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You are here: Home / Inverter Circuits / Low Battery and Overload Protection Circuit for Inverters

Low Battery and Overload Protection Circuit for Inverters

Last Updated on August 3, 2020 by Swagatam 168 Comments

A very simple low battery cut-off and overload protection circuit has been explained here.

The figure shows a very simple circuit set up which performs the function of an overload sensor and also as an under voltage detector.

In both the cases the circuit trips the relay for protecting the output under the above conditions.

How it Works

Transistor T1 is wired as a current sensor, where the resistor R1 forms the current to voltage converter.

The battery voltage has to pass through R1 before reaching the load at the output and therefore the current passing through it is proportionately transformed into voltage across it.

This voltage when crosses the 0.6V mark, triggers T1 into conduction.

The conduction of T1 grounds the base of T2 which gets immediately switched Off. The relay is also consequently switched OFF and so is the load.

T1 thus takes care of the over load and short circuit conditions.

Transistor T2 has been introduced for responding to T1's actions and also for detecting low voltage conditions.

When the battery voltage falls beyond a certain low voltage threshold, the base current of T2 becomes sufficiently low such that it's no longer able to hold the relay into conduction and switches it OFF and also the load.

The"LOAD" terminals in the above diagram is supposed to be connected with the inverter +/- supply terminals. This implies that the battery current from the right side has to pass through R1 before reaching the inverter, enabling the sensing circuit around R1 to sense a possible over current or overload situation.

CORRECTION:

The above shown circuit will not initiate unless the relay is actuated manually through a push switch as shown below:

Parts List

  • R1 = 0.6/Trip Current
  • R2 = 100 Ohms,
  • R3 =10k
  • R4 = 100K,
  • P1 = 10K PRESET
  • C1 = 100uF/25V
  • T1, T2 = BC547,
  • Diodes = 1N4148
  • Relay = As per the specs of the requirement.



Previous: Make This 1KVA (1000 watts) Pure Sine Wave Inverter Circuit
Next: Cheapest SMPS Circuit Using MJE13005

About Swagatam

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

You'll also like:

  • 1.  Class-D Sinewave Inverter Circuit
  • 2.  Arduino Pure Sine Wave Inverter Circuit with Full Program Code
  • 3.  SPDT Relay Switch Circuit using Triac
  • 4.  Solar Boost Charger Circuit with LED Driver Dimmer
  • 5.  Inverter Voltage Drop Issue – How to Solve
  • 6.  Solid-State Inverter/Mains AC Changeover Circuits Using Triacs

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  1. Search Related Posts for Commenting

  2. Mark Rankin says

    Hello,

    Do low battery cutoff circuits wear out?

    I have a 4000 watt inverter and I set the low battery cutoff to energize at 22 volts. This has worked well for 12 years. Lately, the cut off activates when any heavy load kicks in (water pump, refrigerator) even when the battery is fully charged.

    I am trying to figure out if it’s a battery problem or an inverter problem.
    Thanks in advance for any help you can provide.

    Mark Rankin

    Reply
    • Swagatam says

      Hi, no they don’t, because cut off circuits are solid-state circuits, which will hardly ever blow, burn or degrade under normal conditions.
      In your case it could be a battery problem which may be unable to provide the initial high current due to aging or internal degradation, causing the voltage to drop severely below the lower cut-off threshold.

      Reply
      • Mark Rankin says

        Thank you so much for this, Swagatam
        About a month ago I installed a new 1236 amp-hour 24-volt battery (made for solar systems) by GB battery (a forklift battery company). It worked fine for about a month but I have my concerns now.
        The battery seems to go through the inverter’s “3 stage” charge cycles very fast.
        When the battery is at the “top” of a freshly charged cycle and it gets an inrush demand for amperage (blowing the low battery cut off) the voltage will drop way down to below 23 volts after the incident.
        I have been watching my specific gravity which reads (with no load on it) 1.25 per cell when the battery is at the top of the charge.
        Now I am concerned that this battery (even though they say it is made for solar) requires a different charging system.
        It’s like, over the past month of usage, the battery has slowly been depleted even though the charge cycle on the inverter says it’s complete.
        I will be calling the company in the morning. If any of this makes sense I would appreciate your input.

        Reply
        • Swagatam says

          You are most welcome Mark,
          Now it seems there’s actually nothing wrong with the working of the various systems in your controller, and neither the battery.

          The 3 step charger is designed to implement a fast charging on the battery, and for this initially during the first step it has to force a relatively high amount of current into the battery, which is readily accepted by the battery inducing the high surge current. As the battery charges, the charger reduce the current during the subsequent 2nd and the 3rd steps.

          Therefore during the initial 1st step, the surge may be causing a steep drop in the charger voltage and tripping the low cut off.
          To remedy the situation you can either adjust the 1st stage charging current to a lower level, or add some sort of delay feature to the low cut off, so that it doesn’t react to the initial momentary surge.

