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IC 555 Lead Acid Battery Charger Circuit

IC 555 Lead Acid Battery Charger Circuit

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 made a mistake on preset R2 and R5, they should not be 10k but 100k, I just made one and it was a success, thank you.

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

SHARING IS CARING!


About the Author

I am an electronic engineer (dipIETE ), hobbyist, inventor, schematic/PCB designer, manufacturer. I am also the founder of the website: https://www.homemade-circuits.com/, where I love sharing my innovative circuit ideas and tutorials. If you have any circuit related query, you may interact through comments, I'll be most happy to help!



226 thoughts on “IC 555 Lead Acid Battery Charger Circuit”


  1. Howdy, Friend! Interested to Learn Circuit Designing? Let's Start Discussing below!
  2. Sir I want ckt dig with details of components and transformer rating to charge 12v 35 to45 Ah lead acid battery .with auto cut off ckt along with power on and full charge led indication..

  3. Thanks Swagatam for taking time to reply,. Yes I have gone through the internals of 555 and have a fair idea Let me get to my point fast.
    1. Have you/ anyone of your reader tried this ckt
    2. When you add the 100k and 47k to the voltage divider, should not the divider “arm” resistance change
    and give different voltages other than the 2/3 , 1/3 of Vcc
    At least i have tried without success and cant figure out why
    Thanks once again. Yours is great site and have learnt immensely

    • Thanks T Choudhury, As shown in the article one of the readers did try this circuit successfully, however I have myself not yet tried this but I can simulate it in my mind and feel that this would work as stated in the article, however according to my analysis only presets must be employed across the sensing pins so that the thresholds can be perfectly adjusted without the interference of the fixed resistors.

      Thanks and keep up the good work

  4. has anyone really cosnstructed this
    Pl let me know , bec tried and failed.
    the upper and lower cut off voltage is derived from the voltage divider. when you add 100k and 47k , it does not look ok
    So the question has anyone made it to wok ok?
    Thanks

    • To make this work you will have to first understand how the IC pinouts are internally designed to respond.

      When the voltage at pin#6 goes above 2/3rd Vcc of the IC, the pin#3 goes low, conversely when pin#2 potential goes below 1/3rd Vcc, pin#3 becomes high.

      Therefore we have to appropriately adjust the two presets such that pin#6 is set at a voltage near 2/3rd Vcc corresponding the battery’s full charge level….and the pin#2 voltage should be set at 1/3rd Vcc corresponding to the battery’s low level tripping point

  5. Hi swagatam!

    I Have question, what’s the value of relay switch? How much volts and amperes of the relay switch. Have transfo of 12v 500amps for secondary and 220v for primary.

  6. 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?

    • 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!

  7. 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?

    • 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.

    • 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.

  8. 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?

    • 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

  9. 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.

    • 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.

  10. *correction*

    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?

    thanks

  11. 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.

    • 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?

      thanks

  12. 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?

    thanks

  13. 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

    • 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

    • 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…

  14. 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

  15. Swagatam,

    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?

    Ralph

    • 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.

  16. 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!!!

    • 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

  17. 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

  18. 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.

  19. "FOR 19.5 VOLT 3.33 AH RATING OF BATTERY"
    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?

  20. "FOR 19.5 VOLT 3.33 AH RATING OF BATTERY"
    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.?

    • 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.

  21. Thanks sir,

    "FOR 19.5 VOLT 3.33 AH RATING OF BATTERY"

    AS WITH GIVEN FORMULA ,
    BATTERY CHARGING RATE =1/10 X BATTERY AH RATING
    THEREFORE ,
    BATTERY CHARGING RATE =0.333 A
    NOW FOR DIODE RATING
    Id (diode current rating)=2 X BATTERY CHARGING RATE
    THEREFORE
    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) )

    • Nikhil,

      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) …

  22. 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.

    Thanks
    Nikhil

  23. 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!

    • 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!

    • 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.

  24. 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

  25. 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.

    • 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.

  26. 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

  27. 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….

    • 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.

  28. 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?

    • 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

  29. 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.

    • 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.

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

  31. sskopparthy,
    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:

    https://homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html

  32. 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.

  33. dear sir i have 20v5A ups transformer i want to charge 12v45ah battery what are the modifications i should do?

  34. 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……

    • 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.

  35. 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

  36. 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

    • Bhanu,
      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.

  37. 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

  38. 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 ………………………………

  39. 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

  40. 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

  41. 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 ? "

  42. 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

  43. 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 ?

  44. 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 ?

    • 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.

  45. Sir i have done the circuit of 12 V charger given as a reference from this link

    2.bp.blogspot.com/-nwMfNvxGD5g/ULSDWeSQ3qI/AAAAAAAABpU/ZLh413MLJwQ/s1600/40%20watt%20led%20emergency%20light%20circuit.png

    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, 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.

  46. 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.

  47. 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

  48. 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

    • I would suggest an opamp based charger circuit because these are more flexible and simple, i have of plenty of them in this site.

  49. hi,

    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 .

    • 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.

    • 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.

    • 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.

  50. 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?????

    • 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.

  51. The figure is containing in this link sir

    2.bp.blogspot.com/-nwMfNvxGD5g/ULSDWeSQ3qI/AAAAAAAABpU/ZLh413MLJwQ/s1600/40%20watt%20led%20emergency%20light%20circuit.png

    • refer to this circuit and see how the 22k resistors are added with the circuit.

      1.bp.blogspot.com/-VMaUoV6aqLQ/UcsSdjCa6TI/AAAAAAAAEqo/NufzQhrsZeE/s1600/48v%20battery%20charger%20circuit%20automatic.png

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

      ignore the msfet stage and use a relay as given in your circuit

  52. 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………

  53. 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

    • 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.

  54. 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

    • 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

  55. 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

  56. 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

  57. SIR I WANT HELP!!!!!
    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

    • 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

  58. 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 ??''

  59. 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 ) ?????

  60. Sir,,
    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.

    • 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.

  61. 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 ? '

    • 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

  62. 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

  63. 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. '????'''

    • 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.

  64. 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 ???????

  65. 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….

    • 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.

  66. Yes!!!!!!!!! The circuit is functioning sir.
    THANK U VERY MUCH…..

    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 ????'??…..

  67. LED connections will be like this:

    positive———>I——^^^^———-pin6———–>I———^^^^————ground

    the resistor shown are 4k7k each

    yes your current resistor calculation is OK.

  68. 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 ) ???????

  69. 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

    • 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.

  70. 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 ? )

  71. 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 ?????

  72. PLEASE HELP ME SIR,
    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

    • 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:

      2.bp.blogspot.com/-nwMfNvxGD5g/ULSDWeSQ3qI/AAAAAAAABpU/ZLh413MLJwQ/s1600/40%20watt%20led%20emergency%20light%20circuit.png

      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

  73. Dear Majumdar, Will i have to replace the 7805 with a higher voltage regulator IC for charging batteries rates above 50 Ah ?

  74. 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

    • 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.

  75. 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.

  76. 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

  77. 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.

  78. 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.

  79. 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????????

    • 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.

  80. 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.

  81. 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.

    Rashid

  82. 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.

    Rashid

    • 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.

  83. 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.

    Rashid

    • 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….

  84. 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.

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