Monday, July 30, 2012

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.

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



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.



239 comments:

  1. Hi, sir i need a circuit for an 12v battery. so what changes i should make in this circuit to charge a 12 v battery.thanks

    ReplyDelete
    Replies
    1. 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.

      Delete
  2. Hi Sir,
    I need a circuit for DC low cut-off.If the DC voltage goes lesser than 450V it must cut the voltage,if the voltage exceeds 450V,it must turn ON with 5 minutes time delay.Please help.

    ReplyDelete
    Replies
    1. 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.

      Delete
  3. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  4. 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!!

    ReplyDelete
    Replies
    1. 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.

      Delete
  5. Dear sir can I use the 1st circuit to charge the 12 v 7.5 ah using a power supply 12v 0.5 amps. What are the changes I need to make for the above circuit

    ReplyDelete
    Replies
    1. You will have to use at least a 2amp transformer (@14V)for charging a 7.5ah battery, 500mA will not be enough...

      Delete
    2. It market I'm getting 12~13.8v, 1 amp tranformer can I use it. Can I use it as a trickle charger

      Delete
    3. For trickle charging you can use a 1amp transformer...

      Delete
  6. 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! :)

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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.

      Delete
    3. 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.

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

    ReplyDelete
  8. Do you have any other circuits for charging 2.7V 100F ultra-capacitor? The supply Voltage will be 6V 1AH.

    ReplyDelete
    Replies
    1. The IC 317 circuit can be adjusted to produce 2.7V and used for charging a UC safely.... this is what I assume but not exactly sure.

      Delete
  9. How many hours may take to charge 6V 1AH battery by using above circuit?

    ReplyDelete
    Replies
    1. 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.

      Delete
  10. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  11. 0-----/\/\/\/\------0 To battery +
    Rectified DC 10v 1k 1w
    0-------------------0 to battery -

    Hi swagat you mean this will work ?.

    ReplyDelete
    Replies
    1. Yes that's correct, the 1 K resistor can be in series with the positive line....

      Delete
  12. 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!

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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...

      Delete
    3. hello sir,
      kindly help me how to modify upper circuit to charge a 6v 4.5amps
      lead acid battery in just 1 hour.

      thanks in advance

      Delete
    4. 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.

      Delete
    5. 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...

      Delete
  13. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. you can use the upper circuit for charging your battery but remember the above point.

      Delete
  14. Thank you very much Sir i really appreciate it. .im goin to use 12 v and 1 Amp transformer,is it ok?. thanks a lot Sir.

    ReplyDelete
    Replies
    1. Welcome Daj,

      If your battery is 6V you will have to use a 0-6V transformer and not 12V....the current 1 amp is fine.

      Delete
  15. 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.

    ReplyDelete
    Replies
    1. Hi Umar,

      Yes surely you can use it with the circuit shown in this article.

      Delete
  16. Sir,
    Thank you very much for asistance given to us in our world of elctronics.
    With maximum regard and respect.

    ReplyDelete
  17. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  18. 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?

    ReplyDelete
    Replies
    1. 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.

      Delete
  19. can you explain the function of the 3v zener?

    ReplyDelete
  20. 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.

    ReplyDelete
    Replies
    1. You will have to use two separate regulator ICs, a 7805 IC for microcont. and a 7808 IC for the battery...

      Delete
  21. 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. :)

    ReplyDelete
    Replies
    1. 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....

      Delete
  22. 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.

    ReplyDelete
    Replies
    1. All types of 78XX series will generate max 1amp current, so all are suitable for your application.

      AN7808 will also work and is OK to use.

      Delete
  23. hello Sir,
    kindly advise me how to modify lower circuit to charge 6v 4.5amps lead acid battery in just 1 hour,,

    thanks in advance...

    ReplyDelete
  24. 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. :)

    ReplyDelete
    Replies
    1. Helo Daj,

      You'll need to monitor the batt voltage, when it reaches 7.2V you should stop the charging process.

