Thursday, October 11, 2012

Make this 3 Step Automatic Battery Charger/Controller Circuit

Generally it is noticed that while charging batteries people hardly pay any special attention toward the procedures. For them charging a battery is simply connecting any DC supply with matching voltage with the battery terminals.

I have seen motor garage mechanics charge all types of batteries with the same power supply source irrespective of the AH rating associated with the particular batteries.

That's gravely wrong! That's like giving the batteries a slow "death". Lead Acid batteries to a very extent are rugged and are capable of taking on the crude charging methods, however it's always recommended to charge even the LA batteries with a lot care. This "care" will not only increase the longevity but will also enhance the efficiency of the unit.

Ideally all batteries should be charged in a step wise manner, meaning the current should be reduced in steps as the voltage nears the "full charge" value.

For a typical Lead Acid battery or an SMF/VRL battery the above approach can be considered very healthy and a reliable method. In this post we are discussing one such automatic step battery charger circuit which can be effectively used for charging most of the rechargeable types of batteries.

Referring to the circuit diagram below, two 741 ICs are configured as comparaters. The presets at pin#2 of each stage is adjusted such that the output goes high after specific voltage levels are identified, or in other words the outputs of the respective ICs are made to go high in sequence after predetermined charge levels are accomplished discretely over the connected battery.

The IC associated with RL1 is the one which conducts first, after say the battery voltage reaches around 13.5V, until this point the battery is charged with the maximum specified current (determined by the value of R1).

Once the charge reaches the above value, RL#1 operates, disconnect R1 and connects R2 in line with the circuit.

R2 is selected higher than R1 and is appropriately calculated to provide a reduced charging current to the battery.

Once the battery terminals reaches the maximum specified charging voltage say at 14.4V, Opamp supporting RL#2 triggers the relay.

RL#2 instantly connects R3 in series with R2 bringing down the current to a trickle charge level.

Resistors R1, R2, and R3 along with the transistor and the IC LM338 forms a current regulator stage, where the value of the resistors determines the maximum allowable current limit to the battery, or the output of the IC LM338.

At this point the battery may be left unattended for many hours, yet the charge level remains perfectly safe, intact and in a topped up condition.

The above 3 step charging process ensures a very efficient way of charging resulting in almost a 98% charge accumulation with the connected battery.

The circuit has been designed and invented by "Swagatam"



R1 = 0.6/ half battery AH

R2 = 0.6/one fifth of battery AH

R3 = 0.6/one 50th of battery AH.

86 comments:

  1. Hello
    What about the method of using single high capacity silicon diode plus feeding 10-15 percent AC ripple by paralleling the diode with a resistor? This dissolves any sulphide into solution and prolongs the battery life

    ReplyDelete
    Replies
    1. Good suggestion, appreciate it.... but here we are also interested to get the battery fully charged, close to 100%, therefore a step charging process becomes necessary.

      Delete
  2. Yes sure you are right.
    Another thing I wish that I could post you a Grid Tie Inverter circuit that I have got and would like to get your expert opinion about it as its very unconventional. If I can make it work on a lower power scale for our 220V 50Hz AC system it would be nice but I got to understand it first. Would you help me please?

    ReplyDelete
    Replies
    1. I won't say I'm an expert, but i will surely check it and try to understand it if the design is not too complicated.

      Delete
  3. How can I post the schematic figure here so you can study it

    ReplyDelete
    Replies
    1. You can send it to

      hitman2008@live.in

      Delete
  4. hie my mentor i have done this circuit and kept 100ah battery then the both relay coil and transistor was burnt

    ReplyDelete
    Replies
    1. Hi friend,

      You might have done something wrong with the connections, the relay coil and the transistor have no connection with battery AH so they cannot burn because of the battery.

      But anyway, the above charger will not charge batteries above 40 AH.

      Please note that the 4k7 preset must be set to produce 14 volts at the output before charging a 12V battery.

