Computer Switch 100/50/25/strobe/off question

In using the lower output settings does it just dim the light sort of like using a resistor where it won't necessarily use less power from the battery just that the extra current is shunted? For example running a Pyrenees at 25% will it achieve 4 times the runtime of the Pyrenees run at 100%? If not what is the estimation of runtime advantages at lower levels (if any)

I don't know how Peak does it but generally with good driver you will extend runtime significantly. Usually more that current drop. Good driver will have efficiency pretty much the same across load currents (except extremely low and high), so driving LED at 4 times current actually should extend runtime more than 4 times (assuming significant current difference e.g., 1A vs 0.25A, not 40mA vs 10mA). This is because battery can deliver less power with higher current draw.

Even with resistor you will save power. Of course you have significant losses on resistor but your total power will be less.

Assume 3.5V Vf LED and 4V battery. You need 0.5 Ohm resistor to get 1A current, losses on resistor are 0.5W and power on LED is 3.5W, total 4W. Now you want 0.25A, assume Vf will drop to 3V, you need 4 Ohm resistor, losses on resistor are 1W, power on led is 0.95W, total 1.75W, your runtime will be more than doubled. Far cry from switching regulator but you still get more run time.

So, regardless of method used lower output will result in longer runtime. How much will depend on driving method.

That was a great technical explanation, but I am looking for a more practical answer as far as actual runtime.

arek98 said:

I don't know how Peak does it but generally with good driver you will extend runtime significantly. Usually more that current drop. Good driver will have efficiency pretty much the same across load currents (except extremely low and high), so driving LED at 4 times current actually should extend runtime more than 4 times (assuming significant current difference e.g., 1A vs 0.25A, not 40mA vs 10mA). This is because battery can deliver less power with higher current draw.

Even with resistor you will save power. Of course you have significant losses on resistor but your total power will be less.

Assume 3.5V Vf LED and 4V battery. You need 0.5 Ohm resistor to get 1A current, losses on resistor are 0.5W and power on LED is 3.5W, total 4W. Now you want 0.25A, assume Vf will drop to 3V, you need 4 Ohm resistor, losses on resistor are 1W, power on led is 0.95W, total 1.75W, your runtime will be more than doubled. Far cry from switching regulator but you still get more run time.

So, regardless of method used lower output will result in longer runtime. How much will depend on driving method.

Click to expand...