Automatic High Watt LED Emergency Light with Battery Cut-Off

The following post explains a very simple yet an outstanding automatic 10 watt to 1000 watt emergency lamp circuit. The circuit also includes an automatic over voltage and low voltage battery shut off feature.

The entire circuit functioning may be understood with the following points:

Simulation and Working


Referring the below given circuit diagram, the transformer, bridge and the associated 100uF/25V capacitor forms a standard step down AC to DC power supply circuit.

The bottom SPDT relay is directly connected with the above power supply output such that it remains activated when mains is connected with the circuit.

In the above situation, the N/O contacts of the relay stay connected which keeps the LED shut OFF (since it's connected with the N/C of the relay).

This takes care of the LED switching, making sure than the LEDs are switched ON only in the absence of mains power.

However, the positive from the battery is not directly connected with the LED module, rather it comes via another relay N/O contacts (the upper relay).

This relay is integrated with a high/low voltage sensor circuit stationed for detecting the battery voltage conditions.

Supposing the battery is in a discharged condition, switching ON the mains keeps the relay deactivated so that the the rectified DC can  reach the battery via the upper relay N/C contacts initiating the charging process of the connected battery.

When the battery voltages reaches the "full charge" potential, as per the setting of the 10 K preset, the relay trips and joins with the battery through its N/O contacts.

Now in the above situation if the mains fails, the LED module is able to get powered via the above relay and the lower relay N/O contacts and get illuminated.

Since relays are used, the power handling capacity becomes sufficiently high. The circuit is thus able to support in excess of  1000 watts of power (lamp), provided the relay contacts are appropriately rated for the preferred load.



The finalized circuit with an added feature can be seen below:



The circuit was drawn by Mr. Sriram kp, for details please go through the comment discussion between Mr Sriram and me.

Need Help? Please send your queries through Comments for quick replies!




Comments

The Knight said…
Simply click on the diagram, it should open in a new tab or your existing tab depending on your browser setting.
The Knight said…
Dear Mr. Majumdar,

Thanks for suggesting this circuit and from the looks of it, it appears that it will work quite well and that too for any load (the benefit of having relays here). I, however, plan to do it with solid state components for two reasons. First, I do not have much faith in relays. Solid state circuits are far more reliable. And secondly I do not plan on using really high number of LEDs'. I would use say, some 50 odd white LEDs'. That should suffice for emergency light purpose.

As I said earlier I do not get much time to design circuits. But I will sure design one such as time permits me which will also incorporate a circuit to stop the light from switching on automatically in the day time.

I much appreciate your quick responses and suggestion. You indeed are very helpful. I wish you all the best in your future endeavors.

NAVIN ROY.
Swagatam said…
Thank you Navin, I wish you all the same!

By the way the second circuit provided in the following article might just suit your application, you can give it a try in your free time:

https://homemade-circuits.com/2012/08/led-emergency-light-circuit-with.html
Swagatam said…
the charging relay is controlled by the IC741 and it's triggering will depend on the settings of the two presets which can be accurately set as per individual preference.
Swagatam said…
In that case you can eliminate the relay and use a mosfet instead, and also remove the 100K preset link entirely, as shown in the following example:

https://homemade-circuits.com/2013/07/motorcycle-headlamp-battery-over.html
Swagatam said…
A relay could be actually more compact and powerful than mosfets or transistors, as these can handle higher current without getting hot...mosfets and transistor could require heatsink and have limited current range compared to a relay
Saeed Abu said…
If i want to use cell phone battery 3.7v and Nokia standard cell phone charger instead of that transformer then what things i have to change in above circuit please feedback me urgently.
Swagatam said…
you can try the last circuit given in the following article:

https://homemade-circuits.com/2013/04/1-watt-led-emergency-lamp-circuit-using.html
Hanish Kumar said…
Dear sir,
many thanks sir.May i know which battery(Ah) i have to use for 500mA transformer with 50 watt LED module then?
Swagatam said…
Dear Hanish,
what is the voltage rating of the LED?
Sriram Kp said…
In the above circuit, suppose low voltage value is set as 11.5 and high voltage value is set as 13.5 for a 12v battry in the two presets. Now the battery is fully charged and stopped charging. Now If the AC power gets failure means the battery wil start draining by iluminating the LEDs. Now the battery voltage become 12v and if the AC power comes means, will the battery get charging?? or it wont start charging unless it wil come to 11.5 as set in the preset?? If it wont start charging unless the battery level becomes 11.5 means how to modify the circuit like if AC power is there means the battery should start charging , it should not wait until the battery becomes 11.5 and it have to stop charging at 13.5.
Swagatam said…
The battery will not start charging and will wait until 11.5V is reached.
To avoid this do the following things:
Take a BC547 transistor, connect its collector to pin3 of the IC, its emitter to ground, and base to bridge positive via a 0.22uF resistor and a 220 ohm resistor in series.