      Delete
  25. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  26. sir can i use 7808 in your circuit instead of lm 317 ,so that i may avoide r1, r2,c1

    ReplyDelete
  27. Thanks. But for 7806 regulator is it ok if i always connect it to 6v 4ah battery?

    ReplyDelete
    Replies
    1. I think it is OK, because the charge wouldn't be able to increase to dangerous levels.

      Delete
  28. 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

    ReplyDelete
    Replies
    1. Hi Bibin,

      It's a long explanation, you will have to go through the following article for knowing how an opamp works as a compaarter, as applied in the above circuit

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

      Regards.

      Delete
    2. Ok swagatam,

      Let me go through that article... Nd i'll be back with more doubts.. see ya.. take care..

      Regards,

      Delete
  29. hi swagatam,

    Can we use zener diode and transistor instead of 741 for auto cutoff when battery is fully charged.. ?

    Regards,

    ReplyDelete
  30. i was talking about a circuit like this...

    image

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. so the circuit will not work as said... ?

      Delete
    3. cutting-off means the output of LM338 should become zero, but that will never happen here.

      Delete
    4. swagatam,

      Can u please explain the working of the above circuit with transistor..?

      Delete
    5. 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.

      Delete
    6. 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



      Delete
    7. 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.

      Delete
    8. Thanks yaar... i'll try the second circuit... !

      Delete
  31. 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?

    ReplyDelete
    Replies
    1. 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.

      Delete
  32. Hi,

    I want to Charge a 6 Volt rechargeable battery with a 5Volt 0.5A supply.How can I do that. Some sort of boost is required.

    ReplyDelete
  33. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  34. 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

    ReplyDelete
    Replies
    1. Hi Nikhil,

      anode will go to transistor base, and cathode to the 10 k resistor.

      Delete
  35. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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

      Delete
    3. I think the IC has been damaged...

      Delete
    4. 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

      Delete
    5. 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.

      Delete
  36. 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.

    ReplyDelete
    Replies
    1. Yes you may do as you have mentioned....

      Delete
    2. 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.

      Delete
    3. 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

      Delete
    4. 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

      Delete
    5. 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.

      Delete
    6. 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.

      Delete
    7. 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.

      Delete
  37. 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.

    ReplyDelete
    Replies
    1. 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

      Delete
    2. 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.

      Delete
  38. 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..

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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..

      Delete
    3. 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.

      Delete
  39. 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..?

    ReplyDelete
    Replies
    1. 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.

      Delete
  40. 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

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

      Delete
  41. 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

    ReplyDelete
    Replies
    1. Relay will switch ON when the red led comes on, after the battery gets fully charged.

      Delete
  42. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  43. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  44. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  45. Dear swagatam Majumdar, if i want to use LM393 then how would i wire them in
    1. the first circuit
    2. the second circuit
    3. please email me both diagrams using lm393 ic on khansparty@hotmail.com

    ReplyDelete
    Replies
    1. The wiring would be exactly the same.

      use LM324 instead, because I have tested the circuit using LM324.

      Delete
    2. 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..........

      Delete
    3. 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.

      Delete
  46. 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.

    ReplyDelete
    Replies
    1. I have updated the diagrams correctly as per your doubts, please check them out.

      Delete
  47. Are all these resistors 1/4 watt on the second diagram above?

    ReplyDelete
  48. Dear Sir,
    With 10K RESISTOR and 3v Zener, how much Refrence voltage (Vf) will be set by this ZENER-RESISTOR VOLTAGE REGULATOR.

    ReplyDelete
    Replies
    1. Dear Bashir,

      It will be always equal to the zener voltage, that is 3V.

      Delete
  49. 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?

    ReplyDelete
  50. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  51. 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.

    ReplyDelete
    Replies
    1. Please see this circuit:

      http://homemadecircuitsandschematics.blogspot.in/2012/01/how-to-make-versatile-variable-voltage.html

      R4 = 0.1 ohms.

      adjust the pots to get the desired voltage and current.

      Delete
  52. The feedback resistor connected to PIN-2 and PIN-6 of the IC in the second circuit, is that 100K resistor OR 10K resistor as used in first circuit.

