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LED Emergency lighting

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pirate_edmund

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Hey Guys,
Need your help and advice!

I am trying to make an led emergency lighting system. This is completely disconnected from the regular wiring circuit at my house.

I live in India and have been having a few issues with the electricity supply! More demand less supply not enough rains...blah blah blah...!

Either ways..

I wanted to build this..View attachment 67525 (sorry about my paint skills)

I want to be able to control the intensity, know how much juice is left, and turn on/off the panels individually.


I plan to use 5 BL-5C (http://www.cpkb.org/wiki/Nokia_BL-5C_battery ...we get them for cheap here like $4 or 5) in series so that the Voltage is around 18V (3.7V*5) and 5100mAh (1020mAh*5) (hope my math is right!)

I plan to use 36 piranha 4 pin leds (Datasheet: https://www.electro-tech-online.com/custompdfs/2012/09/9263.pdf)
on each panel in a 6x6 configuration. And there are a total of 6 panels (I plan to put each panel in a separate location like at 4 corners and 2 in the middle of my gallery and hall)

now.. here is where i need help and advice from you guys !

Question 1) I need to chose an led driver.. which can work 18V and has PWM..

Question 2) Do you think that these batteries will suffice? i was thinking 3 would be enough! i was hoping to get about 8 odd hours of lighting

Question 3) I need help in putting together a charging circuit since this will be fixed.. i prefer using the Nokia male and female connectors here.. have a few lying around the house.

Question 4) Also an LCD or LED panel (easier the better) where i can read how much battery charge is left.. so that i can turn off the panels which i don't need and get more battery life

Question 5) Will a 10k Pot do for input for intensity control?

Question 6) What value resistors will I be needing on the panels?

I would really appreciate your help in helping me figure this out!
Thank you!
 
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The battery cells "average" 3.7V but are fully charged to 4.2V each and are discharged to about 3.2V each. Then the minimum voltage for 4 in series is 12.8V.

If you connect 3 LEDs in series then they need a maximum voltage of (3.8V x 3=) 11.4V leaving a few volts for the current-limiting resistor and driver saturation loss.

For a panel of 36 LEDs you csan connect 12 groups of 3 LEDs in series, each using 20mA or 25mA then their total current is 240mA or 300mA.

Then the 720mAh battery will last 3 hours or 2.4hours for one panel and the LEDs will be slowly dimming as the battery voltage runs down.

The battery needs a fairly complicated "balancing" charger that charges each cell separately and limits their charging current to about 360mA and charging voltage to 4.20V each.

The battery needs a low voltage disconnect circuit to protect it since a low voltage destroys it.

You can use a voltmeter circuit to measure the battery voltage when it is powering the LEDs to see how close to low voltage disconnect the voltage is.

I cannot tell you an LED driver since it is probably not available in your country.
 
Hey Audioguru, thanks for the reply!!

I can get the required parts in India or I can get them shipped so it wont be a problem...

Secondly I was thinking that if the BL-5C batteries are gonna be a pain then i can get https://www.hobbyking.com/hobbyking/store/__8932__Turnigy_2200mAh_3S_20C_Lipo_Pack.html

or

https://www.hobbyking.com/hobbyking...gy_nano_tech_2200mah_3S_45_90C_Lipo_Pack.html

or any one of those shown in the hobby king store and the required charger.. they do the balancing on their own!! (https://www.hobbyking.com/hobbyking/store/__216__408__Battery_Chargers_Acc_-Battery_Chargers.html)

so could you suggest any one from the above store for batteries? i was hoping to get at least 5 hours on all panels..! ( the reason i say 5 is because that's how long the longest power outage lasted)

and yes 12 groups of 3 leds in series is good.. I will be using that..

So to reiterate..
1) Please suggest batteries and chargers from the hobby king store as I can get them here..
2) Please also suggest the led driver I can get them here or I will have them shipped..!
 
The Noika battery has a protection circuit built-in. The cell phone has the charger and the circuit that disconnects the battery inside when the voltage drops too low.