This will break the hysteresis every time the power returns and forec the IC to resume the charging instantly.
Sriram Kp said…
0.22uF resistor or capacitor? bridge positive means where?, which spot??
And what is the purpose of that LED which is connected to the transistor?? If I dont want that LED means, can i remove the LED and the 10K resistor and directly connect the pin6 to the transistor??
Swagatam said…
0.22uF is a capacitor.
the white band of the filter capacitor in between the 4 diodes is the positive point or line.
LED indicates battery full.
you can remove the LED but not the 10K
Sriram Kp said…
I will use 13.5v adapter instead of AC line. so i wil remove the transformer and the diode part fully. So can i connect the 0.22uF capacitor to the positive point of the adapter?? And what should be the volt for the capacitor?? 25v is enough??
Swagatam said…
yes you can do it.

0.22uF is a disc capacitor which are all 50V rated always.
Sriram Kp said…
Hai,
I did some modification in the above circuit. Pls check whether the circuit is correct or not. Link for the circuit diagram : imgur.com/J7kbOAz
Working of the Circuit:-
The LED module is connected to the "common" of the relay 1. The +ve of the adapter is connected to the coil , NO of the relay 1, and NC of the relay 2. The NO of the relay 2 is connected to the NC of the relay 1. The battery's +ve is connected to COMMON of the relay 2. So if AC power is ON means, the battery will be in charging state and also the relay 1 wil get the activated and the LED module will iluminate by connecting COMMON to the NO of the relay 1. suppose if AC power failure occurs means , the COMMON and the NC get connected in relay 1. The COMMON and the NO get connected in the relay 2. So the LED module will start iluminating...
One doubt, At what state ,the COMMON will be connected to NO and NC of relay 2 ??
As u said, I added a transistor along with resistor and capasitor in the left side of the circuit. check whether I placed that correctly in order..
And Do a diode is need at relay 1 and current limiter??
And check whether anything is needed in the circuit...
Thank you...
Swagatam said…
hi, it looks OK to me, you can connect a diode across the first relay for safeguarding the smps electronics.

when relay is powered or energized it's common will be at N/O and vice versa.
Ajish said…
Sir can you mention the model name of the mosfet?
Sriram Kp said…
Is the arrangement is correct in my left side of the circuit for transistor part??? I.e From adapter +ve to capasitor, capacitor to resistor, resistor to base of the transistor... Just need a clarification...
Swagatam said…
yes it's correct, use 1k instead of 10k.
Sriram Kp said…
One doubt, Is preset and potentiometer are same or different??
Swagatam said…
preset is adjusted with screwdriver, pot with a shaft/knob rotation.
Swagatam said…
...the functions are the same.
Sriram Kp said…
For a 12v relay, If I supply 13.5v across the coil means will this create any heat or problem in the relay??
Swagatam said…
very slight proportionate heat will be generated, but won't cause any problem.
Sriram Kp said…
Oh. can i use resistor in between the 13.5v power supply and 12v relay to supply a 12v to relay?? If possible means what value resistor I can put??
Swagatam said…
although not required, you can try putting a 10 ohm resistor in series.
Sriram Kp said…
I am having 12v, 7.5A lead acid battery. How much volt need to charge the battery??? 12v or 13.5v??
Swagatam said…
14.3V to be precise at AH/10 current rate.
Sriram Kp said…
I am not able to get a 14v adapter in the market. 15v smps adapter is available. So shall I connect 15v to charge the battery and adjust the 10k preset to activate the relay at 14.3v??
Sriram Kp said…
Or is there any possibility to reduce 15v ,1A to 14v 1A using zener diode or something else??
Swagatam said…
use a 1N5402 diode in series with the positive of the 15V, it will drop to 14.4V
Sriram Kp said…
Anode of the diode should be connected to positive of the 15v and negative of the diode to battery. Am I correct??
And If I want to futher reduce the the 15v to 13.8v means can I connect two diodes in series to the battery??
Check whether this is correct...
imgur.com/7Xr4SST
Swagatam said…
yes the mentioned doubts are correct.
Unknown said…
Sir give me good circuit diagram for inverter 12v to 230 v 1000w is out put give full details for doing my final project very good
Alihan Güllü said…
Hello Swagatam,

I have some questions regarding the circuits above.

