Simple Low Battery Indicator Circuit Using IC 741

The proposed circuit was requested by one of the avid readers of my blog. It is a low battery warning indicator circuit and can be used for monitoring a particular low battery voltage threshold.
The circuit may be understood with the following points:
The entire configuration is wired around the IC 741 and it becomes the heart of the circuit.
Basically it is configured as a comparator with one of its inputs clamped to a fixed reference level while the other input used as the sensing terminal.
Here as can be seen in the diagram, the non inverting input is provided with a fixed reference voltage through a resistor zener network.
This input is fixed to about 5 volts.
The other inverting input pin #2 is wired via a preset to sense the input supply voltage from the source.
The preset is adjusted such that the voltage level at this input becomes lower than the fixed reference voltage at the other pin of the IC as soon as the source voltage becomes lower than the desired threshold level.
When this happens the output of the IC immediately becomes high, illuminating the connected LED.
The illuminated LED immediately provides the indication of a low voltage situation so that the required actions may be initiated.
Optionally, the output may be replaced by a piezo buzzer instead of the LED for getting an audible response of the above situation, eliminating the headache of monitoring the LED condition every now and then.


The above circuit can be modified by adding a relay stage for controlling a articular stage which may be relevant to the low battery cut of actions.



The above low battery indicator circuit can be even further improved in the following manner for controling both lower and the upper charging thresholds:

Initially keep the 100K preset link disconnected.
Apply a 14.4V source across the shown "12V input" and adjust the 10K preset such that the upper relay just activates, confirm the triggering by subsequently moving the preset to-and-fro. Glue it once fixed. The LED will respond by switching ON to the fixing of this preset.
Now reconnect the 100K preset feedback link, and reduce the input supply to about 11.2V.
Next, adjust the 100K preset such that the relay just deactivates. Confirm by flipping the preset as above.
Ignore the lower relay as it will switch ON as soon as the input supply is switched ON, so its operation is obvious.

That's it, the low battery warning circuit is all set now and will accurately respond to the above settings or any different setting that may be preferred and implemented by the particular user.




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187 comments

comments
January 27, 2012 at 6:51 PM delete

hai.. can u help me.. i have created a blog comutronics.blogspot.com/ for Ece students.. how can i get it by google search

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January 28, 2012 at 9:00 AM delete

Make it big....Google won't notice your blog unless you have put 50 to 70 thousand words into it and also you need to keep it updated from time to time.

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femmy
March 19, 2012 at 5:19 PM delete

this circuit is wrong,since the pin 6 is positive,then ur resistor(2.2k) should be there before your led i.e the positive led.Can buzzer be use in place of led?

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March 19, 2012 at 5:37 PM delete

A resistor is used only for limiting the current through the LED, it does not matter whether it's at the anode or the cathode end of the LED, it works OK as long as it's in series with the LED.
A buzzer can be used in place of LED, the resistor will not be required if a buzzer is used.
Regards.

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April 25, 2012 at 4:37 PM delete

Hi,
i am wanting to make a low battery indicator, when the battery voltage comes down to 2 volts
how should i go about it?

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April 25, 2012 at 8:31 PM delete

Hi,
Please search this bog, I think I have already published one article with this topic using the IC 741

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Anonymous
May 1, 2012 at 1:35 AM delete

Hey there,

What part of the circuit is it that actually sets what voltage the LED should be lit at? I assume its just whatever the ZENER diode is rated at?

Thanks

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May 1, 2012 at 11:11 AM delete

Hi,
The output voltage to the LED will be equal to the supply voltage....the zener voltage is there to provide a reference voltage level to the IC

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Ajay
August 12, 2012 at 4:12 PM delete

Hi,

How can the above circuit be modified to work with 12v 7ah Sealed Lead Acid battery. The LED should light up when the battery voltage reaches 12.5v

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August 12, 2012 at 6:00 PM delete

Hi,
Just interchange the pin connections in the diagram, keeping rest of the things as it is.
Now apply a measured 12.5v at the +/- terminals, adjust the preset such that the LED just lights up.
Next remove the above source and connect the battery to the shown position, when its voltage reaches 12.5v, the led will instantly light up, giving the required indications.

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Ajay
August 13, 2012 at 7:23 PM delete

Hi Swagatam,
Thanks for your reply. I didn't understand which pin connections your asking me to interchange. And moreover i don't have a variable voltage source to calibrate the preset. Is there any other way to do it. Please help me.

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Ajay
August 14, 2012 at 2:53 AM delete

Hi Swagatam,
I thank you for posting this circuit. I got it working finally by allowing the battery voltage to fall till the required voltage(12.5v) and then calibrated the preset(10k). Though there exists a small issue. The LED glows faintly when the above circuit is connected to battery which is fully charged . I tried to increase the resistor value to 4k but still there is a dim light glowing. Other than that the circuit works absolutely fine and the LED glows brightly when it is 12.5v.

I also wanted to know if the above circuit can be modified to cut-off the load when the battery is low along with the LED glowing.

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August 14, 2012 at 11:48 AM delete

Hi Ajay,

You are welcome,

You can add a 3v zener diode in series with the LED, this will stop the parasitic glow of the LED.

The circuit can be modified by adding a relay driver stage in place of the LED, and the relay contacts may be wired with the battery terminal for getting the required cut-off feature.

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Sandip
August 15, 2012 at 4:45 PM delete

Dear Swagatam,

Can the above circuit be modified to place a piezoelectric buzzer after the LED?

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August 15, 2012 at 6:41 PM delete

Dear Sandip,

Yes, just replace the led resistor with the buzzer....

be sure to connect a 3 v zener in series with them.

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Sandip
August 15, 2012 at 11:57 PM delete

Thank you for your reply. I tried it works great.

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Ajay
August 21, 2012 at 4:28 PM delete

HI Swagatam,

I wanted to your help in building an automatic charger for my 12v battery. So i have a rough idea of building it like this:

The output of op-amp in the above circuit is connected to the base of npn-transistor, where the collector and emitter of the transistor are connected to Output & Adjust pin of LM317 voltage regulator.

I adjust the preset in the above circuit such that the output of op-amp is high when the voltage of battery rises to 12.5v.

So when the battery voltage rises to 12.5v, the op-amp output is high and it makes the npn transistor to conduct. Thus LM317's output and adjust pin are grounded and the charging process is cut-off.

As soon as the voltage of battery falls below 12.5v the op-amp output is low and the transistor is off. This will result in switching on the charging process.

Will this work? Please help me out.

