# Half the power lasts 4x as long!?

#### Katherine Alicia

##### Enlightened
Maybe I`m having a senior moment but I seem to be missing something here?
I have a night light that uses 2x 3000mah `D` cells, so 3v @ 3Ah and it`s running a 6V 500ma bulb, it makes lovely ~2000k ish color and plenty of it, I get just over 12 hours runtime.

Now if I use 4 of these D cells to make up 6v @ 3Ah I get about 6 hours runtime, which is correct for the 6v 500ma bulb.

my problem is that I get 6 hours at full brightness with 4 batteries, and 24+ hours at half brightness using 2 batteries twice, surely it should be twice as long and not 4x as long runtime?

what am I missing? I`m sure it`s something really obvious! LOL

#### thermal guy

##### Flashaholic
That's WAY to much maths for me😁 are you sure your getting half the brightness? Might be 1/4 of full output and then it kinda makes sense. "I think"

#### Katherine Alicia

##### Enlightened
That's WAY to much maths for me are you sure your getting half the brightness? Might be 1/4 of full output and then it kinda makes sense. "I think"

well, I can`t be 100% as I don`t have a lumen meter yet (it`s on the way though), I`m sure i`m doing something wrong that I just can`t see, my maths says that: 4x D cells at 3000mah should only be able to deliver 18 watts of power no matter How you wire them up, 4x D = 6 volts, and 6v x3 Amps= 18W

even if I wire them as 2s2p, I get 3v at 6A and that still makes 18 Watts.

yet I`m getting 24 hours of what I assume is 1.5w (based on halving the voltage to the 3W bulb) which makes 36W?

in this instance our 6v .5a bulb is 3 Watts (6v x .5a=3w) Therefore (3v x .5A=1.5W)

if we but 4x 3000mah D cells in series, we get 6v at 3a, so that will run out bulb for 6 hours (3000mah / 500ma).
so 6 hours at 3 watts is what we get (18W total)

Now...

wire those same D cells as 2 parallel and 2 series, to make 3v at 6000mah. now hook up the same 6v 500ma bulb, we now get 24 hours of runtime at 1.5 watts.
but 24 hours of 1.5 is 36W total.

How!???? it`s the same D cells and bulb! :duh2:

#### thermal guy

##### Flashaholic
Efficiency of the bulb at that lower level might just be more "better " then you think?

#### xxo

##### Flashlight Enthusiast
You need to measure the current draw to see what is going on. Even with NiMH's incan bulb's output drops off quickly with even small fading of Voltage (most incan's don't have drivers to adjust themselves). Test both lights current every hour or half hour and see how much they are actually drawing throughout their runs.

##### Flashlight Enthusiast
LED efficiency goes down as current increases. Depending on the type and manufacturer of the LED, that logarithmic scale can differ but are generally the same. So it's very possible that the lamp you're using requires 4x the amount of current just to produce twice the lumens.

You also mention "twice the brightness". Is that what you're perceiving or is that what the manufacturer of the lamp says? If it's perception then you have to factor in that your eyes also don't scale linearly either.

#### thermal guy

##### Flashaholic
Lots of maths. Lots of maths😁😁😁

#### Katherine Alicia

##### Enlightened
You also mention "twice the brightness". Is that what you're perceiving or is that what the manufacturer of the lamp says? If it's perception then you have to factor in that your eyes also don't scale linearly either.

Twice the brightness based on the fact that it`s twice the wattage, I have no objective way of testing brightness yet though, just that it`s Brighter on 6 Volts than on 3 Volts.

#### wrf

##### Newly Enlightened
in simple case, half the voltage results also in half the amperage
so
(3v x .5A=1.5W)
would actually be closer to
(3v x .25A=0.75W)

#### archimedes

##### Flashaholic
None of the functions you reference are truly linear, and how accurately are you monitoring each of the variables (volts, amps, lumens, etc) throughout the entire run ?

Alkaline cells will sag significantly, and that will not be consistent under different loads. That's just for starters.

Put more simply ... what and how are you measuring versus what are you calculating ?

Last edited:

#### Katherine Alicia

##### Enlightened
None of the functions you reference are truly linear, and how accurately are you monitoring each of the variables (volts, amps, lumens, etc) throughout the entire run ?

Alkaline cells will sag significantly, and that will not be consistent under different loads. That's just for starters.

Put more simply ... what and how are you measuring versus what are you calculating ?

the only thing measured is the runtime, the rest is worked out from the voltage and mah of the D cell(s) and the rating of the bulb 6v .5A those are the only things I have to play with at the moment.

#### Katherine Alicia

##### Enlightened
in simple case, half the voltage results also in half the amperage
so
(3v x .5A=1.5W)
would actually be closer to
(3v x .25A=0.75W)

Hmmm... now That`s interesting! and may very well explain what`s going on

#### archimedes

##### Flashaholic
So, for the (single) C cell test I linked, @HKJ measured ~ 8.4Wh at 0.1A test current ... but 0.9Wh (!) at 3A

The third graph from the top, I believe, most directly addresses your question here (titled as - discharge, time scale)

Last edited:

#### Katherine Alicia

##### Enlightened
So, for the (single) C cell test I linked, @HKJ measured ~ 8.4Wh at 0.1A test current ... but 0.9Wh (!) at 3A

and couple that with the fact that a bulb isn`t a fixed load and changes it`s resistance as it gets hotter and it really isn`t a simple calculation at all!
I don`t feel so bad now LOL

#### archimedes

##### Flashaholic
Yes, I had left heat aside, but it is also a non-trivial factor.