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

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Sunday, January 1, 2012

How to Make Simple Low Battery Indicator Circuit Using IC 741

The proposed circuit was requested by one of the avid readers of my blog. It is a low battery warning indicator circuit and can be used for monitoring a particular low battery voltage threshold.
The circuit may be understood with the following points:
The entire configuration is wired around the IC 741 and it becomes the heart of the circuit.
Basically it is configured as a comparator with one of its inputs clamped to a fixed reference level while the other input used as the sensing terminal.
Here as can be seen in the diagram, the non inverting input is provided with a fixed reference voltage through a resistor zener network.
This input is fixed to about 5 volts.
The other inverting input pin #2 is wired via a preset to sense the input supply voltage from the source.
The preset is adjusted such that the voltage level at this input becomes lower than the fixed reference voltage at the other pin of the IC as soon as the source voltage becomes lower than the desired threshold level.
When this happens the output of the IC immediately becomes high, illuminating the connected LED.
The illuminated LED immediately provides the indication of a low voltage situation so that the required actions may be initiated.
Optionally, the output may be replaced by a piezo buzzer instead of the LED for getting an audible response of the above situation, eliminating the headache of monitoring the LED condition every now and then.


The above circuit can be modified by adding a relay stage for controlling a articular stage which may be relevant to the low battery cut of actions.



The above low battery indicator circuit can be even further improved in the following manner for controling both lower and the upper charging thresholds:

Initially keep the 100K preset link disconnected.
Apply a 14.4V source across the shown "12V input" and adjust the 10K preset such that the upper relay just activates, confirm the triggering by subsequently moving the preset to-and-fro. Glue it once fixed. The LED will respond by switching ON to the fixing of this preset.
Now reconnect the 100K preset feedback link, and reduce the input supply to about 11.2V.
Next, adjust the 100K preset such that the relay just deactivates. Confirm by flipping the preset as above.
Ignore the lower relay as it will switch ON as soon as the input supply is switched ON, so its operation is obvious.

That's it, the low battery warning circuit is all set now and will accurately respond to the above settings or any different setting that may be preferred and implemented by the particular user.




147 comments:

  1. hai.. can u help me.. i have created a blog comutronics.blogspot.com/ for Ece students.. how can i get it by google search

    ReplyDelete
  2. Make it big....Google won't notice your blog unless you have put 50 to 70 thousand words into it and also you need to keep it updated from time to time.

    ReplyDelete
  3. this circuit is wrong,since the pin 6 is positive,then ur resistor(2.2k) should be there before your led i.e the positive led.Can buzzer be use in place of led?

    ReplyDelete
    Replies
    1. A resistor is used only for limiting the current through the LED, it does not matter whether it's at the anode or the cathode end of the LED, it works OK as long as it's in series with the LED.
      A buzzer can be used in place of LED, the resistor will not be required if a buzzer is used.
      Regards.

      Delete
  4. Hi,
    i am wanting to make a low battery indicator, when the battery voltage comes down to 2 volts
    how should i go about it?

    ReplyDelete
    Replies
    1. Hi,
      Please search this bog, I think I have already published one article with this topic using the IC 741

      Delete
  5. Hey there,

    What part of the circuit is it that actually sets what voltage the LED should be lit at? I assume its just whatever the ZENER diode is rated at?

    Thanks

    ReplyDelete
    Replies
    1. Hi,
      The output voltage to the LED will be equal to the supply voltage....the zener voltage is there to provide a reference voltage level to the IC

      Delete
  6. Hi,

    How can the above circuit be modified to work with 12v 7ah Sealed Lead Acid battery. The LED should light up when the battery voltage reaches 12.5v

    ReplyDelete
    Replies
    1. Hi,
      Just interchange the pin connections in the diagram, keeping rest of the things as it is.
      Now apply a measured 12.5v at the +/- terminals, adjust the preset such that the LED just lights up.
      Next remove the above source and connect the battery to the shown position, when its voltage reaches 12.5v, the led will instantly light up, giving the required indications.