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August 21, 2012 at 9:58 PM delete

Hi Ajay,
This concept has been covered in the following article:

http://homemadecircuitsandschematics.blogspot.in/2012/02/how-to-build-automatic-6-volt-12-volt.html

However this concept cannot be very effective because the voltage tends to restore within a fraction of second after switching off the output, due to the absence if the required instantaneous triggering, and the voltage instead of cutting of completely, switches rapidly regulating the voltage upto a certain point but never cutting the voltage completely.

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Ajay
August 22, 2012 at 1:24 AM delete

Thanks for the info and the link Swagatam.

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Ajay
August 22, 2012 at 4:34 PM delete

Hi Swagatam,

Can you please suggest a better circuit as you are saying that the above concept can't be very effective.

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August 22, 2012 at 6:18 PM delete

Hi Ajay,

May be this circuit can be used in conjunction with a LM338 IC for complete cut off:

http://homemadecircuitsandschematics.blogspot.in/2011/12/high-current-10-to-20-amp-automatic.html

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Anonymous
September 8, 2012 at 3:39 PM delete

what is a 10k preset?please post a list of equipment and what battery to use?(battery spec.)

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September 8, 2012 at 4:22 PM delete

10k preset is a variable resistor, the circuit will monitor all types of battery voltages....

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Anonymous
September 9, 2012 at 9:05 AM delete

Can you post the link of your own version of low battery indicator using and ic741,I've tried finding but it did not show up,Thanks in advance!

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September 9, 2012 at 10:49 AM delete

The above circuit is the only way of using a 741 for low battery indications...

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Anonymous
September 9, 2012 at 10:06 PM delete

Hey swagatam----1kohm,2kohm resistors,a 10k potentiometer,an led,a 741 ic,a 4volt zener diode,a 9 or 12 volt battery...are these all we need or is there anything else? Thanks.

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Anonymous
September 9, 2012 at 10:16 PM delete

Thanks for the reply Swagatam, After fixing all the connections what to do...I'm a novice,I would like you to explain about its working in basic language Thanks a lot.

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September 10, 2012 at 9:20 AM delete

That's all you would need, the battery to be monitored is connected at the +/- points.

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September 10, 2012 at 9:22 AM delete

I have explained everything in the article in simple words, pin#2 is the inverting input and pin#3 is the non inverting input.....
Please read the article once again.

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Anonymous
September 10, 2012 at 9:31 PM delete

hi swagatam,I bought all the above instruments and tried connecting the circuit but i feel its too complicated,....the battery's getting heated but the LED ain't glowing...:(((

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September 11, 2012 at 9:56 AM delete

The shown design is a standard configuration, and will surely work if everything is done correctly as per the diagram.

The battery will become hot if there's a short circuit, meaning something is wrong with the wiring.

LED will light up as per the setting of the preset at the threshold levels only.

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Anonymous
September 11, 2012 at 7:43 PM delete

Thanks man,Works like a charm,so the basic principle is that
The lighting of the LED depends on the battery voltage which is adjusted by a variable resistor eh?

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Anonymous
September 17, 2012 at 9:13 PM delete

Hi there,
Will this circuit work in a 24V setup (2 X 12V batteries in series) if i apply the following changes:
1. Use a 12 V Zener Diode as a reference at the positive input.
2. Use a 1k pot at around the 50% level to the negative of the OP AMP.

I would require it to be a LV indicator for a 24V system setup.

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September 17, 2012 at 9:50 PM delete

Hi,

I am afraid the shown design cannot support 24V, because the IC has a tolerance of a max 16V across its pins 4 and 7....the circuit will require major modifications for working with 24v supply.

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Anonymous
September 18, 2012 at 12:06 AM delete

I was looking into placing a voltage regulator at the IC's input (pin 7)...could this be used to reduce the input Vcc

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September 18, 2012 at 7:50 PM delete

Yes that will do, also make the zener resistor to 10K and add a 1K resistor across pin#2 and ground.

The zener value can be kept as it is in the diagram.

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Anonymous
September 18, 2012 at 11:13 PM delete

The LED at the output was rated at 30mA (2V). I used a 1k pot at the input to pin 2, Zener resistor at 10k at first but the simulation did not light the LED. I changed it to a 1k( and 500ohms) and it worked like a charm. The voltage regulator indeed was the key to the use of the OP AMP at this voltage!

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Anonymous
November 11, 2012 at 7:43 AM delete

Hi Swagatam,

for the relay version? what is the value of resistor connected to LED and what type of transistor do i use?

Regards,

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November 11, 2012 at 9:21 AM delete

resistor can be a 10K 1/4 watt, and the transistor BC547...

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Anonymous
November 13, 2012 at 8:49 PM delete

Hi,

can i integrate this circuit to this blog? http://homemadecircuitsandschematics.blogspot.com/2012/11/automatic-dual-battery-changeover-relay.html

Regards,

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November 20, 2012 at 7:29 PM delete

how is indicator different from voltage regulator???

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November 20, 2012 at 8:03 PM delete

An indicator will only indicate while a regulator will control the voltage as well.....

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Anonymous
December 6, 2012 at 11:57 PM delete

Hi.
Can you please tell me what do I have to do , to make this circuit works with 1 lithium shell battery (3.7v)
Thanks

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December 7, 2012 at 8:57 AM delete

Hi,

Just replace the zener with two 1n4148 diodes in series, cathode toward ground. No further modifications would be required.

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Anonymous
January 18, 2013 at 7:53 AM delete

dear sir , how can i modify as over voltage protection for battery

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January 18, 2013 at 9:57 AM delete

just interchange the input pin numbers of the IC.

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Anonymous
January 18, 2013 at 10:16 AM delete

please explain sir,

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January 18, 2013 at 1:46 PM delete

put pin#2 at zener, and pin#3 at 10K preset.

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Anonymous
January 18, 2013 at 3:14 PM delete

ok sir i will try and tell u later . thanks

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Anonymous
January 21, 2013 at 11:46 PM delete

dear sir , i am bala, as per your suggestion, i did the circuit .it is working fine ,please tell me how can i use for 21 volts ?. bcos iam planing to use for solar panel change over. panel is giving 21 volts in the peek time.

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January 22, 2013 at 11:18 AM delete

for over 20 volts, change the 741 opamp, replace with one of the LM324 opamps.

LM324 has 4 opamps, use any one of them.

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Anonymous
January 23, 2013 at 2:54 PM delete

sir, how can i construct 12v inverter with at least 500watt without transformer.

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Anonymous
January 23, 2013 at 3:02 PM delete

sir, help me i want construct 500watt inverter without transformer, what do i do?

