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Looking to make brightest 1 metre of LED's with Sony NPF lipo battery

I'm basically trying to make the brightest LED bar light for video that I can which runs off of the most commonly used NPF 8v batteries.

I've been experimenting with these 2835 LED strips. I have cut it to 1 metre which has 240 LED's and to my understanding wants 24v and apx 1.5amps or 36W (180WattsPer5meters/5=36/24v=1.5a). I chose these because they seem to be the most efficient LED's out there and are low heat (which I assume is related to efficiency)

This NPF battery claims to have 8000mAh and while that's probably a vast exaggeration, in my understanding there should be enough juice to power these lights at full power for somewhere between 1-1.5 hrs. (8amps X 8v = 64Wh). Now I've also read that these NPF batteries have a max 5A discharge auto cutoff which is just above my calculated needs.

I've been using this Boost Converter to adjust output current/voltage which is rated for somewhere between 35-50W. I rigged it all up and adjusted the voltage to 24 and connect the LED with a low current. This low current kept the starting voltage below turning on point. As I raised the current I eventually got to 23 volts and 1.5 amps output and a very bright light strip. Any higher and the circuit would shut off. I saw that the battery started drawing more than 5amps and my guess is that it shut itself down. I ran the circuit in this setup and it lasted.....30 minutes :/

I noticed that the Boost converter was getting quite hot and I wonder if this is where there is great energy loss.

I also happen to own this NPF powered 176 LED light and I've found that it draws 1.5 amps from the battery and powered the LED's which are set up in two parallel sets with a total of only 5.8V without any resistors. The light claims to output 1320 lumens and while it is indeed quite bright, is about 3 times dimmer than the theoretical output of my strip. However when I let it run on full power using the exact same 8000mA NPF battery it lasted 330 minutes! This suggests to me two things:

1. That my NPF battery is actually close to 8000mA capacity because with a 1.5A draw it would be calcluated to last 320 minutes (8Ah/1.5A=5.3hrs x 60minutes).

2. Therefore my setup definitely has some serious inefficiencies.

I tried running my setup at slightly lower voltage/current of 21V 1A and it lasted 90minutes which is certainly better but again the Booster gets hot and not very close to the calculated 140 minutes I could theoretically get with these settings.

It seems that the makers of the 176 LED light I purchased were smart in deciding to setup the LED's in an array that requires LESS voltage than the batteries native output and therefore only requires a slight step-down rather than a large boost. Do you think this could be where my efficiency is lost? I can't find any sub 9v LED strips, and while I've done a fair amount of research I don't exactly see why not.

Sorry for my long and probably rather uneducated post and thanks for any help!
 
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Fish4Fun

So long, and Thanks for all the Fish!
Hey alex ezorsky-lie !

Most LED strips use resistors to limit the current from a nominal voltage (typically 12Vdc or 24Vdc commonly found in cars and boats). In these applications "saving power" with respect to LED lighting is meaningless, and the (rather significant) losses added by resistive current limiting are deemed acceptable because the implementation of resistive current limiting in these applications is vastly easier to manage than more efficient means of current limiting.


Assuming you have white LEDs, the typical forward voltage is ~2.8V-3.4V per LED. For a 24V "string" the LEDs are likely arranged in groups of 6 in series giving us a nominal forward voltage 18.6V but in reality might vary from 16.8V to 20.4V. Assuming an input Voltage of 24V the minimum resistor used would be calculated as follows:

24V - 16.8Vf = 7.2V

Assuming the LED current is 60mA ...

7.2V = 0.060 * R
R = 120 ohms


The power consumed by the resistor would be:
P = i * E
E - i * R
So,
P = i^2 * R >> 0.060^2 * 120 = 0.432W


The power consumed by the LEDs (in this case) would be:
16.8V * 0.060A = 1.008W


Total Power = 1.008W + 0.432W = 1.44W

Efficiency = 1.008W/1.44W = 70.0%

Now, lets look at the case where the Vf = 20.4V:
24V - 20.4Vf = 3.6V
3.6V = 120 Ohms * i ---> i = 0.030 Amps


Power Consumed by the resistor:
i^2 * R = 0.030^2 * 120 Ohms = 0.108W


Power Consumed by the LEDs:

20.4V * 0.030A = 0.612W

Total Power = 0.72W

Efficiency = 0.612W/0.72W = 85%

The way LED "strings" are constructed, the nominal voltage determines the number of LEDs in a "Series Unit", the length of the string is then an integer number of these "Series Units" ... 1M of LEDs = 240 LEDs ==> 40 "Series Units" like the above. So ...

40 * 1.44W = 57.6W

40 * 0.72W = 28.8W

The "Nominal Rated Power" of 1M of the LEDs you linked is 180W/5 = 36W, which is likely based on a nominal forward voltage of 3.25V ...
Vf = 19.5Vf
24V - 19.5Vf = 4.5V


4.5V = 120 * i >>
i = 0.0375A


P => i^2 * R = 0.16875W lost in each resistor

19.2V * 0.0375A = 0.73125W

Total Power = 0.90W

0.90W * 40 = 36W

Assuming your DC-DC Converter is ~80% Efficient then your actual power consumption should be:

57.6W/.8 = 72W

36.0W/.8 = 45W

28.8W/.8 = 36W

The battery you linked clearly states it has a nominal voltage of 7.2V, NOT 8V ; it also clearly states the capacity to be 57.6WH.

From a forum dedicated to batteries discussing the exact type of battery you linked to:
These batteries are often faked and I never seen any non-oem production with actual capacity larger than 6000mAh.


72W/7.2V = 10A
45W/7.2V = 6.25A
36W/7.2V = 5A


Assuming the 72W case, it would appear your battery has a "real" capacity of ~5000mAh ...
Assuming the nominal case of 45W battery drain, your battery likely has a "real" capacity of ~3125mAh ...
Assuming the 36W case, it would appear your battery has a "real" capacity of 2500mAh.


As you pointer out, it is possible that the pack was not designed for high current discharge; if this is in-fact the case, then it is quite likely that heat build up from an excessive discharge rate could substantially reduce the available capacity.

So, in short, is the glass half Full or is the glass half Empty? I am thinking the glass is just twice as big as it needs to be ;-)

Good Luck!

Fish
 
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