          However, if your battery is showing depletion, that means its condition is deteriorating, perhaps due to the high current charging, in that case it would be better to change the 3 step charger into a regular 1 step charger which will enable a moderately slow charging and ensure longer life for the battery and also stop the low voltage cut off tripping

          Reply
          • Mark says

            Swagatam,
            You are a rare find, thank you so much for sharing your knowledge. Just being able to bounce these problems off someone of your caliber is such a gift for me right now.
            I will attempt to make the changes you suggested, I hope you and your family have a Merry Christmas.

            Mark

            Reply
            • Swagatam says

              It is a pleasure Mark, always happy to help! Wish you all the best, and Merry Christmas to you too!

          • Mark Rankin says

            I think I found the problem!!!

            I did a gravity test on all twelve cells tonight, this battery is only two months old. To my surprise, I found one cell that did not even float the hydrometer indicator, it read 0 volts on the multimeter. Not sure how I missed this before. I am contacting the company about a replacement cell (hopefully under warranty) this week.

            I just wanted to pass on the good news.
            Thanks again, Swagatam!

            Reply
            • Swagatam says

              Sounds great, Glad you could find the fault so quickly!

  3. jay says

    hello sir , i’am from philippines and a beginner electronic hobbyist. I have finish your circuit ‘low battery cutoff and overload’ installed on my oscillator circuit ( pwm inverter ) sg3525 ic (not on the battery side of the inverter ) .. I set the low battery cutoff to 11 volts and it is working perfectly on a small load like 60 watts electric fan, and 100 watts soldering iron , but when my load is 60 watts 21 inches crt television the relay will cutoff immediately .. I’am using 10 ampere relay and .02 ohm resistor for r1 .. please i need your help .

    Reply
    • Swagatam says

      Hello Jay, 60 watt fan/iron and 60 watt TV are similarly rated? May be the TV rating is much higher than 60 watt…But anyway 0.02 is too small and will NOT provide any current limit or over current protection below 30 amps, and 30 amps looks impractical for your application

      Reply
  4. Okanama says

    I really appreciate this, I will only ask for a circuit that changes from the power station to inverter and vise versa.
    When done, charging the batteries begins.

    Reply
    • Swagatam says

      You can probably try the following designs:

      https://www.homemade-circuits.com/automatic-inverter-supply-and-mains/

      https://www.homemade-circuits.com/how-to-convert-inverter-to-ups/

      Reply
      • Joshua says

        hello sir, im Joshua. i designed an inverter with sg3524 and it has been working perfectly but i have been having serious problems with the rate my batteries overcharged. the circuit does not have overcharged protection feature. could you please help me out?

        Reply
        • Swagatam says

          Hello Joshua, you can try the following concepts:

          https://www.homemade-circuits.com/opamp-low-high-battery-charger/

          or simply use a fixed regulated 14.1V as the input supply for the battery, without the need of any cut off circuit.

          Reply
  5. Robert Obiorah says

    Sir, I have been having this constant problem with repair of Indian inverters. And it has to do with overload, no load, etc fault. Please, an you give me a clue as to how to tackle this issue. I have no idea where to start

    Reply
    • Swagatam says

      Robert, all inverters have different board configuration and component settings, so it can be difficult to judge their faults commonly…

      Reply
  6. ben says

    Sir help me for 24v battery full cut off using opamp lm358
    Please is ulgent

    Reply
    • Swagatam says

      Ben, you can try one of the designs explained in the following article:

      https://www.homemade-circuits.com/opamp-low-high-battery-charger/

      Reply
  7. Saeed Abdullah says

    Hi sir please i my circuit diagram and if i built it blow the mosfet and if
    i can send my circuit diagram to check it if there is any problem then please you help me. Thanks sir

    Reply
    • Swagatam says

      there’s no mosfet in the diagram, it is a simple and a tested design and will work definitely, but only if you connect and set the circuit correctly

      Reply
      • Saeed Abdullah says

        hi sir, pls i have build inverter with ic TL494 plus mosfet irf3205 and if i connected with 12v battery it blow mosfet and i used ferrite core transformer and sir if you can help me with this issue I will be glad. Thank you

        Reply
        • Swagatam says

          Hi Saeed, if it is a ferrite based design then it can be difficult to troubleshoot for me, because ferrite trafos require strict calculations for the winding, and a wrongly built trafo can cause instant burning of the devices.

          Reply
  8. Emmanuel John says

    good day sir, i hope you’re doing well? back in a months, I ask you so many questions about inverter and the challenge which i had with the one I built. with your help, i had solve the challenge,my mosfet are not getting heat anymore ,tanks for the help. please Sir I am back with another question. question is that,can at the same time the inverter out put still charge the battery while is still working or ho can I make it work constant that is, without removing the battery to charge separate before used again?

    Reply
    • Swagatam says

      Hi Emmanuel, that’s not possible, your battery will get discharged and then you will have to switch OFF the inverter and charge the battery from an external source until its fully charged again

      Reply
  9. itar eric says

    please sir, can I use the above circuit with my home made power inverter , will I connect the protection between the load and inverter or how do I connect it?if I can’t please I need an over load and short circuit protection circuit for 12v inverter and how to connect it, thank you sir

    Reply
    • Swagatam says

      Hi Itar,

      yes you can use it for inverter overload protection. THe “LOAD” here is the inverter, so connect the inverter DC supply terminals across these points, and connect the battery on the right hand side as shown

      Reply
  10. Yusuf Abdlazeez says

    Hi sir. Thanks for d job well done. The calculation of R1 is my prolem. You said R = 0.6/Trip current and TC = Wattage/Voltage. I get 0.00xx Can you explain more. Thanks and God bless.

