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Low Battery Cut-off and Overload Protection Circuit.

Last Updated on by 248 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 the Circuit 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.

 

 

QUIZ = Explain the introduction of R4 and C1.

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.

Reader Interactions

Comments

  1. 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

  2. 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

  3. 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

    • 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

  4. 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

    • 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

  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?

    Reply

    • 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

  11. 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

  15. 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

    • 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

  17. 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

  18. 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

  19. 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

    • 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

    • 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

  27. Hi Swagatam,

    I submitted the comment above.
    I'm now using two 12 volt, 15 amp-hour AGM deep cycle batteries in series. The load will be a 350 watt DC motor.
    Shall I still use a 2N2222 for T2? The automobile relay should be rated up to 24V correct?
    You said the wattage for R1 should be 10 watts. Should all the resistors be rated at 10 watts?
    Once I have the circuit set up do I just use the potentiometer to adjust the cut of voltage?

    Reply

    • Hi anonymous,

      an 30 amp automobile relay coil might require heavier current to operate therefore 2N2222 should be used since it is rated to handle upto 1 amp.

      yes the relay coil should be rated at 24V

      a 24V 350 watt motor will require around 14 amp current, therefore the sensing resistor should be

      R = 0.6/14 = 0.042 ohms

      wattage = 0.6 x 14 = 8.4w or 10 watts for max safety.

      at 14 amp your 15 ah battery will become flat within 1/2 an hour or even less.

      Reply

  28. Hi Swagatam,

    I want to use two 11.1 volt, 12 Amp-Hour LiPo batteries in series. I'm guessing the cutoff voltage will be at 18 volts. The load will be a 500 watt DC motor. What values of the circuit diagram should I change to do this? What trip current will I use to calculate my R1?

    Thank you

    Reply

    • Hi Anonymous, if two are connected in series the cut off voltage should be 24V or somewhere near to it… how can it be 18V?

      for current limiting R1 will be = 0.6/12 = 0.05 ohms using two 0.1 ohm resistors in parallel,wattage will 0.6 x 12 = 7.2 watts or simply 10 watts for better protection.

      the relay will need to be replaced with a heavy duty automobile relay and T2 will need to be replaced with a 2N2222

      Reply

  29. Hello Swagatam,

    I want to use two 11.1 Volt, 12 Amp-hour LiPo batteries in series. I'm guessing the voltage cut-off will be at 18 volts. The load will be a 500 watt DC motor. What values in the circuit diagram should I change? What trip current would I use to calculate R1?

    Thank you

    Reply

  30. 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

  35. 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

    • 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

    • 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

    • 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

    • 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

    • 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

    • 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

    • 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

  37. 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

    http://youtu.be/AshLT5g010M

    Reply

    • 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

    • 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

    • 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

  43. 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

    • 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

    • 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

  46. 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

  47. 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

    • 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

  49. Hi!
    I found your circuit very interesting.
    Is there problem to make it work at 24V?To protect 2 SLA.
    Exept relays what else do i have to change?
    Thanks in advance.
    Joca.

    Reply

  50. 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

  51. 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

    • 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

  52. 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

  53. dear swagtam
    can i use bc 557 to get +ve signal at emmiter to give pin no 10 of sg 3525 to stop action when voltage goes below the cuttoff level.and what will be the changes in the circuit.
    thanks

    Reply

  54. swagatam,
    Im trying to implement your circuit into a battery powered project i have to use it only as a low voltage disconnect. I have a smart charger to control over voltage so Im searching for a small simple circuit i can build to put into my projects to stop the batteries discharging to far. Your circuit seems perfect for this.

    i have built the following circuit in circuit lab does it look ok for a simple LVD? this is my first circuit design so any help will be appreciated.

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

    Also i am wandering if i can implement a set reconnect voltage to stop the circuit reconnecting immediately when the load is removed hence increasing voltage which would continually occur i think this is called hysteresis but its a bit above my level?

    Also i believe the bottom outlet is the protected (switched) outlet is this correct if so where does the top outlet go to ground?

    Reply

  56. sir when i connect the 25watts load through this circuit to 12volt 7.5AH battery the load is running well. but when i replace battery with 60watts solar pannel the relay is vibrating why? what can i do. value of R1 i used is 0.1ohm 2watts.

    Reply

  57. sir i connected this circuit to 20ah 12volt battery. when i shorted the output
    the spark is appeared in circuit and then the circuit is not working why.(but the circuit connection is correct)

    Reply

    • What is the value of R1 that you have selected? It is the main component which decides the functioning of the circuit.