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January 23, 2013 at 9:39 PM delete

you can try the following circuit:

http://homemadecircuitsandschematics.blogspot.in/2012/05/how-to-make-transformerless-inverter.html

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Anonymous
January 29, 2013 at 11:24 PM delete

dear sir ,
i did the circuit for automatic cutoff for solar panel charging and A.C mains charging to a battery connected inverter.my idea is if 15 volts comes to this ciruit from the solar panel the relay will on and cut the A.c mains charging to the battery in the day time . so we can avoide parellel charging . but the problem is the realy is not on .i used 200 ohms relay to cut the A.C mains. but lm 3914 I.C 10 th pin led is glowing when i gave 15 volts dc to circuit .the relay is not operated. please help me to solve the problem

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January 30, 2013 at 9:41 AM delete

It will be difficult for me to check the fault without actually seeing the circuit board or the diagram.

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January 31, 2013 at 9:50 PM delete

it should work actually, not sure what's wrong with your circuit.

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Anonymous
February 11, 2013 at 11:16 PM delete

dear sir ,
iam bala. i did the above relay circuit without any changes .working well. but little "dick" "dick" sound is come from the relay when it is going to disconnect at 11.1 volt.i am using 12 volts d.c 250 ohms relay. please help me to solve this problem.

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February 12, 2013 at 11:15 AM delete

Connect a 1N4148 diode across pin3 and pin6 of the IC.

anode to pin3....cathode to pin6

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Anonymous
February 12, 2013 at 2:03 PM delete

ok sir,i will do it and give you feedback. thanks.

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Anonymous
February 14, 2013 at 11:39 AM delete

sorry sir. After connected the 1N4148 the cut off voltage comes to 9 volts( the relay on at 9 volts). not able to set 11.1 in my preset (10 k). also doubted with my relay and changed . not able to set at 11.1 volts. please help me.

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February 14, 2013 at 12:07 PM delete

Replace the diode with a 10K resistor and check...

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Anonymous
February 14, 2013 at 2:20 PM delete

u mean the 4v7 zenar diode ?

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Anonymous
February 15, 2013 at 1:57 AM delete

dear sir , connected 10 resistor to the 1N4148 diode. the circuit is working perfectly.now preset set to the relay operation at 11.1 volts. thanks for your valuable co-operation.

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February 16, 2013 at 11:14 PM delete

Hi,
i want to make 54V low battery indicator, when the battery voltage comes down to 46 volts
how should i go about it?

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February 17, 2013 at 11:36 AM delete

Hi, this circuit can be used for upto 24V application only, not above that...

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February 22, 2013 at 1:55 PM delete

SwagatamAugust 12, 2012 at 5:30 AM

Hi,
Just interchange the pin connections in the diagram, keeping rest of the things as it is.
Now apply a measured 12.5v at the +/- terminals, adjust the preset such that the LED just lights up.
Next remove the above source and connect the battery to the shown position, when its voltage reaches 12.5v, the led will instantly light up, giving the required indications.

Sir, in respect to this explanation Can you tell me which pin have to be interchanged (is it that the preset goes to 3 pin and the zener goes to pin 2) and should the zener also be changed for 12v battery applications

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February 22, 2013 at 9:40 PM delete

Yes, pin#3 will go to the preset and pin#2 to zener for making it a high voltage cut off circuit. zener value will not change.

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Anonymous
February 28, 2013 at 8:11 PM delete

can you tell me the name of the box after the pin#6

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February 28, 2013 at 9:28 PM delete

Instead of the LED or piezo can this be used to switch a 12V relay off so the battery wont fully discharge?

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March 1, 2013 at 6:22 PM delete

Hi Mike, see the second circuit, it has a relay.

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March 1, 2013 at 8:16 PM delete

What is that red thing on pin 6 of the 741? Is that a resistor and LED?

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March 2, 2013 at 8:01 AM delete

yes it's a resistor and a LED, 10K resistor and a red 5mm led.

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March 2, 2013 at 9:23 AM delete

Another question is say I want the load to be switch off the battery when the battery level reaches 11.9V would I be changing the zener or the 10K preset?

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March 2, 2013 at 8:52 PM delete

everything is very flexible with such comparators, you may adjust any of the two.

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March 7, 2013 at 7:21 PM delete

Iam a little confused please, help, I have constructed the circuit (1) without the relay using all the parts according to your specifications, and I have two batteries 10.9v and 11.7v, I used the 11.7v to adjust the preset to set the red LED glowing, now when I connect to a battery of 12.6v it still glows and when I connect to the battery rated at 10.9v it does not glow, is it correct what i've done, or am I wrong some where, should the circuit lamp glow only at 11.7v and low, or does it go off after 11.7v please explain as I am trying to build a Low battery indicator.
Thanks in advance

Prem Antony

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March 8, 2013 at 9:35 AM delete

the preset should be set such that the voltage at pin#2 gets just a little lower than the reference voltage at pin#3. This will make the output of the IC high, illuminating the LED.

Now if the battery voltage increases, the set pin#2 voltage will also increase, and will tend to get higher than pin#3 voltage, this will instantly make the output of the IC to become low, switching OFF the LED.

So I think you might have wrongly adjusted the preset, adjust it so that the LED just lights up at 11.7V, and stops glowing even if the voltage increases marginally to 11.8 or 11.9 V and above.

the LED should stay illuminated at voltages which are lower than 11.7V in your case, and vice versa.

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March 8, 2013 at 10:18 AM delete

Thank you for your reply sir, I used 741 first and it did not work, may be there was some problem with the IC so I tried with LM324, I have connected Pin 4 to +, pin 11 to Negative, Pin 1 output to RED lED through a 2K2 resistor 5v zener and 10K resistor from ground and + respectively to PIN 2 and PReset to PIN 3, Please help me to adjust it to work for low voltage cut off.

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March 8, 2013 at 1:49 PM delete

apply 11.7V and adjust the preset so that the LED just lights up at 11.7V, and stops glowing even if the supply voltage is increased marginally to 11.8 or 11.9 V and above.

below 11.7, the led should continue to remain illuminated.

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March 8, 2013 at 10:19 PM delete

Thank you sir, one more doubt please tell me the output of the IC through a resistor goes to the cathode or anode of the LED, I've connected the anode to the output and cathode to ground, am I correct or wrong sir.

