Help me understand runtime.

I have a quark turbo that can run on 2xcr123 or 18650. The voltage is 3.0-6.0v. If it can run 3 hours on high on 2xcr123, will the runtime be better on an 18650 (3400mah)?

Thanks!

18650 will give you about twice the runtime.

Although the voltage is lower on the 18650?

The 123 batteries used in series gives the higher voltage that the light can handle (always check that!). But the light is running at 3.7V nominal voltage despite which batteries you use (in THIS light, up to 6V). And… the 18650 has vastly more mAh (capacity) than the two 123’s combined capacity- and so it runs longer.

It’s actually not a simple question. Many variables come into play. Also, mAh are not the only story. Energy (Wh) in this case should be the metric compared.

The Quark turbo uses a buck driver, this is a circuit that exchanges volts for amps, in simple terms. There are efficiency losses that aren’t necessarily constant between the two battery configurations, but for simplicity we’ll assume it is equal. The reason you can’t only look at mAh here is that the CR123s are going to have ~30% less current demand than the 18650 due to them having about 30% higher voltage.

Cells in series add voltage, capacity is unchanged. Cells in parallel add capacity, voltage is unchanged. Energy though combines these values and it all makes sense.

Energy = Capacity * Nominal Voltage >>>> Wh = Ah * V

At a 3hr rate, a single CR123 provide roughly 1.5Ah at ~2.4V or ~3.6Wh each. On the other hand, a high capacity 18650 such as a Sanyo/Panasonic NCR18650GA, will deliver ~3.2Ah at ~3.6V or ~11.5Wh. >>> 11.5Wh/7.2Wh = 160%

From this we can see that even assuming equivalent efficiency the 18650 will deliver about 2/3rds more runtime. So in high it should run about 5hr on the 18650.

If you want to read more about this exact topic, check out the awesome writeup HKJ provided years ago CR123A and rechargeable substitutes.

This will also depend on what you consider runtime to actually be.

I don’t know anything about this light or its driver. However I have a few lights that support similar configurations.

And some of these lights will have better regulation on the higher voltage and possibly offer higher output too when using the 6v power source. In such instances total runtime until the batteries are flat is less on 2xCR123a. But the higher output and flat regulation mean the High output runtime may be considerably longer compared to the 18650.

Some of my other lights however that use both of these power sources show no difference in output or regulation on either battery, in these cases the 18650’s offer much better runtime at all outputs.

Thanks for the info guys.

This is a real good explanation.

For comparison, here is a shot of the package on a Streamlight Protac HL 5-X

On this light, 4xCR123s give less runtime and lower brightness (on high) than a pair of 18650s.
This light is very well regulated and is a current controlled light. So on medium and low, the output is the same, but the run time is less with the CR123s.
All the Best,
Jeff

maybe, it will be better, but it also may not. simple numbers are , 2 cr123 are 4,5wh each, 1 8650 at 3500mah is about 12wh. but even those numbers wont answer the question, all depends on a current draw, and how well your cells handle it, cells capacity is very dependent on the load, so is runtime

i would guess 1 18650 will run about 8 times longer than 2 cr123

it is worth it, seriously

wle

8 times??? only if 9wh is 8 times less than 12 wh

oh
i;m wrong

50% more energy

now look at cost per hour :slight_smile:

cr123a —— 800 cents / 10 watt hours = *about 80c/watt-hour
*
18650 is about :
cell cost $5.00, reuse 500 times, 12 watt hours = 6000 watt hours
——————–500 cents / 6000 watt hours ==* about .08 cents*

so roughly 960 times cheaper !

Another question.

Which has better runtime at 30 lumen?

A quality 1000 mah 14500 4.2v or Eneloop 2000 mah?

A quality 1000 mah 14500 should run much longer. You have the voltage in the right spot and the current to run it. On the Eneloop 2000 mah your voltage is going to have to be raised up to run most LED’s. It’s going to have to raised up to typical forward voltage of between 1.8-3.3 volts. Even if only doubled in a perfect circuit your Eneloop 1.2 volts 2000 mah becomes 2.4 volts at 1000 mah. Perfect circuits don’t exist you’ll lose a bit of that 1000 mah in the boost. Even more to get just usable wattage as the boost driver raises the voltage higher. In a setting listing 30 lumen output you may find that a lithium-ion cell gives you 45 lumens and the Eneloop gives you 20 lumens. Then the Eneloop might run longer.