    Reply
    • Swag says

      Thanks Yusuf, what trip current or max overload current are you trying to apply? please let me know I’ll solve it for you.

      Reply
  11. Abioye says

    sir i have a 3000watt inverter built from a 650watt inverter with your honourable advice and guidance but i want to make a overload cutoff for it my maximum drain current from the battery is 125amp when i divide 3000watt by the batteries voltage which is 24v and it give me the R1 value of 0.0048 how many 0.1resistor do i wire in parrallel to achieve this value

    Reply
    • Swag says

      You will need 20 resistors in parallel. Each must be rated at 5 watts

      Reply
  12. Godson Ugo says

    Hello sir Swagatam, I asked for an overload protection circuit for an inverter with auto-power off feature in the comment session of one of your posts and you referred me to this post. Having read through the post and comments, I concluded that I may not be able to use the circuit for the intended purpose. You explained in the comments that the inverter should me made the load but I’m afraid that I won’t be able to get the relay that can handle the huge current drain from the battery. So what I need is circuit that will be connected to the output side of the inverter transfo and which can operate a small 12V relay such that I can use the relay contacts to power off the inverter driver when overload is sensed at the inverter output. I’ll be glad if you could help me with such circuit. Thanks a lot sir.

    Reply
    • Swag says

      In that case you can try the concept explained in the following article:

      https://homemade-circuits.com/2014/06/simple-current-sensor-circuit-modules.html

      use the NPN version, and connect the collector with the mosfet gates via separate diodes or simply connect it with the “Ct” of the IC,…. for SG3525 you can use the PNP version, and connect the collector with pin#10 via a 1N4148 diode

      Reply
      • Godson Ugo says

        Ok sir. Thank you very much. I’ll check it out.

        Reply
  13. moses kwagam says

    Thank u Swagatam and every one participating in this blog. Pls how do I determine the trip current and/or the value of R1 if I want to make 1KVA inverter and I want the inverter to cut off when it senses 800 watts of load

    Reply
    • Swagatam says

      thanks moses, it can be calculated by the formula:

      R = 0.6 / trip current

      trip current = wattage/Voltage…in your case it will be 800 / battery voltage

      Reply
    • Dady says

      Sir I need overcharged protection circuit diagram and low battery cutoff circuit diagram. I built my inverter with SG3524 and its 24v. I would prefer the link sent to my mail: zuredu@gmail.com.
      Thanks

      Reply
      • Swag says

        Dady, you can find many options under this category:

        https://www.homemade-circuits.com/category/battery-cut-off-switch-diagrams/

        Reply
  14. Faith Jumbo says

    in the case of inverter the load is were the ac is connected but in on one of the comments i read u said the entire circuit is connected to the dc side so how do we monitor the load since the load are the appliances since the appliances are connected to the ac side

    Reply
    • Swagatam says

      the load cannot be an AC load here, the circuit is designed for DC loads, if an inverter is connected as the load then the appliances consumption would be ultimately drawn from the battery, and therefore the circuit would sense it and trip if an overload from the appliance was detected.

      Reply
  15. i.c.v. young king says

    Pls sir can u send me the circuit diagram of low battery cut through this mail ,youngkingicv@gmail.com pls sir l'll be waiting thanks.

    Reply
    • Swagatam says

      ICV, you can refer to the following article for making a low battery cut off circuit

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

      Reply
  16. dawar jan says

    How does the circuit work.esplain step by step. How can i change it to only cut off at low voltage.reply plz

    Reply
  17. dawar jan says

    What do u mean when the voltage cross 0.6v

    Reply
    • Swagatam says

      it is the triggering base voltage for T1

      Reply
    • Wise Essien says

      Hello sir I have challenge on how to build h_bridg inverter

      Reply
  18. MORRIS CERULLO says

    What should i do for this circuit to cut off at 8volt when using 12v 70amp-hour battery?

    Reply
  19. Usa Juana says

    Hello sir . am building the bubba osilator inverter and am adding the overload and no load detection circuits , please I have these question on those circuits .1, I have finished the oscilators and I am getting 1.4 Volts ac , is that OK?.2, for the sensing resistor of the overload circuit RESISTOR X I calculated 0.00909 ohm , so 12 0.1ohm in parallel will solve it but what wattage of resistor will be OK will 1/2 be OK as it is many?. 3,the no load detection circuit have 0.05 ohm so 2 0.1 ohm resistor will solve it , now this concern me more,what watts do I need to use so that it can with stand the 800watts inverter at its full operational power?

    Reply
    • Swagatam says

      Hello Usa,

      please comment separately under those articles, so that I can refer to those diagrams and address the issues accordingly.

      the bubba oscillator output needs to be processed to SPWM first in order to achieve sine wave output.

