Sunday, May 27, 2012

Low Battery Cut-off and Overload Protection Circuit.

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.

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 = 10 PRESET

C1 = 100uF/25V

T1, T2 = BC547,

Diodes = 1N4148

Relay = As per the specs of the requirement.

185 comments:

  1. 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

    ReplyDelete
    Replies
    1. Hi Peter,

      For handling higher loads, simply upgrade the relay contacts and dimension R1 accordingly.

      As explained in the parts list, R2 may be adjusted for making the circuit switch at any desired voltage levels.

      Regards.

      Delete
  2. 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

    ReplyDelete
    Replies
    1. R1 may be calculated with the help of the given formula, it may be made using a small length of iron wire, thickness should be selected by some experimentation such that the wire does not heat up at tripping currents.

      Sorry R2 does not have any role, it's R3 that needs to be replaced with a preset so that the low voltage tripping point may be set by adjusting it.

      Regards.

      Delete
  3. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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?

      Delete
    3. 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?

      Delete
    4. 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?

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

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

      Delete
    7. 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.

      Delete
  4. what will be the value of the preset

    ReplyDelete
  5. I think this what you said.[img src="http://i1243.photobucket.com/albums/gg546/vinodvinu2/th_lowbatteryandoverloadbatterycutoffcircuit.png"]

    ReplyDelete
    Replies
    1. sorry, I did not understand your point.

      Delete
  6. 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!

    ReplyDelete
    Replies
    1. Namaste Ji,

      The circuit shown in the above article is relatively crude, so I won't recommend the circuit, there are many simple highly reliable battery charger circuits which I have discussed in this blog.

      You may find them through the search box given above.

      Dhanyawaad!!

      Delete
  7. hi swagatham
    plz design a simple low voltage cutoff with buzzer..which can be used in between a charger and inverter

    ReplyDelete
    Replies
    1. Hi Sandy,

      I'll try to do it soon...will inform you if it's done.

      Delete
  8. 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

    ReplyDelete
    Replies
    1. Thankss!

      R1 = 0.6 divided by the max specified current limit to the output

      Delete
    2. Thanks for your quick respond,

      If current will 10A, R1= 0.6/10 = 0.12oms
      what about the wattage of R1 is it 100W(w=12/.12) ?

      Delete
    3. I think your calculations are absolutely correct...

      Delete
    4. Hi, I'm probably missing something, but I think the calculations are wrong. For maximum current 10A R1 should be 0.06 Ohm (not 0.12). And the maximum voltage on R1 is 0.6V when the current is 10A, so the resistor must be able to handle just 6W (not 100W).

      Delete
    5. That's correct, may be i didn't check the previous calculations properly.

      Watts = V x I = I.R x I = I^2 x R = 10x10x0.06 = 6 watts (correct)

      Delete
  9. 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??

    ReplyDelete
    Replies
    1. Please try the following circuit, it's much more effective and accurate:
      http://homemadecircuitsandschematics.blogspot.in/2012/04/how-to-make-automotive-electronic-fuse.html

      By the way 0.3 Ohm is correct, you have to employ a resistor of this value for controlling current above 2 amps.

      Delete
  10. overload short and low battery protected is it work for 1kva inverter

    ReplyDelete
    Replies
    1. Yes, just use RL1 with appropriate contact current rating.

      Delete
  11. overload short and low battery protected , if i want inverter to be cut off at exacly 1kva what can i do?

    ReplyDelete
    Replies
    1. divide 1kva with input voltage, you'll get the trip current, now solve the following to set the trip at 1000 watts.

      R1 = 0.6/Trip Current,

      Delete
  12. 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?

    ReplyDelete
    Replies
    1. Hi Vinod,

      If you are using it in the DC side then it's correct....

      Put two 0.1E in pararallel

      Delete
  13. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  14. 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

    ReplyDelete
    Replies
    1. You are welcome Toludan,

      The entire circuit should be connected at the DC side

      Delete
  15. 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

    ReplyDelete
    Replies
    1. You can try this circuit:

      http://homemadecircuitsandschematics.blogspot.in/2012/04/how-to-make-incubator-timer-egg.html

      Delete
  16. 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!

