrives the LED when the battery dies.
- a buck puck is constantly busy working but provides its target amperage (= constant LED brightness) all the way till the (battery's) end.

That said I would guess that a resistor solution might provide more run time, but only with severely reduced brightness during the last third
-> Does that answer your question or raise even more ?

[Sifu John]

Thanks Racona Nova,
I didnt realize the sound board had a constant drain as well.
 

In a sound saber it's not only the LED that has a constant drain - the sound board has it, too.

If I'm correct, due to excessive heat development in the Resistor setup it would be logical that you have a reduced runtime of your Li-Ions because a certain amount of the voltage and mAh provided by the batteries aren't used for the LED but transformed into heat. So a BuckPuck setup, which has a decreased heat development and thus increased efficiency, would last longer than a Resistor setup because the energy that would otherwise become heat can be used for the LED.