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March 12, 2013 at 12:29 AM delete

Hi Sir, I want to use this cutoff circuit for 6v 4.5ah SLA Battery and the cutoff voltage will be 5.7v so should i use:
1) 5v Relay or 6v Relay
2) secondly can 5.1v .5w zener can be used instead of 4v7 zener

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March 12, 2013 at 10:56 AM delete

Hi Bashir,

1) use a 5v relay.
2) zener should be much lower than the supply voltage, use a 3 or 4.7 volts zener

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Anonymous
March 13, 2013 at 11:13 PM delete

I need to measure 12v battery voltage using pic16f877 and display in lcd...if the charge in battery decreases it should get automatically charged from external battery ..while charging i should measure power in it and display..for that i need current(I) of the battery to measure power..can you provide the circuit for measuring current...

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March 14, 2013 at 10:40 AM delete

current can be measured by an ammeter simply, but there should be some load connected to battery....or may be i did not understand your question properly.

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March 27, 2013 at 10:03 PM delete

plz tell how to modify this circuit and turn it into a over voltage cutoff ,
basically the opposite of low voltage cutoff

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March 28, 2013 at 10:36 AM delete

in the last circuit, adjust the preset so that the relay just activates at around the high cut off threshold.....adjust the 47K resistor value so that the relay just deactivates at the lower cut off threshold.

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April 8, 2013 at 7:43 PM delete

Hi, i had completed the 3rd design and fixed it for the invertor for battery low voltage cut-off. Initally the power supply for this circuit was taken frm the inverter board, at times this circuit when on inverter mode it terminates the power even when the battery is at full charge or when switching off the inverter o/p load. After then powersupply to this cut-off circuit was taken directly frm the battery terminals now the issue is reduced but rarely happens. kindly advice

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April 8, 2013 at 7:58 PM delete

It might be happening due to instantaneous transients in the power line, try adding a capacitor (10uF/25V) across pin#2 and negative, and across pin#6 and positive, this might help to control the stray disturbances in the power line.

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April 8, 2013 at 8:44 PM delete

Thank u......i'll try this and give a feedback

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April 12, 2013 at 7:07 AM delete

this worked out...........but a drawback is there like, if we disconnect and reconnect the supply to this circuit the status of the o/p is changed it does not remain the same as earlier

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Anonymous
May 21, 2013 at 4:55 PM delete

Hi..I want to make low battery voltage indicator which can show the voltage level from 3v to 24v. Can it be possible to make the same with this ckt??? what changes would be required to make this ckt?

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May 22, 2013 at 10:35 AM delete

yes the above circuit can be used, but the IC741 opamp will need to be replaced with opamp LM324 (use any one out of the four available in the IC)

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Anonymous
May 27, 2013 at 8:34 PM delete

Hi Sir..I have few doubt
LM324 is 14 pin IC having 4 inputs .
Can these i/p be given to different voltage levels like 3V,9V,12V,24V?

I want to select one i/p at a time what should be done for that?

Also in low battery indicator ckt if IC replaces with LM324 then what about value of other components like resistor,zener diode etc do we need to change their value if yes then what will be value of components that to be sustain upto 24V level?

i want to show the indication in different leds like green for live battery,yellow for medium and red for dead.for this is their need to change the circuit ?If yes then what??

please do reply...

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Anonymous
May 31, 2013 at 8:02 PM delete

Hi. Do you know how costly in terms of power consumption this probe is? I would like to spend at most 10 mA for such functionality.

I will use it in a solar energy system. I do not need a LED, but will use a relay to inform the controller about low voltage condition.

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June 1, 2013 at 2:08 PM delete

With the relay switched ON, the power can be no less than 40mA (maximum)

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Anonymous
June 28, 2013 at 8:06 PM delete

Ecellent!!!
So you are connecting the relays in series and both have to be charged in order for the load to function. Am I right?

Does this also mean that in case of low battery level, both relay coils are not energized (i.e for minimum consumption in case the battery is very discharged)?

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June 29, 2013 at 9:00 AM delete

Thanks!
The relay contacts are in series, and the lower relay is wired in such a way that the load gets power from the charger DC source as long as it's present, and reverts to battery power during the absence of the source voltage.
However the battery power is available to the lower relay only if its fully charged otherwise the load gets no power.

The charging source can be a solar panel voltage or some other renewable option.

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Anonymous
June 29, 2013 at 5:31 PM delete

I see...

I don't really mind the upper threshold thing but the rest of the circuit is perfect! So would you be kind enough to upload the previous version of the third circuit again please?

I also wonder if there can be a hysteresis implemented in this circuit, preferably adjustable.

Lastly, is there a particular reason you are using an opamp instead of a voltage comparator (i.e why not use LM393 instead of LM324)? I know it is not wrong, but I trust you might have a reason for your selection.

Thanks again,
Pete

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June 30, 2013 at 11:25 AM delete

The present circuit can be also used for just monitoring the lower threshold, simply eliminate the second relay and the associated connections and wire the N/O contact of the upper relay with the load, I am assuming that you want to disconnect the load from the battery when the voltage reaches the lower threshold.

Apply a sample lower threshold voltage to the circuit and adjust the 10k preset such that the relay just deactivates at this voltage, while doing this keep the 100k preset link disconnected, once the setting is done you may connect it back.

The 100k preset link will provide the required hysteresis which can be adjusted by adjusting its value.

LM324 is rated at 30V which allows it to be used with 24V supplies, that's why I preferred this IC.

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Anonymous
June 30, 2013 at 7:03 PM delete

So, if I need only the lower threshold version, I can use the second diagram and just a 10K fixed and 100K variable between pins 3 and 6 for the hysteresis. Right?

Thanks again,
Pete

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Anonymous
June 30, 2013 at 7:09 PM delete

Also, assume that pins 2 and 3 are interchanged for use as a high voltage indicator. Can u still add hysteresis in this case which will be a little lower than the high threshold voltage?

This blog does amazing work mate, congratulations for running it!

Pete.

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Anonymous
July 1, 2013 at 2:44 AM delete

Hi I need to build an LED low battery indicator for a 7.2v battery. The low battery indication would be around 5 volts.
How would I modify the circuit? would it work for my application.

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July 1, 2013 at 1:38 PM delete

Interchanging the pins would make things complicated because we have the hysteresis loop involved....instead of thinking much, very simply the same circuit which is explained above can be used for the high voltage cut off also, just by pushing the preset setting at the required higher threshold, so when the battery voltage reaches say a set 14.3V mark, the relay actuates and cuts off the external power to the circuit and the battery...after this the circuit monitors the battery voltage and restores the relay back by switching it off once the battery voltage falls below the set lower threshold, determined by the hysteresis loop preset.

You are most welcome....we all are enjoying your interesting feedbacks.

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Anonymous
July 1, 2013 at 2:05 PM delete

Just when I thought I had it all figured out, I am confused again!