      Delete
    2. thanku friend i mean the vcc 12v to relay coil is connected by the battery.

      Delete
    3. Yes, but that will not do any harm to the relay coil, as long as the voltage does not go far beyond 12V

      Delete
  5. Hello Swagatam,
    What if i am using a DC power supply instead of transformer?
    What input DC voltage should I use just at the input of LM338? Like if I want to use 20V or 24V DC, can i use that directly from power supply?
    And then how would the value of capacitor will change? And would this thing affect other component values?

    Thanks.

    ReplyDelete
    Replies
    1. Hello Osama,

      You may connect any DC at the input of LM338, it should not exceed 30 volts though.

      With a DC input you may reduce the input capacitor to 1uF/50V.

      The input voltage changes will not affect the other circuit components.

      But with 741 ICs you cannot charge a 24V battery, for charging 24V batt you will need to change the 741 opamps with IC324 opamps.

      Delete
  6. 1. What is the connection of resistor R2? Where is it connected from top terminal?
    2. Why have you left an open terminal just below 100 ohm resistor at left side of battery? What is that connection? Is it a mistake?

    Kindly clear these issues.

    ReplyDelete
    Replies
    1. R2 is connected with the relay RL1 contact (N/O point)), the terminal below 100 ohms can be removed....

      Delete
  7. And secondly, how have you made the relay connections below the battery?
    Can you explain that? I can't understand that.
    I have to make the simulation and so I don't know which wires to connect where with the relay wires?

    ReplyDelete
    Replies
    1. Those are the relay contacts, please read the following article to know about relay contacts and their symbols:

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

      Delete
    2. Are the two relays connected to each other too?
      The two contacts (N/O) of both the relays are connected to each other? Is that correct?

      Delete
    3. Relay poles are connected with each other, their N/O contacts are connected to R2 and R3 and not with each other..

      Delete
  8. I wanted to confirm the TWO terminals of R2. One terminal is connected to N/O contact of relay RL1. But where is the other contact connected? You have showed in your circuit diagram that it is connected to resistor 4K7, but how is that possible? Is that a mistake? Where should be the upper terminal of R2 be connected to?
    Kindly clear that.

    ReplyDelete
    Replies
    1. It's connected to the ground line. Please enlarge the diagram by clicking on it.

      Delete
    2. Is 4k7 a fixed resistor or variable resistor? I mean should i ignore the wire coming out of middle of resistor 4k7? it is just useless right?

      Delete
    3. 4k7 is a preset as indicated by the symbol, the center pin is the most important connection...the upper pin goes to ADJ, the lower and the center goes to ground, these are the preset pin connections of 4k7

      Delete
  9. One more thing to ask, I have to implement this circuit on hardware, so all these resistors are simple resistors or high-watt resistors? What should be the rating of these resistors i mean in sense of wattage?

    ReplyDelete
    Replies
    1. all resistors except R1,R2,R3 are 1/4watt rated, R1,R2,R3 can be 2 watt rated or more....

      Delete
  10. What are the two LED's indicating? When will the green LED lit up and when will the red LED lit up? At what voltage levels? They are dependent on voltage levels right? and what do these 2 LED's exactly show in this circuit?

    ReplyDelete
    Replies
    1. upper led ON indicates "charging"

      lower led ON indicates "full charge" and battery cut off (upper led shuts off).

      The second result depends on the setting of the 10K preset of the lower opamp, which should be set for full charge cut off (trickle charge initiated)

      Delete
  11. So upper LED (green) will be always ON once i connect the battery charging circuit until the cut-off?
    do you know any EXACT Voltage levels when upper and lower LED will lit up and when will it be OFF?

    ReplyDelete
    Replies
    1. upper led will be ON when power is switched ON, with a discharged battery connected.

      As mentioned earlier, the lower led will light up when battery reaches full charge which must be set by the adjusting the 10K preset. for 12V battery it should be about 14.3V and for 24V batt it should be at about 28V.

