Make a 6v 4ah Automatic Battery Charger Circuit without Using a Relay

The following 6 volt 4 AH battery charger circuit has been designed by me and posted here in response to the request from Mr. Raja, let’s learn the whole conversation.

The Request

“Dear sir, please post a circuit to charge 6 volt 3.5 ah lead acid
battery from 12 volt battery. The charger should automatically stop
charging as the battery is fully charged. 

Please use transistor
instead of relay to stop charging, and also tell me how to use 12 volt
relay for the same circuit. 

Explain Which is safe and durable either
relay or transistor to cut off charging. (At present i am charging my
above said battery by simply using LM317 with 220 ohm and 1 kilo ohm
resistors and a couple of capacitor) i’m awaiting your article, thank you”.

The Design

The following circuit shows a simple automatic 6 volt 4 to 10 AH battery charger circuit using a 12 volt relay.

6V 4AH, battery charger circuit with relay and variable voltage LM317

 The following circuit shows a simple automatic 6 volt 4 AH battery charger circuit without using a relay, rather directly through a transistor.

Update (Jan 05, 2017)

The above transistorized 6V charger circuit has a mistake, and needs to be corrected as illustrated in the following image:

How to Set Up the Circuit

Initially, keep the pin6 feedback resistor disconnected and without connecting any battery adjust R2 to get exactly 7.2V across the output of the LM317, for the IC741 circuit.

Now simply play with the 10k preset and identify a position where the LEDs just flip/flop or change or swap their illumination.

This position within the preset adjustment may be considered as the cut-off or the threshold point.

Carefully adjust it to a point at which the RED LED in the first circuit just lights up……but for the second circuit it should be the green LED that is supposed to get illuminated.

The cut-off point is now set for the circuit, seal the preset in this position and reconnect the pin6 resistor across the shown points.

Your circuit is now set for charging any 6V 4 AH battery or other similar batteries with an automatic cut-off feature as soon as or each time the battery becomes fully charged at the above set 7.2V.

Both the above circuits will perform equally well, however the upper circuit can be altered to handle high currents even up to 100 and 200 AH just by modifying the IC and the relay. The lower circuit may be made to do this only up to a certain limit, may be up to 30 A or so.

The second circuit from above was successfully built and tested by Dipto who is an avid reader of this blog, the submitted images of the 6V solar charger prototype can be witnessed below:

6V, 4ah battery charger prototype breadboard image

341 Replies to “Make a 6v 4ah Automatic Battery Charger Circuit without Using a Relay”

    1. Swagatam

      Hi Tauseef,

      You can use the first circuit, no changes would be needed in the components, just adjust the 2k2 pot so that the output voltage from the IC becomes 14 volts, now adjust the 10K preset such that the green LED comes ON and the relay gets triggered.
      The setting is done, now switch OFF the input supply, connect a discharged 12 battery, switch ON the supply…..the battery will start getting charged and switch OFF automatically when fully charged.

      Reply
    1. Swagatam

      Hi Karthik,

      450V DC is quite high, so presently I don't have any circuit for handling this much voltage, if i get any suitable circuit I'll surely let you know.

      Regards.

      Reply
  1. Muhammad Rizwan

    Hi Bro i have clear my basic concepts i have a question as we know that all step down transformer produce 6,12,24….. volt AC. but we need DC voltage to charge a battery. Question is that how to convert 6,12 colt AC to DC? you Circuit is starting from 12 volt DC Input plz add Converting device or AC Diode any thing suite able in the Circuit . Thnaks

    Reply
    1. Swagatam

      Hi Bro,
      Making a DC power supply is the simplest of all, therefore I didn't feel it necessary to show it here….just type "how to design power supply" in the search box and you will be directed to the article which comprehensively discusses making of AC to DC power supplies…

      Regards.

      Reply
  2. Anonymous

    Very, very useful circuit. Simple as it may seem but it takes patience and time to do it. Nice work sir! I also have a question, if my supply source is already 7.5V DC coming from a rectified 7.5V transformer output, I may not need the LM317, right? So would it still be right to work if I take out the LM317, 2.2k pot, 0.1 cap, and the 240ohm from the circuit? Thanks!!

    Reply
    1. Swagatam

      Thanks!!

      Yes the IC 317 is for restricting the voltage to the desired charging voltage, if the input supply is around 7V, IC317 and the associated R1, R2, C1 won't be needed.

      Reply
  3. Anonymous

    I'm not very clear in second circuit. What is green and red diode works for? Can anyone have idea? Can someone explain detail about how IC 317 works in second circuit diagram. Thank you! 🙂

    Reply
    1. Swagatam

      RED and GREEN diodes are the indicator LEDs. LM317 is a voltage regulator IC, it produces a fixed predetermined voltage at its output terminal, as set by the pot R2.

      Reply
    2. Anonymous

      Is there any circuits diagram for 6V 1A battery charger? Can I put the 6V 1A battery instead of 6V 4AH? Oh… what I mean LEDS about is that when red led light up and when Green LED light up? RED LED will light up when charging or Green LED turn on when charger is full? Isn't it? Thank you for your response. Coz this circuit may useful for my Final year project.

      Reply
    3. Swagatam

      6V/1AH battery can be used, just make sure the transformer is rated at 200 mA or at the most 500 mA.

      Red led lights up when the battery is fully charged, while green indicates the battery is being charged.

      Reply
  4. Anonymous

    Thank you for your immediate response,Sir. I'm really appreciate it. I think your circuit can help me alot. 🙂

    Reply
    1. Swagatam

      charging time is standard and should be done for 10 hours ideally for lead acid batteries. use a 200 mA transformer for this.

      The input power decides the charging time, the circuit only controls the voltage and the over charge.

      Reply
  5. Vinod KC

    Hi Swagat,
    I need to make 6v 4.5Ah battery trickle charger. I do have a 10v 0.6A transformer. Is it possible to use LM317 with this transformer?. I mean the 3volt difference between in and out. Please help.

    Reply
    1. Swagatam

      Hi Vinod,

      You may use a 317 IC for making a trickle charger using the said transformer.

      The voltage difference can be adjusted using the preset.

      However I think for trickle charging, you can simply rectify the 10V and connect it directly with the battery with a 1K 1 watt series resistor, that will take care of everything.

      Regards.

      Reply
  6. Anonymous

    Hi Sir,
    Thank you very much for your circuit. I have something which produce 13 volts, and I want to use it to recharge a battery. The battery is 6V 20Ah Lead Acid. Can I use your upper circuit? What have I got to change to supply the battery with 2Amps?
    Thanks in advance!

    Reply
    1. Swagatam

      Hi,

      You can use the upper circuit. You will have to do just a few modifications:

      Replace LM317 with LM338
      Adjust R2 to produce 7 volts at the output.

      Also connect a 470K resistor across pin#6 and pin#3 of the IC.

      Reply
    2. Edison Alejandro

      hello sir,
      can I use upper circuit to charge my 6volts 4.5 ahmps lead acid battery, but I need the battery to be charge in 1hour only,
      kindly help me what will I have to modify on the circuit?
      thanks in advance…

      Reply
    3. Swagatam Post author

      Hello Edison,

      The ideal and safe rate is to charge a battery for 10 hours at 1/10th of its AH value, lead acid batteries cannot be and should not be charged at quicker rates, that would cause permanent damage to the cells.

      Reply
    4. Edison Alejandro

      Sir, thank you very much for your quick reply,,

      but still i would like to learned how to make a fast charger, I hope you can help me,,, this is my project at school, were going to use it for electronic experiment…

      thank you in advance…

      Reply
  7. Daj

    Hello Sir. im planning to make a power supply for arduino microcntroller at the same time while supplying the microcontroller i want to put a switch to charge a 6V 4Ah battery . but i dont know what is the better transformer capacity for this.? Hoping for your kind consideration Sir.

    Reply
    1. Swagatam

      The ideal rate of charging all batteries is 1/10th the AH of the battery, so in your case it's 4/10 = 400mA.

      A 0-6V transformer connected to a bridge rectifier and 1000uF/25V capacitor may be used, the current can be around 750 to 1Amp from the transformer.

      Reply
  8. Umar Abubakar

    Hi Sir, your circuit is very interested. I have a 12V solar panel and I don't know it specifications. Can I use it as my power source for your battery charger circuit?
    Thanks.