Let me put down all I need to do and see if it can be done:
I need the circuit to keep the relay off (deenergized) for voltages bellow 22V and turn it back on at 24V.

Do you think it can be done with this circuit (a little modified maybe)?

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July 1, 2013 at 7:28 PM delete

yes you can achieve the results using the discussed circuit, 10k preset is for the high level (24V) and the 100k preset is for the lower level 22v.

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July 1, 2013 at 7:37 PM delete

make the first circuit, don't modify anything, use it as is.

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Anonymous
July 1, 2013 at 11:19 PM delete

Dear Swagatam,

Your assistance during the past few days has been incredible, thanks to which I think I have managed to get close to the circuit I need.

Having combined and the above info, a selector circuit I found in a different source and by adding my own ideas I came up with the following circuit:

http://domoarchitecture.eu/scan.jpg

I know there are issues which with my knowledge I cannot solve, but I think I am close (I will list these issues at the end of this post).

My knowledge is limited (I am neither an expert nor a novice in electronics), so I would like to hear your opinion and recommendations.

To my understanding, what I expect it to do is as follows:

LM324 will monitor the voltage and turn off the relay at the bottom of four scales: 18.5-20V, 20-22V, 22-24V, 24-28V adjusted by the various 10K presets.

The circuit starts (battery full) with the relay armed (load connected) and the 24-28V LED on (left hand side set of LEDs). The selector circuit (top part) assures that only the selected OPAMP arms the relay although all of the OPAMP outputs will be high at the same time (this I need for a later function, see end of next paragraph).

During discharge, when 24V are reached the equivalent output (1) will go low and the relay will disarm, disconnecting the load. At the same time BC337 (NPN) will conduct and light up the three right hand side LEDs, to indicate scale options that are available.

Once a button on the selector is pressed for a lower scale (i.e 22-24V), the relay at the lower circuit will arm by the second OPAMP (output 7) and the load will be connected. Same goes for the equivalent left hand side LED. The same applies for the 20-22V scale, but only 2 LEDs will light from the right hand side set).

Now, if the last scale (18-20V) is selected, when 18V are reached the 4th OPAMP of the LM324 will disconnect the load once again but at the same time the SPDT relay at the top left of my picture which is connected as a latching relay will disarm and take power completely off of the circuit to minimize consumption as the battery will be very deeply discharged now. In order to start the system again, manual reset will be required via the push button near the latching relay, once the battery is charged again.

The purpose of this circuit is for a solar system used to power an emergency communications station. This is the reason it cannot be set in a completely automatic way. Although deep discharging of the batteries is not recommended, in such stations the operator will have to judge whether he must deep discharge the battery or not, in case the type of emergency makes it inevitable. Under normal situations, the power will cut off at 24V automatically protecting the batteries, but it is important that the operator will have the option to continue drowning the batteries if necessary.

So here are my issues:
1. I have a fear that in the way the circuit works, the outputs go high, not low when thresholds are met. If it is so the circuit will not operate in the intended way, especially the latch relay function.

2. I have a feeling that there will be a lot of small and big flaws in my additions to the design. I am confident in building it when it is all designed, but designing of circuits is a skill I only have to certain extend.

3. I wonder if I can substitute the relays of the top part of the circuit with some kind of low power MOSFET. If so, please give me a clue as to how to connect them.

4. I am also thinking that the disconnection of the load will cause the system to do some chasing effect, so I wonder if some small hysteresis can be added.

Please get back to me with you thoughts; I believe this can be a useful circuit if ever completed for many people (maybe by adjusting the thresholds to higher values, not everybody needs to drain their batteries so much).

My gratitude once again for the tones of information you have helped me to collect during the past week.

Regards,
Pete

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July 2, 2013 at 1:58 PM delete

Hi Pete,

Please give me some time to think, because the circuit is quite elaborate and a bit puzzling too, I'll revert once I grasp all the details about its functioning.

Regards.

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Anonymous
July 2, 2013 at 3:04 PM delete

Sure my friend, I hope my long description gives you a good clue. Please get back to me if you need more info...

Good day,
Pete

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July 3, 2013 at 11:16 AM delete

Hi Pete,

Please refer to the following article, I have presented a simplified version of your circuit at the end of the page, I think it will do the job, but not entirely sure:

http://homemadecircuitsandschematics.blogspot.in/2013/07/selectable-4-step-low-voltage-battery.html

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July 9, 2013 at 7:02 PM delete

I tried the circuit by using LM741. It is working Perfectly thanks Swagatham

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Anonymous
November 21, 2013 at 1:53 PM delete

hello i had constructed your second circuit http://homemadecircuitsandschematics.blogspot.in/2011/12/highly-accurate-mains-high-and-low.htm
there doesnot occur any cutoff either in higher range or in lower range ...what problem might had occur......would you plz solve my problem
mine lower voltage level is 170 and higher is 245

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November 21, 2013 at 7:50 PM delete

connect LEDs at the outputs of A1 and A2 and ground with 1k separate resistors.

Now apply 170V and adjust P2 such that A2 LED just lights up.
Next apply 245V and adjust P1 such tha A1 LED just lights up.

Your circuit is set now, the relay connected at the collector of the transistor will also respond correspondingly.

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November 24, 2013 at 6:30 AM delete

Swagatam, sir,

I think I've cracked it, Instead of the 10k preset, I used a voltage divider: a 12k in series with a 27K. From between these two, I was able to obtain a voltage of 3.86V at Pin 2 - Pin 3 had 4.49 VDC - and, which triggered the transistor via Pin 6 and tripped the relay, and the lights, connected to the normally open contacts, cut out.

Now to assemble all three circuits.

Stephen Gard

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November 25, 2013 at 9:57 AM delete

OK Stephan, that's great, no issues, if you have solved it you can go ahead with it.....

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November 25, 2013 at 10:15 AM delete

(Reposted without links)

Swagatam,

Thank your for your well-conducted and informative site. I have learned much from it. I am trying to adapt your 741 Low Battery Cut-Off circuit for use with a 6VDC lighting set-up, and I am having some difficulty. I hope you can advise me. Here is what I wish to do.

1. I have four high-power white LED spotlights. They each require 3.2VDC @ 34mA.The four lights, wired in parallel, draw 82mA at 3.2 VDC. So that is the load.

2. The supply is a single 6.5VDC SLA battery. I attach a 33 Ohm 1 Watt resistor in series with the negative supply to the 4 x LEDS, as a current-limiter. I wish to use ONE battery only for the entire circuit, no separate supplies. It will be installed in a friend's garden and not get much maintanence! So it must be fully 'automatic'.

