Lead Acid Battery Charger Using IC 555

The following article explains a simple, versatile automatic lead acid battery charger circuit. The circuit will allow you to charge all types of lead acid battery right from a 1 AH to a 1000 AH battery.

Using IC 555 as the Controller IC

The IC 555 is so versatile, it can be considered the  single chip solution for all circuit application needs. No doubt it's been utilized here too  for yet another useful application.

A single IC 555, a handful of passive component is all that's needed for making this outstanding, fully automatic lead acid battery charger circuit.

The proposed design will automatically sense and keep the attached battery up to date.

The battery which is required to be charged may be kept connected to the circuit permanently, the circuit will continuously monitor the charge level, if the charge level exceeds the upper threshold, the circuit will cut off the charging voltage to it, and in case the charge falls below the lower set threshold, the circuit will connect, and initiate the charging process.

How it Works

The circuit may be understood with the following points:

Here the IC 555 is configured as a comparator for comparing the battery low and high voltage conditions at pin#2 and pin#6 respectively.

As per the internal circuit arrangement, a 555 IC will make its output pin#3 high when the potential at pin#2 goes below 1/3 of supply voltage.

The above position sustains even if the voltage at pin#2 tends to drift a little higher. This happens due to the internal set hysteresis level of the IC.

However if the voltage continues to drift higher, pin#6 gets hold of the situation and the moment it senses a potential difference higher than 2/3rd of supply voltage, it instantly reverts the output from high to low at pin#3.

In the proposed lead acid battery charger circuit design, it simply means that, the presets R2 and R5 should be set such that the relay just deactivates when the battery voltage goes below say 11.3V (for 12V batts) and activates when the battery voltage reaches above 14.2V.

Nothing can be as simple as this.

The power supply section is an ordinary bridge/capacitor network.

The diode rating will depend on the charging current rate of the battery. As a rule of thumb the diode current rating should be twice that of the battery charging rate, while the battery charging rate should be 1/10th of the battery AH rating.

It implies that TR1 should be around 1/10th of the connected battery AH rating.

The relay contact rating should be also selected as per the ampere rating of  TR1.

How to set the battery cut off threshold

Initially keep the power to the circuit switched OFF.

Connect a variable power supply source across the battery points of the circuit.

Apply a voltage that may be exactly equal to the desired low voltage threshold level of the battery, then adjust R2, such that the relay just deactivates.

Next, slowly increase the voltage up to the desired higher voltage threshold of the battery, adjust R5 such that the relay just activates back.

The setting up of the circuit is now done.

Remove the external variable source, replace it with any battery which needs to be charged, connect the input of TR1 to mains, and switch ON.

Rest will be automatically taken care of, that is now the battery will start charging and will cut off when its fully charged, and also will  get connected to power automatically in case its voltage falls below the set lower voltage threshold.

IC 555 Pinouts

IC 7805 Pinout

How to Set Up the Circuit.

The setting up of the voltage thresholds for the above circuit may be done as explained below:

Initially keep the transformer power supply section at the right hand side of the circuit completely disconnected from the circuit.

Connect an external variable voltage source at the (+)/(-) battery points.

Adjust the voltage to 11.4V, and adjust the preset at pin#2 such that the relay just activates.

The above procedure sets the lower threshold operation of the battery. Seal the preset with some glue.

Now increase the voltage to about 14.4V and adjust the preset at pin#6 to just deactivate the relay from its previous state.

This will set up the higher cut off threshold of the circuit.

The charger is now all set.

You may now remove the adjustable power supply from the battery points and use the charger as explained in the above article.

Do the above procedures with lot of patience and thinking
Feedback from one of the dedicated readers of this blog:

untung suharto January 1, 2017 at 7:46 AM

Hi, you have make mistake on preset 2 and 5 not 10k but 100k, I just make one and succesfull...thank you.

As per the suggestion, the above diagram may be modified as shown below:

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


Swagatam said…
Thanks! you can use any zener between 3 and 9V
Swagatam said…
I have added a transistor for the relay, please do the given modification.
Swagatam said…
any npn transistor will do.
Swagatam said…
please explain how you did the setting up procedure?

5.1V zener should also work.
Swagatam said…
I did not test this circuit, but according to the datasheet of 555 this circuit is perfect.

provide all the details of the procedures that you have done with this circuit, then I'll try to troubleshoot.
Swagatam said…
You must do exactly as explained in the article, only then it will produce the intended results.
Rashid Ansari said…
Hello Sir,
I have a question for you. I have a 12V, 130AH lead battery and an inverter (not smps). I run a pedestal fan on it which is of 150W 220V. It runs only for about 2.5 (two and an half) hours.

According to formula W/V=A which is here 150/12= 12.5
and Time = Battery Amp/Amp of appliance which is 130/12.5= 10.5

Question is it should run for atleast 10 hours, why it is only run for about 2.5 hours ??

Hopefully you will help in detail.
Thank you.

Swagatam said…
Hello Rashid,

Your result is as per ideal conditions and nothing in this world will generate ideal results, even a new battery here will not give 10.5 hours of back up.

Therefore either your battery has become old and inefficient or it's not being charged correctly and fully. Try using a step charger circuit for charging it optimally....
Rashid Ansari said…
Thank you for your reply Sir,

In fact my battery is just brand new and I use a PWM charger to charge it. And when it becomes full, indication LED emits on its panel.

Any way thank you again for your reply.

Swagatam said…
You are welcome Rashid! How can you be sure that your battery is getting 95% charged?

Only a step charger circuit ensures that much possibility.
Rashid Ansari said…
Dear Sir,

Yes you are right, i am not sure about the battery either it is fully charged or not. I only assume that by the indication LED on the charge controller.

Sir, can you explain what is the step charger, how it works. And if possible can we have it's circuit diagram.

Thank you.

Swagatam said…
Dear Rashid,

You can go through one example circuit given below, however it is not meant for high current batteries, I will hopefully try to produce a similar circuit for higher current applications soon:

Swagatam said…
yes you can do it.
gurpreet said…
Sir can i use mixture of 1/2 and 1/4 watt resistance. And can i charge 12v computer ups battery using this circuit .
Swagatam said…
yes will do.
you can charge any type of battery with this circuit.
Swagatam said…
I have no idea regarding microtek operating specs so can't say much about it, but anyway it's never recommended to connect two batteries from a source specified for a single battery.
Hello swagatam,
I tried this circuit with the present modification, but it didn't work. i used 7.5v zener instead of 6. still the circuit activates at around 7.6v. i again removed bc547, c1, c2 and used a 12v zener still the same, i.e., it activated at around 7.6v regardless of variation in pot. please help. i used a variable dc power source controlled by lm317.

one more que, disconnecting transformer section means to remove only transformer or along with bridge rectifier????????
Swagatam said…
Hello AME,

I have modified the diagram by replacing the zener with a 7805 IC, please check it once gain with the modifications, I am sure it would work now.

