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Low Battery and Overload Protection Circuit for Inverters

Low Battery and Overload Protection Circuit for Inverters

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.

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.

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About the Author

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



140 thoughts on “Low Battery and Overload Protection Circuit for Inverters”


  1. Howdy, Friend! Interested to Learn Circuit Designing? Let's Start Discussing below!
  2. 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

    • 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

  3. 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.

  4. 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

  5. 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.

  6. 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

  7. 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

    • 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.

  8. 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.

  9. 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?

    • 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.

  10. 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

  11. 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.

  12. 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

  13. 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

    • 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

  14. 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

  15. 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.

  16. 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

    • 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.

  17. 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.

  18. 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)

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

  19. 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…

  20. 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.

    • 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.

    • 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).

    • 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

    • 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)

    • 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.

    • 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?

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

    • 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?

    • 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..

    • 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.

  21. Hello Sir,

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

  22. 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

    • 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.

  23. 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?

    • 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.

    • 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

    • 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 ?

    • 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.

    • 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

  24. 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.

  25. 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 ………

    • 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.

    • 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…..

    • 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.

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

  26. 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.

  27. 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..

    • 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

  28. 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.

  29. 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?

  30. 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?

  31. 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?

  32. 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.

  33. 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.

    • 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.

  34. 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.

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