    Reply
  9. Daj

    is it okay if i used 12v transformer and 7808 voltge regulator and have a series resistor having voltage drop of 2v causing and 0utput of 6v.Thanks.

    Reply
    1. Swagatam

      No that won't do, you can apply 8V directly to the battery by using a transformer whose current is 1/10th of the battery ah.

      However the 7808 is not suitable for charging batteries above 4 ah.

      Reply
  10. Mike

    good day sir
    can you explain what function the 3v zener has? also should the circuit diagram have a second zener in series with the collector of the xsistor to ground? I am having trouble finding a 3v zener locally, can i uns a different rating ie. 12v diode with a resistor in series?

    Reply
    1. Swagatam

      good day Mike,

      the zener provides a reference threshold for the opamp. the zener can be of any value between 3 and 8, but it cannot be 12V because it has to be a lot lower than the supply voltage.

      Collector zener is not required.

      Reply
  11. Daj

    do you have any suggestion that i can what are the materials im gonna used. 12v iamp transformer. . what will be the good regulator i will use?. so that it can charge my battery and the same time supply my 5v microcontroller.

    Reply
  12. Daj

    thanks you very much Sir. .so I need to use 1 amp transformer i brigde rectifier ,1 1000microfarad capactor, and parallel 7805 and 7808. .thanks man. or else i will use two power supply with one transformer. hehe. thanks for the idea Sir. 🙂

    Reply
    1. Swagatam

      That's correct Daj, put the IC pins in parallel except the output pin which should be separately terminated so that respective voltages can be received from them….

      Reply
  13. Daj

    thanks Sir..im just little confuse sir. you that 7808 is not suitable for greater 4Ah. i just want to clarify that is it ok to use AN7808 direct to 6v 4aH?. .what is better?. thanks Sir.

    Reply
  14. Daj

    Hello Sir, Good Day!
    is it okay if i will always connect the 6V 4AH battery from my 7808 regulator? Hoping for your valuable reply Sir. 🙂

    Reply
  15. Menanti

    Hi swagatam, me again.
    1. Could you explain the function of 10K pot?
    2. If the voltage output from 317 change, i.e 3.7v. How about upper and lower value of the cut off? Or it can be set using 10K pot? How to set it?
    3. Is there any part should be change if i use battery with different value?
    Thanks a lot.

    Regards,
    Menanti

    Reply
    1. Swagatam Post author

      Hi Menanti,

      The 10K preset is for setting the high voltage cutoff threshold, voltage level at which you want the battery to be disconnected from the supply.

      With different output voltages from the IC317, the 10k preset will need to be re-adjusted accordingly.

      A low voltage restart has not been included in this design.

      The last circuit given here is better equipped:

      http://homemadecircuitsandschematics.blogspot.in/2012/07/making-simple-smart-automatic-battery.html

      Regards.

      Reply
  16. Bibin Edmond

    hi swagatam,
    Can u explain the working of this circuit.. ? The second circuit without Relay.. ! How the red light works while battery is fully charged.. ? And how the circuit knows whether the battery is fully charged or not.. ? and please explain the cut off function also.. !

    Regards,

    Bibin Edmond

    Reply
    1. Swagatam Post author

      This circuit will not cut off and hold fully, it will siwtch rapidly at the cut off voltage threshold which will keeping supplying the battery with the charging voltage, though at lower current.

      Reply
    2. Swagatam Post author

      Hi Bibin,

      The circuit needs a little correction, the input pins must be interchanged for the second circuit.

      Initially when the battery is not fully charged, pin2 voltage stays below that of pin3, so the output begins with a logic high, and the transistor conducts charging the battery.

      Once the battery gets fully charged, pin2 voltage goes higher than pin3, the output now reverts and goes low, switching of the transistor and also the supply to the battery.

      Reply
    3. Bibin Edmond

      swagatam yaar…

      i'm so confused yaar… i already bought three 6v 5ah sla batteries… i need a good charger with following features.

      1. Battery full Indication (LED)
      2. Battery full Auto cutoff ( Battery overcharge protection)

      In Kerala we got 1 hr power cut( Loadshedding)daily, so i need those battery charged for my lighting backups…

      so brother please send me a suitable circuit with above features..

      Regards,

      Bibin Edmond
      Bibinedmond@gmail.com

      Reply
    4. Swagatam Post author

      Hey Bibin,

      Don't get confused, you can make the second circuit shown in the above article, it's perfect in every respect as per your needs.

      The input to the circuit can be fed from a standard 12V 1 amp adapter.

      To set up the circuit initially do not connect any battery.

      Feed 12V input, adjust the 2K2 pot to get 7v across the battery charging terminals.

      Next, adjust the 10K preset such that the green LED just lights up fully and the red LED shuts off.

      Finished, your circuit has been set.

      Switch OFF power…. connect a discharged battery….and switch ON power, the circuit will do the rest….it will cut off as as soon as the battery voltage reaches 7V.

      Regards.

      Reply
  17. Daj

    How about the reverse current going to my power supply?. is it ok? if i will put diode, 6v will become 5.3v? does this affect my battery?

    Reply
    1. Swagatam Post author

      true….you will have to include a diode just after R1, at the right side……

      use 9 volt supply at the input and adjust the preset R2 to get 7V after the diode.

      Reply
  18. Anonymous

    Hey Swagatam,

    I have designed the circuit with the relay…Whenever I 'ON' the circuit and try to adjust the 10k pot of the LEDs properly swaps its places on adjustment of pot. But my relay stays on irrespective of the state of LED's. When i check the voltage at the PIN6 of OP_AMP 741, it shows around 12V, but at the Base of transistor BC547 it shows 0.78V approx. The whole 12V gets drop at 10k resistor between PIN6 of OPAMP and Base of transistor. I tried changing the value to 5k but was unsuccessful. Can you guide me on this issue pl.

    Regards

    Nikhil

    Reply
    1. Swagatam Post author

      Hi ikhil,

      Connect a 3V zener diode at the base of the transistor, in series with the existing 10K resistor, this will switch ON the relay only when the RED led is ON.

      Reply
  19. Anonymous

    Hi Swagatam..

    Thnx for the reply what how should i connect the zener.. Kathode side towards PIN6 of OPAMP or show i place it opposite…

    Regards
    Nikhil

    Reply
  20. Anonymous

    Ok.. Because I tried placing the zener between the PIN6 OF OPAMP and 10K resistor…My OPAMP IC just got heated up..Will try ur suggestion. Also Can I know why this BC547 transistor is behaving so Odd and why the voltage is getting drop across 10k resistor.

    Regards

    Nikhil

    Reply
    1. Swagatam Post author

      you can place it between pin#6 and 10K also…no issues, it will still work.

      the 10K has been placed to control the voltage so that the transistor gets around 0.7V when its conducting, theerfore it's showing a drop in voltage at the base.

      The IC should NOT heat up….something's wrong with your connections.

      Reply
    2. Anonymous

      I tried getting the zener inbetween the said points but not to much help… But I noticed one thing the LED i connected to the transistor instead of relay, gets dimmer when i change the pot value of Pin3 of OPAMP

      Regards
      Nikhil

      Reply
    3. Anonymous

      I check with two more IC's… But the condition is same.. Any other Idea. Because initially even the RED LED at the output used to glow nicely, but with ur modification it atleast gets dim.

      Regards
      Nikhil

      Reply
    4. Swagatam Post author

      If the LEDs are swapping positions while adjusting the 10K preset it means the IC is responding properly, it seems the transistor is faulty or connected incorrectly……. it can be only reason.

      Reply
  21. Anonymous

    Hi sir, can i use LM393 instead of IC741, similarly 4.7K POT instead of 2.2k POT and 3.6 .5w zener instead of 3v 400mw zener in the first one circuit. Also i dont want the lEDs so if i remove them then what will be further modification.

    Reply
    1. bashir ahmad

      Hi Sir, i want to ask a few question about the first circuit using relay….Plz explain me in detail because i am confused.
      1) you have used a RELAY on the right most side of the circuit, while a relay diagram is also there on the top of the battery. Is this the same relay of the right most side of the circuit OR another relay has been used.
      2) If this circuit has only one RELAY been used on the right most side, then please make clear the CONNECTIONS coming from LM317T.
      3) thirdly is this 5 pin relay been used.