If still it doesn't work then you could probably opt for a 741 IC circuit which is a thoroughly tested design.
Swagatam said…
Hi Superbender

yes an smps can be used here as the power supply, no issues, just make sure it's current is not above 1/10th of the battery AH which is being charged.
Swagatam said…
Hi Superbender,

Yes a single common power supply source is just what is normally required to charge two batteries or even more numbers or for any other application where more number of recipients are needed to be operated.

I would be addressing your above question in the form of a new post soon where I would provide all the required circuit details.
Swagatam said…
Hi Superbender,

I have published your design here:

Jinnamuru Xunte said…
can this charge any capacity?
Gurpreet said…
Hi sir,
what are 22u and 0.1u cap sir it would be helpful if you give the exact volts with those cap. Can i use 16v 220u instead of 22u and 50v 2.2u instead of 0.1u. thanks
Swagatam said…
Hi Gurpreet,

I have removed the capacitors from the design, please check the new diagram.
vizer40 said…
I'd like to use a PC power supply (volt moded 12V to 14,4V) instead of transformer and diodes. Is it possible? I'm concerned because the PC PS can deliver 28A on 12 V rail. Is there a solution, I don't want to buy a 300W transformer. I want to use this circuit for automatic charging a 50Ah lead acid battery.
Swagatam said…
The above circuit does not include a current control feature so it might not be safe, I would recommend you the following design, it has the required current control feature so would be entirely safe for your application even with your PC 14.4v/ 28amp power supply.

See the LAST design in this post:

Swagatam said…
pls try the following circuit, the above might not be suitable for 24v

prithviraj singh said…
Dear swagatham
Is there any problem using this circuit in co-operation with a medium power ( about 350 W )inverter ? I will have to draw more currents from the battery while discharging .
And another serious doubt to clear is, whether this circuit can be used to charge all types of deep cycle lead accid batteries with normal charging cycle fulfilled effectively ? Plz reply me soon......... Sry 4 the bad english
Krishna moorthy reddy said…
Dear Majumdar, Will i have to replace the 7805 with a higher voltage regulator IC for charging batteries rates above 50 Ah ?
Swagatam said…
Dear Prithviraj,

The circuit can be used in conjunction with all types of inverters because the charging procedure has no connection with discharging procedure.

The circuit can be used with any AH battery, provided the trafo, diodes and the relay are dimensioned as per the load current.
Swagatam said…
Dear Krishna,

the 7805 is only for powering the IC and has no connection with the battery, so does not need to be changed.
Prithviraj Singh said…
My 555 ic takes too much time to respond.......
I was tuning the pot at terminal #2 for lower threshold after sustaining the variable voltage to 11.4 V. But initially no response was there although the pot undergone complete rotation. I tried one more time and realized that when pot is rotated at the maximum position such that the resistance measured between the ground and its adjusting terminal becomes 8.2 ohm ( in this situation i think that pot has contributed complelte resistance towards its Vcc side ) and when the power supply ( 11.4 V ) is directly connects, then the relay activates at the same moment. In brief i couldn't efficiently adjust the pot since the relay took too much time to respond.
I just disconnected the relay from the circuit and put a piezo buzzer at the pin#3 of IC to know whether the ic responds to the adjustments made to the pot, but the result made me so much unhappy. I found that the pin#3 becomes active only after 35 seconds when the pot adjusted maximum. Because of this i can' t adjust the pot properly..

Sir what is the reason ? How could it Be solved ?

I used a 0.1 uF CERAMIC CAPACITOR at pin#5 as in the diagram. A TRIM POT was used as R2 and a normal one as R5. I had to use a SERIES combination of resistances to substitute a single higher one in the circuit due to lack of the exact one specified.

Will these effect the overall working.
Anyhow expecting a quick suggestion from your side
Krish Sudarshan said…
Good day sir,
I want to Clear something. I have noticed from the comments given that there are 4 capacitors in this circuit also a zener diode too. But i couldn' t see any such components except c4. That means you renewed the old one composing all these components with a later edition circuit consisting of only C4 compared with the other ?????
Swagatam said…
I saw the above idea in some other site and to me the explanation looked quite OK and therefore posted it here....it said that when pin2 voltage dropped below 1/3rd Vcc the output pin3 goes high and when pin6 goes above 2/3rd Vcc the output returns back to zero....this looked perfect for achieving the battery high and low cut offs, however many are finding it difficult to set up the above circuit so I would recommend you to temporarily quit this circuit and make an opamp circuit which is far more reliable and has been tested by me. you can try the following design:


ignore the lower relay, it's not required. this circuit will work perfectly....adjust the 10k preset for the high cutoff and the 100k preset for the lower cut off.....remember to disconnect the 100k preset temporarily while setting up the 10k preset
Swagatam said…
Good day Sudarshan,

Yes, I have removed the caps as they were of no relevance with the design
Prithviraj Singh said…
Thank You sir for the quick reply
But a bit of confusion i have, with the image you given.
Can i use this circuit to be used for both lower and higher thresholds?
How can i set the threshold as in the previous case. I am asking you this, coz it seems constant power supply from SMPS is used here.
Can i use 4k7 and 47k pots instead of the ones given here, coz i don't have a 100k pot with me( it is possible to get a range upto 100 K by periodically adjusting the 10 k pot with a series of resistances in multiples of ten upto 90 , but will it turn the behaviour of the circuit ? )
Ramswamy Agarwal said…
If there is no temperature compensation, will it aftect the battery life sir ? WHat to do ?
Swagatam said…
The 10k preset is for setting higher threshold, the 100k for the lower threshold, so both can be set in this circuit.

smps can be replaced with any 15V DC source, but the current should be 1/10th of batt AH rating.

you can try 47K preset, I think it should work.
Swagatam said…
....while setting the 10k preset, REMEMBER TO KEEP 100k preset link disconnected, connect it back after the 10k setting is done.

the circuit link, for others to follow:


Swagatam said…
yes it will affect.
Prithviraj singh said…
Sir as per your suggestion i had tried caliberating the circuit under the two thresholds. On adjusting the 10k pot first ( keeping 100 k pot disconnected ) after setting the voltage to higher threshold 14.4 V, i found the LED glowed suddently at a point also the relay activated. Upto this everything was perfect.
But the problem is that, when i am adjusting the 100k pot for lower threshold 11.4 V ( KEEPING 10K POT CONNECTED ), nothings seems happening, no LED glow even the relay sound couldn't heared.
Actually what should be the indication for lower threshold setting ?
What may have occured to ma circuit ?