      Delete
    2. Hi Swagatam,
      Thanks for your reply. I didn't understand which pin connections your asking me to interchange. And moreover i don't have a variable voltage source to calibrate the preset. Is there any other way to do it. Please help me.

      Delete
    3. Hi Swagatam,
      I thank you for posting this circuit. I got it working finally by allowing the battery voltage to fall till the required voltage(12.5v) and then calibrated the preset(10k). Though there exists a small issue. The LED glows faintly when the above circuit is connected to battery which is fully charged . I tried to increase the resistor value to 4k but still there is a dim light glowing. Other than that the circuit works absolutely fine and the LED glows brightly when it is 12.5v.

      I also wanted to know if the above circuit can be modified to cut-off the load when the battery is low along with the LED glowing.

      Delete
    4. Hi Ajay,

      You are welcome,

      You can add a 3v zener diode in series with the LED, this will stop the parasitic glow of the LED.

      The circuit can be modified by adding a relay driver stage in place of the LED, and the relay contacts may be wired with the battery terminal for getting the required cut-off feature.

      Delete
  7. Dear Swagatam,

    Can the above circuit be modified to place a piezoelectric buzzer after the LED?

    ReplyDelete
    Replies
    1. Dear Sandip,

      Yes, just replace the led resistor with the buzzer....

      be sure to connect a 3 v zener in series with them.

      Delete
    2. Thank you for your reply. I tried it works great.

      Delete
  8. HI Swagatam,

    I wanted to your help in building an automatic charger for my 12v battery. So i have a rough idea of building it like this:

    The output of op-amp in the above circuit is connected to the base of npn-transistor, where the collector and emitter of the transistor are connected to Output & Adjust pin of LM317 voltage regulator.

    I adjust the preset in the above circuit such that the output of op-amp is high when the voltage of battery rises to 12.5v.

    So when the battery voltage rises to 12.5v, the op-amp output is high and it makes the npn transistor to conduct. Thus LM317's output and adjust pin are grounded and the charging process is cut-off.

    As soon as the voltage of battery falls below 12.5v the op-amp output is low and the transistor is off. This will result in switching on the charging process.

    Will this work? Please help me out.

    ReplyDelete
    Replies
    1. Hi Ajay,
      This concept has been covered in the following article:

      http://homemadecircuitsandschematics.blogspot.in/2012/02/how-to-build-automatic-6-volt-12-volt.html

      However this concept cannot be very effective because the voltage tends to restore within a fraction of second after switching off the output, due to the absence if the required instantaneous triggering, and the voltage instead of cutting of completely, switches rapidly regulating the voltage upto a certain point but never cutting the voltage completely.

      Delete
    2. Thanks for the info and the link Swagatam.

      Delete
    3. Hi Swagatam,

      Can you please suggest a better circuit as you are saying that the above concept can't be very effective.

      Delete
    4. Hi Ajay,

      May be this circuit can be used in conjunction with a LM338 IC for complete cut off:

      http://homemadecircuitsandschematics.blogspot.in/2011/12/high-current-10-to-20-amp-automatic.html

      Delete
  9. what is a 10k preset?please post a list of equipment and what battery to use?(battery spec.)

    ReplyDelete
    Replies
    1. 10k preset is a variable resistor, the circuit will monitor all types of battery voltages....

      Delete
    2. Can you post the link of your own version of low battery indicator using and ic741,I've tried finding but it did not show up,Thanks in advance!

      Delete
    3. The above circuit is the only way of using a 741 for low battery indications...

      Delete
    4. Thanks for the reply Swagatam, After fixing all the connections what to do...I'm a novice,I would like you to explain about its working in basic language Thanks a lot.

      Delete
    5. I have explained everything in the article in simple words, pin#2 is the inverting input and pin#3 is the non inverting input.....
      Please read the article once again.

      Delete
  10. Hey swagatam----1kohm,2kohm resistors,a 10k potentiometer,an led,a 741 ic,a 4volt zener diode,a 9 or 12 volt battery...are these all we need or is there anything else? Thanks.

    ReplyDelete
    Replies
    1. That's all you would need, the battery to be monitored is connected at the +/- points.