3. I have a solar-panel (5W 17 volt) that charges the battery and a darkness detector. Now I wish place your 741 Low Charge Cut-Off circuit (with relay) in between these two circuits, to prevent the 6.VDC SLA from discharging too far. The solar charger supplies only a small amount of current, and there will be no-sun days, of course.

If I understand the operation of your circuit, when the battery voltage drops, the 741, via Pin 6, allows current to flow through Q1 (the BC547), tripping the relay coil. My 4 x LED load is wired to the NORMALLY-CLOSED contacts of the relay (NEC EA2-4.5VDC), and when the coil is energised via Q2, they disengage the load by switching over to the NORMALLY-OPEN contact. So no current flows, and the 4 x LEDS go off.

That is what I wish to do, but I cannot make your circuit do this. The only change I made to your circuit was to the 10K pre-set pot. My pre-set over-heated. I measured the R, and replaced it with a 12k ohm 1 watt fixed resistor


b. This arrangement delivers 5.65V to pin 2 of the 741. Pin 3 is set at 4.52 V, from the junction between the 4.7V Zener and the 1K resistor connected to the +ve terminal of the battery.

c. To test, I connected one side of a diferent white LED load to the NC pins of the relay, and the other side to the battery ground. This load draws 2.8VDC, at 22mA.

d. I connect a tired old dry-cell battery to this circuit, and on my DMM, I watch the supply voltage drop.

4V99. 4V96. If I insert the DMM at Pin 2 and try to take a reading, the voltage goes down slowly.

e. So I connect a 2MegOhm pot in between the 12K and Pin 2, and bring the voltage down.

But now I notice that the voltage on Pin 3 is droppng as well. For example, from 4.351 to 4.348V.

It seems that the two pins will never achieve the diffential that will trip the relay.

When I connected the fixed resistor, I did not take it to ground, as the original circuit did, after the 10K preset. If I connect this fixed resistor to ground, Pin 2 goes low and the relay trips, but I don't think that is correct.

Can you please advise what I am doing wrong here?

Sorry for such a long letter!

Stephen Gard






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November 25, 2013 at 11:30 AM delete

Swagatam,

I am still testing to get exact values for my requirements. I am also drawing up the whole device in TinyCAD, and can send you a png of the completed three-part circuit, with test-points and voltages marked, if you wish.

Stephen Gard

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November 25, 2013 at 12:58 PM delete

Yes, why not Stephan, I'll be extremely happy to publish your efforts in my blog so that other can also benefit from it.

I appreciate your positive thoughts...kindly do send them.

And whenever you do something new in electronics, please make sure you share them here for us.

We all will love to see them.

Let me know if you need any further assistance?

Thanks very much.

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December 23, 2013 at 3:07 PM delete

Swagatam,

I added this circuit to my solar lighting set-up, but there is a problem. The relay must be activated to cut off the current to the LEDs. But this drains the battery even further. Finally, the battery is drained so much that it cannot activate the relay, which then returns to the NC condition, powering the LEDs, and draining the battery even further. After one dull day, when the solar panel could not recharge the battery, the lights do not work at all, because the battery was drained so low. Perhaps a timer would be better? I am running the whole set-up from just one 6VDC SLA battery, and don't want to add any other power sources. What do you suggest?

Stephen Gard

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December 24, 2013 at 1:42 PM delete

Stephen, the solution is to simply swap the pins of the IC, meaning connect the preset pin to pin#3 and the zener to pin#2.
In the above situation the load must be connected via the N/O contacts, so the LEDs stay illuminated as long as the relay is activated (voltage normal condition), as soon as the voltage drops below the lower threshold the relay and the LEDs all shut off without causing any further discharging issues.

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December 28, 2013 at 7:47 AM delete

Thank you, Swagatam, I carried out this mod, and now it all seems to working as I require, The set-up is currently undergoing a long-term (1 week or so) outdoor test, and after this, I will construct the final (weather sealed) installation and send you a schematic and some photos, as promised. I note that the circuits (there are three) still draw 12mA, even when the LED lights are off, but perhaps the solar-panel will keep ahead of this. The weather for the next week includes sunny and cloudy periods, so my solar-charged set-up should have a good test period,

Stephen Gard

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December 28, 2013 at 1:20 PM delete

That's great Stephen! I will certainly love to see the pics of the finished design, and also post it here for the viewers.

The quiescent consumption should not be more than 5mA according to me, i think something may be leaking through the output of the opamp, a 3V zener diode in series with the transistor could possibly fix this issue too.

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January 9, 2014 at 3:03 PM delete

hai swagatam!!! I had connected this as your circuit.. but when the circuit cut off the output..the battery voltage increases then the relay deactivates and the output works and it again activates and cut the output ...this happens simultaneously ..how to cure it..

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January 10, 2014 at 12:53 PM delete

Hi Dharma,

Please connect a feed back resistor link from pin6 to pin3 as shown in the last diagram, this will prevent the circuit from oscillating at the threshold points.

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January 14, 2014 at 10:06 AM delete

HI
Can you please explain what is the purpose of using a bipolar transistor in the 3rd circuit?
why are we using it?

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January 19, 2014 at 9:47 AM delete

Swagatam,

I got this all installed and working, but by that time, I had learned a lot more about solar lighting, and realised that my approach to this project was wrong: all I needed was a Joule Thief! However, the experience was very helpful, and your low-voltage cut-off circuit works fine. I plan to use it for a 6VDC solar-powered water pump... wilol report back when that's done.

Thanks for your help and advice,

Stephen Gard

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January 19, 2014 at 12:07 PM delete

That's great Stephen, I congratulate you on that. Best wishes to you.

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January 20, 2014 at 11:29 AM delete

hey
can i used 3rd circuit diagram for 12 volt 7.2 to 45 AH battery? And in this circuit there is two battery supply for one is relay operate & another is for load?

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January 20, 2014 at 6:07 PM delete

Hello sir,
Plz I need clarification on ur last circuit dat has two relay. I built d circuit for my 12v inverter as charge control but d problem is i got d first conrol using 10k VR for 14.5v auto cuttoff but.for low batteery sensing I could not get what I want to deactivate at 10.5v. All I got was it deactivate at above 11v and even that I have to tuned d 100k VR toward zero ohm so what do i do to achieve low battary dectation at exactly 10.5v instead of above 11v

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January 20, 2014 at 8:02 PM delete

you can use it...there's only one battery supply that connects with the load

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January 20, 2014 at 8:12 PM delete

Hello Beelal,
try a lower value pot in place of 100k, may be a 33k pot and reduce the series 10k fixed resistor with 1k

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January 24, 2014 at 1:37 PM delete

Hello,
i see 2nd diagram 10 k preset connect @ 2 pin & in 3rd diagram 10K preset connect @ 3pin of IC. Another question is that I have UA 741C IC can i use it or there is difference IN UA 741 & LM741 please guide as i used it for 12 volt 7.2 AH battery.