      Reply
  20. Shuddhatam Jain says

    pl tell me where i purchases lm196 or 78h12a

    Reply
  21. vhafuwi says

    Hi Swagatam

    I require your assistance, I have installed an off-grid solar system for student residents ,just for them to use light , laptops and charge their cell phones , it is becoming difficult to control them since they have other high wattage appliances such as electric iron , boilers in their rooms, i require a help in designing a device that will shape the amount of watts or amps a students is allowed to use per room plug , the unit must only disconnect just the student who plug a overload appliance , i will appriciate your assistance , my email vhafuwi@gmail.com

    Reply
    • Swagatam says

      Hi vhafuwi,

      you can try the following circuit

      https://homemade-circuits.com/2015/03/mains-over-load-protector-circuit-for.html

      install one such unit in every room after the mains DPDT switch

      Reply
  22. adelusi oluwatosin says

    Thank you so much sir. sir how can i select R1 for 1.5kva inverter? i.e the appropriate current sensor resistor for 1.5kva. Thanks.

    Reply
    • Swagatam says

      adelusi, divide 1500 with the battery voltage, and then divide 0.6 with this result.

      Reply
  23. Swagatam says

    Hi Robert, it's possible only through P1, that is through the end terminals of the preset

    Reply
  24. Robert Random says

    Hi Swagatam,
    Would there be a drain of the battery through P1, D1, R1, T1 when the circuit is off due to battery depletion? Thanks. ~Robert

    https://www.circuitlab.com/circuit/227r5y/screenshot/1024×768/

    Reply
  25. kamran abbas says

    Hi How are you
    Sir
    How can I make digital numbers display for example I have inverter and i want show inverter output voltage on display and also battery voltage

    Reply
    • Swagatam says

      Hi Kamran, you can make and use the following circuit and apply it in your inverter for the results:

      https://homemade-circuits.com/2013/05/make-this-simple-digital-voltmeter.html

      alternatively you can buy a readymade digital voltmeter module and use it for the same.

      Reply
  26. rob wiseman says

    Hi Swagman.

    The P1 preset in your low voltage cutout says (10 preset) how many ohms is this component,,
    regards Rob.

    Reply
    • Swagatam says

      Hi Rob, it's 10k, I have corrected the typo

      Reply
  27. Olaleye Kolade says

    Thanks for this simple circuit.
    Am having little problem about the circuit, I build just the circuit and set it, everything is working, but my question is where will I connect the terminal mark load? Is it directly from inverter transfo output
    2. The terminal mark battery is it the battery for the inverter or another battery. Please explain
    Thanks

    Reply
    • Swagatam says

      The terminals indicated as "load" is connected with the inverter input supply terminals which is normally supposed to be connected with the battery directly.

      and the terminals marked as "battery" should be connected with the inverter battery….not anything external.

      with the above set up the battery power now enters the inverter through the above circuit's relay so that the required cut-off actions can be executed

      Reply
  28. ataull hakim says

    Sir, can i use it as a solar charge controller for LVD circuit?

    Reply
    • Swagatam says

      yes it can be used

      Reply
  29. Shuddhatam Jain says

    Respectiv sir 3 problems come me for making circuit Low Battery Cut-off and Overload Protection Circuit
    1. this circuit work for this battery 18650 li-ion 11.1v 6.6Ah battery ?
    2. what is R1 = 0.6/Trip Current and how many watt?
    3. what is setting of pre set is mark for this battery?
    i use this battery pack in hero honda(super splender) biky for the self starting
    pl help me

    thankyou

    Reply
    • Swagatam says

      Shuddhatam, I have already answered to the same question in the other article….please check it

      Reply
  30. Angelous Chavez says

    Sir, i interesting with this circuit. I am planning to make this circuit to add in my project. Can you give me s circuit for battery cut-off and overload protection for dc to ac inverters please if you have.

    I am angelous from the philippines i am an electronics student.

    I am hoping for your reply…

    Thanks in advance.

    Reply
    • Swagatam says

      Angelous, I'll check and try to find one…

      Reply
  31. Shuddhatam Jain says

    respective sir my name shuddhatam jain i likes your circuit digrame i need li-ion battery protection circuit i make Low Battery Cut-off and Overload Protection Circuit according to you in this circuit i don't understand what is meen by QUIZ = Explain the introduction of R4 and C1.
    pl give mr reply

    thankyou very much

    Reply
    • Swagatam says

      Hi Shuddhatam, thank you for appreciating my site.

      the circuit required by you is already present in my blog, please use the search box above to find the preferred one of your choice.

      R4 is for locking the relay when it clicks, and C4 is for preventing the chattering of the relay during the changeover action.

      Reply
  32. Sina Alimorady says

    Hi Swagatam,
    is this circuit suitable to use with https://homemade-circuits.com/2013/04/automatic-micro-ups-circuit.html in order to make the Micro UPS auto cut off the load from battery ??

    if yes what changes i have to make to support 4 or 5AMP load like Micro UPS does ?

    if not what do you suggest to use with Micro UPS ?

    thank you again and again @

    Reply
    • Swagatam says

      Hi Sina, the above circuit is not quite efficient with its detection, and also it could make the entire design too lengthy, so it's better to use the last circuit from this article:

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

      this will take care of the all the required safety conditions.