      Reply
    2. Swagatam Post author

      Hi Bashir,

      1) it's a single relay unit, the relay coil is connected to the transistor while the relay contacts are wired with the battery supply.

      3) yes it's a 5 pin relay

      Reply
    3. bashir ahmad

      Sir thanks alot for your brilliant support and quick response…some more questions sir if you dont mind…

      1)i am using 6v 4.5ah sla battery so will it require any further changes
      2)the connection from LM317 is connected to the N/O (i.e OPEN PIN OF RELAY), am i right. sir why not to use the N/C pin of relay to avoid any current lose
      3) should a diode be placed after R1 to avoid backward flow of current to the IC from the batter, if so which one 1N4007 or 1N4001 or 1N4002
      4) can 12v @ 1A transformer be used

      Reply
    4. Swagatam Post author

      Hello Bashir,

      1)no changes would be required for 6v 4.5 ah battery,
      2)if you are using the N/O of the relay for charging, use an PNP transistor (BC557)
      3) ysea diode must be placed for avoiding reverse current flow.
      4)use a 0-12V or 0- 9v 500 ma transformer ideally, but 1 amp will also do.

      Reply
    5. bashir ahmad

      Dear Sir, with 470 Ohm resistor connected from LM317 Vout to battery positive Terminal, Only 330uA curretly flows to the battery when battery voltage is 6.5v. Sir when i remove the 470oHM resitor then the current flow becomes 49MILIAMPERE which the right flow i think. So sir please help me why there is limited current flow to the battery when 47o ohm resistor is connected.

      Reply
    6. Swagatam Post author

      The 470 ohm resistor has not much importance, it's introduced for trickle charging the battery after the relay has switched ON (when battery gets fully charged). As long as the relay is switched OFF, the battery would get the supply directly through the relay contacts (470 ohms would be bypassed by the relay contacts).

      I think there's something wrong with your circuit connections or you haven't understood the circuit operations properly.

      Reply
  22. bashir ahmad

    sir Thanks for Reply. Sir i have made so far the LM317and voltage divider section before proceeding to the next portion. sir in your circuit you have connected the two PINS of THE POT and connected it with the negative rail and the other pin with the ADJ pin of LM317.
    while i have done it otherway round, i have connected the two PINS of the POT with the ADJ of LM317 and the one pin left of the POT with the negative rail. will it make difference.

    Reply
    1. bashir ahmad

      Dear Sir,
      1) In the first circuit you have connected 470 Ohm resistor to the positive terminal of the battery and in the second circuit you have connected it to the negative termninal of the battery. please explain..
      2) In the second circuit you have connected 100K resistor to the PIN 2 and PIN 6 of LM741 IC. Please explain
      3) In both cicuits, which one is the most reliable and accurate in functioning…
      4) You have posted "PLEASE CONNECT A 3V ZENER DIODE IN SERIES WITH THE EMITTER OF THE TRANSISTOR AND GROUND, FOR ACCURATE RESPONSE.
      " please explain its function and will this zener be needed in both circuits

      Reply
    2. Swagatam Post author

      Dear Bashir,

      the 470 resistor or any resistor in that position needs to be bypassed by the relay or transistor. In the first circuit the relay contacts are connected with the positive, therefore the resistor is also connected to positive, in the second circuits it's the opposite, the negative of the battery is connected to the transistor, and therefore the resistor is connected to the negative.

      The 100k resistor is for hysteresis control in order to latch up the output when the battery gets fully charged, otherwise the output will vibrate at cut off point. It is required in the both the circuits.

      Both the circuits are identical and same.

      zener is required for both the circuits.

      While drawing it becomes difficult to include and check everything therefore these mistakes are present in the diagrams.

      Reply
  23. bashir ahmad

    dear sir, when put protection diode (1N4007)in series with OUTPIN (PIN No:02) of LM317 then the current drops dramatically. will the internal resistance of the diode is limiting the current. if so then plz suggest me the diode which have lowest reistance..

    Reply
    1. Swagatam Post author

      Dear Bashir,

      the diode is there only for preventing the battery voltage from entering the circuit so that it cannot discharge.

      It does not have any relation to your issues. or may be your diode is faulty.

      Reply
    2. bashir ahmad

      Dear Sir, i am really thanful to you for your brilliant support and you are great person. Sir, have some more questions if you dont mind…
      1) In the Second Circuit you have used 1N5402 diodes so can i use 1N5408 instead of the one mentioned
      2)you have connected 1N5402 with the EMITTER of transistor in the second circuit. should that diode be connected in the first circuit also..

      Reply
    3. Swagatam Post author

      A diode at the emitter of the transistor may be required for preventing leakage voltages from the output of the IC, which might trigger the transistor early.

      use a 1n4007 for the relay circuit, and 1n5402 for the transistor circuit.

      Reply
  24. micho dalanon

    we've tried making the circuit.. but the green
    LED always lights up even if theres no charging happening.. is that ok? what do u think is the problem if there is..?

    Reply
    1. Swagatam Post author

      the 10k preset should be set to keep the red led switched off until the battery is fully charged.

      when the set threshold is reached, the red led should illuminate and the green should shut off.

      remove the feedback 10k resistor (if you have included) across pin6 and pin2 and then test the results.

      Reply
  25. bashir ahmad

    Dear Sir, you have connected 10K POT to PIN-No 03 in the first circuit while you have connected the same POT to PIN-NO 02 in the second circuit… plz explain, the diagrams has confused me

    Reply
    1. Swagatam Post author

      both the circuits should have the same components and connections except the relay….I'll correct the diagrams soon so that they become identical.

      Reply
  26. bashir ahmad

    Dear sir, green LED comes on Successfully, but the RELAY is not triggering what could be the problem..Although i have checked the Relay alone and its working . plz help

    Reply
  27. bashir ahmad

    Dear sir, thanks for your reply and support. some more questions if you dont mind.
    1) you have mentioned in your posts that while setting the 10K POT the GREEN and RED should swap, and when GREEN LED comes completely on and RED LED goes off the RELAY should trigger. But in my case the LEDs does not swap, only the GREEN LED comes ON and and RED LED remains OFF and RELAY doesnot trigger in the initial setting of 10K Pot before connectiong the battery.
    2) In which circuit (First or Second) the IC 741 has correct PIN connection. Because in the first circuit you have connected the 10K POT to PIN-3 of the IC while in the second ciruit you have connected the same POT to the PIN-2 of the IC. So i am confused whether 10K POT should be connected to PIN-2 or PIN-3

    Reply
    1. Swagatam Post author

      Dear Bashir,

      There might be some problem with your connections or the IC, because adjusting the preset should toggle the output condition.

      Both the circuits are correct, in the first we want the transistor to switch ON after the battery gets fully charged, while in the second we want to do just the opposite, here we want to switch OFF the transistor when the battery gets fully charged.

      Reply
  28. bashir ahmad

    DEAR SIR,
    1) When i configure the 10K POT the GREEN LED come ON successfully, but the RELAY sounds buzzy (like bee buzzing), where i am making the mistake, please correct me.

    2) Are the two LEDs connected to the junction of the PIN-6 and 10K resistor of Transistor as shown in diagram. if not then please tell me because i have connected them.

    Reply
    1. Swagatam Post author

      Dear Bashir,

      1) that should not happen, absolutely not.

      2)yes the LED junction is connected to pin#6 of the IC……. do one thing make the LED resistors = 10K and then check the results.

      Reply
  29. bashir ahmad

    Dear sir,
    1) I have made the second circuit. I checked each and every connection according to the diagram every connection went well. But sir, when i am going to adjust the 10K POT, the GREEN LED lights up and RED LED negligibly glow also, very very little glowing, is there a chance that IC 741 might be a faulty one.

    2)Also sir, when i switched off MAIN POWER supply, the GREEN LED Still glow taking current back from the battery, how should i resolve this problem as well.

    3) How to check IC 741 alone whether is in working condition or a faulty one.

    Reply
    1. Swagatam Post author

      when you adjust the preset, the leds must swap from red to green and vice versa. The low glow may be due to parasitic leakage, it's not a fault…. but the leds must swap while adjusting the preset.