      Delete
    2. hi swagatam,I bought all the above instruments and tried connecting the circuit but i feel its too complicated,....the battery's getting heated but the LED ain't glowing...:(((

      Delete
    3. The shown design is a standard configuration, and will surely work if everything is done correctly as per the diagram.

      The battery will become hot if there's a short circuit, meaning something is wrong with the wiring.

      LED will light up as per the setting of the preset at the threshold levels only.

      Delete
  11. Thanks man,Works like a charm,so the basic principle is that
    The lighting of the LED depends on the battery voltage which is adjusted by a variable resistor eh?

    ReplyDelete
  12. Hi there,
    Will this circuit work in a 24V setup (2 X 12V batteries in series) if i apply the following changes:
    1. Use a 12 V Zener Diode as a reference at the positive input.
    2. Use a 1k pot at around the 50% level to the negative of the OP AMP.

    I would require it to be a LV indicator for a 24V system setup.

    ReplyDelete
    Replies
    1. Hi,

      I am afraid the shown design cannot support 24V, because the IC has a tolerance of a max 16V across its pins 4 and 7....the circuit will require major modifications for working with 24v supply.

      Delete
    2. I was looking into placing a voltage regulator at the IC's input (pin 7)...could this be used to reduce the input Vcc

      Delete
    3. Yes that will do, also make the zener resistor to 10K and add a 1K resistor across pin#2 and ground.

      The zener value can be kept as it is in the diagram.

      Delete
  13. The LED at the output was rated at 30mA (2V). I used a 1k pot at the input to pin 2, Zener resistor at 10k at first but the simulation did not light the LED. I changed it to a 1k( and 500ohms) and it worked like a charm. The voltage regulator indeed was the key to the use of the OP AMP at this voltage!

    ReplyDelete
  14. Hi Swagatam,

    for the relay version? what is the value of resistor connected to LED and what type of transistor do i use?

    Regards,

    ReplyDelete
    Replies
    1. resistor can be a 10K 1/4 watt, and the transistor BC547...

      Delete
    2. Hi,

      can i integrate this circuit to this blog? http://homemadecircuitsandschematics.blogspot.com/2012/11/automatic-dual-battery-changeover-relay.html

      Regards,

      Delete
  15. how is indicator different from voltage regulator???

    ReplyDelete
    Replies
    1. An indicator will only indicate while a regulator will control the voltage as well.....

      Delete
  16. Hi.
    Can you please tell me what do I have to do , to make this circuit works with 1 lithium shell battery (3.7v)
    Thanks

    ReplyDelete
    Replies
    1. Hi,

      Just replace the zener with two 1n4148 diodes in series, cathode toward ground. No further modifications would be required.

      Delete
  17. dear sir , how can i modify as over voltage protection for battery

    ReplyDelete
    Replies
    1. just interchange the input pin numbers of the IC.

      Delete
  18. please explain sir,

    ReplyDelete
    Replies
    1. put pin#2 at zener, and pin#3 at 10K preset.

      Delete
  19. ok sir i will try and tell u later . thanks

    ReplyDelete
    Replies
    1. dear sir , i am bala, as per your suggestion, i did the circuit .it is working fine ,please tell me how can i use for 21 volts ?. bcos iam planing to use for solar panel change over. panel is giving 21 volts in the peek time.

      Delete
    2. for over 20 volts, change the 741 opamp, replace with one of the LM324 opamps.

      LM324 has 4 opamps, use any one of them.

      Delete
    3. dear sir ,
      i did the circuit for automatic cutoff for solar panel charging and A.C mains charging to a battery connected inverter.my idea is if 15 volts comes to this ciruit from the solar panel the relay will on and cut the A.c mains charging to the battery in the day time . so we can avoide parellel charging . but the problem is the realy is not on .i used 200 ohms relay to cut the A.C mains. but lm 3914 I.C 10 th pin led is glowing when i gave 15 volts dc to circuit .the relay is not operated. please help me to solve the problem

      Delete
    4. It will be difficult for me to check the fault without actually seeing the circuit board or the diagram.