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January 25, 2014 at 10:09 AM delete

Connect the pinouts as shown in the third diagram.

all types of 741 will work for this circuit

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January 27, 2014 at 10:02 PM delete

Hi Swagatam,

You have recommended this third version for my hi/low contacter.
You suggested that I don't require the second relay.
Which one do you mean and what was it's original function?

Thanks for this
Yours
Carl

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January 28, 2014 at 12:31 PM delete

Hi Carl,

The lower relay is not required for your application.

The circuit is a battery over charge/discharge controller.

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February 23, 2014 at 12:05 AM delete

Hi swagatam,
For the third circuit, I would like to know the exact values of the potentiometers, because i am only going to put resisters instead of potentiometers.my circuit will be using 12v input, 12v battery and the load current will be 3a.

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February 23, 2014 at 8:37 PM delete

Hi Anthony, the simplest way of implementing it is to first set the thresholds with pots and then measure the pot resistances to get the exact values of the fixed resistances.
The pot can be removed thereafter and replaced with the measured equivalent resistances.

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March 16, 2014 at 5:43 PM delete

can i use this circuit for 24v gel battery with 180 ah??

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March 16, 2014 at 9:40 PM delete

yes can be used..use a 324 IC for the opamp.

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May 9, 2015 at 9:38 PM delete

hai sir,
I made this circuit it is working perfect with sugar cube relay but it cant pull the coil of a 6 amp 200 ohm 12v relay,I am making a 800 watt relay,since the current has to pass through the relay I have to use a high amp rating relay,how to solve it.

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May 10, 2015 at 9:41 AM delete

Hi Vijeesh,

according to me a BC547 transistor should be able to pull a 200 ohm relay easily, try reducing the base resistor to 4k7 and check the response, or you can also try a 8050 or a 2N2222 in place of the BC547 for the same....

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June 3, 2015 at 7:08 AM delete

Swagatam Majumdar ...sr....
My ups is 40W ..
Battery is fully charged is there ..
But my ups is showing battery low why sr...

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June 3, 2015 at 2:30 PM delete

sharanabsava, check the battery voltage while the inverter is operating and without any load....if the battery volatge drops then that would indicate an incorrectly charged battery or a faulty battery...also check the current with an ammeter in series with the batt positive....if it indicates high consumption without any load then the fault could be in the inverter, not the battery

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July 18, 2015 at 1:33 AM delete

hello
i used the above circuit to shutdown my inverter when battery voltage is 9.5v but inverter goes on and off severally at the set voltage.

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July 18, 2015 at 10:15 AM delete

that can never happen....use the second circuit, remove the 100k resistor and adjust the 10k preset such that the relay just cuts off at 9.5V and activates at around 9.7V....once this is done the circuit will implement the same accurately each time the battery voltage falls to this level.

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August 24, 2015 at 11:17 PM delete

Dear Sir
Thanks for your prompt reply to my other query and referring this circuit for my purpose. Kindly help me with the following queries regarding this circuit
1. To use a relay can the LED and its resistance (as in first diagram) be simply replaced by the relay OR is it mandatory to do the other changes of 2nd diagram like addition of transistor.
If I want to use <12v as the voltage to activate the relay will the required zener diode be of 12v OR as shown in diag 4v is needed.
3. What will be time required by the relay to act. (will the relay be immediately energized as soon as voltage falls below set 12v Or will some time be required and Vice Versa when the voltage again resumed to normal).
Thanks

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August 25, 2015 at 10:49 AM delete

Dear Ankit,

the IC will not be able to handle a relay directly, therefore a transistor stage will be required as shown in the second diagram.

The zener diodes should have a much lesser value than that of the supply voltage level, so for a 12V supply any value between 3v and 6V would be OK for the zener diodes.

the relay will act almost immediately as soon as the voltage crosses the set threshold limit.

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August 25, 2015 at 10:52 AM delete

....if the 100K feedback resistor is used then the restoration will not be immediate rather as per the 100K response...lower values will produce relatively quicker restoration....as the resistor value is increased the restoration point will move backwards (will be delayed proportionately)

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August 29, 2015 at 9:13 PM delete

Dear Sir
Kindly advise if 4.7V zener diode is not available then can 3.3v or 7.5v zener diode be used.

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August 30, 2015 at 5:15 PM delete

Dear Ankit, yes any zener below 8V will be fine...

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September 4, 2015 at 8:09 AM delete

Dear Sir, I made the circuit exactly according to second schematic and it is working fine. Just one query, In this Initially at normal voltage the relay is activated and when the voltage drops below the low level then the relay gets deactivated. CAN this function the OTHER way ie at normal voltage the relay remains deactivated but when voltage drops it activates.
Further what change may be required if low voltage cut is to be set at around 12V (currently it is 5V probably)
Thanks

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September 4, 2015 at 9:39 PM delete

Dear Ankit, what you are asking is not recommended because if you allow the opposite to happen that would mean at the lower critical level the relay coil will be energized and this would lead to even further consumption of the battery which will in turn discharge the battery even faster........ this won't be a favorable situation for the circuit and the battery.


by the way you can do it by simply changing the NPN transistor with a PNP...and connect the relay diode stage across its collector and ground.

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September 5, 2015 at 6:46 AM delete

Thanks Sir. As per your suggestion of using PNP, will the ground be emitter or the base.

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September 5, 2015 at 7:33 PM delete

emitter will go to the positive....

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September 5, 2015 at 9:13 PM delete

Dear Sir, I did all the changes with pnp BC 557. But now the circuit is not working. Kindly advise

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September 6, 2015 at 5:44 PM delete

Dear Ankit, you might have done something incorrect with the connections for sure...otherwise there's no reason why it wouldn't respond....relay should be connected between collector and ground....emitter to positive supply....base to opamp resistor/LED......relay diode cathode must go to transistor collector...anode to ground

the LED polarity needs to be reversed.