Sir i have to say onething that, I HAVE NO 100K POT WITH ME. So i HAD USED THE 49.2 K RESISTOR ( 47k+ 2k2 ) in series with the adjusting terminal of the 50K POT which gave a variation of 50 to 100 k at one full rotation, and the remaining was taken as usual.
IS THERE ANY PROBLEM IN REPLACING THE ORIGINAL 100 K POT with this configuration. Will it affect the result ?
Anyway i am waiting for your valuable suggestions about this problem to keep ma project go ahead
Prithvirai singh said…
Dear sir
Where could i connect a Green LED in the circuit of opamp to indicate charging progress ?
If i am using a 12-0-12 V inverter transformer as the power supply, then if i only want maximum of 6 A for safe charging, What will be the value of current limiting resistor which is to be used prior to the bridge rectfier section ? Will it be 2 ohms, since by ohm's law r= v/i ( 12/6 ) ???????
Swagatam said…
Hi Prithviraj,

Everything is centered around the reference voltage set by the zener voltage at pin#2.

While setting the higher threshold, the LED lights up because the pin#3 voltage exceeds the zener reference voltage.

Now in order to switch OFF the LED the 100K must be adjusted to a point where pin#3 voltage is again dragged below the zener voltage of pin#2.

Measure this with a meter while adjusting 100K pot.

If this is not happening then probably either we'll have to increase the value of the 100K preset to some higher level or connect the 100k preset with reference to ground.

You may confirm this first then we can proceed further.
Swagatam said…
LED connections will be like this:


the resistor shown are 4k7k each

yes your current resistor calculation is OK.
+iF@ Kir@n@ said…
What components should I replace? when I use a 6 volt battery. thank you
Prithviraj Singh said…
Yes!!!!!!!!! The circuit is functioning sir.

Now i just connected the power supply section. As i said earlier, an inverter transformer rated 25 A at secondary is being used. I had followed the transformer with a bridge rectifier ( formed using four 1N5402 diodes ) and a 1000 uF/25V capacitor in parallel. But when i had just finished the circuit and powered up, a bursting occured in the center tap or the transformer seconrary, then i suddently disconnected the device. I have only tested the power section ( without the charging portion ) to avoid any awaiting damages.

I am using a 25 A rated 230/12-0-12 inverter transformer with 5 inputs( primary ) and 4 outputs ( secondary ). The center tap had been taken BY PAIRING THE TWO CENTER WIRES IN THE SECONDARY TOGETHER and fed this tap to the -ve terminal of the capacitor forming ground. The other two terminal wires have been connected to the respective junctions in the bridge......n but the system FAILED.
I have noticed several circuits in which this bridge configuration is employed, but in all these, the transformers used were not having a center tap, having only two outputs..........
THen, was my connection wrong ?
How can i solve this issue sir ????'??.....
Prithviraj Singh said…
Sir one more thing to ask to you.
On noticing the envelope of the battery, it has been written there that , the battery has to be operated by a charger with the feature of constant current and constant voltage charging. So any problem will my battery face operating with the opamp circuit sir ? Will this simple circuit be able to provide the enlisted features here ?????

I have found one of your schematic named ' automatic 3 step charger '.
But as one of your followers said in that blog, i am also not able to collect such low value resistors r1, r2, r3 ( of the order of 0.01 ohm ) from the local shops in our area. So what will i do to charge my battery with the safety features and without the lack of such triffle components ????

Give me a better suggestion sir....
Swagatam said…
Your bridge connection seems to be incorrect. Your method would be suitable for dual supplies as shown in this article:


It's better to use the half winding, that is 0-12V winding and connect the bridge to it in the conventional manner, this method would create full wave rectification and also eliminate confusions.

Swagatam said…
you can use the same components.
Swagatam said…
Three step charger would be too complicated for you to build, however if the transformer is rated at current 1/10th of the of the battery AH and the voltage at around 15V. then no additional circuit would be required...the opamp circuit would be enough as it will do the cut offs at the set moments.

With a higher current trafo, yo would certainly require a current control circuit along with the opamp circuit.
Prithviraj singh said…
Very good morning sir.
I have tried the 0-12 tap, eventhough spark and the bursting sound along with smell comes again. I am very much afraid of this and not so much confident to connect this into the opamp circuit. I think the spark comes due to the sudden high current flow towards the rectifier section. I have used a 2 ohm resistor to reduce the current to 6 A. But nothing happens as i wish. I should have to use this transformer to follow the inverter design along with this. Can you please suggest me a better modification which can be applied to the secondary terminal with 4 NO. OF OUTPUT LEADS ???????
Prithviraj singh said…
Sir i have noticed another thing that, no such problems ( sparking and busting ) occurs when the transformer is used with half wave rectification mode. The two center wire twisted together connected to the -ve terminal of capacitor forming ground and the other two wires connected in normal manner.
No problem yet.
But i don't wanna this, coz i think it will not be efficient for my purpose.
Then what may have occured to ma bridge config. '????'''
Swagatam said…
Good morning Prithviraj,

Using a meter set at AC range, check the voltage of the taps, and find the one which gives the required 12V, and connect these taps to the bridge.

I think the diodes might have got shorted and burnt....you can check them to confirm or use fresh diodes to make a new bridge rectifier with correct polarity and connections.
Sooryanarayanan said…
Sir your blog is very interesting. It helps a lot of people who are interested in electronics. Therefore hearty CONGRATS for being this blog a very success.

I have a small doubt. Will you clarify it??
I have three sealed lead accid batteries of 12 V each. All the 3 having same Ah rating. I had been using the batteries for 3 or4 months before. But since then it hasn't been ever used. But no problem..they are zero maintenance batteries.
Now i have to charge all of them. But the problem is that each batteries shows different open circuit voltages on checking with a voltmeter. My doubt is whether they can be used in parallel to charge the whole setup or not?
I prefer a parallel set up, becoz it suits my purpose ot getting the required higher Ah keeping voltage constant..
I am very sorry for the large describtions which may have wasted your time much more. Again sorry for the bad english.
Hoping a better suggestion from your side
Swagatam said…
Thank you very much Soorya,

You can connect them in parallel with each individual high current rectifier diodes in series with their positives. The diodes will automatically balance outthe difference in voltage levels of the batteries and charge them uniformly, you mayu refer to the following post:


Be sure to use diodes rated with current twice that of the charging rate of the batteries.
Sooryanarayanan said…
Thank you very much sir. Now i want to know whether a 1N5408/02 is sufficient for maximum charge current of 6 A ????

What about 6A4 diode ?

What word i have to search for the maximum current rate of a diode in it's datasheet ? Is it 'maximum forward current ? '
Swagatam said…
Thanks Soorya,

You can use two 6A4 diodes in parallel for each battery positive.