      Read this article to know how an opamp works like a comparator, then you may understand the actual principle behind it:

      http://homemadecircuitsandschematics.blogspot.in/2012/03/how-to-use-ic-741-as-comparator.html

      add a 10k resistor with the LED so it will not consume much power from the battery, because we cannot isolate the 741 from the battery otherwise it won't sense the low voltage threshold and revert the charging process.

      Reply
    1. Mohammad Jasim

      I have a 12v 1 amp transformer, I want to 6v 4.5ah and 12v 7.5ah two battery how to charge these battery individually. if I use the second circuit which component is change for those battery individually. plz help me for……….

      Reply
    2. Swagatam Post author

      no modifications would be required, just adjust the preset for getting 7V for the 6V battery and 14V for the 12V battery before connecting them.

      Use 1 amp transformer for both.

      Reply
  30. jennifer

    One confusion: You mentioned, "PLEASE CONNECT A 3V ZENER DIODE IN SERIES WITH THE EMITTER OF THE TRANSISTOR AND GROUND, FOR ACCURATE RESPONSE" and you also mentioned both circuits require this zenner to someone's response above but there is one already in your both circuit connected in series with 10K resistor. Is this the second zenner we need to complete the circuit? I am trying to make the second circuit, for that can I place this zenner in between diode 1N5402 and the emitter of Tip122? Another question is there any specific reason you preferred 1N5400 series diodes over 1N4000 series diodes as you replied Bashir, "use a 1n4007 for the relay circuit, and 1n5402 for the transistor circuit." And what about the one that is attached at LM317 out on second diagram? Can that be replaced with 1N4000 series? I am just trying to understand. Thank you.

    Reply
  31. bashir ahmad

    Sir,
    1)In the 1st Circuit you have connected the 10K POT to NON-INVERTING INPUT OF IC (PIN-03), because here you want to switch ON the Transistor after battery get full charge. am i right?
    2) While in the 2nd Circuit you have connected the 10K POT to INVERTING INPUT OF IC (PIN-02), because here you want to switch OFF the Transistor after battery get full charge. am i right?

    Reply
  32. bashir ahmad

    Sir,
    You have connected two parallel 1N4007 diodes in series wiht LM317 for purpose of preventing reverse current. but i have always seen one diode being used in most of circuits. Using two diodes in parallel, will it make some difference in efficiency or performance. why not to use one diode.

    Reply
    1. Swagatam Post author

      two diodes would provide more current handling capacity, around 2 amp.

      you can also use a single 1N5402 diode in place of two 1N4007.

      1N4007 will get hot at 500mA.

      Reply
  33. bashir ahmad

    dear sir,
    I want to use 5 watt high power LED, as LED needs constant current source so please post a circuit which can be adjusted through POT to desired value of constant current and which has also brightness option and which can be integrated with the above circuits for delivering constant current to LED.

    Reply
  34. Amrit Mandal

    sir i have successfully made the circuit using tip122. but during boxing the circuit i had replace the RED led i used previously with long RED led. after that the circuit not working unfortuately. all the connections are okay and the ICs are perfect. sir is there any problem regarding the Voltage/power of the newly one used led in my circuit?….pls help me sir.

    Reply
    1. Swagatam Post author

      You might have shorted the IC pins, not sure exactly what may be wrong.

      Clean the board with thinner and check for shorts or leakages.

      Reply
    1. Swagatam Post author

      yes it can be used, adjust R2 to get 4.2V across the battery terminals without any battery connected.

      use a 0-12V 500mA ac/dc adapter for the input source

      Reply
    1. Swagatam Post author

      Did you finish building the circuit? Please make ti and then I'll explain how to proceed. you will need a digital multimeter for setting up the circuit.

      Reply
    2. Swagatam Post author

      if you are using the second circuit, do it in following manner.

      initially do not connect any batteryand also disconnect the 10K at pin2 of the IC

      keep volt meter prods across the points where battery would be connected.

      adjust 2k2 to get 7V here.

      now adjust 10k preset such that the green LED just lights up.

      setting is complete.

      now reconnect the 10k resistor at pin2.

      now you can connect the battery for charging, it will be automatically cut off when its voltage reaches 7V

      Reply
    3. Swagatam Post author

      for 10K, connect the center lead to the IC pin, connect the outer two leads to positive and negative supply points.

      for 2k2, short the center and any of the outer leads of the pot, connect this joint to ground while connect the other free lead to the ADJ of the IC LM338

      Reply
    1. अमृतमन्थन

      sir,
      i checked the circuit on bread board which was fine with 12 volt dc but on pcb board i use 9 volt 200 mA transformer and the battery charging (1000 mah 3.7volt mobile battery) not happening. Was it insufficient input?
      I ve set 5 to 6 Volt Across battery terminals. And also both LEDs also not illuminating afer adjusting 10k pot
      Please guide

      Regards

      Reply
    2. Swagatam Post author

      If you are using the second circuit, do it in following manner.

      initially do not connect any battery and also disconnect the 10K at pin2 of the IC

      keep volt meter prods across the points where battery would be connected.

      adjust 2k2 to get 7V here.

      now adjust 10k preset such that the green LED just lights up.

      setting is complete.

      now reconnect the 10k resistor at pin2.

      now you can connect the battery for charging, it will be automatically cut off when its voltage reaches 7V

      since it's a i-ion cell, you can try using a higher 1amp transformer if 200mA is not providing the required charging current.

      Reply
    1. Swagatam Post author

      Helo Tanmoy,

      connect the power supply input to the IC LM317 and connect the emergency light transistor emitter directly to battery positive and also connect the LM317 output to the base of the transistor.

      Reply
  35. Anonymous

    I am not able to understand what is N/C in the upper circuit, strait above the
    battery's +ve terminal.

    Reply
  36. Tanvir Ahmed

    Dear Sir,
    I want to use 2nd circuit for 12v 7.5 ah battery charger using source from 12v solar panel. so please help to which component need to modify?

    How to connect 5 pin 12 relay in 1st circuit?

    Reply
  37. Anonymous

    sir can u suggest easiest usb mp3 player circuit diagram without micro-controller even without display

    Reply
  38. Anonymous

    Sir
    Thanks a lot for making this awesome thread. I learnt a lot from all these conversations.
    I have made the second circuit and want to charge a 3.7V Li ion battery @1500mAh. Though I have a power supply of 9V I thought how about using a computer smps for the input. So here is my question, can I use the 5V from smps as input voltage for my charger? The smps has a rating of 45Amp max on 5V and this is worring me. So can i use smps or should i stick to the 9V 1amp power supply?

    Reply
  39. Anonymous

    Hi sir
    I have a question not directly related to this circuit but out of curiocity. We know the lm 317 is rated at 1.5A. I was thinking if we could use two lm 317 in parallel ( solder pin 1 of 1st ic to pin 1 of 2nd ic and so on for the other pins then using both ic as a single ic ) to get more output current like 2.5A? If it is not possible can u explain?
    And if possible then how many ICs can we put together in parallel?

    I found LM 350 costs 40 rupees against LM 317 @ Rs 10 only. So two 317 in parallel would produce power same as a 350 but at half its price.

    Thanks
    Rahul

    Reply
    1. Swagatam Post author

      Hi Rahul,

      Yes technically it looks possible, you should mount both the ICs over a common shared heatsink plate, this will ensure uniform current dissipation and avoid abnormal functioning.

      Reply
  40. Anonymous

    Bro, regarding the second circuit you asked to disconnect the 10k resistance connected to pin 2 of ic and then do necessary adjustment. But I cant find any 10k attached to pin 2 of ic. A 100k is attached to pin 2 of ic! So which resistor to remove during adjustment?

    Abishek

    Reply
    1. Swagatam Post author

      The diagram was modified later on may be…. It's the 100K resistor which needs to be disconnected before setting up the 10K preset.

      Reply
  41. Anonymous

    Bro, regarding the second circuit you asked to disconnect the 10k resistance connected to pin 2 of ic and then do necessary adjustment. But I cant find any 10k attached to pin 2 of ic. A 100k is attached to pin 2 of ic! So which resistor to remove during adjustment?

    Another question is, I have a 6v 2Ah lead acid battery. How can I understand that the battery needs recharging? Or in other words what would be the battery voltage when it is low and needs recharging?