      Delete
  12. I appreciate ur circuit...
    I just want to know that can I use 15-0-15 V , 2A T/F for this circuit

    ReplyDelete
    Replies
    1. Thanks!

      Yes it can be used, use the 0-15 tap if the battery to be charged is a 12V battery

      Delete
    2. Can't I use 15-0-15 all three taps so that I can eliminate the bridge rectifier and left over with only two diodes.

      Delete
    3. Though it's not a recommended configuration, will do the job......bridge is always the best.

      Delete
  13. good day sir! how can i make this circuit capable to charge a battery which is 60ah up? i am also excited if you could share any modifications. im planning to build one using your design. im carrem from the philippines your avid follower. thank you.

    ReplyDelete
    Replies
    1. good day carrem,

      Add a TIP36 transistor over the IC LM338 in the manner shown in the following post:

      http://homemadecircuitsandschematics.blogspot.in/2012/11/high-current-transistor-tip36-datasheet.html

      Feel free to post your work, I'll publish it with your name here.

      Delete
  14. Good day sir! This is Carrem. If I use 60AH battery then the value of R1=2 ohms R2=50 ohms R3=500 ohms pls correct me if Im wrong. I need your expertise regarding this matter. Thank you.

    ReplyDelete
    Replies
    1. Hi Carrem you cannot use a 60 ah battery with the above circuit, please see my previous comment....you will need a tip36 transistor for up-gradation.

      Delete
  15. Good day sir! Last month I have already tried and built the circuit using http://homemadecircuitsandschematics.blogspot.in/2012/11/high-current-transistor-tip36-datasheet.html as reference and it worked well. Now I'm planning to incorporate your suggestion on upgrading the circuit on the 3 step battery charger. That is the reason why I asked you about my computation on R1=2 ohms R2=50 ohms R3=500 ohms. Are the figures I arrived at correct by using 60 AH battery? I really need your help on this I'm not so sure about myself when it comes to numbers.
    Thank you so much and Godspeed! carrem

    ReplyDelete
    Replies
    1. Good day!

      R1 should be selected to pass max current, may be at AH/2 rate, therefore 60/2 = 30amp.....therefore R = 0.6/30 = 0.02 ohms.

      Similarly R2 may be selected as 0.6/15 = 0.04 Ohms

      and R3 = 0.6/7.5 = 0.08 ohms......finally R4 = 0.6/3 = 0.2 Ohms

      All these are approximate assumptions though, changes can be made as per personal preferences.

      Delete
  16. sir, thank you so much for sharing your expertise. i just open your blog spot today. we have no connection for a few days. ill keep you posted regarding this circuit as soon as i finish building this project. more power! carrem

    ReplyDelete
  17. Hi Swagatam,

    can i use 6.2volts for the zener diode?

    Regards,

    ReplyDelete
  18. good day sir,can this charger use as an automatic on and automatic off charger and can explain how?thanx

    ReplyDelete
    Replies
    1. The circuit automatically increase and decrease the current to the battery as per the charge levels of the battery.

      Delete
  19. Hi Swagatam
    In reference to the TIP36. Could I use 2 across the LM338 to charge 210Ah?
    The next question is ….Is the a way we could get rid of the relays? Use transistors, opto couples or cmos gate’s to do the switching?
    Regards
    Louwrens

    ReplyDelete
    Replies
    1. Hi Louwrens,

      Yes TIP36 may be used for increasing current capacity of the circuit, however replacing with other devices would make the circuit unnecessarily complicated, so here relays are more preferable....you may use the low current type relays, example 5V/500mA type.

      When it comes to sensing voltage thresholds, opamps work better, so CMOS gates wouldn't be a good choice here.

      Delete
  20. Hi Swagatam,

    What modifications I have to make on your circuit diagram in order to charge 24V batteries of 200 AHours

    Many thanks in advance

    Regards

    ReplyDelete
    Replies
    1. Use 24V relays and replace the 741 opamps with IC324 opamps.