    ReplyDelete
    Replies
    1. 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.....

      Delete
  17. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  18. Hi . i need low battery procted output with current limet output. Circuit . anyone help me. Email : joynal_307@yahoo.com. send me this emil address.

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

    ReplyDelete
    Replies
    1. Please provide full specifications....

      Delete
  20. 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 ?

    ReplyDelete
    Replies
    1. Dear Michael,

      booster circuit for what application?

      What should be the voltage of it?

      You can try the last battery charger circuit given in this article:

      http://homemadecircuitsandschematics.blogspot.in/2012/07/making-simple-smart-automatic-battery.html

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

    ReplyDelete
    Replies
    1. Do You want to boost a low watt inverter power to 1000 watts??

      Delete
  22. 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)

    ReplyDelete
    Replies
    1. 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.

      Delete
  23. 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?

    ReplyDelete
    Replies
    1. Hello Olayemi,

      I have made some changes in the diagram please do it according to it....

      Delete
  24. Why are u afraid about booster circuit. It can not work with inverter or what?

    ReplyDelete
    Replies
    1. It's not possible to increase watts of an inverter by using external circuits.

      Delete
  25. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  26. 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.

    ReplyDelete
    Replies
    1. By Feedback circuit do you mean a regulated output inverter circuit??

      Delete
  27. 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?

    ReplyDelete
  28. 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)

    ReplyDelete
    Replies
    1. 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....

      Delete
  29. 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.

    ReplyDelete
    Replies
    1. Yes it will work if R1 is selected correctly.

      Delete
  30. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  31. 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

    ReplyDelete
    Replies
    1. Nothing needs to be changed, just adjust P1 for a 5.7V cut off

      Delete
    2. thanks a lot Mr. majumdar

      Delete
  32. Replies
    1. P1 is a variable resistor for adjusting the low voltage cut off limit.

      Delete
  33. 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.

    ReplyDelete
    Replies
    1. apply 11.1 volts to the circuit and adjust the preset such that the relay just deactivates.

      Delete
  34. 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

    ReplyDelete
    Replies
    1. Hi Prem,

      I couldn't understand the connections properly, anyway you can add one more diode in series with D1 and check.

      Delete
  35. sir what is the relay specification i need for 12 volt battery.

    ReplyDelete
  36. DEAR SIR
    WHAT IS THE SELF CONSUMPTION OF THIS CIRCUITE AND CAN I TAKE ONE BATTERY VOLTAGE FROM 40 SERIES BATTERY BANK FOR BATTERY SENSE.

    ReplyDelete
    Replies
    1. 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.

      Delete
  37. Dear Sir,
    I want a Only Battery Low voltage circuit,which is consume very low loss itself. Please hepl me.
    Thanks & Regards

    ReplyDelete
    Replies
    1. Dear Vijay,

      what is the battery voltage that you want to monitor??

      Delete
    2. Thank you very much for your valuable time.
      I want to monitor 12VDC battery
      And second is 40 battery bank voltege, it is 480 VDC & i want 435 VDC cutoff, Is this posible?

      Delete
    3. 1) 12 VDC
      2) 480 VDC (40 battery require cut off 435 vdc)

      Delete
    4. For 12V, you may try the second circuit explained in the following link:

      http://homemadecircuitsandschematics.blogspot.in/2011/12/how-to-make-simple-low-battery-voltage.html

      For 480V, I'll have to design a specific circuit, may take some time, I'll inform you when it's posted.

      Delete
    5. Dear Sir
      Any circuit available for 24
      VDC battery low?

      Delete
    6. Dear Vijay,
      you may try the following circuit, just replace 741opamp with IC324 opamp

      Delete
  38. Replies
    1. Sir I used 324 and circuit working good, but 1k resistor getting hot.what can I do?

      Delete
    2. yes sir,

      You are the best.
      Thanks a lot.

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

    ReplyDelete
    Replies
    1. yes you can connect it without problems.