    Ankit

    Reply
  42. Amlan

    Sir this is the best work i ever seen! I have a question? i make 1st circuit. i used 6-0-6 600 mA transformer. when i power on both led are act. i didn't understand why both led are acted?

    Reply
    1. Swagatam Post author

      If your circuit is correctly build then that shouldn't happen, because the opamp output can either be positive or zero, under these conditions only one LED will be lit, never both. Recheck and confirm for any issues with the connections…..

      Reply
  43. Amlan

    Brother Swagatam,
    Before the power on i connect the battery and green led is lit.Is it OK or not? i used 6-0-6 600mA transformer and and 6v 4.5AH battery. battery is not heat up. i didn't understand is it charging or not? relay is energized also. i get low voltage 5v from battery terminal. please give me some advise what should i do now? I designing pcb for it. i will send it to oyu later. Good Night…

    Reply
    1. Swagatam Post author

      you should first set up the preset. Remove the battery and adjust the 2k2 pot to produce a 7V input to the circuit (assuming the first circuit on this page).
      Now adjust the 10 preset such the relay such switches OFF (deactivates)

      While doing this keep the feedback 10k across pin6 and pin3 disconnected, reconnect it when the adjustments are done.

      Reply
    2. Swagatam Post author

      …after this you can connect the battery, it will start charging, and once its voltage reaches 7v, the relay will disconnect it and stop any further charging.

      Reply
  44. Tanvir Ahmed

    Dear Sir,
    I have made 2nd circuit. when output -v connect to battery from transistor collector but unfortunately couldn't charge the battery. after when input negative voltage connect to the battery and its charging to battery with heating LM317.
    I think transistor couldn't be on. already replace the TIP122 transistor but same problem exit.

    Reply
    1. Swagatam Post author

      Dear Tanvir,

      the transformer should be rated at 1/5th or 1/10 of the battery AH.

      you should first set the Lm317 pot to get the required charging voltage and then set the 10K preset to switch off the transistor precisely at that voltage. This setting must be done without connecting the battery.

      Reply
  45. Amlan

    I finished the circuit successfully. Thanks my dear friend.
    I designed a pcb for it. I want to send it. Please give your email address.
    Thanks

    Reply
  46. Tanvir Ahmed

    Dear Sir,
    As per your instruction I have repeat the
    previews adjustment, first adjust 2K2 POT
    for required output charging voltage & swap
    the 10k variable resistor with successfully
    Green LED on and shut off red but not fully
    off with slide lighting . but could not
    charging when connect negative voltage from
    TIP 122 (Collector). after input negative
    voltage connect to battery terminal & it
    charging to battery with heat to LM 317.
    please help me to find-out the problems.
    for your information I have attached picture
    file of circuit.
    1. How to off & on TIP 122 transistor with
    which voltage need to done this job?
    2. please explain the function of base
    Zaner diode of TIP 122 transistor?

    Below the voltage status of circuit
    Input voltage 11.94 with 5A
    output regulated voltage 7.77 after adjusting the 2.2k variable.
    IC PIN 2 voltage =7.77
    IC pin 3 voltage =7.08
    IC pin 6 voltage =2.17 after swap the 10k
    variable & before swap 10kV voltage was V6.74
    TIP122 Base voltage after swap 10k variable
    140 mV and 0.657V before swap the 10k
    Variable.

    Thanks & Regards
    Tanvir Ahmed
    Bangladesh
    http://facebook.com/cse.tanvir

    Reply
    1. Swagatam Post author

      Dear Tanvir,

      You have done just the opposite of what was required to be done.

      Without connecting any battery, set the 2k2 pot for the required output, next adjust the 10k preset to just switch ON the RED LED.

      After this switch off the circuit, connect the battery, and switch ON power. Now you will find red LED switched OFF and the green LED coming ON…once the battery gets fully charged, the RED will again switch ON indicating that the battery is charged and disconnected.

      While setting 10K preset remember to disconnect the 100k feed back resistor….connect it back after the setting is done.

      Reply
  47. apudiu1

    Hi,
    I am sure this circuit is very good. But I need your help here. This circuit is for 6v 4ah battery but I have 6v 4.5ah (1) battery. and I also have 4v 3ah (1) and 3v 2ah (1) total 3 batteries that I bought few days ago.

    Please help me build charger for the battery 6v 4.5ah battery. and also please tell me can I use Nokia charger (5v 350 ma) for two of my other batteries directly to charge?

    I will be very happy if you help me.

    Regards,
    MD

    Reply
    1. Swagatam Post author

      Hi,

      You can use the above circuit for charging all the three batteries that you have bought by suitably adjusting the given presets in the circuit.

      You can use the 5v charger supply with the above explained circuit for charging the 3V batteries.

      Reply
  48. apudiu1

    Thank you very much for you very useful information. But can you please tell me which circuit should I built for my case ? the first one or the second one ?

    And Kindly please tell me which presets I should change to charge 3v & 4v batteries ?

    I understand that I can use the Nokia charger to power the circuit to charge the 3v & 4v batteries, but in this case I need any parts change on the circuit ?

    Waiting for your information.

    B. Regards,
    MD

    Reply
  49. Anonymous

    dear swagatham,what is the maximum voltage and current which can be fed to 6v4.5ah emergency battery for charging.kindly give the details.

    Reply
  50. Anonymous

    Hi Swagatam, thanks a lot for posting this schematic. I'm a bit confused adjusting the 10K pot and the function of the relay.

    1. I adjusted the 2K2 pot to get the 7V then turned to 10K till I heard the relay coil works but the Red LED glows while Green lights up pretty weak. I moved the pot to the max and the red dims while the green illuminates kinda weak. I verified the wiring thrice for two days and I still can't figure out what's the problem. Could it be defective LEDs? I tested them before soldering.

    2. Since the relay is connected only to the coil and not thw switching part how will it shut down when the battery charged?

    Reply
    1. Swagatam Post author

      1) It's due to parasitic leakage from the IC output, it can be corrected in the following manner.
      Swap the positions of the transistor base resistor and the zener, next shift the LED joint in between the resistor and the zener, this will clean up the issue.

      2)The relay contacts can be seen connected with the battery positive, so as soon as it activates the battery gets disconnected from the supply at the full charge conditions.

      Reply
    2. Anonymous

      Hi Swathigam, I did the change as you've mentioned but still it looks the same. Plus the relay always activates regardless of the 10K pot's position. I checked the part numbers to verify and found out that instead of 1N4007s my electronic parts seller has given me 1N5819s the next day when I asked him why he did that he said it wouldn't make much of a difference although I seriously doubt that now. Could diodes be the cause of the problem?

      Just to verify if I got the circuit wiring right ; you two 1N4007s to the 240 resistor and one between the relay coils right?

      Reply
    3. Swagatam Post author

      It means your opamp is faulty or there's some other kind of mistake, because altering the 10K preset should certainly make the relay flip.

      I think it would be better if you refer and learn how an opamp functions, it will help you to troubleshoot the issue immediately from the core. there are many online sources dealing with this subject

      By the way did you remove the feed back 10k resistor while doing the testing, it is important otherwise the relay will not respond properly.

      The relay N/C contact connects with the diodes, diode should be a rectifier diode rated at least at 1amp, any type will do.

      Reply
  51. Anonymous

    I want to use 5v ,800mah cellphone charger to charge the 6v 4.5AH battery, i changed the zener diode of the circuit,6.2 volt zener to 7.5 volt,but the transformer of the smps charger got blasted every time. Please bro help me…what else must i do to modify the circuit to charge 6V, 4.5Ah battery

    Reply
    1. Swagatam Post author

      you can try a readymade 12V smps 1amp adapter, and regulate the output using an LM317 circuit…that would be much clean and easy.

      use a good heatsink for the LM317

      Reply
  52. Anonymous

    Sir, in the second circuit does the GREEN and RED LEDs are connected to PIN-06 of IC —-OR—- Green LED is directly connected to 1K resistor of RED LED.

    Reply
  53. Anonymous

    sir, i used 50K POT instead of 10K POT in second circuit. so will it make difference. secondly, when i adjust the the POT the both LEDs swap, but during adjustments the When the Green LED lights up the RED led slightly glow.