6A4 can handle 6amps so two would provide a safety margin of 12 amps

I haven't update diode datasheets here yet so at present you won' t be able to find any relevant info in this blog
Prithviraj singh said…
How do i measure the current drawn from my high power transformer using multimeter???
I want to ensure that the current limiting resistors are working properly to avoid any awailing damages.
I have heard that it is possible to measure the current only under LOAD. But except ma batteries i don't have any other devices functioning at that much current with me. I don't wanna to take risk on ma batteries too.
Swagatam said…
Hi Prothviraj,

If you check your meter features you will find a slot for 10Amp or 20 amp range, you will have to plug the red prod into that slot, set the selection knob to 10Amp position and atsrt checking current by inserting the prod terminals in series with the transformer output. Use DC range if you are connecting them after the bridge, and AC range if it's before the bridge.
nazriya said…
Hellow sir, Good morning.
I have 5 no.s of 20 Ah batteries connected in parallel forming a 12 V 100 Ah battery bank. What will be the required charging rate of the whole setup ?
Is it 2 A max ( 1/10 th of the individual battery ) or 10 A max ( 1/10 th of the whole battery capacity ) ?????
Sooryanarayanan said…
Sir my previous comment was a wrong assumption. I individual batteries are 20 Ah rated. SO may i use a diode rated above 2A for each battery for parallel charging?????
1N4508 is enough ??''
Prithviraj singh said…
When i am turning on my dc power supply the resistors connected in series with the +ve secondary of the 300 W transformer is getting burned. I have tried a 2 W resistor, eventhough the burning effext didn't change. what to do sir ?
My power supply connections are everything perfect
Swagatam said…
Hello Nazriya,

It'll be 10amps and not 2 amps because parallel connection adds up current.
Swagatam said…
As I have told many times in this blog, the charging rate should be 1/10th of the battery AH, is your trafo generating this much current?? Pls confirm this first.

Secondly the resistor will need to be dimensioned according to the current limit that's being used at the output...if you can tell me the charging rate of your system then I'll tell you the correct resistor dimensions
Swagatam said…
for 20 ah you can use 6A4 diodes, or any other type 6 amp diode
Vikram vaidee said…
Hellow sir,
I have noticed a battery charger circuit using 741 opamp which you had given as a reference circuit to one of your readers. In this circuit when the battery voltage exceeds a higher cut off ,opamp output becomes high due the increase in voltage at terminal 3 compared to zener reference voltage at pin#2.
But i want an extremely opposite action to this. I mean, when the voltage rises above the zener voltage, output of opamp should go low and vice versa. ie., opamp in -ve config.
Can you please help me sir...n
Swagatam said…
Hello Vikram,

Just reverse the pin arrangement, meaning connect the zener to pin3 and the preset to pin2
Prithviraj singh said…
Sir sorry for disturbing you again at this time. I was busy with some other works, that is why i hadn't noticed your last comment on my request. Now i am giving here the exact requirements i need.

I have a 300 W inverter transformer with me ( bought in order to make a medium power inverter ). From basic calculations it could be understood that anyhow it draws maximum 25A current under full load at secondary.
I have connected the rectifier section along with the transformer for the first time, but a fuzzing sound could be heard. Therefore, i decided to add a calculated resistor of value 2 ohm ( 12V/ 6 A) in series with the +ve tap of the secondary to limit the current. But it also got failed. Resistor burned.
Here i am using a 60Ah battery. Therefore it should draw maximum of only 6 A from the transformer secondary, but the thing is that the secondary terminal supplying very high current.
I am employing your automatic single transformer inverter/charger/changeover circuit with the charging circuit using opamp which i got from this blog. So all the above problems came from this single transformer design such that current limiting and power handling problems.

BUT I DON'T EVEN WANT TO PAY ANY MORE FOR ANOTHER TRANSFORMER and i have have to implement the specified charger block only since i prefer the charging of my lead accid batteries with complete safeguarding

So in brief ( talking from a beginners level ) i want to clarify the below things-
1). What should i do to limit the current for charging section???? Will it be a single resistor alone with high power rating????
2). Any modification to be carried out for remaining sections ( such as oscillation section, relay coils ) which deal directly with transformer secondary ???
3). Can the limiting resistor be used in series with transformer secondary??? Will it be able to limit AC current???
And the last thing to ask to you ' Whether i have to use a LM338 IC or Traic circuit as a current limiter???????'''''''

Now i think you got the exact thing that i meant before and asking you several times. So if you can i want your valuable suggestions and help on this. Please don't neglect ma request as a triffle. I am expecting a satisfactory reply
Prithviraj singh said…
Sir can i use a seperate 1:1 5A transformer for the charging section??? ( since i am not able to find higher watt resistances from our local shops )
If it can, i need help in modifying other sections of the inverter handling with high current delivered by high power transformer
Swagatam said…
Hi Prithviraj,

That would be the right approach, since you are new in the field you shouldn't try adventurous circuit concepts, it could be risky.

I would also suggest that you spend some more money and buy another transformer and make a separate charger,using IC LM196 in place of IC338.

By the way which inverter design are you trying right now?

NTC will not work as current limiter, because it's function is to limit surge current and not continuous current
Prithviraj singh said…
The single transformer inverter/charger/changeover circuit that you posted before sir.
Sir i want to clarify something on that inverter circuit.
I need the circuit to be function as follows.......
I am connecting the output of the inverter directly to an ac wall outlet with a plug in a room. My purpose is that during power failure, suitable instruments (to the inverter power) can be turned on which is situated at any room in the house powering from the ac fed by the inverter at the other end through that plug.
One of my cousin had done a similar inverter and he plugged the inverter directly to the wall socket and swithed on it after turning the mains switch off during power failure. It worked well. But all the thing he had done was changing a normal PC UPS into an inverter.
On searching the internet, i am confused about this output connection, because all of them saying that the instruments to be powered during power failure are directly connected to the inverter output. So will there be any problem in my strategy.
And also I want the inverter to be connected permanently to the ac wall outlet and ni need to TURN OFF THE MAINS SWITCH during power failure to make it operating.....mTry giving me best reply considering my cousin's situation
Swagatam said…
If you connect the referred single transformer inverter/charger output to wall socket, then all the appliances in the house would get accessed with it as normally happens with utility mains.

But you should first confirm the operations using smaller transformer and without wall socket connection, otherwise you may end with a possible fire hazard.
yaswanth reddy said…
Sir i found an automatic charger circuit using an opamp ic 741 referred in one of the comments in this article. But i have only a 24V 20 Ah battery to charge. I am very interested to implement this circuit. Can u plz help me in modifying the given circuit to charge 24V battery or batteries rated upto 48V.........
yaswanth reddy said…
The figure is containing in this link sir

Swagatam said…
refer to this circuit and see how the 22k resistors are added with the circuit.


For your circuit use 10k resistors in place of 22k.

ignore the msfet stage and use a relay as given in your circuit
Daty said…
pls, Mr. Swagatam i would like you to post me the modified Automatic Lead Acid Battery Charger using NE555 IC.keep up the good work either on the blog or send it to my Email: datepaka@gmail.com
Swagatam said…
thanks Daty, I'll try to do it soon.
Govindh shenay said…
Can the two 12V lead accid batteries connected in series be charged with this diagram???
I want to use the series config. for my need. i.e, as 24V bank always.
Can you tell me which are the lower and higher threshold levels for this combination and how can they be set?
I also want to know whether any damage will occur to the individual batteries with this charging method?????
Swagatam said…
The above diagram is only for parallel connections not for series connections.