    ReplyDelete
  53. sir will you pls provide the parts list in detail? like the capacitor type,it's rating, resistors wattage.

    ReplyDelete
    Replies
    1. all resistors are 1/4 watt rated, capacitor is disc type, or any other standard type will do.

      Delete
  54. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  55. Can this circuit will be used to charge 3.7 volt 1000 mah Li-Ion Battery or any changes should be made in this circuit

    Thanks

    ReplyDelete
    Replies
    1. 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

      Delete
  56. Sir,

    Please explain me how to fix the 2 pots (10k and 2k2) in 2nd circuit which have three connections in it.

    Thanks & Regards

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. Sir
      I built the circuit but confused about the pots please explain how to fix the pots(10k 2k2)

      Delete
    3. 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

      Delete
    4. Sir U dont understand my problem i want explanation about the pots which have 3 connections each. I dont know which connection has to fix at which end of pot in circuit

      Delete
    5. 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

      Delete
  57. Sir

    I dont have 3V zener diode Can i use 5V zener instead

    ReplyDelete
    Replies
    1. Sir,
      Do i need specifically 400mW zener Diode

      Delete
    2. 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

      Delete
    3. 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.

      Delete
    4. Sir
      As a input source can i use your transformer-less power circuit (12V- 500mA output) so that whole unit can be cased in a compact mode.

      Regards,

      Delete
    5. use a cell phone charger or a readymade 12v smps for the input supply, don't use a capacitive type non-isolated circuit.

      Delete
    6. how to use 5v cellphone charger to charge battery of 6v ??

      Delete
  58. Swagatam Majumdar PLEASE DO NOT STOP THIS BLOG...

    ReplyDelete
  59. Hello Swagtam Da,
    I want to make an automatic emergency light circuit using this battery charger circuit ant the concept you use in "http://homemadecircuitsandschematics.blogspot.in/2011/12/how-to-make-efficient-led-emergency.html)". So, can I connect a PNP transistor before the battery or it will affect the charging process. Please Help.
    Thanks in advance.

    Tanmoy Singha Mahapatra

    ReplyDelete
    Replies
    1. 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.

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

    ReplyDelete
    Replies
    1. it's the position of the relay contact terminal which is "Normally Closed" denoted as N/C

      Delete
  61. 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?

    ReplyDelete
    Replies
    1. Dear Tanvir,

      you won't have to modify anything, just adjust the settings to suit 12V battery charging.

      you may refer the following post to know about relays in details:

      http://homemadecircuitsandschematics.blogspot.in/2012/01/how-to-understand-and-use-relay-in.html

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

    ReplyDelete
    Replies
    1. mp3 player cannot be made without using mcontroller, that's not feasible.

      Delete
  63. 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?

    ReplyDelete
    Replies
    1. It is better to use the 9V adapter as it's safer than the 5V/45a option.

      Delete
  64. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  65. 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

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

      Delete
  66. Can i use 12V 3A transformer for supply 12v dc??

    ReplyDelete
    Replies
    1. Amps will depend on the battery AH, should be 1/10th of battery AH

      Delete
  67. 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

    ReplyDelete
    Replies
    1. At 5.6V the battery can be assumed to be fully discharged.

      Delete
    2. And my 1st question??

      Delete
    3. I have answered it in the earlier comment....posted by Abhishek.

      Delete
  68. 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?

    ReplyDelete
    Replies
    1. 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.....

      Delete
  69. 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...

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. ...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.

      Delete
  70. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  71. 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

    ReplyDelete
    Replies
    1. You are welcome my friend!

      Please send it to homemadecircuits@gmail.com

      Delete
  72. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. ....the input current must be rated at 1/10th of the battery AH, otherwise the above procedures will not work.

      Delete
  73. I'm awaiting your helpful feedback.

    ReplyDelete
  74. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  75. 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

    ReplyDelete
    Replies
    1. You can try the second circuit.

      No change in parts would be required.

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

    ReplyDelete
  77. 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?

    ReplyDelete
    Replies
    1. 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.

      Delete
  78. 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

    ReplyDelete

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