    Reply
  54. P Ramamurthi

    Sir
    I am interested to make a battery charger for 12 volt battery.But when I went shops to get parts they could not understand relay for 12 volt battery charging.What specifications like any series number should i tell so that I can get it
    Thanking you
    P Ramamurthi

    Reply
  55. Amlan

    hello, i tested the ckt some month ago. after that i was an accident and do not finished. now i again test. but something is wrong. when i switch on the power of ur ckt my relay is energized and green led is lit. i bought new 6V 4.5A battery and connect to the ckt. but my battery do not charge. plz give some advice to me. i use 3.3v 1w zener diode.

    Reply
    1. Swagatam Post author

      Without any battery connected the input voltage should be 7V, at this voltage first adjust the 10k preset such that the relay just cut-off or deactivates.

      Once this is done, then connect the battery and switch ON power, now the relay will not activate until the battery is charged.

      While adjusting the 10k preset remember to keep the pin6 10k resistor disconnected…..connect it back after the adjustment is finished.

      Reply
  56. Anonymous

    2.2k pot is not available in my locality. What can i replace it with. I have a 5k ceramic variable resistor, can i use it.

    Reply
  57. Skl Dj

    please clear the confusion, 2nd design tip122 transistor emiter and collector are joined to 470R resistor with ground,i think collector should go to positive. what is the purpose to use 3v zener at the base of tip122?
    if i want to use 2nd circuit to charge cellphone battery (3.7v,800mah) with USB power source, what modification should be done?

    Reply
    1. Swagatam Post author

      the TIP122 connection is correct, rest assured.

      It will be difficult to explain all the modifications for a USB 3.7 charger, because there are many changes to be done.
      I'll post the compete design in my blog and provide you with the link soon.

      Reply
  58. Anonymous

    Hi Swagatam, I have assembled your first circuit that with relay in project board. Everything seems to work as per your description but a there is a slight problem which I wanna state below.

    Actually I want to combine this circuit with one of your emergency led light circuits in which the white leds are controlled by LDR. is it possible to combine this charger and that circuit ?

    I want to set up the final circuit along with battery and white leds for 24×7 in my room like UPS. I don't wanna touch the system and the system will provide auto charging and LDR based output. And I also want the relay to cut off mains supply (220VAC) from the transformer when battery is full.

    Brother, please reply to me. I am badly needed the solution. Advance thanks to you!

    Reply
  59. Yeshwanth Kumar

    Hello Swagatham thanks for your help, I believe i can use the same for charging 12V battery. with suitable corrections of resistors?? Thanks for your great help
    Regards
    Yeshwanth kumar J

    Reply
    1. Swagatam Post author

      Hello Sumantri,
      yes, it's to keep the battery trickle charged even after cut-off.

      you can eliminate the 470 ohms if you don't want this to happen.

      Reply
  60. Kolley

    Thanks so much, you are indeed angel sent from heaven to solve human electronic hobbyist problem, more grease to your elbow.
    Sir , am having problem getting TIP122 pls can I replace it with BC or mosfet? If yes which BC or mosfet can I use? Thanks in advance

    Reply
  61. Kolley

    Thank you so much for this great circuit, I always like to build your circuit.
    Pls am having problem here getting TIP 122 pls can I use BC transistor or Mosfet instead of TIP122, if yes which BC or mosfet can I use?
    Thanks so much

    Reply
  62. Richard Fernandes

    Sir,I need to charge a 11.1 volt 2200 mah lion battery, i have read your above reviews and understand that i will have to use a 14-15 vdc source. I am not too sure about the Current rating of the dc source, should be be around 250 ma(1/10 of the mah) or how much more? as there would be current draw from the circuit itself. thanks and warm regards

    Reply
    1. Swagatam Post author

      Hi Richard, I might have missed your comment, so I am replying it now, for Li-ion you can use a current source that may be equal to the batt AH rating, therefore in your case you can charge it at the the rate 2.2 amps.

      Reply
    1. Swagatam Post author

      Hello Dipto, for charging a 6V batt you will need at least an input of 9V in the above circuit, so probably a 9V or a12V panel may be required.

      yes it can be used in conjunction with the referred changeover circuit….

      Reply
    2. Dipto

      thanks you sir for your quick reply. I will be doing my final year project & your blog is helping me a lot.

      another question I have sir,
      what should be the current rating of the solar panel? and how long will it take to charge the battery 6V 4.5AH with a 10W 12V solar panel?

      Reply
    3. Swagatam Post author

      withe the given solar/batt specs it could take around 6 to 8 hours for the battery to get fully charged

      current rating of the panel may be calculated as follows:

      10/12 = 0.833 amps

      Reply
    1. Swagatam Post author

      thanks for the image Dipto,

      remove the 3V zener from the transistor base and change its position by connecting its cathode directly with pin6 of the IC and the anode with the junction of the two LEDs and the transistor base resistor.

      Reply
    2. Dipto

      Sir, I have followed the procedure you have mentioned the red LED is turned off indeed however the green LED always stays on even after varying the 10k pot.

      One thing I did in the original circuit was connecting a 10k resistor is series with red LED (between red LED and 1K resistor) and the red LED seemed to be almost turned off. So it is okay if attach a 10k resistor with red LED?

      Reply
    3. Swagatam Post author

      Dipto, the procedure is given as under:

      disconnect the feedback persist which comes from pin6 of the IC temporarily.

      without any battery connected, set the output to 7V

      now adjust the pin2 10k preset until the green LED just lights up.

      seal the setting with glue.

      reconnect the 10k resistor between pin6 and pin2.

      the setting is done!

      if you have done the zener modification as suggested by me, then a single resistor with each LED will be enough….do not manipulate in any other way.

      Reply
    4. Dipto

      Sir, after shifting the position of the zener diode as per your suggestion the problem seems to be gone now. The LEDs are swapping without any issues now. I have also implemented the first circuit using relay and attached a solar panel it is also working as well. Here are some pictures,

      https://drive.google.com/file/d/0ByOP1BiJWDUab2k5UGotTWtBYm8/view?usp=sharing

      https://drive.google.com/file/d/0ByOP1BiJWDUaMVpWaWRDR2V0Smc/view?usp=sharing

      https://drive.google.com/file/d/0ByOP1BiJWDUaZGRQekk3OTdFMmc/view?usp=sharing

      Reply
  63. Dipto

    Hello Sir, I have built the 2nd circuit again but I am facing one problem. When ever I vary the 10k preset the light shifts position but the other LED glows dimly. For example if red LED is glowing brightly then green is glowing dimly vice versa.

    I interchanged the position of the resistor (100 ohm) and zener at pin 6 and placed LED junction in between them as you have instructed earlier, but it is not working. Is there any other modification I can do?

    Reply
  64. VIJAY AJ

    Thank you very much sir for the above circuit it works perfectly for me.I need one more feature on the above circuit because the above circuit didn't cutoff the load when the battery voltage is to low this will affect the life of a battery.please say any idea that battery should be automatically isolated from the load when battery voltage is to low.I am waiting for your reply sir.

    Reply
    1. Swagatam Post author

      Hi Vijay,

      the above two circuits will never allow the connected battery to go below a predetermined level (as set by the value of the feedback resistor from pin6 to pin#3)…as soon as a low battery is detected the circuit will revert and begin the charging process for the battery

      Reply
    2. VIJAY AJ

      OK sir.my supply voltage is from solar panel.during night if the voltage from a panel is to low the charging process get stopped on such period of time my battery voltage reaches to low it will affect the life time of battery .can u get my point sir????.pls suggest an idea for me.thanks for your reply sir

      Reply
  65. Louis Nieuwoudt

    Hi Swagatam,

    I am looking to build the second of the two circuits you listed above. Going through the list of explanations you provided in the comments I think I have a fair understanding of how it works, except for the tip122 on the output of the op-amp. Could you please explain what the purpose is of this component ?