      Use 30V/30 amp transformer.

      Connect an outboard transistor with LM338 for increasing current limit, you may refer to this article:

      http://homemadecircuitsandschematics.blogspot.in/2012/02/how-to-make-solar-inverter-circuit.html

      Delete
    2. Good morning Swagatam,

      Thanks very much for your quick reply.
      Isn't it better to replace LM338 with LM350 ( increased current ) than introducing an outboard transistor.What is your opinion?

      Thanks very much

      regards

      Delete
    3. Good morning Anonymous,

      LM338 is rated at 5 amps while LM350 is rated at 3 amps, so it's lower in its rating.

      Delete
    4. Good morning Swagatam,

      You are correct about LM350.
      What about LM196? Do you think I can use this regulator?
      My aim is to construct a 24V battery charger to be able to pump 20-25A to the 200AH battery bank which I have and supplying an inverter 24V 2000VA.
      Since you are the designer of this charger do you think that it can be modified to these requirements?

      Thanks very much

      regards

      Delete
    5. Good morning Anonymous,

      yes you can use LM196, and get maximum 10 amps current and 15V as per the datasheet, they haven't mentioned anything regarding how to use the IC with higher voltages.
      if you want to use the above IC with the above circuit, you can replace the IC with it and adjust the 4k7 preset accordingly.

      R1,R2,3 will also need to be modified by using the formula.

      Delete
    6. Good morning Swagatam,

      Many thanks for your quick respond and your advice.

      Delete
  21. Good morning Swagatam,

    I am coming back to the problem of charging 20-25A to the 24V battery bank. By using the LM196 I can only pump 10Amps as you correctly said. Could you please guide me to modify your circuit to be able to pump 20-25A at 24V

    Best regards

    Tsitsios

    ReplyDelete
    Replies
    1. Hi Tsitsios,

      Since there's no single chip device which can handle so much current, we would require an outboard transistor to reach the required level, you may refer to the circuit design provided in the following article, it might help you:

      http://homemadecircuitsandschematics.blogspot.in/2012/11/high-current-transistor-tip36-datasheet.html

      you may replace 7812 with 7824

      Delete
  22. hi sir , i like u site , and im newbie about electro , but i want to make this project , can u tell me ,what components are used, r1, r2, r3?
    and which place to put the 3 lamp signal?
    Can you present an image of pcb?
    thanks sir, sutan.rajosati@yahoo.co.id

    ReplyDelete
    Replies
    1. Hi Sutan,

      thanks, you can calculate the values of the r1, r2, r3 as given in the formula with the help of the battery AH rating.

      two lamps are already shown for the two steps after the initial high current stage, the initial lamp may not be required so it's not shown.

      I am sorry, there's no PCB drawing for this presently.

      Delete
  23. Hi! Sir. I have a 15v 5a transformer. I want to use this circuit for charge 12v 50ah car battery. Can you guide me which components should be replace.

    ReplyDelete
    Replies
    1. Hi Seok,

      you won't have to replace anything, you may proceed with the shown design.

      Delete
  24. Thank you sir! I will try to build this circuit..

    ReplyDelete
  25. Hai sir, I am a sound engineer working in malayalam film industry, but an electronics enthusiast too.
    It seems your project be very much helpfull to me in order to make my own inverter which is rated 300 w. but i have some doubts regarding the circuitry
    1) First of all i am asking you that, will there be any problem in using a 300 W 12-0-12 V transformer in the circuit since it produces much current (25 A) than that given in the schematic.

    2).' Installing a 60 Ah battery needs the addition of a TIP136 over LM338 '. I found this advice in this blog. So please help me to identify this modification correctly by providing me the schematic. Please help me quickly, becoz i am using three 12 V 20ah lead accid batteries in series.....

    3).Is there any other modifications needed for complete safeguarding of the battery??