      Reply
  33. Ben says

    How is the protection circuit connected to the inverter circuit?

    Reply
    • Swagatam says

      the "LOAD" is the inverter supply input….

      Reply
  34. Swagatam says

    R4 is for making sure that the relay gets latched after activation when any of the situation is detected…C1 is for preventing the relay from chattering during the operating thresholds.

    Reply
  35. Swagatam says

    yes that's correct

    Reply
  36. Swagatam says

    10k ohms is the right value

    Reply
  37. Swagatam says

    …ok got it, you meant to say the lower cut off voltage, yes that's right, 18V is correct.

    Reply
  38. Shankar ks says

    hi sir, can i connect load cell directly to arduino. if no, please give the interface circuit to connect load cell to arduino(supply voltage should not more than 6v)

    Reply
    • Swagatam says

      Hi shankar, please provide the info in more detail regarding the application so that I can understand and design it better.

      Reply
  39. Yel Bordallo says

    Hello sir can i used overload protection only in this circuit…what parts of this circuit i will remove…thank you hope for your quick response…

    Reply
    • Swagatam says

      Hello Yel, you'll have to use the entire circuit as shown even if you want only the overload to work….

      Reply
    • Naveenkumar T N says

      what is the maximum load it can handle

      Reply
    • Swagatam says

      will depend on the relay contact…..can be upgraded to any limits.

      Reply
  40. HuynhNguyet says

    Hello Sir,
    Intermediate relay only endure current 10A, how do we use for current 30A?
    Thank you very much.

    Reply
    • Swagatam says

      hello Huyn, what's intermediate relay? relays are available in all shapes, sizes and ratings, 30amp are also available.

      Reply
  41. HuynhNguyet says

    Hello Sir,
    Where do load and battery connect?

    Reply
    • HuynhNguyet says

      Hello Sir
      I'm sorry if my comment made you discomfort.
      I change BC547 by C1815, that is ok.

      Thank you very much for your design.

      Reply
    • Swagatam says

      OK that's great!

      Reply
    • HuynhNguyet says

      Hello Sir,
      I don't understand: When the battery is 13V, there is not power to B pin of BC547 (T2), T2 can not active, so relay can not active.
      Would you like to explain that? Thank you very much.

      Reply
    • Swagatam says

      hello Huyn, if you have selected 13V as the triggering threshold then you must adjust the preset until the relay gets activated at 13V…..

      Reply
  42. HuynhNguyet says

    I'm sorry sir
    I can not find any error. T2, T1, relay … are ok. I don't know why.

    Reply
    • Swagatam says

      there's no way a relay won't activate if the transistor is good, connected rightly, base getting 0.6V and the power supply current sufficient…

      remove the emitter diode and check, may be it's faulty.

      Reply
    • HuynhNguyet says

      Hello Sir,
      I made new board with new components, the relay does not active. When I check the load voltage is 1.3V at 13V battery.
      Thank you very much for your support!

      Reply
    • Swagatam says

      Hello Huynh, try the following basic relay driver stage first, and check the response:

      https://homemade-circuits.com/2012/01/how-to-make-relay-driver-stage-in.html

      Reply
    • HuynhNguyet says

      Hello Sir,
      Does R1 relate this problem?

      Reply
    • Swagatam says

      No, R1 has no connection with this problem.

      Reply
    • HuynhNguyet says

      Hello Sir,
      Does N/C pin of relay connect to R4 and NO pin connect to R3?

      Thank you for your support!

      Reply
    • Swagatam says

      hello Hunnh, yes that's correct.

      Reply
    • HuynhNguyet says

      Hello Sir,

      If I only use low bettery cut-off function, What component's I remove?
      Thank you very much.

      Reply
    • HuynhNguyet says

      Hello Sir,
      If I choice C1 = 100uF/50V, is it ok?
      Thank you for your support.

      Reply
    • Swagatam says

      yes C1 = 100/50 will also do.

      Reply
    • Swagatam says

      you can try the following design which is without overload cut off

      https://homemade-circuits.com/2011/12/simplest-smf-automotive-battery-charger.html

      Reply
    • HuynhNguyet says

      Hello Sir,
      My equipment use 12V battery. I want to isolate between the equipment and battery when the battery voltage is 11.5V in order to protect battery. When battery voltage is 12V, it will supply power for the equipment again.
      Thank you for your help.

      Reply
    • Swagatam says

      Hello Huyn, a 12V batt must be charged until it reaches 14V, therefore once your battery gets discharged to 11.5V, it should be charged to 14V before using…

      the above linked design will do the above mentioned operations appropriately

      Reply
  43. HuynhNguyet says

    Hello Sir,

    I assembled this circuit. However, the relay do not active when the bettery is 13V (0V in the load). When I remove T2, I connect direct from relay->diode-> mass, the relay actived (the battery is 13V, 13V in the load). I don't known why.