Parallel charging would be better using the above explained method, series charging will not give optimal charging results compared to parallel so it should be avoided.
Admin said…

I need a little info about tabular acid battery.
I purchased a new 100 AH battery(not branded). I am using the battery @ of 10Amps load using a mosfet inverter battery is lasting 5hrs and 10min.
according to my knowledge It should give backup of 9+ hrs @10 Amps discharge.

dealer claims he sold the correct item of 100AH but I feel being cheated.

I want to know how to determine the correct AH of the battery. It's only 15 days old .
Swagatam said…
hi, your dealer is not cheating, if it's printed as 100ah then it should be a 100ah battery.
Possibly your battery might not be fully charged....to charge it fully keep it connected with a 14V 10amp charger for about 14 to 18 hours and then you can expect the desired results from it.
Admin said…
hi, thanks for reply,
1. battery has nothing printed as it is not branded dealer build it at his shop assembling the battery plates.

2. Battery gets charged using the same transformer ( inverter + charger) @10 -15 Amps automatically it does not get over 15 Amps during charging. and has 14V at the battery terminals. After switching off the charger battery has 13.93 V at the terminals.
(Purchased this ready made kit from market based on Mosfet and SG35xx series)

3. I had fully charged the battery like you are saying for 14-18hrs before testing , and as the charging is automatic I don't need to switch off the charging as it gets charged to it fullest level charging automatically kept @ 14V.

4. Tested the 100AH battery 2-3 times in past days after fully charging but it never gives more than 5Hrs backup @10 Amps discharge.

5. Lastly I have another 32AH battery (1 year old) which gives a backup off 2Hrs 40min at 10Amps discharge rate using the same inverter for charging . This is the reason I highly doubt there is something wrong with the 100AH one.

Swagatam said…
In that case you can the expect the quality to be not up to the mark, obviously you would have got it at a rate much lower than the standard rates.
A hand assembled battery can never produce results equivalent to branded ones due to the low tech manufacturing procedures involved.

mexzony said…
can i build this charger and it will work
Swagatam said…
I would suggest an opamp based charger circuit because these are more flexible and simple, i have of plenty of them in this site.
arasan san said…
sir i have completed circuit.but connect across the voltmeter -ve(from battery) & IC 2 pin this time p2 adjust relay active,remove voltmeter not work. what i do it?i am try to many times but same error how to build it? thank you
arasan san said…
sir i have completed circuit.but not work relay. i will try to many times, change a set of new component. no reaction,tell me solution
Swagatam said…
arasan, the adjustments involved with the above circuit needs a good knowledge of the IC555 functioning, if do not know them could be difficult for you to set them.
It's better you try the first circuit from the following link, which is much easier:

Ajish said…
Sir I am just a hobbyist. The specified component VR 7.5v 1W 1N4737A does not appear anywhere in the circuit diagram. Am I missing it anywhere? Kindly help me out.
Swagatam said…
Ajish, it has been removed from the circuit for simplifying the design
Deva RajaIAS said…
Sir i have done the circuit of 12 V charger given as a reference from this link


Everything was successfull and the charging too. But a bit of confusion still exists with charging.
I am using a 15V 5A old Thoshiba Laptop charger to give charging voltage to this circuit.
Initially on connecting a battery having a voltage of 12.5 V the voltmeter shows nearly about 14.5 V or higher in charging phase, thereby having a confusion in making out the exact higher cut off voltage. I have chosen the cut off voltage to be 14.0V approax., but when i am connecting the battery to be charged with the circuit the relay activates and being in discharging mode automatically disconnecting from charging mode, thereby i am not able to charge my battery.
Also i can't select a threshold voltage at 13.8 V or below 14.0 V. I am saying about this threshold level, because i didn't get apperance of voltage more than 13.9 V on charging similar types of batteries for more than 10 hours. So i am affraid of overcharging if the threshold is set to above 13.9 V and think that they willn't reach above 13.9 V while charging.
Could you please help me in taking the exact cut off for the 12 V 20 Ah selaed lead accid battery sir?????
I have also an another doubt that,
When i am seeing one of my batteries is having a voltage of even 12.6 V i get little brightness in a 12 V car lamp but when this is checked with an another battery having same voltage the brightness is awesome. The batteries are expected to be good. Then what may be the reason for this sir ?
Deva RajaIAS said…
Also how can i get know that the battery is FULL CHARGED ?????
Swagatam said…
Deva, you should use a transformer or input voltage with a current rating 1/10th of the battery AH value, and connect the battery before switching ON power.

Once the battery voltage reaches 14.3V, gently adjust the preset so that te relay just cut off.
If the relay activates in between, keep adjusting the preset to deactivates th rleay until the voltage 14.3V.

Keep the feed back 100k feedback preset disconnected while the above setting is being done.
Connect LEds across the opamps pin6 ad positive and pin6 and ground with individual 1k resistors, for the required indications.
Arun Dev said…
Sir a simple dout...........
It can be seen that in any ceiling fans in India, the capacitor is placed near the joining portion of leaves and the motor cover . My question is whethrr it is possible to place the capacitor near the switch of the fan in the electrical switch board on the wall parallel to P and N so that we can easily replace the capacitor if damaged avoiding the risk of climbing over height to gain access to the leaves ?
Swagatam said…
Arun, yes you can put the capacitor anywhere and at any distance from the fan, but that would mean you'll have to manage two long wires from the fan upto the capacitor.
Arun Dev said…
Why it requires two wires sir???? I could see the capacitor terminals are connected in parallel to the following junction pairs
AC phaseline - Possitive line of fan ( For preferred rotation )
AC neutral line - Negative line of fan ( for preferred rotation ).
So could it be possible to connect between switch of fan and AC neutral. That was my question.
Also could you please tell me, why we use capacitors for fans ?????
Should i use a capacitor of atleast 24 V while operating a 12 V DC fan ?
Arun Dev said…
And what will occur if it is kept disconnected ???
Swagatam said…
Arun, the capacitor will be in series with the "start" winding of the fan, its one end could be connected with the live output of the switch but the other end must be in series with the "start" winding of the fan coils.
DC fans do not require a capacitor
Arun Dev said…
Ok sir.. Thanks for the reply. I will now correct myself.
You didn't give the answer for last question
" why the capacitors are necessary ? "
Swagatam said…
it's for giving the initial start current to the "start" winding without which the fan cannot begin rotating
chintu teja said…
Sie can i use this circuit to charge my 12v 1.3ah sealed lead acid battery for battery full and low conditions? My application is portable audio amplifier, i have 12v 3amps adaptor, one maintainence free sealed lead acid battery, i want to use battery in the amplifier and the charger indicator circuit so can i use this circuit while same time my amplifier should work and battery should be charged up, and when i unplug, my amplifier should run from battery, please please help me
Swagatam said…
chintu, I think you should make the last circuit given in the following article. It's a better design and has been tested by me and will fulfil the conditions properly.