    Kind regards
    Louis

    Reply
    1. Swagatam Post author

      Hi Louis,

      TIP122 acts like a switch, it remains switched ON and connects the negative of the battery with the negative supply line as long as the battery is charging and is below the full charge mark…it switches OFF the negative line to the battery the moment it reaches the full charge,

      Reply
  66. Muhammad Faizan

    Hi swagatam, thanks very much for posting these circuits. I want to ask 2 questions about your second circuit.
    1) Can i charge a 6V 6.8Ah battery from this circuit? Any modifications?
    2) What is the function of the 10K potentiometer used at the right side of the battery with op amp. How much should i adjust both the potentiometers to charge 6V 6.8Ah battery

    Thanks in advance

    Reply
    1. Swagatam Post author

      Hi Muhammed, yes you can charge the mentioned battery, just adjust the LM338 pot and ensure a 7V across the battery terminals without connecting any battery.

      the 10k preset is for adjusting and setting the cut off threshold at the battery full charge level, set this such that the green lights up and red led shuts off at 7V across the battery terminals

      Reply
  67. Goran Radkov

    Hello Swagatam,

    Can i ask you for help for this circuit, please… i made the first one and i want to use it for 12V battery; the problem is, that when i'm adjusting the output voltage, and when it exceed 1.5V the relay is triggering, so the charging LED is lighting, and when i try to adjust the other pot – on the IC741, it is always flicking

    PS
    I used resistor R1 – 220 ohm, instead of 240 ohm;
    and also i saw in the first coment, that initially, the pot under LM317 has been 2.2K

    Reply
    1. Swagatam Post author

      Hello Goran,

      initially keep the preset slider arm towards ground, completely…so that the pin#3 of the opamp is at a zero potential.

      Now adjust a 14.4V from the LM317 and slowly adjust the preset slider arm upwards until the red LED just lights up (green LED shuts off)

      220 ohm is OK….

      2.2k pot is not proper…it must be around 5k to 10k

      Reply
    2. Goran Radkov

      Yes, initially the preset is giving 0.0V at pin#3, but adjusting the preset is just slightly increasing the light of the RED LED, and the relay is getting triggerd when i'm adjusting the output voltage, just when it exceed 1.5V;
      Also the RED LED is always very slightly flickering, even when the preset is giving 0V at pn#3,
      maybe some resistor must be bigger…?

      Reply
    3. Swagatam Post author

      No, everything's perfect in the circuit and the resistors don't matter much, even if you were to use 100k resistors at the input of the opamp, it would provide the same sequence of results.

      make sure that you remove the feedback resistor across pin6 and pin3 while setting-up the pin3 preset.

      did you include the zener at the output pin6 of the opamp?

      there should be no battery connected while the setting up procedures are being carried out.

      Reply
    4. Goran Radkov

      I wasn't removed the reedback resistor while setting up… so i did it with a swich, but there was no difference, because the voltage at pin#6 is always about 2.8 – 3V for 14.4V output;

      what do you mean about the zener? i have it on the place, like on the circuit diagram;

      yes, there's no battery while setting up;

      But i have used C1 – 0.1uF ceramic? Maybe it must be electrolitic…?

      Reply
    5. Swagatam Post author

      that means your IC is faulty or you have done something wrong with the connections, the output pin6 must instantly turn equal to the supply voltage as soon as pin3 potential is pushed just above pin2 potential….this will happen while you adjust the preset….but if that's not happening indicates that something's not right in the assembly.

      Reply
    6. Goran Radkov

      I remade the circuit entirely from the beginning, but it's happening the following:
      1 – when the voltage exceed ~2.3V the relay is triggering
      2 – the voltage at pin#3 is 0V but at pin#6 it is not the same, but about 2.6 (which i think is because of the 3V zener…)
      3 – at the situation of "2" the both LEDs are lighting equally, and when the input voltage (at pin#3) exceed ~2.3V the voltage is jumping to about 9V at the both input and output, and afterwards it's increasing while adjusting… but in that very moment the RED-Batt Full LED starts shining stronger, and the GREEN LED shines slower
      4 – all the time the relay is triggered and the LEDs are lighting, no matter of the brightness

      I didn't understand what happened, but in one moment the IC exploded along with the input C1, as the last one set on fire, so i wanted to ask you, do you have that circuit made and all working properly…?

      Reply
    7. Swagatam Post author

      All these indicate that your IC is duplicate or of poor quality….with pin3 grounded the voltage at pin6 should be not more than 1V or max 1.5V (leakage volts)

      the 3V zener is introduced to block the pin6 leakage voltage from reaching the BC547 base, it won't create the volts as you are assuming.

      did you remove the 10k feedback resistor across pin6/pin3, while adjusting the 10k preset??? it needs to be removed and later on reconnected

      and under no circumstances the IC can explode unless the pin7 voltage is increased above 22V

      please go through all the comments above to get a better idea regarding the procedures.

      Reply
  68. Angelous Chavez

    sir what should be the resistance value of 10k potentiometer that is connected to pin2 of ic1 if im going to charge a 6v lead acid battery…

    i have maked the circuit above but the two leds are glow.

    Reply
  69. Angelous Chavez

    i have already make this circuit and i connected correctly all of the components based on the circuits above without using a relay when i connect the 6v bat and a 9v supply the green and red leds are glown.. it supposed to be the green led will glow first then when the bat is full the red led will glow…

    Reply
    1. Swagatam Post author

      Measure the voltage at pin6 with reference to ground, it must show zero volts or supply voltage depending on the 10k preset setting.

      the red LED is glowing due to leakage voltage at pin6, but the 3V zener is supposed to stop this, make sure you have connected the zener at pin6 correctly.

      I think I have explained the entire setting up procedure somewhere within the comments please search it and set the 10K preset accordingly.

      Reply
    2. Swagatam Post author

      Angelous, did you you use the 3V zener diode at pin6, it's supposed to block this leakage 2V…or may be your IC is duplicate quality and faulty, change it with a good one and check again

      Reply
    3. Angelous Chavez

      sir, it should be set the pot preset at pin 6 to 0v first then you can charge your battery now? then when the battery being charge reach the reference voltage lets say the ref voltage is 7.2v then the red led will on and the green led should be in off state when the battery gets full charge?

      Reply
    4. Swagatam Post author

      Angelous, the 3V zener at pin6 is supposed to block the leakage voltage across pin6 and ground….if you wish you can use a preset also across pin6 and ground for the same but that can be risky and damage your IC if accidentally the preset is grounded while adjusting.

      Reply
    5. Angelous Chavez

      sir i think there is something wrong with your circuit because i have already made it base on your circuit diagram with the right components as your circuit indicates. please check it again and make some improvement.

      thanks

      Reply
    6. Swagatam Post author

      Angelous, I have myself tested and used this circuit many number of times for different applications, and it worked perfectly….may be you are not able follow my instructions, and so not able to troubleshoot your circuit.

      and if you keep trying without understanding you are never going to succeed with any circuit…so attempting after understanding the details is the most important thing.

      Reply
  70. Angelous Chavez

    sir, in this circuit i used a 9v rectified as supply in this charger circuit so it is okay to use 9v or to supply a 9v? cause as far as i know a 9v supply is good enough to charge a 6v and 4.5ah lead acid battery.

    Reply
  71. Angelous Chavez

    sir thank you for responding! i have read the article about the opamp. and i learned more about it.

    sir the 10k resistor and a 3v zener diode juction in which the pin 3 is connected across it is a "REFERENCE VOLTAGE" if so, what voltage value or how much voltage is your ref voltage?

    and if i am not mistaken the 10kpot preset is the sensing voltage. Meaning when the voltage across it falls or increase above the reference voltage the output at pin 6 will change to high state then it will be "on" the transistor TIP122.
    Sir if i wrong please correct and guide me so that i can learn much about this circuit.

    Reply
    1. Swagatam Post author

      Angelous, you understood it correctly, an opamp output will always go high if its (+) input pin attains a higher potential than its (-) input pin, regardless of which pin may be connected as the reference.

      The reference voltage can be any voltage below the actual supply voltage for the IC.

      when the output is high tip122 will conduct and vice versa

      Reply
  72. Angelous Chavez

    sir one more question, why did you used a 3v zener diode in series with the 10k resistor and that is connected at pin3?

    and what is the function of a 470ohms resistor that is connected to negative terminal of the battery. sir please explained it to me so that i can understand more.

    Reply
    1. Swagatam Post author

      as explained above, it's for the setting up a reference point for the other pin.

      the 470 will keep the battery charging continuously at a very low rate even after the actual charging is cut off…in order to prevent the batt from self discharging.

      the 100k is for latching the IC in the cut off position so that it does not oscillate while the battery voltage drops slightly after the cut off

      Reply
  73. Angelous Chavez

    sir i try again to make this circuit with a new ic, as you have said a few days ago, the voltage at pin6 should be in 0v but in my reading the lowest voltage whenever i adjust the preset is proximately 2v but i connect the zener diode correctly. sir what do you thinks is the problem?