    Reply me as earlier as possible to ma email id

    dev.arun999@gmail.com

    ReplyDelete
    Replies
    1. Hi Arun,
      If you are referring to the above circuit you can replace LM338 with LM196 and upgrade the circuit to handle upto 10amps.

      Rest will remain as is, but the limiting resistors will need to be calculated and fixed appropriately.

      Delete
    2. Thank you very much sir for replying fast. I am now gonna take care of it. But a doubt still remains that whether the 60 Ah battery can run that much load ( 300 W ) in the inverter. If it can how much hours of back up will i get ?
      And using TIP36 over the LM316 means replacing the regulator ic 7812 of the battery charging circuit with TIP36 given in the link

      http://homemadecircuitsandschematics.blogspot.in/2012/11/high-current-transistor-tip36-datasheet.html

      with the regulator ic LM196 ?


      And also i want to know whether 0.01 ohm with 5w is sufficient instead of a similar one with 2 W, becoz 5w resistors are only available in our area. Can you please suggest me a site to buy these cheep resistance online ?

      Delete
    3. Hi Arun,
      60 AH will not support 300 watts full load, the discharging rate should not exceed 10amps for a 60AH battery, meaning 10 x 12V = 120watts max.

      With LM196, TIP136 will not be required, just replace it with LM338 in the above step charger circuit, it will provide you with all the required protections and cut-offs.

      Delete
  26. Thank you sir for the advice. but a bit of confusion in it.
    I am using three 12 V 20 Ah lead accid batteries ( 12EC25L SEALED ZERO MAINTENANCE RECHARGEABLE BATTERIES ) in series which have been taken from my old electric scooter. As per your suggestion 60 Ah can't drow current beyond 10 A. Then how it is possible for the four 20 Ah batteries in the scooter can run a motor rated above 600 W ?

    ReplyDelete
  27. Thank you sir for the advice. but a bit of confusion in it.
    I am using three 12 V 20 Ah lead accid batteries ( 12EC25L SEALED ZERO MAINTENANCE RECHARGEABLE BATTERIES ) in series which have been taken from my old electric scooter. As per your suggestion 60 Ah can't drow current beyond 10 A. Then how it is possible for the four 20 Ah batteries in the scooter can run a motor rated above 600 W ?

    ReplyDelete
    Replies
    1. Is your scooter an electric bike? even if it's so if you check the current at any instant, it won't be above 6 to 10amps....the ideal rate of charging or discharging any lead acid battery is at 1/10th its AH rating.....you can draw higher currents from it, but that would only mean shorter backups from the battery. discharging current is inversely proportional to backup time

      Delete
  28. Dear sir,
    Are those resistors marked as R1 to R3 are necessary in the circuit ? What are their real functions in this circuit ?
    The 0.01 ohm resistors are not at all available in our area. Can you please suggest me a website from which i can buy that in smalk quantity ?

    WHAT TYPE OF 0.01 OHM RESISTORS I HAVE TO BUY ?
    I am asking this because i have seen different types of this resistance in internet such as wirewound, thick film, current sense resistor, surface mount type etc. Which should i prefer most. Please help me

    ReplyDelete
    Replies
    1. Dear Devraj,

      R1, R2, R3 determine and confirm the 3 step charging process, without them the circuit will lose its purpose.
      You can put 10nos 1 ohm 1/4 watt resistors in parallel for getting the 0.01 ohm value, by the way R1, R2, R3 will need to be calculated as per the battery being used....did you calculate them as per your battery specs, everything's comprehensively explained in the article, pls read it carefully.

      Delete
    2. But sir, one doubt whether power will get added on paralling those resistors. I mean whether i will get 2 or more wattage rate for the entire resistor bunk by placing them parallel. Anyway that is very crazy to have such a connection. Anyway i am going to order for 0.01 ohm 2W current sensing resistors. Will that enough ?

      Delete
  29. Sir a small request is there, please carry on reading.......