Vickey Singh said…
hiiiiiiiii I need 12volt and 24 volt 180Ah to 200aH auto cut battery charger circuit plz help me and send me in my mail thanks
Vickey Singh said…
Hello Sir,I need 12 volt 180 Ah to 200 Ah auto cut battery charger circuits plz help me and send me in my mail. Thanks ....................................
Swagatam said…
You can try the following circuit:

chintu teja said…
Sir i have designed one portable amplifier, i have one 12v 1.3ah sealed lead acud battery, i also have 12v 3amps adaptor, i need one circuit which in which if i connect the adaptor, my amplifier should run and parallaly charge my battery, the circuit should be in a manner that even if i connect the adaptor continuosly to amplifier the battery should not be affected or damaged, do you have any such circuit please provide me the parts values and ratings, m email is chintu.shiva27@gmail.com
Swagatam said…
Chintu, you can try the second circuit from the following link:


Replace R2 with a 10k pot in order to be able to set 14v output for the battery.

You can power the amp from the battery while it's charging.
Carl Corbeau said…
Hi Swagatam,

I can't read the numbers on this schematic.
Can you increase the resolution?

Carl Corbeau said…
Sorry Swagaatam,

I must have printed the wrong one.....I have a readable print now.

Swagatam said…
Hi Carl, sometimes our computer responds a little late, may be that might have caused the problem:)
Sir me bhanu...
i have assembled all the components but but sir my relay is not tripping , according to your description i have connected variable dc power supply & adjusted preset.
But sir i am in doubt i think the preset should be adjusted by giving supply in both side means in the battery side we should connect variable supply & at the same time we should connect the bridge to AC supply through transformer...... is it the correct procedure
Thank you...sir
Swagatam said…
The presets should be adjusted by supplying voltage only to the battery side.
first connect a fixed 11V to the battery side and adjust the pin2 preset for activating the relay.
Then increase the battery side voltage to 14.3V and adjust the pin6 preset to just disconnect the relay.
I have not tested this design yet, so can't provide confirmed opinions.
Yes sir thanks ....Me bhanu
I have already tested this procedure but I am not getting the desired result….
1) One big problem in this circuit is I am not getting output in the 555 IC #3.
But while I am testing the same IC in breadboard by giving VCC & ground I am getting the desired output..
2) I think we should not use voltage regulator in pin 4&5 , it should be connected directly to battery side. Actually I can’t understand why this voltage regulator is used here….what is the logic behind this..
3)Another problem I am not getting voltage in relay coil side…
Plz sir help me
Swagatam said…
The voltage regulator is at pin4 and pin8
The internal opamps of the IC require a fixed reference voltage for sensing and comparing the output variations at pin2 and pin6, if 7805 is removed the IC won't be able to get the fixed references and will produce entirely wrong output results.

If you are difficulty wit the above design you can try an opamp based circuit,
an example circuit may be seen here:


Thank u sir
I have visited this link this is totally different from this circuit but technology & logic is same
Sir still i am trying to get output from the old circuit
1) Today i connect a variable power supply and followed the same step but while testing when i am touching the prob of multimeter to pin 2 of IC 555 the relay is tripping but when i am living the prob again the realy gets off ..... means i am not getting output in pin 3.....what the proble here...
2)Sir i checked the datasheet of IC 555 there it is given voltage across #2,,#3,,#4,,#6.... according to the datasheet particular voltage i am not getting in this following particular pin can you please help me sir......
Swagatam said…
you should adjust the pin2 preset until its voltage reaches below 1/3rd of pin4/8 voltage. as long as this is sustained the output will be high.

connect 10uF capacitor across pin2 and ground, similarly across pin6 and ground for stabilizing the effects.
dear sir i have 20v5A ups transformer i want to charge 12v45ah battery what are the modifications i should do?
sskopparthy said…
Sir, the circuit is working but the relay begins to drifts to N/C and N/O very fastly at certain point of charging(12v, 7ah batt, 1amp transfo used). The led connected to relay coil with 1k resistor switches on and off with relay but it doesn't seem to be doing so.(I wanted to tell how fast the relay os switching on and off). I'm afraid that this behavior would damage the relay. Please guide me how to stop this.......thank you very much sir.
Swagatam said…
I would like to know how much did you understand this circuit?
just by making it correctly is not all, you must know how to tweak it until it works.
did you check the voltages at pin2 and pin6 while confirming the results?

whether it's 100k, 10k, or 1k, for presets and pots it doesn't matter because the variable ratio across them will be always the same. what a 100k will produce, 10k will also produce the same results.

The circuit may be difficult for you but that doesn't mean the circuit doesn't work.

Instead of saying "it doesn't work" it's more important to find why it doesn't work, and this may be done by finding the possible faults in the design....can you do it?

If you having difficulties with this design, you can try the second circuit given in the following link:

Swagatam said…
the preset value change that you have done won't produce any difference, as I have explained in the above comment.
Swagatam said…

you can try the second circuit from this link:


I have tested it practically.
Partha P Baruah said…
Sir, can you help to build a very simple electronic magneto phone that can connect one room to other .In very simple ,with ringing and communicating with the other end facility.
Swagatam said…
Partha, do you mean a simple intercom system??
Antero Duarte said…
Hello there, great post here, and great blog you have going on. I've been looking around other posts and this one seems very interesting for it's simplicity and relatively low cost. But i really like the idea of using an LED driver to indicate battery status like you've shown here: homemadecircuitsandschematics.blogspot.pt/2013/08/3v-45v-6v-9v-12v-24v-automatic-battery.html

How could i incorporate that part in this circuit? Although your explanations are perfect, i dont seem to understand how the LED driver part works on the other schematic, so i need some help in incorporating that part in this schematic, would it be difficult?
Swagatam said…
Thanks so much!
You can do it by connecting one LED across pin4 and pin3 and another across pin3 and ground.
each of these should have there own individual 1k series resistors.
the pin4 led will show battery is full while the ground led will indicate battery is charging
Partha P Baruah said…
Sir not exactly the intercom,in intercom speaker is used and the communication can be heard by others also. I mean a telephone with transmitting and receiving coil.The two similar phones are connected to communicate between two places like the magneto phones of earlier days.In magneto phones we use hand generators for ringing purpose .This hand generator is to be replaced with electronic circuit for ringing and for talking purpose also circuits to be designed.Please help.
Swagatam said…
You can make the first circuit shown in the following link:


You will require two sets of these for the communication.

The headphones for the two circuits will need to be exchanged across the locations.