      Delete
  40. 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.

    ReplyDelete
    Replies
    1. connect a 47uF/25V capacitor across the relay coil

      Delete
    2. when i connect 47uf capacitor the frequency vibration decreases and not completely eliminated(i also tried 470uf capacitor) plz tell solution.

      Delete
    3. this is the only correct solution, if it's not still not working then there might be some other abnormal issue with your circuit.

      Delete
  41. 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)

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. value of R1 i used is 0.1ohm .but i check the value of resistor in multimeter it shows as 0.3ohm.

      Delete
    3. 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.

      Delete
  42. 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.

    ReplyDelete
    Replies
    1. what is the load wattage at the output??

      Delete
    2. load is OK....voltage input to the circuit should not exceed above 14V, it should be near about the relay voltage, please check it.

      Delete
  43. hello sir, how many 0.1 ohms resistor i will connect in paralle to get 0.09 ohms? Thank u sir.

    ReplyDelete
    Replies
    1. there's hardly any difference, both are approximately the same.

      Delete
  44. 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/1024x768/

    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?

    ReplyDelete
    Replies
    1. Hi Chris,

      The diagram looks OK to me although the setting up procedure for such crude circuits is always time consuming.

      the circuit which you have shown already consists enough hysteresis so you won't have to add anything more, it would work just fine the way it's shown.

      Delete
    2. correction: the relay contact shown toward the arrow mark shouldn't be connected to ground, this will short circuit the battery and instantly damage everything.....you can leave it open.

      Delete
  45. 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

    ReplyDelete
  46. 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.

    ReplyDelete
  47. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
  48. 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.

    ReplyDelete
  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.

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

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

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

      Delete
  51. 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..

    ReplyDelete
    Replies
    1. 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

      Delete
  52. 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.

    ReplyDelete
    Replies
    1. 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

      http://homemadecircuitsandschematics.blogspot.in/2012/07/making-simple-smart-automatic-battery.html

      Delete
  53. Thanks for this circuit.

    How can I modify this to use solid state relay?

    ReplyDelete
    Replies
    1. 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.

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

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

      Delete
  54. sir i need circuit for cutoff voltage with 2millivolt.

    ReplyDelete
    Replies
    1. use any of the following circuits, replace 741 with LM311

      http://homemadecircuitsandschematics.blogspot.in/2011/12/how-to-make-simple-low-battery-voltage.html

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

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

      Delete
  55. 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 .........

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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.....

      Delete
  56. hello sir, can this kind of cct can be modified to trip low standby current? thank you :)

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

      Delete
  57. 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.

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

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

    ReplyDelete
    Replies
    1. 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

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

    ReplyDelete
    Replies
    1. 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.

      Delete
  60. 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?

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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

      Delete
    3. No that won't happen because of the transistors hysteresis properties...

      Delete
    4. ok that's good

      but 'm facing a problem with relay connections

      this is the diagram :
      http://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 ?

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

      Delete
    6. this is a video I've recorded :
      http://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)

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

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

      Delete
  61. 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

    ReplyDelete
    Replies
    1. 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.

      Delete
  62. Hello Sir,

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

    ReplyDelete
  63. 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.

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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).

      Delete
    3. 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

      Delete
    4. 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)

      Delete
    5. 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.

      Delete
    6. 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?

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

      Delete
    8. 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?

      Delete
    9. 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..

      Delete
    10. 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.

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

    ReplyDelete
    Replies
    1. 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.

      Delete
    2. 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!

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

      http://homemadecircuitsandschematics.blogspot.in/2012/01/how-to-make-relay-driver-stage-in.html

      Delete
    4. Hello Sir,
      Does R1 relate this problem?

      Delete
    5. No, R1 has no connection with this problem.

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

      Thank you for your support!

      Delete
    7. Hello Sir,

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

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

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

      http://homemadecircuitsandschematics.blogspot.in/2011/12/simplest-smf-automotive-battery-charger.html

      Delete
    10. 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.

      Delete
    11. 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

      Delete

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