      Delete
    5. it should work actually, not sure what's wrong with your circuit.

      Delete
  20. sir, how can i construct 12v inverter with at least 500watt without transformer.

    ReplyDelete
  21. sir, help me i want construct 500watt inverter without transformer, what do i do?

    ReplyDelete
    Replies
    1. you can try the following circuit:

      http://homemadecircuitsandschematics.blogspot.in/2012/05/how-to-make-transformerless-inverter.html

      Delete
  22. dear sir ,
    iam bala. i did the above relay circuit without any changes .working well. but little "dick" "dick" sound is come from the relay when it is going to disconnect at 11.1 volt.i am using 12 volts d.c 250 ohms relay. please help me to solve this problem.

    ReplyDelete
    Replies
    1. Connect a 1N4148 diode across pin3 and pin6 of the IC.

      anode to pin3....cathode to pin6

      Delete
    2. ok sir,i will do it and give you feedback. thanks.

      Delete
    3. sorry sir. After connected the 1N4148 the cut off voltage comes to 9 volts( the relay on at 9 volts). not able to set 11.1 in my preset (10 k). also doubted with my relay and changed . not able to set at 11.1 volts. please help me.

      Delete
    4. Replace the diode with a 10K resistor and check...

      Delete
  23. u mean the 4v7 zenar diode ?

    ReplyDelete
  24. dear sir , connected 10 resistor to the 1N4148 diode. the circuit is working perfectly.now preset set to the relay operation at 11.1 volts. thanks for your valuable co-operation.

    ReplyDelete
  25. Hi,
    i want to make 54V low battery indicator, when the battery voltage comes down to 46 volts
    how should i go about it?

    ReplyDelete
    Replies
    1. Hi, this circuit can be used for upto 24V application only, not above that...

      Delete
  26. SwagatamAugust 12, 2012 at 5:30 AM

    Hi,
    Just interchange the pin connections in the diagram, keeping rest of the things as it is.
    Now apply a measured 12.5v at the +/- terminals, adjust the preset such that the LED just lights up.
    Next remove the above source and connect the battery to the shown position, when its voltage reaches 12.5v, the led will instantly light up, giving the required indications.

    Sir, in respect to this explanation Can you tell me which pin have to be interchanged (is it that the preset goes to 3 pin and the zener goes to pin 2) and should the zener also be changed for 12v battery applications

    ReplyDelete
    Replies
    1. Yes, pin#3 will go to the preset and pin#2 to zener for making it a high voltage cut off circuit. zener value will not change.

      Delete
  27. can you tell me the name of the box after the pin#6

    ReplyDelete
  28. Instead of the LED or piezo can this be used to switch a 12V relay off so the battery wont fully discharge?

    ReplyDelete
    Replies
    1. Hi Mike, see the second circuit, it has a relay.

      Delete
    2. What is that red thing on pin 6 of the 741? Is that a resistor and LED?

      Delete
    3. yes it's a resistor and a LED, 10K resistor and a red 5mm led.

      Delete
    4. Another question is say I want the load to be switch off the battery when the battery level reaches 11.9V would I be changing the zener or the 10K preset?

      Delete
    5. everything is very flexible with such comparators, you may adjust any of the two.

      Delete
  29. Iam a little confused please, help, I have constructed the circuit (1) without the relay using all the parts according to your specifications, and I have two batteries 10.9v and 11.7v, I used the 11.7v to adjust the preset to set the red LED glowing, now when I connect to a battery of 12.6v it still glows and when I connect to the battery rated at 10.9v it does not glow, is it correct what i've done, or am I wrong some where, should the circuit lamp glow only at 11.7v and low, or does it go off after 11.7v please explain as I am trying to build a Low battery indicator.
    Thanks in advance

    Prem Antony

    ReplyDelete
    Replies
    1. the preset should be set such that the voltage at pin#2 gets just a little lower than the reference voltage at pin#3. This will make the output of the IC high, illuminating the LED.