For the ringer, you can make separate buzzer circuits and connect them with push-buttons across the desired locations

you can find plenty of simple buzzer circuits online.
sskopparthy said…
Sir, sorry for disturbing you again. I wanted to ask you if you have a circuit for detecting faulty earth(leaking earth) and give a buzzer with indicating led? I am facing a lot of problems with faulty earth in my house.....can you please help me in designin this circuit....if possible....
Swagatam said…
SS, you can do it simply by measuring the volts across LIVE/Neutral and LIVE/EARTH
both should be exactly identical, if the live/earth shows lower volts than the live/neutral value, would indicate a bad or insufficient earthing.
ak style said…
I'm not into electronics but I am really in need of making this circuit. Can you tell what does the arrows pointing from R7 to R5 and R8 to R2 denotes and N/O and N/C part. I sorry but I really don't know much about electronics.Thanks in adavance
Swagatam said…
Those are pots or variable resistors, N/C stands for Normally closed contact and N/O for normally open contact
ak style said…
I have also found that there are no 8k resistors and in your diagram it says "8k2", is it some type of error.
Swagatam said…
where is 8k written?
ak style said…
R3 the one below the first potentiometer
Swagatam said…
yes it's 8k2, so where's 8k that you were referring to?
Kishore Cp said…
Hi Sir,
i would like to charge 12v 7.5 amp battery with the circuit, i have chosen 15v 5amp transformer.

please let me know the diode and capacitor value should i use for bridge circuit. Thank you.

Please correct me if my transformer value is too high or ok to charge the battery or drive this circuit. Thank u so much.
Swagatam said…
Hi Kishore,

a 15V transformer will produce about 24V after bridge rectification which could be too high for the battery.
use a 0-12V transformer, which will be just enough for charging a 12V battery, but 5amps could be again too high for a 7AH batt, use a 0-12V/1amp transformer.
ainsworth lynch said…
I built this battery charger but the relay keep makes a huming sound and I only get 1.3 Volts out
Swagatam said…
not possible to troubleshoot without seeing it practically
I made your circuit .it is not working perfectly.give me suggestion please
thavamani84 said…
Sir Can send. T1 secondary? How much volt possible
umesh said…
hello, for the first time iam visiting here found very good information. can i too get some ideas regarding increasing of voltage level in a ready made 12v 1A smps adapter
Swagatam said…
glad you liked my site....you can modify any smps for getting a desired voltage from it by followign the instructiosn as give here

Unknown said…
hi i want to make a circuit which can detect ultrasound at 37.5 khz. please suggest some circuit. e-mail neerajair12@gmail.com
Swagatam said…
Hi, please explain the application and the requirement in detail, so that I can design it correctly.
Goran Radkov said…
Hello Swagatam!

Can you please help me, because i didn't understand, what is the difference between: "How to set the battery cut off threshold", and: "How to Set Up the Circuit" it's one and the same, i think?

And the other question - saying "variable power supply", means DC, which can be varied by voltage value, but not AC, right?

Thank you!
Swagatam said…
Hello Goran,

I would recommend an opamp besed automatic battery charger circuit instead of the above 555 circuit because opamp circuit is much easier to adjust and get accurate results.

you may try the following circuits

Goran Radkov said…
I already made this, and also made a voltage regulator with LM338, adjusted the voltage to 11.5V but in the whole range of the R2 the relay didn't deactivate, also with R5, so it turned out, that these 11.5V applied to the battery terminal are loosing, because at pin#2, pin#6, pin#4 and pin#8 there is 4V, pin#5 - 2.8, at pin#3 - 0 if R2 gave 4V and 0.36V if R2 gave 1.76V...
So R1-33K, R4-22K and R10-10K should be smaller?

Thank you!

Swagatam said…
OK, I think you can try removing the R1, R3, R4, R6, these are not relevant.

As per the rules, when pin#6 voltage goes below 1/3rd of the Vcc, then pin3 is supposed to go high, and when pin#6 goes higher than 2/3rd of the supply voltage then the pin3 should revert to zero.

please check the circuit by referring to the above instructions.
R said…
hi sir, what charging circuit can i use for the main supply that made by piezoelectric transducer to charge 12 volts lead acid battery?
Swagatam said…
Hi R, sorry I could not understand your question corrector, please elaborate
Nikhil Sonar said…
hi sir,
I want to charge battery of 19.5 V / 3.33 A and 65 W.
and also i want tolerance for wattage of 20 %.
does this circuit is sufficient or any changes are required.

Swagatam said…
Hi Nikhil,

you can use the above circuit, just remove R1, R3, R4, R6...these are not required
Nikhil Sonar said…
Thanks sir,


Id (diode current rating)=2 X BATTERY CHARGING RATE
Id = 2 X 0.333 A
Id = 0.666 A
as ,with 230 VAC primary and 24 VAC volt secondary transformer
diode rating will be
p=v x i (24 x 0.666)
15.98 wattage of diode will be sufficient ,does it ok?
(please tell me diode and transformer specification for this circuit(19.5v 3.33 ah) )
Swagatam said…

for the diodes you can use 1N5408, for transformer you can use 0-15V/1amp, or a 7.5-0-7.5V/1 amp trafo (ignore the center tap) ...
Nikhil Sonar said…
I am using relay of 24 volt coli voltage.
is it correct?
And also form circuit diagram i can see only two capacitor ,i.e. C4 and C5,
what about C1, C2, C3.?
Swagatam said…
what is the voltage across C5?? if it's 24v then 24V relay will be OK, at 19.5V a 24V relay will not operate.

I removed the other capacitors from the circuit as those were not important.
Nikhil Sonar said…
for this battery (19.5 v) lower threshold is 17.5 and upper threshold is 21v .

SIR , across c5 there is 24 volt dc.
so 24 volt relay is ok,

but sir during calibration of circuit ,

for setting lower threshold level, relay is activating by adjusting preset at pin #2.
but for upper threshold level ,relay is not activating by adjusting preset at pin #6.
Note : for upper threshold level setting ,voltage at pin 6 is varying from 0 to 4 volt.
Why so?

Swagatam said…
that's strange should not happen...did you increase the voltage to 21V while adjusting the pin 6 preset?
Swagatam said…
also connect an LED in series with the transistor base for getting quick indications of the changeovers
Nikhil Sonar said…
sir ,
i have checked circuit again ,
but still same results .

i set value of variable power supply to 17 vdc ,and then by adjusting the preset at pin#2 such that the relay just activates.relay is activating ,because output of timer ic is high at pin no 3 is high,
by keeping same arrangement but now value of variable power supply is increased to 21 volt dc.
but now, when i adjust preset at pin 6 ,voltage varying at pin number 6 from 0 volt to 4 volt something , but output of ic is high ,even vtg at pin 6 goes above 2 3rd of supply.

#activating of relay is working rightly but deactivation is not working as suggested circuit.
Swagatam said…

what is the voltage at pin2 when the supply voltage is increased to 21V?

it should be above the 1/3rd only then the output will flip.
thanks for sharing..
idowu victor said…
Sir pls what will be d output current of this cicuit
Swagatam said…
current will depend on the relay contact rating, it could be as high as 25 amps
adam basha said…
please any one give me the 48 volt,10amps charging circuit diagram
Sir could you suggest an efficient circuit to charge a 12V 7Ah battery properly.. The circuit should include the following features

- automatic charging cut off when it is fully charged
- short circuit protection
- using minimal but efficient components..