The Hobbyking batteries have the protection circuits built into the model airplane and the charger. Using their charger then you must design and build a low voltage disconnect circuit.
 
@ Audioguru - I have found a local shop here which sells batteries and chargers with protection.. could you advise me on what batteries i should pick up? He can custom make my batteries too..! So what Li polymer battery voltage and amp do i pick up for here?

@ronv- I dont want to use lead acid batteries as im not a big fan of how dirty and heavy they are so i will stay away from them..!
 
@ Audioguru - I have found a local shop here which sells batteries and chargers with protection.. could you advise me on what batteries i should pick up? He can custom make my batteries too..! So what Li polymer battery voltage and amp do i pick up for here?
We talked about using a 3-cell Li-po battery that averages 11.1V.
Make sure that the "protected" battery shuts off the load when its voltage drops to 9.6V.
Make sure that the charger is balanced.
Make sure the charger limits the charging current to the amount recommended by the battery manufacturer.
Make sure the charger limits the voltage to 12.60V.
Make sure the charger disconnects the battery when the charging current drops to about 1/40th the limited current.
 
pirate_edmund;1085246[quote said:
I dont want to use lead acid batteries as im not a big fan of how dirty and heavy they are so i will stay away from them..!

The ones I posted are sealed so they are very clean. They are heavy, but for a fixed installation I don't see that as a problem - but it is only money to use LiPo.

Please note your white LED's have a forward voltage of 3.3 to 3.8 volts.
 
@ Audio Guru Yes Sir.. 11.1V 3 cell Li-po battery.. sorry to ask this but would 2000mAh do? or do i get 5600mAh ? And yes will make sure all the precautions and safety systems mentioned are present.

@ Ronv.. Sir I understand and I will go for Li-po and yes they have a 3.3- 3.8 V forward and that is the primary reason for a heavier voltage battery... i wish i could run them of the normal 9V D cells! :D
 
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pirate_edmund;108558511.1V 3 cell Li-po battery.. would 2000mAh do? or do i get 5600mAh ?[/QUOTE said:
Did you calculate how many LEDs can be in series and their maximum current?
A 3-cell Li-Po battery has a minimum voltage of only 9.6V.
The LEDs have a maximum forward voltage of 3.8V then 3 in series need 11.4V plus maybe 2V for the current limiting resistor.
So only two LEDs can be connected in series.

The battery is 12.6V when fully charged. The maximum current for an LED is 30mA when it is cool but 36 grouped together will get VERY HOT so figure on 25mA for each string. Then each group has 18 strings and the total current of a string is 450mA. You want it to work for 5 hours so a battery for one group should be 2250mAh.
One battery for 6 groups must be an expensive 13500mah monster!
Of course one battery charge will last longer when the LEDs are dimmed.
 
@Audioguru Here is what i calculated...

Parameters:
Input Voltage from battery 12.6 V
LED forward voltage 3.8V (max)
LED forward current 30mA
Number of leds in array 36..

Solution:
3 leds in series... and 12 of these in parallel to have 36 leds per panel.
So that should be 360mA from the battery..
I will be using 47 ohm 1/4 watt resistors

Now 6 of such panels 2160mA
Now 5 hours so 10800mAh battery!

So i would have to get one of these!

**broken link removed**
or
https://cpsbattery.manufacturer.glo...ium-ion-battery/1039923461/Li-ion-Battery.htm
or
https://zeefocharger.en.made-in-chi...ina-Portable-Charger-for-Laptop-U-N1083-.html
 
THINK!
Don't you want to have light emitting diodes and not dark emitting diodes?

Look at my post #9.
3 LEDs at 3.8V each need 11.4V plus a couple of volts across the current-limiting resistor.
But a 3-cell Lithium battery averages only 11.1V and is 9.6V when it should shut off. Then 3 LEDs in series will be dark.
That is why I calculated with 2 LEDs in series.