    Reply
    • Swagatam says

      hello Huynh, either your transistors are not good, or the preset is not correctly set or T1 is interfering due to some incorrect connections, pls check everything again.

      Reply
    • HuynhNguyet says

      Hello Sir,
      Thank you very much for your reply.
      I will check T1 and some connection again. When I use Proteus software to sumilate, the result is the same my assembled circuit (the relay does not active when battery is 13V).

      Reply
    • Swagatam says

      check the base voltage of T2 at 13V, it should be 0.6V to activate the relay…the relay will not activate if its coil resistance is very low in that case R3 will need to be reduced for triggering the relay….1K could be tried for R3

      Reply
    • HuynhNguyet says

      Hello sir,
      Thank you very much for your support.
      I already check the base voltage of T2 at 13V, it is 0.6 V. I change R3 by 1k resistor, then I set P1 (P1=10k). The relay is not active (I am sure the relay is ok)

      Reply
    • Swagatam says

      Hello Huyn, connect a LED with a series 1k resistor parallel with the relay coil and check the response…if the LED glows/shuts off in response to the preset adjustments would mean the relay has problems or is not correctly matched.

      Reply
    • HuynhNguyet says

      Hello Sir,
      Thank you very much for your support.
      I already connect a Led with series 1k resistor parallel with the relay coil. The Led light. That it's mean the relay has problem?

      Reply
    • Swagatam says

      hello Huynh, short the collector/emitter of T2 manually, if the relay clicks would indicate a good relay but a faulty transistor…

      Reply
    • HuynhNguyet says

      Hello Sir,
      Thank you very much for your support.
      When I shorted the collector/emitter of T2 manually, the relay clicked. That it's mean T1 or T2 faulty?

      Reply
    • Swagatam says

      OK now remove the collector of T1 from R3/P1 and check again, this time if the relay functions normally through T2 would indicate a faulty T1….otherwise it's T2 that may be doubtful..

      Reply
    • HuynhNguyet says

      Hello Sir,

      I already remove the collector of T1 from R3/P1, the relay does not active at 13V. May be I will change T2 tomorow.
      Thank you very much.

      Reply
    • Swagatam says

      OK!

      Reply
  44. HuynhNguyet says

    Hello Sir,

    I assembled this circuit. However, there is not the voltage in the load with any level battery voltage (11V – 13.5V).

    Reply
  45. Mouhammad Alnajjar says

    well , I've made the circuit and put a red led to detect the passing of the current , and R1 = 3 resistors 2watt : 2x(1 ohm) and 2.7 ohm
    I've tested it with 9v and 12v battery
    when I change the preset the circuit doesn't deactivate !
    although I've connected each point of the relay to the true point
    can you help me please ?
    thanks

    youtu.be/AshLT5g010M

    Reply
    • Swagatam says

      for setting up the low voltage cut off, you will have to connect the 9V source to the "load" marked terminals and then adjust the preset until the relay just trips, after this the source may be removed from the "load" terminals and the actual 12V battery connected to the "battery" marked terminals for normal operations

      R1 has not related to the above setting. It determines the overload threshold trip point for the relay.

      Reply
  46. Mouhammad Alnajjar says

    hello sir

    if the circuit current is 1.5A and i wanna cut-off at 10.5 volts which relay I have to use ?

    if the cut-off volt is changed what is the rule to choose the suitable relay?

    Reply
    • Swagatam says

      hello Mouhammad,

      the relay has no relation to amps and cut off voltage of the design….you just have to choose a relay whose coil voltage is matching with the supply voltage of the circuit and adjust the preset such that relay just cuts off at the specified lower threshold….the adjustment will need to be made by supplying this lower threshold through an external power supply while setting up.

      Reply
    • Mouhammad Alnajjar says

      thanks so much sir

      there is something else :
      is this circuit re-activate automatically after it is deactivated ?
      because if it cuts-off at (e.g 11 volts) , the voltage will be more than 11 volts with after the loading ends, so the circuit will be activated then deactivated many times

      Reply
    • Swagatam says

      No that won't happen because of the transistors hysteresis properties…

      Reply
    • Mouhammad Alnajjar says

      ok that's good

      but 'm facing a problem with relay connections

      this is the diagram :
      imagizer.imageshack.us/v2/280x200q90/537/zaHcTY.jpg

      with the meter I've found the points (2,1) in the place of (3,5) and (3,5) in the place of (2,1) I don't know how or why

      because I've found connection between the points 4 with 2 not 4 with 5 !

      how can treat this please ?

      Reply
    • Swagatam says

      If you can show me the relay image I can try to help, without seeing the relay image it would be difficult to locate.

      Reply
    • Mouhammad Alnajjar says

      this is a video I've recorded :
      youtu.be/bi4bgS_6YEU

      in the website of the store I've bought from , there is the diagram of points connections the same I told you
      http://www.matni.com/Arabic/Relay/RELAY.htm

      the relay model is T-73 and in the website I think it's the same JQC-3F(T-73)

      Reply
    • Swagatam says

      the connections for this relay have been elaborately explained in this article, please check it out:

      https://homemade-circuits.com/2012/01/how-to-understand-and-use-relay-in.html

      Reply
  47. Vishnu Menon says

    How to make resistors of such low value 6miliohms etc here I am stuck do u have any idea pls let me know

    Reply
    • Swagatam says

      use a couple of inches long non-plated iron wire, or a meter long copper wire wound on a former, tweak and adjust the lengths by verifying the ohms through a suitable multimeter.

