Running 24*2=48 white LEDs in series with a 6V 4.5AH lead acid battery

[mcGeek]

I'm running 48 leds with a 6v 4.5AH lead acid battery, there are 20 leds connected in parallel each parellel line has got 2 leds making them 48 leds. the leds are white leds. please i need to know the total current and watt of the leds and how long the battery would last. please check the attached picture

 

[KlausST]

Hi,

Without knowing the LED type it is impossible for us to know the LED voltage/current/power.

Without knowing the efficiency of the LED driving circuit is impossible to calculate the lifetime of the battery.

Klaus

 

[mcGeek]

it is a white LED with 3.2Vf

 

[mcGeek]

don't mind the picture, its just a quick proteus schematics to show how the leds are arranged.
48 5mm 3.2Vf white LEDs

 

[KlausST]

Hi,

farnell has over 250 different white LEDs types with 3.2Vf ...
We can not know wich of them you use.

*****

20 LEDs x 2 = 40 LEDs , you say 48. What is correct?

Klaus

 

[mcGeek]

sorry its a typo, i meant 24*2=48 and i meant a standard white 5mm 20mA led 25cd.

 

[KlausST]

Hi,

obviously if you have 24 strings with 20mA each the total current is 24 x 20mA.

And obviously when you connect two 3.2V LEDs in series there is a total voltage of 2 x 3.2V.

Multiply both values to get total power.

**
Your battery is 6V nominal. But your LED needs more than this. Therefore you need a driver.
I asked you about the driver before. (It´s really annoying to get the data piece by piece and asking for every value twice)

Klaus

 

[mcGeek]

sorry for the stress, I'm using a 15R/10W resistor.
from your reply 24*20mA=480
2*3.2=6.4
therefore total power should be 480*6.4=3072w ?
or is there something else i'm missing?

 

[KlausST]

Hi,

or is there something else i'm missing?

Yes, the "m" of mA.

If you use the "m" then you get 3072mW = 3.072W

***
You see you need 6.4V but you nominally have 6V only. No resistor can "generate" the missing voltage. You need a LED driver. "Boost" mode, "step up" mode or "SEPIC" mode.
In short: with a higher output voltage than the input voltage.

Klaus

 

[mcGeek]

Thanks alot, what if i use a standard 5mm green led with 2.8Vf and 20mA connected with th 15R/10W resistor to limit current would it be OK?

 

[KlausST]

HI,

I don´t think so, because th 6V is the nominal voltage.
Expect a dicscharged voltage of 4V or less and a full charged voltage of 6.8V or 7.2V depending charging method.

Again: Maybe there is a solution, but I think there is no way around a LED driver.


Klaus

 

[chuckey]

Your circuit is very unsatisfactory because the current through any pair of LEDs could be 30 or more percent higher then for another pair of LEDs. So to make sure that you do not over dissipate any of the low Vf pairs is to run them at a design current of 15 mA. So with a fully charged battery of 7.3V you get your 15 mA per LED pair. Of course as the battery runs down so will the current and brightness of the LEDs.
The proper way to do this is to wire them all in series so you need a constant current supply of 20mA and with 48 LEDs, it must be able to output at least 48 X 3.2 ~160V.
You could try buying binned LEDs - matched for Vf.
Frank

 

[Audioguru]

Nobody sells a "3.2V" white LED. Some will be 3.0V, some will be 3.2V and some will be 3.4V and you do not know the voltage of your LEDs unless you test then sort each one. That is why we use a supply voltage that is high enough that low voltage LEDs and high voltage LEDs use almost the same current from the same resistor value and look almost the same brightness.
Expensive high power LEDs are "binned" into voltage groups but ordinary cheap LEDs are not binned.

But you do not have "extra" voltage, you do not have enough voltage. Two 3.2V LEDs connected to a 6.0V supply will be very dim and two 3.0V LEDs will be so bright that they quickly burn out, and two 3.4V LEDs might not produce any light.

The maximum current for most 5mm diameter LEDs is 30mA. They are spec'd with 20mA. Two 3.0V LEDs need 6.0V and the current-limiting resistor can have 2.5V so the supply needs to be 8.5V.
The resistor for 25mA is (8.5V - 6V)/25mA= 100 ohms. If both LEDs are 3.4V then they need 6.8V and the resistor will have across it 8.5V - 6.8V= 1.7V and their current will be 1.7V/100 ohms= 17mA which is a little dimmer than 25mA but it is fine. Since some LEDs will be 3.0V and others will be 3.4V then you cannot parallel them unless you measure and sort them. Then you can calculate a resistor for each voltage group.