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September 7, 2015 at 8:05 PM delete

Dear Sir, Thanks for the reply. After reversing the LED polarity, the modified circuit is working fine. As a final piece of advise regarding this circuit, kindly let me know the following-
1. What is the appox consumption of the circuit (on 12V battery) (when the relay is in deactivated stage), and
2. Is it OK to keep the circuit connected 24X7 or should it be switched on only intermittently (ie will continuous operation reduce the life of circuit)
Thanks

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September 7, 2015 at 9:45 PM delete

Dear Ankit,

increase the value of the perset to 100k, and same for the zener resistor, this will allow just a few mA current to be consumed by the circuit (only as long as the relay is not active)

the circuit can be kept powered ON infinitely without any harm

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September 14, 2015 at 9:11 PM delete

Dear Sir, The circuit is working as intended (using PNP so that relay is activated when voltage is low). However on borderline voltage, the relay chatters. Can something be done to avoid this eg adding delay with IC 555 or some other way. Thanks

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September 15, 2015 at 1:50 PM delete

Dear Ankit, just connect a 100uF/25V cap in parallel with the relay coil, this will take care of the issue

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September 18, 2015 at 7:20 AM delete

Dear Sir, I connected 100uF/25V cap in parallel with relay. Though this works fine when I test it at my work bench with a dc adapter, but as soon as I use the circuit in car (12V 70 Ah battery), the relay again starts chattering at borderline voltage.
I maintained the polarity of caps (+ to + of relay)
The slight modification that has been done to your original circuit (in addition to using PNP) is following
1. I am using 2 relays in parallel (as I need 2 switches and double pole relay not available) (diode put across both relays)
2. The 100uF 25 V caps have been connected to each of the relays separately
3. I have connected a buzzer in parallel to the LED (buzzer spec 3V to 30 V)
Kindly advise to prevent relay chattering
Thanks

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September 18, 2015 at 1:47 PM delete

Dear Ankit, do you have the feedback resistor in place? (between pin6 and pin3)

anyway, try increasing the value of the cap to 1000uF

or try placing the 100uF across the base and emitter of the BC557

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September 18, 2015 at 6:54 PM delete

Yes, the resistor between pin 6 and 3 is there.
Which option will be better increasing value of each cap to 1000 or placing cap between E and B

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September 19, 2015 at 8:22 PM delete

first try the E,B, option if it doesn't then you can try the 1000uF method

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October 19, 2015 at 3:26 PM delete

Great circuit , i wanna use it with Automatic Micro UPS , but what AMP does it support ? i need to use with 5 AMP Micro UPS

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October 20, 2015 at 11:29 AM delete

thanks, it will support any desired amp, 5 amp can be easily used with the shown circuits

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October 20, 2015 at 2:49 PM delete

Thank you ,
it's really good circuit , but i can not understand , in wanna use 2nd circuit with your great Auto Micro UPS circuit , but what is charging voltage input ?? is this circuit has built in charger ?!! i'm really confused !!

i wanna Auto Micro UPS to cut off the load when 12volt battery is going down and reaches its critical voltage

who to use this circuit with Auto Micro UPS ?

thanks in advanced

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October 20, 2015 at 5:45 PM delete

This diagram is correct:

https://drive.google.com/file/d/0B2VgP4YQwP4ETzczQ2Y3VExKX00/view?usp=sharing

you can use the second circuit from the above article and use it as indicated in your diagram.

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October 31, 2015 at 7:00 PM delete

Hi Swagatam,
I'm back again.... :-) .... I'm working on a digital thermometer using ICL7106 and an LCD display. The LCD has an icon of BAT for low battery indication. The main power supply is a 9V battery, reduced in 5V by a 7805 regulator. I need to have the icon BAT light on (which is terminal 40 of the LCD and was not included in the original project) whenever the battery goes below the minimum of the regulator, which is 4V. The LCD is common Anode. How can I interface this circuit with the digital thermometer and what changes do I have to do?

Best Regards.
Nélio

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November 1, 2015 at 7:20 PM delete

Hi Nelio,

you'll need to first confirm regarding what logic level pin40 is assigned for (zero or high?) After that you can use an opamp based low voltage detector circuit for activating this pin with the relevant logic level when the low voltage is detected....for converting it into a digital thermometer you'll need to use a temperature to volatge converter stage and integrate this with the voltage sensor input of the IC 7106

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November 2, 2015 at 3:01 AM delete

Hi Swagatam,
The part of the thermometer I have. All I need is to trigger the BAT icon in the LCD when the battery goes low. How to I know what is the logic level of pin 40 of the LCD?

Best Regards,
Nélio

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November 2, 2015 at 11:05 AM delete

Hi Nelio,

you can manually inject a 5V supply through a 100k resistor into this pin, and/or do the same by connecting the pin to ground via a 100k resistor....and check the response on the low batt icon...this will let you know the logic assignment for this pin.

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November 10, 2015 at 7:12 PM delete

Pls sir i have made the first circuit nd it's working fine...my question is(1)is the second diagram used for cutting the relay when voltage increases to 14.4v cos it does not seem like a low battery circuit since preset is connected to pin 3 and zener to pin 2 and i want to build this second circuit for battery full cutoff at 14.4v .(2)i want to add the first circuit to my sg3524 inverter without relay driver stage so i'm planning on connecting pin6 of the op-amp to pin10 of sg3524 so that it will shutdown my inverter at the low voltage level will it work like that?(3)what will be the effect of adding a 2.2uf 400v fan capacitor in series to the load at the output of my inverter?

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November 11, 2015 at 9:00 AM delete

victory, the second circuit has both the features, it will switch ON the relay when the battery reaches the full charge value...and will switch OFF the relay when the battery reaches the low charge level...the low charge cut-of setting is determined by the value of the feedback resistor connected across pin6 and pin3 of the IC.

yes it will work in tat way with SG3524 shut down pin....

capacitor should be added in parallel to the load, not in series...it will help to modify the square wave to a sine wave.

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February 12, 2016 at 12:51 AM delete

If the second diagram has both features then what is the extra feature of the 3rd diagram with the second relay

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February 12, 2016 at 11:35 AM delete

the 3rd diagram allows the load to access both the SMPS voltage as well as the battery voltage depending on the upper relay switching, and the battery condition.

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March 12, 2016 at 11:39 PM delete

Hello,

I have a 28v dc power supply and I am looking for a circuit to monitor voltage drop below 10v in the load. I want to have led light up when load voltage is above 10v and another led to light up if voltage drops below 10v. I also want the led to draw power from the power supply instead of the load. How can I achieve this please? Thank you.

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March 13, 2016 at 11:02 AM delete

you can use the circuit, and connect one LED from positive to pin6 and another LED from pin6 to ground....both these LED must have a 10K resistor in series.

Use LM321 IC instead of 741 since LM321 can work with upto 32V supplies

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March 13, 2016 at 11:02 AM delete

...and use 10k for the zener resistor

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