Boot Diver 88% efficient. is that good?

Flashlight Modding and DIY Parts
1

I have a 7 XP-L light that is running from 2s2p 18650 battery pack. Nominal voltage is 7.4V

I am not getting the runtimes out off the battery pack and I am trying to figure out where the problem is. Suspect 1 is the battery pack but it got me to measure the driver efficiency. So i have a few questions that will follow.

Setup
So on my bench power supply I have set the voltage to 8Vdc.
on full power i measure 1.93A

driver is a boost driver and the 7 leds are in series:

Output is 20V x 0.676A

Eff = 20x0.676A / 8x1.93 = 87.5%

1. Is that good or bad?

2. Would a step down driver not be more efficient? 95-98%?

3. Is it ok to put 7 XP-L in parallel? Will one or two not take all the current and the rest idling along?

2

Anything over 85% is efficient. Something like 95% you can say it is very efficient.
A step down driver could be more efficient at around 92% realistically
You can't put 7 XP-L in parallel to the boost driver and inputting 2S2P, the voltage of the load (7 XP-L in parallel) is lower than the input voltage so there can't be no boosting.

However runtimes are not entirely related to efficiency of the driver but also due to weak battery pack (capacity, age, current capabilities) or voltage drop in the wires (you need to calculate it both ways, because where are 2 wires).

3

I know this is a bit off topic but do you have a link with

That 7 xpl light? Most 7 xpl light i see here are fake.

4

Efficient enough but very underdriven…

5

I want to change the driver and the led configuration to step down(Buck) and put the leds all in parallel. will that work? will the 7 all havce the same Vf?

6

boost drivers can also be very efficient, taskled hyperboost is like %93-%94 at certain voltage in\out ratios, usually the less voltage difference is between in \out, the more efficient they are, same for buck drivers.

7


You can do that, the Vf will not be technical identical but well you should not worry about it, make sure the current path is not of different resistances (wires from the driver to the LEDs)

But you first need to sort out the voltage drop in wires, no matter how efficient a driver is you still need to have that sorted out, over long and thin wire you basically "lose" power, 87% efficiency is not by any means a bad efficiency which should make you worried. What is the cell model in the packs? What is the wire length and gauge from the pack to the driver?

8

Ok I Will make sure the lengths of the tracks are the same on the PCB.

The length is about 100mm but the wires are thick about 2mm diameter. I am not sure what that gauge that translate to.

I am using LISHEN 2600mAh 18650 cells. but I am starting to doubt my choice.

9

2mm is about 14 awg. 100mm lenght, (4”) of that guage is plenty for 2A, more than enough.

assuming you meant 2mm for copper, not copper and insulation

10

I assume they are the wires with the insulation? something like 18AWG, but anyway at 2A and 2meters back and forth there is only a voltage drop of 0.085V for 18AWG.
Unless the wires have an additional insulator, then the conductors may be thinner.

11

Another VERY important consideration regarding the efficiency of boost drivers is the actual input voltage. Simply put, the higher the input voltage, the higher the efficiency. Boost drivers have a bad reputation for low efficiency in the flashlight world because their most common application is for 1xNiMH lights. The electronic components of boost drivers have certain, fixed voltage drops associated with them. So if all you are feeding them is 1.2V, these voltage drops burn off LOTS of your input voltage. Couple this with the need to boost voltage by a factor of three, and you can see that efficiency will be low.

2xAA lights both start off with a higher voltage AND don’t have to boost it by as large of a factor as 1xAA lights. So the efficiency of the converter is higher than in a 1xAA light. The boost driver in the Fenix PD40 is even more efficient. It starts off with 1xLi-Ion (3.7V) rather than 1.2V. And it boosts voltage by a factor of less than two. So the result is a MUCH more efficient light than your typical 1xAA or even a 2xAA.