 

[BradtheRad]

Since I like to create simulations, this one illustrates the suggestions that you boost 6V to a higher voltage. A boost converter can do this.

Post #12 suggests 160V to drive 48 bright whites in series. However if that seems like too high a voltage to work with, then you can run 80V to two strings of 24 led's each.

LED strings are driven at 20 mA. Each led represents 12 led's.



The battery provides about 1/2 A average current draw. Theoretically the battery can last 9 hrs (= 4.5 / 0.5).

 

[marce]

Having done some automotive LED lights recently, I would recommend an automotive LED driver IC and associated circuitry to drive the leds...
Driving the Leds was the simpler task, controlling the junction temperature of the Leds was more problematic. Minimum would be a double sided PCB with additional heatsinking added to the rear, look at LED manufacturers sites for information on thermal considerations.
To avoid spacing problems and to comply with SELV (Safe Electrical Low Voltage directive)(I know its a hobby circuit but safety is important) during layout keep voltages below 50V, 48 is good...

 

[mcGeek]

I have wasted some many LEDs because of this. I first blew up 24 Leds then i doubled the leds to 48 same thing happened. i've connected it with a 15R/10W resistor, it seems to be working pretty fine now. I just want to know my calculations like
* How long the battery will last?

 

[Audioguru]

You need to know the exact voltage of the LEDs, the exact resistance of the "15 ohms" resistor and the average voltage of the discharging battery to calculate the exact current. Or you can measure the current by measuring the voltage dropped by the resistor and use Ohm's Law to calculate the current. As the battery voltage runs down you need to decide when the LEDs are dim enough to end the duration test.

 

[c_mitra]

A fully charged lead acid battery will have 2.2V (per cell; you will get around 6.6V) but this will quickly drop to 2V with use and will be ready for charging when the voltage will drop to 1.8V (in your case 5.4V). The battery will be damaged if you discharge it below 1.6V so you will need a circuit to cut off the battery when it reaches around 5.5V. The battery has a rather low internal resistance and can give high currents when shorted- they must never be shorted.

The forward drop of all diodes (and this includes LEDs) depend on the forward current and as the battery runs down the intensity will decrease because of the decrease in the forward current. At high current the LED will be damaged irreversibly and you will also need something to cut off high current. The LEDs are current devices and you need to stick to the rated current. The typical value will be around 20mA per LED device and this will drop around 3-3.5V (varies) but this means that each LED will dissipate around 70mW (approx; per device) which is not much. With 48 LEDs you will have less than 7W of heat coming out of the LEDs.

You need to seriously think about a suitable driver to supply 20mA current at a voltage of around 160V or so. It should shut off if the current drops below 15mA. This should not be a difficult task.
You have a 6V 4.5AH battery that has enough energy to run this configuration for about 4 hours. But this battery DOES not have enough voltage to drive a pair of LEDs in series for most of the time (the LEDs will turn off as soon as the battery voltage gets below 6.4 (which will be very soon). So you will need a circuit to increase the voltage and provide a (approximately) constant current to a string (of series connected; not connected in parallel) of LEDs. This will avoid the use of series resistances to control the current in individual LEDs.

 

[Easy peasy]

24 leds (assuming back to the white ones) x 3.2V x 20mA = 1.54 watts if you have perfect low loss drive to every led (i.e. they were all in series and you had a perfect step up constant current driver 80V)

so 6V x 4.5Ahr, is 27 watt hour for a 450mA discharge (C10) or lesser, lets say the driver is 80% efficient so you need 1.25 x 1.54W = 1.925 watt out of the 6V batt, this is 320mA, so you should get near to rated amp-hours out of the batt, giving 4500mA-hrs / 320mA = 14 hours best case with a suitable driver and series connected LEDs.

Green leds @ 2.8 volts (20mA) would give 14% more time, i.e. 16 hours. (2.8 x 24 = 67.2V, so an 80 volt driver still needed)

 

[c_mitra]

so 6V x 4.5Ahr, is 27 watt hour for a 450mA discharge (C10) or lesser, lets say the driver is 80% efficient so you need 1.25 x 1.54W = 1.925 watt out of the 6V batt, this is 320mA, so you should get near to rated amp-hours out of the batt, giving 4500mA-hrs / 320mA = 14 hours best case with a suitable driver and series connected LEDs.

You computed for 24 LEDs but I did it for 48 LEDs. That will give about 7 hours (half of what you calculated).

It is not good to drain lead acid batteries down to their rated capacity and manufacturers usually recommend stopping at 90%; I computed around 60%.

Your reported value of 7 hours (14/2) is the best case; my guess value of 4 hours is worst case. Actual value will be something in between.