      Reply
  48. Vishnu Menon says

    Sir ,
    I didn't understand the preset part correctly can u explain its working in the given circuit

    Reply
    • Swagatam says

      Vishnu. apply the lower battery cut off voltage to the circuit through an adjustable power supply and adjust P1 until the relay just deactivates, once this is setting is done the circuit would automatically switch OFF whenever the battery voltage reaches this level

      Reply
  49. adelusi oluwatosin says

    Thanks for time spending on this site. Sir can this circuit work for 3kva inverter? If yes kindly do the adjustment more so I need to know total calculation of R1. Thanks.

    Reply
    • Swagatam says

      Yes it will work with 3kva inverter, just modify the relay contact accordingly and use 8050 for T2.

      Reply
  50. Shahirah MNY says

    hello sir, can this kind of cct can be modified to trip low standby current? thank you 🙂

    Reply
    • Swagatam says

      hello shahirah,yes it can be done by setting the preset appropriately.

      Reply
  51. sskopparthy says

    sir, you told that "https://drive.google.com/file/d/0BytEbOgq6mqeQU5mdzU4Yl9kUDA/edit?usp=sharing" is the circuit low battery cut off(to walkabout in above comments). i would like you to help me to modify this circuit to suit my needs. i am not a genius and just a beginner. so, i dont know whether my idea is correct and it can be done or not. i wanted to tell that when load is applied to circuit, and when its voltage decreases to preset level, it cuts off the relay. but when load is disconnected, i observed that the battery regains its voltage to some extent. so, then the circuit may connect the load again……so, i request you to modify this circuit such that, once it cuts the load, it should stay in cutoff position only until a reset button is pressed….can this be done to this circuit, sir? im just a beginner, so please help me ………

    Reply
    • Swagatam says

      ss, there are two faults in the shown link, first the N/C should not be connected to ground, second, R3 other end must be connected to ground, rest everything looks OK.

      it's highly unlikely that the relay would oscillate, due to the presence of the transistor hysteresis, so probably no latching feature would be required.

      Reply
    • sskopparthy says

      Sir, I connected the circuit in the correct way only as you said……before making this comment. But I failed to check the circuit diagram before giving the link(I erased n/c and connected r3 other end to ground but didn't save it but I thought I did). And thank you very much sir…now the circuit seems to be good working…..

      Reply
    • Swagatam says

      OK, no problem, that's great!

      Reply
  52. Shankar ks says

    sir i need circuit for cutoff voltage with 2millivolt.

    Reply
    • Swagatam says

      use any of the following circuits, replace 741 with LM311

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

      Reply
    • Shankar ks says

      will that circuit need any other changes for cutoff voltage as 2 millivolt.

      Reply
    • Swagatam says

      no changes would be required…it will respond even to the minutest changes between its inputs.

      Reply
  53. Rolando Avecilla says

    Thanks for this circuit.

    How can I modify this to use solid state relay?

    Reply
    • Swagatam says

      It may be done wit the following mods:

      Remove the relay and R4 entirely.

      replace the relay coil connection points with the battery poles.

      T2 may be upgraded as per the load amp specs.

      Reply
    • Lester Johnson says

      Sir . Will you low battery cut off circuit, turn on the inverter when the battery is recharge?

      Reply
    • Swagatam says

      it may be done by incorporating another relay parallel to the existing one and by wiring its contacts appropriately for the intended chageover

      Reply
  54. Oluwaseyi Onasami says

    Hello Swagatam,

    I have tried working with the Overload, Short Circuit and Load Battery Cut-Off Circuit but having challenges setting the cut-off voltage (10.5V) and getting the right value for R1. I want to use it for 2KVA inverter.

    Secondly, can you help with a 12V, 100AH charger circuit with constant current charging and overcharge protection at 13.5V.

    Thanks in this regard.

    Reply
    • Swagatam says

      Hello Oluwaseyi,

      if you are using a battery then the settings should be easier to implement, if the input is through a power supply then it would need to be regulated.

      If you find it difficult then it could go for an IC based design.

      You can try the last charger circuit for your 100 ah battery from the following link, just replace the shown LM338 with LM396 or LM196

      https://homemade-circuits.com/2012/07/making-simple-smart-automatic-battery.html

      Reply
  55. Muhammad Junaid says

    Hello Sir I want to make an overload protection circuit which can sustain upto 600Watts or as per requirement by use of Potentiometer. Please help me out how to modify this circuit for this purpose..

    Reply
    • Swagatam says

      Hello Muhammad, you can do it by replacing R2 with a 1k pot. the ends of the pot will go to transistor base and R1, while the center will touch the ground

      Reply
  56. Debashees Sikder says

    Sir,
    Can I use this circuit with sg3525 without relay?I am thinking that I can use a PNP transistor , am I right?