      Don't short the output to check….instead use a 2200uF/50V capacitor, add a 10K resistor across leads of this capacitor, then connect this assembly at the output for simulating a short for a second, use it as many times you like. This will not damage the battery.

      Reply

    • with 0.1 ohms, the short circuiting should pass about 7 amps of current at the output only then the relay would conduct.

      remove the base capacitor and then check again.

      also connect around a 33uF/25V capacitor across the relay coil.

      Reply

  58. sir when short circuit is made at output the relay is not holded in normaly closed position but it is vibrating like buzzer what can i do. even if R4 is used.

    Reply

  59. can i connect this circuit directly to 12volt 100AH battery,will there any damage to circuit in such high current.

    Reply

    • The self consumption is around 30-40 mA if a 400 ohm relay is used.

      If your batteries are connected in series, each might have a slightly different discharge rate…..I think that won't be advisable.

      Reply

  64. Res Sir,
    I've used for Low cut off circuit a preset 2nd leg connected to the base of BC547(No resistor in between), first and sec and third to +ve & -ve respectively, BC547 emmiter through a diode to the -ve, collector to Relay with Diode, now when I apply 11.5 and adjust preset to deactivate the relay, but later even If I give 14v also the relay does not activate in this scenario, how will the circuit work , please help sir, where am I gone wrong, should I add some components to make it work.

    Prem

    Reply

  65. dear sir , i am going to use 150 Ah battery and 150 watts D.C load the battery should cut off at 11.1 volts .please give me your opinion.

    Reply

  67. Hello,
    thanks for the circuit( low cut only ) now i want to know how to change the circuit for 6 volt battery.thanks in advance

    Reply

  68. Res Sir,
    Thank you for this circuit, Please explain to me how to set this circuit for low voltage protection when I am using a 12v battery 30 Ah battery and using a 30w 1 amp solar panel, what to modify and how to calculate R1 please reply to my Query

    Reply

    • Adjust the preset such that the relay just deactivates at 11.5V.

      The circuit will work for your application, please read the article I have explained everything there.

      Reply

  69. When an inverter is overload can this circuit above cut-off the inverter? Becuase you are said that it work with an inverter. Thank u sir.

    Reply

  70. Yes sir, i need feedback circuit diagram for an inverter to regulate output inverter. Pldase sir help me to send it to this E-mail (Olayemi.michael@yahoo.com)

    Reply

    • OK Michael, I'll try to design it, I'll post it in my blog after a few days because already there are quite a few circuits requested by other readers which are to be updated….

      Reply

  71. What u are saying is that it will not work with an inverter? Which circuit i can use for overload cut off of an inverter?

    Reply

  72. Please sir, i need feedback circuit diagram for an inverter and also sin wave inverter circuit diagram. Please sir, help me to send it to this E-mail (olayemi.michael@yahoo.com) . Thank sir.

    Reply

  73. This circuit, is it work with an inverter for overload cut-off? If i connecte two of 0.1 ohm's resistor in paralle for R1 in order to cut -off the load at exacly 160watt. Is it this two of 0.1 ohm's afect the seting of low battery cut-off. Thank u sir.

    Reply

    • According to me this circuit will work for all DC applications which requires overload sensing and tripping.

      R1 will not affect the low voltage cut off setting.

      Reply

  75. Hello sir, i build this circuit above, i used 10k variable resistor for R3 and also i connect two of 0.1 ohm's in paralle for R1, by the time am turning variable resistor to set low voltage cut-off that 10k was cash fire. What can i do to set low voltage cut-off at 10.5v?

    Reply

  76. Yes sir, want booster circuit diagram to boost low watt inverter to 1000watt. Please sir, help me to send it to this E-mail (Olayemi.michael@yahoo.com)

    Reply

    • I am afraid that would be only possible by upgrading the transformer of the existing inverter to 1000 watts and by using a battery capable of supplying 1000 watts of sustained power, no external circuit can work for implementing this.

      Reply

  77. Thank u sir, i need 1000watt booster circuit diagram for inverter at 220v ac- 240v ac(olayemi.michael@yahoo.com).

    Reply

  78. Thank u sir, for your response to me. Sir, i need 1ooowatt booster circuit daigram. My E-maill (Olayemi.michael@yahoo.com) and also i build battery charger for lead acid battery, the battery is 100AH,12V, I whant battery to cut-off at 13.6v what can i do ?