1- Would it be ok if I use 1000uf/25V capacitor instead of 100uf/25V ?
2- On Sriram'a drawing, we will be using 1K resistor instead of 10K in series with the capacitor, right?
3- On the drawing, I see that LM338 adj pin is not connected anywhere, correct?
4- After LM338, I see that there is a 1,78R 1W resistor, as there is no 1,78R in local market here, I will use a parallel resistor with a 10R and 2.2R. Questions is do I need two 1W resistors or two 1/2W resistors?

Thanks in advance for your help..
Swagatam said…
Hello Alihan, the answers are as follows:

1) 1000uF/25V can be used instead of 100uF
2) 1K will work quicker so it is recommended instead of 10K
3)The ADJ is mistakenly omitted, please connect it with the LED positive (after the limiting resistor).
4) any equivalent form will work for the indicated value, so you can adjust it as per the availability and convenience.
Swagatam said…
4)....using two 1 watt resistors will be better.
Alihan Güllü said…
Thanks a lot for the answers..hopefully I will get this working soon..will update you with the results :)
Swagatam said…
sure!...wish you all the best!!
Alihan Güllü said…
Hello again...

I built the circuit mixing your and Sriram's diagram and came up with the following circuit

https://drive.google.com/file/d/0B54a7D9qm1YQT25UTTRzX0ZKZ0k/view?usp=sharing

I kept 10K and 100K presets at zero resistance in the beginning, and when I apply 14.3V through the test points at the bottom, the first relay activates as it is supposed to do.

When I increase the resistance of 10K, the second relay starts buzzing somehow. I'm thinking it might be broken, didn't try with an alternative. And the 100K preset doesn't have any effect on the relay. Here is a video how it goes:

https://drive.google.com/open?id=0B54a7D9qm1YQRzVjcHVMQS1nQlU

To apply the voltage, I used a 12V 5A LED driver with adjustable voltage. The relays are rated 12V coil - 30VDC contact - 10A - 400R.

I couldn't figure out what could be the problem, I was hoping you could give me an idea.

Also, I didn't try connecting the battery and led only, to see if the battery would light up the leds.
Swagatam said…
hello, thanks for updating your questions, your drawing looks great, however testing procedure needs some corrections.

The 100K preset is for setting the lower threshold at which the opamp output is supposed to switch OFF the transistor and the relay so that charging can resume again.

Initially the 100k preset must be kept ENTIRELY disconnected from the loop, and the 10K preset at pin#3 must be set for the higher threshold triggering of the relay.

for this to happen you must apply 14V to the circuit from the "battery" points.

then slowly shuffle the 10k preset to ensure that the transistor base LED just lights up and the relay clicks.

That's it your upper threshold is fixed now...seal the preset in this position.

now without switching off the supply reconnect the 100k preset in the loop and adjust its resistance to 0V...this will instantly latch up the opamp, and will now not allow switching off the relay even if the voltage is dropped from 14V to any lower voltage level.

Now we want to set the lower restoration threshold for the relay by adjusting the 100K preset, therefore slowly bring down the supply voltage to the desired lower level, say to 11V, or 10.5V, and very gently begin adjusting the 100K preset until the relay just switches OFF (latch breaks)

if suppose the relay is not responding would indicate that the 100K value may be too low.

In that case you can replace the 100K preset with a 220K or a 470K or may be a 1M preset.
Alihan Güllü said…
Thanks for your quick reply..as you said I removed the 100K preset and applied 14.3V on the battery points.

I arranged the 10K preset to light up the led, also switching off one of the relays and switching on the other one automatically.

BUT, when I put the 100K (also tried with 470K) preset, with being set to 0 resistance, the led and the relay switches off instantly, and doesn't do anything when I lower the voltage and mingle with its resistance. Also, when I restart and connect the preset with max resistance, the relay doesn't switch off at that moment, but a single drop of voltage switches off the relay again.

Swagatam said…
when the LED lights up itmeans that pin#6 is switched at the supply voltage level, so when this supply level is fed back to pin#3, this pin#3 will get more positive and become further higher than pin#2 voltage which will make sure that pin#6 gets latched up with its positive and the transistor becomes permanently ON and locked.

you are saying pin#6 gets zero that's impossible unless there's something incorrect in your connections.

pin#6 will continue to remain "high" as long as pin#3 is held higher than pin#2...this is the basic logic which you must check and verify while setting up

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