I don't want to use bulky expensive transformers for this circuit.. so a circuit similar to that used in a laptop charger is required which uses small transformer
Swagatam said…
RT, At the moment I do not have an SMPS charger circuit with auto cut off, if possible I'll try to design it
biswajit das said…
hello sir,I have a question, that is, if the charging voltage is greater than upper threshold(14.2 volt) then it will always be deactivate the relay or not?I don't have too much idea on battery charging, so if you explain on it would very helpful for me.
Thank you!!!
Swagatam said…
Hello Biswajit, yes that's true but as soon as the 14.2V gets connected with the discharged battery it is supposed to drop down to the lower battery level

I am just starting to get into electronics.

I want to keep using my 18v cordless drill. I cannot get the batteries and the charger has stopped working.

I want to run three PS640 batteries in series to get the 18v. However, I want charge them as a pack not as individual batteries.

I have a 19V, 3.42amp laptop charger. Will the circuit described above be suitable for the task?

How can it be modified to suit my parameters?

Swagatam said…
ralph, actually no circuit would be required for your case, just connect a 20 ohm 5 watt resistor in series, and connect the 19V supply directly with the battery, keep monitoring after every 2 hours, and when it reaches 19V you can disconnect it from the charger.
bursach said…
Hello Swagatam, can you tell me what will be a values of the components(resistors) in this circuit for controlling and cut off 24V batteries?
Kind regards
Swagatam said…
Hello Bursach, I have not tested this design practically so wont be able to suggest much...I would recommend you to try an opamp design instead as shown in the following article

untung suharto said…
Hi, you have make mistake on preset 2 and 5 not 10k but 100k, I just make one and succesfull...thank you.
Swagatam said…
thank you for updating this....If you remove R1, R3, R4, R6 then the preset value will not matter whether it's 100K or 10K all will work...
Iyan Pasqualine said…
Thanks for sharing the circuit
if I want to charge 3,7v lithium ion battery then how much voltage need from power supply?
and how much voltage from ic 780x that I need?
and how to set up pin 2 and pin 6?

please answer my question sir
Swagatam said…
for 3.7V batt you can use the 7805 output in common to the IC pin#8/4 and also with the battery....but connect a diode in series with the battery positive to make it around 4.4V

the setting procedures will be as given in the article
Candra said…
Hi Swagatam, I have several questions..

I have 5A transformer do I have to change anything else beside omitting R1,R3,R4,R6 in above circuit?

is it save to use TIP127 as mentioned in conjuntion with 5A transformer?

last, what is the use of R7 and R8, how to calculate R7,R8 and R10?

Swagatam said…
Hi Candra, the transformer current must be 10 times less than the AH rating of the connected lead acid battery.

the transistor should be rated well above 1/10th current rating.
Candra said…
thanks for reply swagatam,

I have four 6v4.5Ah batteries and plan to desulfate it by following your article, i have tested and need about 1.5 hours to fully charged with single battery using 5A transformer by watching voltage down to 7.2v from 9v, now I want to automate it on and off when the voltage touch their limit, any suggestion?

Candra said…

thanks for reply swagatam,

I have four 6v4.5Ah batteries and plan to desulfate it by following your article, i have tested and need about 1.5 hours to fully charged with single battery using 5A transformer by watching voltage down to 7.2v from 9v, now I want to automate it on/off when the voltage touch their lower/upper limit, any suggestion?

Swagatam said…
OK, it will do, or you can also use the following opamp based cirucit instead


just replace the LM317 with a LM338
Hiram Kumado said…
Sir please thanks for the circuit because it is working but my problem that is giving me headache is when I connect the output of the circuit to the battery it the diodes which I'm using for the rectification begin to become HOT and if I disconnect the from the circuit it cools down, pls help me cos I'm using 16 volt transformer with a bridge rectifier without name. Plss i need help.
Swagatam said…
Hiram, the diodes will burn if there's a short circuit in your circuit or if the bridge rating is lower than the load current....so please make sure that the bridge rectifier's current handling capacity is twice that of the connected load current specs.
Sribasu said…
Setting relay driven cutoff levels by adjusting potentiometer is a tough work. I am also confused with the high and low cutoff settings. Suppose the battery is at 12.5v level and lower and upper cutoffs are 11.4v and 14v respectively. Will it start charging the battery when connected to the charger? I mean will upper cutoff take presedence and start charging or due to lower cutoff's presedence charging won't start until the battery drains to 11.4v?
Swagatam said…
Yes at 12.5V if the battery is connected, then the circuit will initiate by charging it...and cut off when it reaches 14V

if you are finding it difficult to set this design, in that case you must try any 741 based circuit from this website...you will find many such circuits here and you can select anyone as per your choice
Kesav.N said…
555 can be handle 12v...
Then y using 7805 regulator in this circuit.
Swag said…
here IC 555 must be applied with a constant input for initiating correct cut off operations, therefore 7805 was used, whether its 5v or 12V IC 555 will work equally well for both, but 5V will allow a cooler condition and lower consumption for the system.
Kesav.N said…
Thank you sir
manash Athparia said…
I built one simple charger using 7815 which provides output of around 1.2amps but when I charge the battery to 14.4 volts and remove the charger then the battery voltage goes down to around 13v. Is the battery fully charged or it requires more charge if yes upto how many volts I should charge the battery?
Swag said…
It is OK, battery will never stay at 14V after the charger is removed and will return to 12.8V or 12.5V and it can be considered as fully charged....however just make sure you charge the battery at the correct ampere rate as specified by the manufacturer, and for the correct duration of time.
Michael Angelo said…
Awesome! I will build one, but i have question what transformer do I need if i have a lead acid battery with a 12V 5Ah?
Swag said…
Thanks, you can use a 12V, 1 amp transformer.
Michael Angelo said…
1.) What if, I use higher value of transformer example 3amps or 5amps. What's the effect?

2.) What's that thing next to the D1? (N/O N/C) is that the value? If not, please let me know ;)

3.) Lastly, what's the wattage of the resistors?

I hope you help me with this project. ;) More powers to you sir!
Swag said…
1)higher value trafo will damage your battery quickly...charging current should be ideally 1/10th of the battery AH, for lead acid battery

N/O...N/C are relay contacts referring to normally closed and normally open contacts

all resistors can be 1/4 watt 5% CFR
Michael Angelo said…
Sir whats the value of the D2-D5?
Swag said…
use diodes whose current is 2 times your battery AH rating.
Michael Angelo said…
I have 12v 3.5Ah, what diode should I use? Sorry for many questions, I really don't have much idea on electronics but I am enthusiastic in it. I really hope you understand.
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