Hi Everyone,

This project sounds alarm when the battery voltage falls below the user defined (low) voltage .It is used in 12 volt battery (It may be any type of Battery- Lead acid ,Lithium -ion, Lithium - Polymer etc.,) and it will produce audio and visual indication to us. It can be connected directly to Inverter Battery.

It uses IC LM358 as main component,which is dual operational-amplifier,used in comparator mode, with some active and passive components.The Battery low voltage can be set by user by adjusting Variable Resistor. For instance, Lead acid battery 10.7 Volt is considered as safe low voltage. Here I am setting the voltage of 10.6 volt for demonstration purpose.

Besides ,Viewers with some basic knowledge in electronics can do this Project.

The total cost of this project would around 3 USD.

The componets used in this project are given below:

1. LM358 IC

2. 1Kilo Ohm Resistor------------------------2Nos.

3. 4.7Kilo Ohm Resistor----------------------1Nos.

3. 10Kilo Ohm Variable Resistor-----------1Nos.( I am using Multi-turn Pot)

4. 1N5234B (6.2V ) ZENER DIODE-------1No.

5. 1N4007 DIODE-----------------------------1No.

6. BC 547 (or) 548 NPN Transistor--------1no.

7. LED (Small -3mm red) -------------------1No.

8. Buzzer 12 volt small ----------------------1No.

9. Two pin screw Header -------------------1No.

10. General Purpose PCB -----------------1No.(or) use my PCB which is 5cm x 3.7cm.(approx.)

Step 1: Assembly

Picture of Assembly

The schematic circuit diagram is shown in step 1. You can use this circuit and you can make PCB circuit as your own or you can use my PCB layout readily available. In my Pcb use jumper wire at J-J. Make Print out and transfer it to copper sheet and do etching as regular process.Then pcb is ready. Make holes by drilling using 0.6 to 1.0mm bits as necessary.etc.,After choosing the PCB, solder all the components. check for any short circuit.

Note: After discussion in this forum, revised schematic, PCB,Components files were updated.

Step 2: Calibration / Setting Low Voltage

Picture of Calibration / Setting Low Voltage

Using DC-DC step down converter or variable voltage source , set the voltage to 10.59 volts. Then connect this 10.59 volt to the Battery low alarm circuit. Adjust R3 multi turn pot in Alarm circuit , clockwise or anticlockwise until buzzer beeps and LED on.Hold on R3. Then increase the votage in DC-DC step down converter or variable voltage source to 10.65volts. Now you can see the Buzzer and LED stops.

Now the Battery Low voltage is set for 10.60 volts.

Step 3: Testing

Connect the Low Battery Circuit to a variable power supply and adjust the voltage slowly upto to 10.75 volts .Here I am connecting 3cell Li-Po battery to dc-dc step down converter . The output of li-po is around 11.1v , which is reduced by step down converter to around to 10.75 volts and it is fed into the Low Battery Circuit. since the set value is 10.6 volts and battery voltage is around 10.75 volts, Alarm doesn't work .Slowly reduce the voltage to 10.61 volts and when voltage reaches 10.59 volts the buzzer and LED becomes ON.


you can use this circuit in many fields . In automotive cars, almost all vehicles up to 24 volts, Inverters,UPS Burglary Alarms, and many more fields , etc.,
Thanks for reading. Any questions/suggestions/improvements in respect of this project, you can mail to


Tomsrig (author)2017-10-16

hi, really like this project. Want to get started but you said an update will come soon. I want to make the printed board but need update. Can you post it ?

arunscuddalore (author)Tomsrig2017-10-16

The Schematic and PCB presented here is Revised /updated one.You can use the PCB to make your project.

Tomsrig (author)arunscuddalore2017-10-17

Thank you for quick response. Just one more question
Some circuits use 12 to 20 ma or so standby current. Eventually it will drain the battery down just sitting there.
Do you know what the standby current is for your circuit?

arunscuddalore (author)Tomsrig2017-10-18


As per calculation, the standby current for the above circuit is less than 2mA.

Omnivent (author)arunscuddalore2017-11-09


How so?

Current through R1: (12V-6.2V) / 1k = 5.8mA

Current through R3 12V/10k = 1.2mA

Op-Amp current: ~1mA

Total stand-by current: ~8mA

(More with noise and oscillations of the unused and untied op-amp)


Omnivent (author)2017-09-26


As your schematic stands, the circuit doesn't work

Assuming you're attempting to use D3 as a ~0.7V voltage reference, you need to turn it 180°, to pass current through it. Further, you need to swap pins 2 and 3 of the LM358 (as is, the LED and buzzer goes off when the voltage falls).

Plenty of other things can be improved in your circuit (and PCB), but due to the response I've often gotten, when trying to help people improve their circuits, I won't comment on those, unless you really want me to.

Have a nice day either way :)

arunscuddalore (author)Omnivent2017-09-27

Thanks for your comment. Have you done prototype of this circuit or you verified with any simulation software.Anyhow once again i will verify both schematic and PCB

Omnivent (author)arunscuddalore2017-09-27


The flaws of the schematic needs neither sim'ing nor prototyping to see, just a bit of common sense. I'll try to walk you through it...

To get an alarm (i.e. LED lighting and buzzer making noise), Q1 needs a positive signal on its base from pin 1 (of the LM358).

For pin 1 to get positive (enough), pin 3 (+ in) needs to see a higher voltage than pin 2 (- in).

As drawn, pin 3 sees a fraction of the battery voltage (via R3, the POT), while pin 2 is set at a voltage derived from R1 and D3. Since D3 is the wrong way around and thus blocking current flow, almost the total voltage of the battery (say 12V minus the ~0.7V of the 1N4001 and a negligible amount caused by reverse leakage in D3)

This way pin 3 cannot ever get higher than pin 2.

