Noob question: Cree P7 at 6~16V, 16V max

Hi guys. Noob here.:laughing:

I bought a diving flashlight with the following specs:

Emitter Brand/Type: Cree
Emitter BIN: SSC-P7-C-SXO
Color BIN: White
Total Emitters: 1
Battery Configurations: 2 x 18650 / 4 x CR123A / 4 x 16340 rechargeable Ii-on batteries
Voltage Input: 6~16V, 16V max
Switch Type: Twist/Twisty
Switch Location: Mid twist
Modes: 3
Mode Memory: -
Mode Arrangement: Hi > Mid > Low
Circuitry: Digital Regulated 1500mA Current Output
Brightness: 900 lumens maximum brightness

So after reading this article, https://www.candlepowerforums.com/threads/247543 , is it correct to say that using 4 x CR123A batteries (@ an estimated 14,8V) instead of 2 x 18650 (@ 7,4V) will not make the flashlight burn any brighter because of the 'driver' regulating 1500mA current output at all times?

The SSC-P7 works from 3.25V to 3.75V depending on Vf bin (I or J not in flashlight spec). Consistently feeding it >4V will make magic smoke. So the voltage of the batteries does not matter as all configurations of the torch you are looking at are above the 4V to 4.5V required by the BUCK driver. The driver will take 6 to 16 volts and convert it to 3.6V 1500 mA.
The driver will do this:
2*18650 - 7.2V 750 mA >> 3.6V 1500 mA.
4*18650 - 12V 450 mA >> 3.6V 1500 mA.
4*16340 - 14.4V 375 mA >> 3.6V 1500 mA.
It just draws different amounts of current from the different battery setups. Feeds the same power to the LED.
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BTW the SSC-P7 C bin is rated 700-800 lumens at 2800 mA.
At 1500 mA you will only get ~300-400 lumens out the front.

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In some BOOST drivers a 3AA light 3.6V is brighter than 2AA 2.4V is brighter than 1AA 1.2V. That is because at 3.6V the amplifier is cruising, at 2.4V it has to work and at 1.2V it has to work really hard. With the better boost drivers output remains the same at all input levels.

Just to point out that there is no such thing as a Cree P7. There is a only Seoul SSC-P7.

Certain retail outlets are so inaccurate with their descriptions and specs that they are close to worthless in terms of relying on.

In order to really understand what is happening in the electronics of this light, you would probably need to take current measurements at the tail and at the emitter using various battery configurations.

I think you have a misunderstanding of battery chemistries.

There are two types of batteries:
Primary, meaning single use, disposable
Secondary, meaning rechargeable

Within this second category, there are a variety of types of rechargeable battery.
IE: Nickel Metal Hybrid (Ni-MH), Nickel Cadmium (Ni-Cd), Lithium ion (Li-ion), and Lithium Polymer (Li-Po).

Ni-MH and Ni-Cd come in a variety of sizes, and are a stable, relatively safe and high capacity rechargeable battery type.

Li-ion and Li-Po also come in a variety of sizes and many different chemistries. They are mostly nominally rated at 3.6/7v, but fully charged are actually closer to 4.2v. There are, however, chemistries that have a lower nominal voltage of around 3.2v.

Li-ion cells have a higher voltage than Primary Lithium cells, such as CR123, which is 3v, not 3.6v. There are Lithium Primary AA's, that are a different chemistry than their CR123 brethren, and are usually rated at 1.7-1.8v.
Whereas most cells are given a certain size (ie: AAA, AA, C, D, 1/2C, 1/2D, etc.), Li-ion cells are described by a standardized numerical system. Ie: an 18650cell is 18mm in diameter, 65mm in length. The final "0" indicates that it is a round cell. So a 16340 is 16mm in diameter, 34mm in length, and round.

So, your math is off, as far as voltages.

The driver you're looking at is a BUCK driver, meaning it reduces the input voltage to an assigned output voltage, in this case, around 3.5v. It also changes it's current draw (from the batteries) depending upon the voltage being supplied.

So current draw will not be the same from 2CR123(@6v), to 1x 18650(@~3.6v, under a high load), to 10xNi-MH (in series, at ~12v). The higher the voltage, the lower the current draw, and vise-versa.