    Sir i am very much tired of searching for those low value resistors ( of the order of 0.01 ohm ) in the internet. I need some few of them, but all of the websites that i searched want me to buy an atleast quantity which is predetermined by them and at the highest rates i have to pay. If anyone is gonna make this circuit, they will probably meet this dilemma for higher values of battery Ah.
    SIR as per your design, the more higher the battery Ah, lower will be the resistance values ( R1, R2 and R3 ). So could you please suggest us a more better circuit utilizing almost all features of this one, but having no such low valued components included also having the feature of charging higer Ah batteries too.

    ( As per your previous reply to ma comment i will not get 0.01 ohm resistor with ten 1 ohm resistors. Please not it....... I will have to take hundred instead of ten. It will not be feasible sir )

    I HOPE A QUICK REPLY SIR.........

    ReplyDelete
    Replies
    1. I just forgot to tell you that LM338 wil not hold more than 5 amps, meaning more ta 50 AH batteries can't be charged with is circuit.

      A transistorized current controller can be replaced for LM338, but even it will require the same low value resistors.

      One solution is to use resistor divider network which will allow us to use higher value resistors.

      I'll try to update the article soon.

      Delete
  30. Good evening, sir, and thanks a lot for this very educative blog.
    I would like to build a charger for a 12V-66AH Automotive battery based on the above schematic.
    I have done the following modifications:
    1. Replaced the LM338 with an LM196 Adjustable Voltage Regulator
    2. Replaced the 4no. Rectifier Diodes 6A4 with a single Bridge Rectifier KBU1000
    3. Installed a 10A05 Rectifier Diode across each of the Relay Coils
    Please advise me if there is anything else I may have overlooked.
    I thank you in advance.

    ReplyDelete
    Replies
    1. Thanks Ronald,

      Everything's rightly done, just be sure you use the correct calculated value resistors for R1, R2, R3.

      Delete
  31. Thank you very much much for your quick reply. I will now embark on building the charger.
    Your help is greatly appreciated

    ReplyDelete
  32. By the way, am I right to assume that the same formulas above are to be used in computing the R1, R2 and R3 Resistor values, i.e.

    R1 = 0.6/ half battery AH
    R2 = 0.6/one fifth of battery AH
    R3 = 0.6/one 50th of battery AH

    Again I thank you.

    ReplyDelete
    Replies
    1. You are welcome.

      Yes, The formulas will be exactly as shown in the article.

      Delete
  33. Sir can i reduce a 25 A dc current to 5 A or less using LM196 IC????
    If yes how can it be done ?????'

    ReplyDelete
    Replies
    1. ...please refer to this post, it's explained there:

      http://homemadecircuitsandschematics.blogspot.in/2013/06/universal-high-watt-led-current-limiter.html

      Delete
    2. Can you please suggest me a NTC THERMISTOR having resistance of atmost 500 ohm and power dissipation of 40 J SIR????'

      Sir how can i depict the Joule rate in terms of wattage?
      I have found many specified with only Wattage not with Joules.
      How much wattage curresponds to a 40J rate. Is it 40 W or 40 mW????

      Delete
    3. Hi Nazriya,

      You will have to check the datasheet of an NTC thermistor manufacturer for getting the required value, I do not have the info right now.

      Here's the formula for converting J to W

      The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
      P(W) = E(J) / t(s)

      Delete
  34. Hi, im interested in doing a simple project of controlling home electrical system fan, tube light through remote control. can you pls help me by your simple circuit diagram so that it is very helpful and easy for me to do.

    ReplyDelete
    Replies
    1. Hi, You can try and use the design which is shown in the following article

      http://homemadecircuitsandschematics.blogspot.in/2013/07/simple-100-meter-rf-module-remote.html

      It will allow you to control 4 different appliances separately

      Delete

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Readers are advised to proceed with the construction of the presented circuits only after understanding the concepts from the core. Not adhering to this can lead to failures and frustrations.