Turning D3 around will create a reference voltage of around 0.7V to compare the instantaneous voltage of the battery against.

However, setting the 1k trimmer potentiometer to reach below the 0.7V reference when it is going flat, only means that pin 3 (+) then goes below pin 2 - turning off the buzzer and LED, which has kept you entertained so far ;) swapping pins 2 and 3 will remedy the "alarm on Low Batt" function.

Again, this is all based on my assumption that you are using D3 for a voltage reference (I'd probably go higher than 0.7V with an
LM358 though, as it doesn't play nice around the power rails - a 2V..4V reference would probably give you less headache down the road).

Using the second op-amp, you could make a vastly improved reference voltage BTW.

The 1k pot alone draws 12mA from the battery (when it's at 12V) and if you turn D3 around, a further 11.3 mA is lost, so you might wanna up the resistor values for less battery drain.

If you add a bit of switching hysteresis, you will counter any tendency to flip on/off/on/off/etc. at the point where it switches - perhaps this is not happening in your circuit though, as the buzzer/LED increases the current draw and lowers the battery further, but it's an option if you see any "hunting" around the switching point.

If you're not using the second op-amp, you should tie its inputs to ground, to kill any chance of oscillations (which could influence the first op-amp in a bad way).

If you like, I can give you a couple of schematics (with and without the second op-amp utilized) to test.

Hope the above is comprehensible :)


arunscuddalore made it! (author)Omnivent2017-09-30

Dear Friend, Thanks for explaining more theoretically. After careful examination of my circuit ,two components were wrongly typed.

1) D3 is 1N5234 (wrongly typed as 1N4148). Since 1N5234 is a 6.2 volt Zener Diode,which has to be placed in reverse biased condition.So, no need to change the circuit.

2) R3 is 10Kilo ohm( wrongly typed as10k/100k)

To get an alarm , Q1 needs a positive signal on its base from pin 1 (of the LM358).For that, pin 1 to get positive (enough), pin 3 (+ in) needs to have a higher voltage than pin 2 (- in).

This condition is TRUE , if we are using LM358 in NON INVERTING COMPARATOR circuit. In this type when Vin is greater than Vref, output will be postive.

According to your statement swapping pins 2 and 3 i.e Vref to pin 3 and Vin to pin 2 will not work properly and leads to a reverse effect or reverse alarm.

But I have designed the circuit in which , LM358 is used in INVERTING COMPARATOR. In this type when Vin is less than Vref, output will be postive.

Here Vref is taken from R1 and D3 voltage divider junction and it is fed to non inverting pin(no.3/+).

Vin is taken from variable resistor R3 junction and it is fed to inverting pin(no.2/-) and accordingly schematic and PCB will be updated soon.

A picture enclosed here will be explaining more theoretically and the values were verified practically.

As per the datasheet of IC, it is not mandatory to tie second op-amp inputs to gnd, if we are not using the second op-amp.

Thanks for making me in depth analysis of the circuit.

Omnivent (author)arunscuddalore2017-10-01


1) D3 is 1N5234 (wrongly typed as 1N4148). Since 1N5234 is a 6.2 volt
Zener Diode,which has to be placed in reverse biased condition.So, no
need to change the circuit.

I'll have to disagree. A schematic that doesn't show the correct components used should be changed ;)

This condition is TRUE , if we are using LM358 in NON INVERTING COMPARATOR circuit.

When talking about a comparator, there's no such thing as inverting and non-inverting. There's just two inputs that is used for the comparison. (A real comparator is a better option by the way).

But I have designed the circuit in which , LM358 is used in INVERTING COMPARATOR. In this type when Vin is less than Vref, output will be postive.

Same comment.

According to your schematic, you feed pin 2 from the R1/D3 node and pin 3 from the wiper of R3 and that's what I said wouldn't work - glad you changed your mind about it :)

The need to tie unused inputs is common knowledge when dealing with high impedance inputs (analog as well as digital), while the datasheet deals with describing inputs in use, but hey, you don't have to take my word for it - I'm not forcing you to learn a thing ;)

OTOH, you could get a wider perspective here. ;)


arunscuddalore (author)Omnivent2017-10-03

Hi, Now almost everything is rectified . Go through the entire project.



Omnivent (author)arunscuddalore2017-11-09


Sorry for the delay!

Yes, your schematic is almost there, but you still need to tie the inputs of the unused op-amp.

Have a nice day :)

arunscuddalore (author)2017-10-07

Thank you for your comment

D6equj5 (author)2017-10-05

A neat and useful instructable. Thank you.

Murugan V (author)2017-09-27

Good one Arun

Ganeshkumar21 (author)2017-09-27

Good one sir

gm280 (author)2017-09-26

Could even use SMD on a lot smaller PC board to make it even smaller in size. That way you can actually install in on the back of the buzzer if you like.

arunscuddalore (author)gm2802017-09-26

It is better idea. But micro soldering station is costlier when compared to ordinary 35 watt soldering iron.Moreover skill is required for soldering SMD components, which is somewhat difficult to ordinary people.

BeachsideHank (author)2017-09-25

R3 value is missing on the PDF sxhematic.

Thanks for pointing out missing of R3 in Schematic . The value is 100 Kilo Ohm which is mentioned in the component list.Soon it will be updated in the Project.

Swansong (author)2017-09-25

Thanks for sharing :)

About This Instructable



Bio: Hi, This is ArunS,hobbiyst of Electronics,Aeronautics,Automation.
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