Explain this tail reading please

Flashlight Modding and DIY Parts
1

I built a C8 using the small 3 - 12V ” buck” driver that I tested and posted in This thread .
It’s working, but I don’t understand the tail reading I’m getting with 2 X CR123 cells.

With 1 X 18650, I’m getting a tail reading of 1.9A
With 2 X 18650, I’m getting a tail reading of 1.0A, which is correct because with two cells the tail reading needs to be multiplied by two, which is 2A for the emitter…fine.

But with 2 X CR123 (3V ea.), I’m getting a tail reading of 2.2A. I don’t understand this. With two cells, does that mean my little XP-E2 is getting 4.4A? How do I read this? Do two 3V cells somehow change the math? Why do I get 1A with two 18650 and 2.2A with two CR123.

Pleas enlighten me.

2

What brand CR123s? Looking at HKJs graph they sag a lot more then 18650s at this current. Is it brighter then the single 18650? If its an 8w driver, that would mean the CR123s are dropping under 2v each under load.

3

I just did a test with the same driver I already had hooked up on my desk.
2xCR123 at the emitter - 2.2A
2xCR123 at the tail. - 2.2A

2x18650 at the emitter - 2.2A
2x18650 at the tail. - 1.1A

So why do the 2xCR123 give me the same current reading at the tail and emitter. Why doesn’t the same double cell math apply to 2x3V cells as compared to 2x4V cells? I’m sure there is a logic to it, but I’m not smart enough to grasp it.

4

Voltage across the emitter stays at a constant 3V regardless of cell type 1-2 X 18650 or 2 X CR123.

5

Its because the two cr123 sag to the point that they have roughly the same voltage as one 18650. There’s not enough overhead for the buck circuit to do much.

6

It does seem like the CR123a are sagging under load,
what are the battery voltages when loaded?

Since the driver is driving the led at constant current (leading to constant power) and from the relation
Power=Voltage*Current:

Emitter_power=3.3V*2.2A=7.3W

—>7.3W=Battery_Voltage*Battery_Current

—> Battery_Current= 7.3/Battery_Voltage

If the voltage drops, the current should rise

7

6V configuration appears to be far less efficient :~

8

I didn’t notice any sag. The readings are constant. The CR123 are brand new and each tested at 3.2V

9

I guess with 2 CR123 at this load voltage at cells will be about 5 volts. With this voltage the driver probably acts as a linear driver, so amps at input and output are equal.

10

Lets say u have an 8 watt driver. At 4v it will pull 2amps to produce 8w, at 8v it will pull 1 amp. That’s why when we have 2 batteries we always double the current to figure out emitter amps. If u had 6v it will pull 1.33 amps to maintain the 8w so you would have to times your tail current by 1.33 to work out emitter amps.

But you didn’t account for the voltage sag, being smaller batteries the volts are dropping more under load then a single 18650 so its pulling more to maintain the wattage. Kind of confusing but I hope this helps.

watts = volts x amps

11

I knew it had to be something like that, but this was the first time I’ve played with CR123s. I guess the main saving grace is that the driver is keeping my voltage constant across the emitter.