    Reply

  79. Thank u for your response to me. Please i need booster circuit. E.g 1000watt to 2000watt (olayemi.michael@yahoo.com)

    Reply

  81. Thank u for response to me. How many 0.1ohm's resistor, i will connect in paralle to get 0.007 ohm's. (2) i build 1kva inverter, i add 260 load, the output voltage inverter droup from 250v to 165v. I used SG3524 with 4N35 optocoupler, and also i used this mosft IRFP26N60K each of them is 22A,600V,300Watt, 4+4 in each saide 12v 100AH battery dry cell, what can i do to make output voltage to being permanet at 250v ? (2) i need booster circuit. E.g 1000watt to 2000watt. E-mail (olayemi.michael@yahoo.com) . Thank u.

    Reply

    • Connect 14 of them in parallel.

      I am afraid you cannot modify your inverter so simply.

      Without studying the fault correctl it won't be easy to rectify the problem.

      Reply

  82. Hi,

    It's not written in the text, but it seams that R4 and C1 are here to create a gap between the cutoff voltage and switch-on voltage (to avoid ON – OFF oscillations), now I want to use the circuit to automatically shut down the supply from a car accessories outlet when the car engine is off and automatically reconnect when the engine is started again. If I adjust R3 for a cutoff voltage at 13 volt, what would be the value of R4 for the circuit to switch on again at 13.8 volt when the alternator is charging? Thank you!

    Reply

    • Hi,

      Yes C1 is placed to create a gap before final switch on, however R4 is placed for latching the system.
      For your application you won't require this circuit. Simply derive a DC by rectifying the alternator voltage and apply it to a suitable relay coil directly, and wire up the contacts of the relay with the battery and the outlet…..

      Reply

  83. pls i wanna ask if it is possible to make timer that will on for 24hour and off for six hours, to control a charger, thanks so much

    Reply

  84. thanks so much for this circuit, i am a fan of ur circuit. please i have a question. for tripping overload with R1, should it be connected to ac load side or dc side that is battery, and what is the arrow of the overload and low battery output of R1 connected to. thanks

    Reply

  85. I =w/v =1000/12 =83.3A therefore R1 =0.6/83.3 = 0.007 ohm's how can i get this value of resistor. (2) i build 1kva with optocoupler 4N35, i used SG3524 as an oscillator, when i add the load of 220watt the output voltage droup from 240v(ac) to 199v(ac) what can i do to it.

    Reply

    • You may have to use many 0.1 Ohm in parallel for achieving that value, or simply a calculated iron wire piece will do the job.

      If the voltage is dropping means either your battery is not sufficiently rated or the the battery and the transformer together are not sufficiently rated.

      Reply

  87. Hi Swagat,
    My inverter's cut-off limit is 120w and my input voltage is 10. So the trip current is 12A. As per your formula,0.6/12 = 0.05 Ohm. Where to find such a low value resistor?

    Reply

  90. hey.. i am building a 0-15v 1 amp DC power supply and need to add an overload/short-circuit protection circuit. how can i modify your circuit to remove the low battery cutoff feature?? ps- suppose the trip current is 2 amp, R1=0.6/2=0.3ohms… 0.3ohm= how is that possible??

    Reply

  91. HI,
    great and simple circuit, thanks for shearing with us.
    can you explain how to choose R1. or more detail about how to make it.
    regards

    Reply

  93. Namaste Swagatam ji,
    I am using AC/DC autocharger 12V to charge Lead acid sealed battery 12V. Can I connect the autocharger parallel to the output ? Will it charge and cut off at high voltage said 14.0~15V and if the battery reach 10.6V will the low voltage cut off and prevent over discharge of the battery. What relay and resistor to use? Can I use variable resistor to adjust the high & low cut voltage ? What resistor to use? And how to adjust the desire voltage? Do I need to separate the high voltage cut off circuit portion (to between the auto charger and battery) and low voltage cut off circuit portion (to between the battery and 10W LED lighting output.
    Thank you for you help in advance. Dhanyavaad!

    Reply

  96. hi Swagatam, i found this so interesting almighty will keep strengthen you. pls i will like you to tell me what to do if what i need from the circuit is just low battery cut-off.

    thanks

    Reply

    • 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

    • 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

    • 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 http://imageshack.com/a/img443/5532/6s0f.png.

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

      Reply

    • 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

  97. hi,

    am very impress with the way you have been responding to my questions and i say very big thank you.

    one more thing sir, in your reply to my questions above you said i should upgrade relay fine but what will be the upgrade for R1 and R2. thank you

    Reply

  98. Thank you very much for your quick response. I have few questions to ask 1. can the over load section of this circuit handle high voltage for instance if the capacity of an inverter is 1kva can the overload protect section protect the inverter from been overloaded beyond 1kva (2) if I want to increase the threshold i.e. want to use it for inverter and want the inverter to shut-down when the battery is drain up to about 10-10.5v, if yes does it require modification if so what are the modification. Thanks a lot

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