Another problem with your lack of thinking is the very high current if the LEDs are actually only 3.3V each and the battery is fully charged at 12.6V. Then the current in the LEDs and in the 47 ohm resistor is (12.6V - 9.9V)/47 ohms= 58mA and the LEDs will quickly burn out!

Another problem I mentioned before is that 36 LEDs together at their maximum allowed current (30mA) will get too hot!
Only a single LED operating at 30mA will probably be at its maximum allowed temperature. 36 together will heat each other much more. The LEDs in the middle of the group might melt.

The first battery you show is 5V, not 11.1V. But a Lithium battery is 3.2V to 4.2V or 6.4V to 8.4V or 9.6V to 12.6V, not 5.0V.
The charger (it is actually just a power supply) does not have a 12.60V output for a 3-cell Lithium battery.
 
@ Audioguru.. sorry seems i forgot the basics..

2 leds 3.8V drop across each of them in series.. so that is 2*3.8V = 7.6V adding 2V for the current limiting resistor we get 9.6V..
As you specified earlier i will need a battery which will give out enough to keep the leds at full glow even at its cutoff charge
Which would mean a battery with 9.6V as the cutoff
Hence u chose the 12.6V battery
the current drawn remains at 30mA across the string of 2 leds in series..
Now 18 strings for 36 leds so we get 30mA*18=540mA
6 panels so we get 3240mA
5 hours so we get 16200mAh or 16.2Ah
For the resistors (12.6V-9.6V)/0.03mA=100ohms
I will be using a 120ohm resistor for this case so i will be supplying 25mA
Hence i have to find a 12.6V 16200mAh battery

if we have 3 leds in series with 3.8V drop across each we need 11.4V +2V for the resistor so we need a battery with 13.4V as the shut off and ~15V at fully charged state
30mA current consumption by each string a total of 12 strings so we need 360mA
6 panels = 360*6=2160mA
5 hours=10800mAh..

I haven't been able to find a battery with 13.4V as the cutoff voltage as you had said "But a Lithium battery is 3.2V to 4.2V or 6.4V to 8.4V or 9.6V to 12.6V, not 5.0V". So I shall stick to the 2 leds in series as described by you.

Regarding the heat i think i will have to use a heat sink which will be built into the back shell of the panel holding the leds and a few holes and maybe try and get a small brush less fan into it.. to keep the leds cool..
 
2 leds 3.8V drop across each of them in series.. so that is 2*3.8V = 7.6V adding 2V for the current limiting resistor we get 9.6V.
For the resistors (12.6V-9.6V)/0.03mA=100ohms
I will be using a 120ohm resistor for this case so i will be supplying 25mA
No.
The minimum voltage for each LED is 3.3V, not 3.8V. The total minimum voltage of 2 LEDs is 6.6V, not 9.6V. The resistor is 200 ohms, not 100 ohms. Use 240 ohms for 25mA, not 120 ohms.

I have to find a 12.6V 16200mAh battery
No.
12.6V is its fully charged voltage. It is rated and advertised at 11.1V which is its average voltage.

Regarding the heat i think i will have to use a heat sink which will be built into the back shell of the panel holding the leds and a few holes and maybe try and get a small brush less fan into it.. to keep the leds cool..
No.
Ordinary 5mm diameter LEDs are impossible to heatsink.
I made a night light with 36 Luxeon SuperFlux LEDs. They have a 4-pins package and the pins conduct some of the heat away to the copper on a pcb. They have a maximum continuous current rating of 70mA so I operated them at "only" 53mA.
I mounted the pcb in the case for an audio cassette tape and drilled many big holes for cooling. I added foot pads to raise the case above the tabletop.
Boy oh boy they got HOT! The case nearly melted and the clear LED packages turned yellow.
 
go to DIODES.COM under APPLICATIONS go to LIGHTING then SAFETY then EMERGENCY.
**broken link removed**
There are many circuits under flash lights that work well.
Safety/Emergency and Garden lighting will give you ideas.

I use many of these parts.
Also look under LINEAR.COM and TI.COM.
 
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