    Reply
    • Swagatam says

      The above circuit is suitable only with a relay, for 3524 IC only an opamp circuit would be suitable

      Reply
  57. Swagatam says

    R3 preset bottom free end should be connected to the negative line.

    the indicated +/- lines are correct. The relay out is positive, the bottom rail is the negative.

    Reply
  58. Walkabout says

    Now i tried google drive, if you can view the image. https://drive.google.com/file/d/0BytEbOgq6mqeQU5mdzU4Yl9kUDA/edit?usp=sharing

    Reply
  59. Swagatam says

    the image link is not opening in my computer.
    It should be OK if you have done as per the mentioned instructions.

    Reply
  60. Walkabout says

    Continuing the discussion of just having low voltage cutoff circuit. You mention(I'm quoting you) "Keep only D1, D2, T2 and the relay, eliminate everything else.

    However R3 now gets replaced by a preset, the center goes to the base of T2, the other two terminals go to the positive and the ground respectively.

    For safety add a 1K resistor with the base of T2 which then can be connected to the center tag of the preset."

    ok following your instructions and basing from the updated diagram, lead me to redraw it to fit for a low voltage cutoff circuit imageshack.com/a/img443/5532/6s0f.png.

    is that the schematic for a adjustable low voltage cutoff circuit?

    Reply
    • Andrew says

      Hello sir, I understand everything else about this circuit except for (R1) please i do not know what am suppose to use, if it,s resistor, fuse or breaker as (0.6/Trip Current) because what i understand is that the negative taminal of the better is suppose go to the inverter negative source which means using a resistor is not going to be a good idea, please i need your help because i have already finished building this circuit with a (0.22 ohm) resistor in place of (R1) but i need to know exactly am suppose to use before testing it because am sure. Thanks

      Reply
      • Swagatam says

        R1 is a resistor, its wattage should be = 0.6 x maximum trip current value.
        You are supposed to do exactly what is shown in the diagram, the negatives have to connect through the resistor R1.

        Here’s another similar design you can try which is thoroughly tested by me:

        https://www.homemade-circuits.com/how-to-make-automotive-electronic-fuse/

        Reply
      • Andrew says

        OK, thanks 4 ur quick respond, i really do appreciate.

        Reply
  61. Swagatam says

    1) In the above updated diagram R3 is a fixed resistor P1 is the preset, yes, any two terminals of the preset can be selected, the center lead being the mandatory one.
    2)the line which joins R1, T1 emiter, C1 negative and D1 cathode is the negative line.
    3) please refer to the above diagram, I am not sure which diagram you are referring to?

    Reply
  62. Walkabout says

    I easily get confused, after I drew the schematic diagram. My questions are;
    1st:There are 3 connections for R3 preset, the 3rd will not be connected, right?
    2nd: Where will the negative polarity of the battery be connected?
    3rd: T2 base is connected to the 1K, while the other two terminals go to the positive and the ground respectively. Is the emitter the ground?

    Could help me redraw the schematic?

    And for just overload cut-off circuit, what would I need in a circuit and what would the schematic look like?

    Reply
  63. Swagatam says

    Make R3 = 22k, and R2 = 1k, that's all, other components will remain as is except the relay which should be 24V rated.

    Reply
  64. tervkal says

    Hi, thank you for your great cutoff circuit. I suppose the cutoff voltage is changing by temperature quite large range in your circuit. A base-emitter voltage of the bipolar transistor has a temperature coefficient at least -2mV/C. That is not a problem because a battery to be charged only in specified temperature range.

    Reply
    • Swagatam says

      you are right, thanks!

      Reply
  65. Allnight says

    New question about this: Could two of these circuits be hooked up to the same relay?

    I have an RV with two sets of batteries: house and chassis. When the engine is running (the chassis battery is over 13V), I want to connect the two sets of batteries. When the house is plugged in (house battery over 13V), connect the two sets of batteries.

    Also, if the voltage of either side goes over 16V, disconnect them (that part may be harder, and not an original requirement, it just occurred to me).

    I was hoping to build two of these circuits, one hooked up to the chassis battery and the other hooked up to the house battery, and have a single relay connect the two batteries. I want to limit the current that crosses the relay to about 30amps (the max output of the house battery charger, as the engine alternator can put out 190amps).

    I'm currently hunting for a relay that will strongly move to open when the coil isn't energized, as I don't want bumps in the road to accidentally flip the relay to the connected state.

    Reply
    • Swagatam says

      Yes that would be possible, you just have to make the collectors of T2 common from the two circuits, and join with the relay coil.

      For 30 amps you would need a powerful relay which would obviously have a tough electromagnetic connection, so hopefully it won't be rattled by the "bumps"

      T2s will need to be appropriately dimensioned as per the relay coil ratings.

      Reply
  66. Swagatam says

    No this circuit cannot be used in any manner for feeding a 3525 input, because the IC requires definite logic level which cannot be achieved from a transistor, you will have to incorporate a 741 IC for it.

    Reply


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