    Reply
    1. Swagatam Post author

      Check the voltage across pin6 and ground….and then across zener anode and ground, if both these voltages are same then probably your zener diode is faulty or is wrongly connected.

      Reply
  74. Uzair Khan

    Hi bro. I want automatic circuit for my 6v 4ah battery which is connected with my wifi router and the source is the dc input of 9v which is the adapter of wifi. I want that 9v 2amp adapter is to charge mai battery when battery get charge then it can automatically stop further charging. Any idea pls

    Reply
  75. Anil Kumar. K

    Hi Swagatham, thanks
    I assembled 2nd circuit for charging 3.7v/800mah Li-Ion battery. Not yet tried hysterisis control and TIP122 section section.Circuit working perfect as you described.

    I have a doubt to clear.
    Input supply is 9v/500mah and charging supply is 4.2v/500mah. Battery rated 3.7v/800mah. So, I think I have to add a current limiting resistor (42v-3.7v/160mah=3.125ohms) 3ohm/1W series with battery to reduce charging current to 160mah (800mah/5=160mah).

    Reply
  76. Waqas

    Thank you so much Swagatam for your valuable share. As I am not an engineer but hobbyist and have enough Know-How about Arduino's and it's circuitry and using ESP8266 WiFi module a lot in my hobby projects. I have bunch of 4v 1 amp/hr lead acid batteries, lot of LM393, 5v dc wall supplies, TIP122 darlington transistors, bunch of IN4148, and off course capacitors and resistances. I want to build an automatic chargers to charge these batteries used in my hobby projects. May you suggest some design? I have few in my mind, including your this project, but want to seek help of a professional one like you. I don't want to buy automatic battery chargers available in market as they are more than what needed. May you help me?

    Reply
    1. Swagatam Post author

      Thank you Waqas, the above designs are the best and universally suited to all batteries for achieving automatic cut offs at the intended thresholds….so I would recommend you to go for one these circuits and try it for your application..

      Reply
  77. K CS

    Hello sir, Chaw is here again !! I want to clarify some of the facts with you in second diagram(using transistor).

    (1) Since I wanted to have an input DC supply of 3-30V, I will use LM338 IC, 480ohm resistor(two 240 ohms in series) and 10k preset to have an output of 27V, calculated based on the calculator u have provided. Am I on the right track ?
    (2) For the input current, let's say 1A is given, the battery terminals will receive the full charging current right ?
    (3) May i know what is the use of 470 ohms resistor connected to the cathode of battery ? Any other value can I use (0.1k,1k,10k etc) ?
    (4) Plus, Is there anything do I need to modify to the auto cut off part of the circuit according to my specs ?
    (5) Instead of TIP122, can I use BC547 instead ?

    Regards,

    Reply
    1. Swagatam Post author

      Hello Chaw, here are the answers:

      1)If you have calculated the figures then it would be fine to go ahead with the mentioned values of the resistors and the port.

      2) yes full 1amp would reach the battery.

      3) It's for providing the battery with a floating or trickle charge current after it has reached the full charge level and the transistor has switched OFF

      4) No need of modifying anything, but with a IC741 you cannot exceed the voltage above 19V across its pin#7 and pin#4

      BC547 will be able to supply only upto 50mA…for 1amp you will have use a TIP 122 or a TIP142

      Reply
    2. K CS

      Noted sir !! But LM741CN i bought from the market has a supply voltage from 10-36V. So I could apply up to 30V right ? By the way, I am going to charge 3.7V 4000 mAh lithium ion battery. I need to confirm with you that
      (1) Before charging, 100k resistor is removed and I know that we need to give an input supply. Is this input voltage is three or five volt higher than the battery like the last time in 6V,12V battery charger circuit ? (for my case would be 6.7V )
      (2) For the input current, is it possible to set at full rate 4000 mA ?
      (3) There is no effect to TIP122 right ? My one has a DC collector current of 5A/
      (3) An input current of five or ten time lesser than the battery is only for lead acid batteries right ?

      Regards,

      Reply
    3. Swagatam Post author

      1) yes the input should be adjusted to achieve around 20% higher than the battery rating…meaning for example in case of a 12V batt it's 12 + 20% = 14.4V…this should be confirmed across the points where the battery needs to be connected.

      2) input current can be 4 amps but that will generate some heat on the batt which will need to be monitored and controlled with a cooling fan.

      3) TIP122 will get significantly hot at 4amps so TIP142 is recommended, which will also require a heatsink.
      3) that's right 10 times less current is recommended for lead acid batts not for Li-ions.

      Reply
  78. K CS

    So, if I used TIP122, either I give lesser input current (1-2A) or 4A together with a cooling fan would do right ?
    If I were to use TIP142, how to make a heatsink ? I am sorry sir since I am still a student and never learnt this term before. Can you suggest me the way to do it ? Thank you for always making my doubts clear 😀
    Regards,

    Reply
  79. The Rocking Time

    Sir why have u used a 470 ohm resistor in between the battery negative rail and the ground rail in the last circuit ( charger without using relay ) ?
    Will it cause the continuation of charging even after the over charge cut off is detected since the current is enable to enter into the ground rail from the battery -ve through the 470 ohm resistor always ?

    Reply
    1. Swagatam Post author

      RT, it is optional, I have included it to enable trickle charging for the battery after the transistor has cut off at the full charge threshold. This is to make sure that the battery is always in the topped-up condition and to eliminate self-discharge

      Reply
  80. Sy Do

    Hi Swagatam,

    I made one circuit without using a relay for battery charger. I using power source with 14v 1A. Because I do not have R1=240 Ohm should I replace it with one resistor 220 Ohm.
    I have set up the circuit in the following way:
    1. Without connecting any battery, disconnect one pin of the 100k resistor to pin6 (IC741). Another pin of 100k resistor keep connected to pin2 (IC741)
    2. Adjust R2 to get 7.2v. At this step, I have put 2 ends of VOM at 2 points on the circuit to get 7.2v:
    – point 1: Behind 1N5408 diode
    – point 2: Behind 10k POT (GND)
    is that right?
    3. Adjust 10k preset and identify a position where RED led just lights up and GREEN led supposed to get illuminated.
    4. Reconnect 100k resistor to pin 6 of the IC 741.
    5. Connect 6V 4AH battery to charging (the voltage of the battery when not connected is 4.8v).
    At the moment, I use VOM to measure amperage passing through battery is 20mA, too small compared to 4/10=0.4A=400mA.
    Could you please help me know where I was wrong?

    Reply
    1. Swagatam Post author

      Hi SY, all the mentioned procedures are correct….now the moment you switch ON power with the batt connected, the 7.2V instantly should drop to the battery level at 4.8V, please check whether this is happening or not?

      If this is happening next you can check the current by connecting ammeter in series with the battery positive line.

      If it's 20mA as you say then either the battery is faulty or the LM338 is faulty…confirm the output current from the LM338 by connecting ammeter across its output and ground line…it must show the maximum available current as per the supply source rating….if not then do the same across the supply source terminals to confirm the current.

      Reply
  81. Sy Do

    Hi Swagatam,
    I was replaced LM317 by another one (not LM338 as you mentioned) but I still get the current is 20mA. Maybe the problem is due my battery.
    Have you any battery charging circuit similar with above circuit (without relay) which can adjust the voltage and current by VR? If yes, could you please give me the link?

    Thanks,

    Reply
  82. Siddharth Talesra

    Dear Sir, I have the 6v, 4.5ah battery and it's charging nicely with the second circuit you provided, and now I want to add one more feature how to know how much charge is present in battery and supply this information to arduino microcontroller and also power the same arduino with that battery. What changes I have to add to make this new feature work.
    Thank you,

    Yours sincerely
    Siddharth Talesra

    Reply
    1. Swagatam Post author

      Thanks Siddharth, your Arduino must be coded such that it is able to read the charging rate and the reduction in current consumption while it's charging and then compare these parameters to figure out the approximate time left.

      A crude method is to attach a 9V incandescent bulb in series with the positive of the battery and monitor its intensity electronically to learn the same.

      Reply

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