      Now if the battery voltage increases, the set pin#2 voltage will also increase, and will tend to get higher than pin#3 voltage, this will instantly make the output of the IC to become low, switching OFF the LED.

      So I think you might have wrongly adjusted the preset, adjust it so that the LED just lights up at 11.7V, and stops glowing even if the voltage increases marginally to 11.8 or 11.9 V and above.

      the LED should stay illuminated at voltages which are lower than 11.7V in your case, and vice versa.

      Delete
  30. Thank you for your reply sir, I used 741 first and it did not work, may be there was some problem with the IC so I tried with LM324, I have connected Pin 4 to +, pin 11 to Negative, Pin 1 output to RED lED through a 2K2 resistor 5v zener and 10K resistor from ground and + respectively to PIN 2 and PReset to PIN 3, Please help me to adjust it to work for low voltage cut off.

    ReplyDelete
    Replies
    1. apply 11.7V and adjust the preset so that the LED just lights up at 11.7V, and stops glowing even if the supply voltage is increased marginally to 11.8 or 11.9 V and above.

      below 11.7, the led should continue to remain illuminated.

      Delete
  31. Thank you sir, one more doubt please tell me the output of the IC through a resistor goes to the cathode or anode of the LED, I've connected the anode to the output and cathode to ground, am I correct or wrong sir.

    ReplyDelete
  32. Hi Sir, I want to use this cutoff circuit for 6v 4.5ah SLA Battery and the cutoff voltage will be 5.7v so should i use:
    1) 5v Relay or 6v Relay
    2) secondly can 5.1v .5w zener can be used instead of 4v7 zener

    ReplyDelete
    Replies
    1. Hi Bashir,

      1) use a 5v relay.
      2) zener should be much lower than the supply voltage, use a 3 or 4.7 volts zener

      Delete
  33. I need to measure 12v battery voltage using pic16f877 and display in lcd...if the charge in battery decreases it should get automatically charged from external battery ..while charging i should measure power in it and display..for that i need current(I) of the battery to measure power..can you provide the circuit for measuring current...

    ReplyDelete
    Replies
    1. current can be measured by an ammeter simply, but there should be some load connected to battery....or may be i did not understand your question properly.

      Delete
  34. plz tell how to modify this circuit and turn it into a over voltage cutoff ,
    basically the opposite of low voltage cutoff

    ReplyDelete
    Replies
    1. in the last circuit, adjust the preset so that the relay just activates at around the high cut off threshold.....adjust the 47K resistor value so that the relay just deactivates at the lower cut off threshold.

      Delete
  35. Hi, i had completed the 3rd design and fixed it for the invertor for battery low voltage cut-off. Initally the power supply for this circuit was taken frm the inverter board, at times this circuit when on inverter mode it terminates the power even when the battery is at full charge or when switching off the inverter o/p load. After then powersupply to this cut-off circuit was taken directly frm the battery terminals now the issue is reduced but rarely happens. kindly advice

    ReplyDelete
    Replies
    1. It might be happening due to instantaneous transients in the power line, try adding a capacitor (10uF/25V) across pin#2 and negative, and across pin#6 and positive, this might help to control the stray disturbances in the power line.

      Delete
    2. Thank u......i'll try this and give a feedback

      Delete
  36. this worked out...........but a drawback is there like, if we disconnect and reconnect the supply to this circuit the status of the o/p is changed it does not remain the same as earlier

    ReplyDelete
  37. Hi..I want to make low battery voltage indicator which can show the voltage level from 3v to 24v. Can it be possible to make the same with this ckt??? what changes would be required to make this ckt?

    ReplyDelete
    Replies
    1. yes the above circuit can be used, but the IC741 opamp will need to be replaced with opamp LM324 (use any one out of the four available in the IC)

      Delete
    2. Hi Sir..I have few doubt
      LM324 is 14 pin IC having 4 inputs .
      Can these i/p be given to different voltage levels like 3V,9V,12V,24V?

      I want to select one i/p at a time what should be done for that?

      Also in low battery indicator ckt if IC replaces with LM324 then what about value of other components like resistor,zener diode etc do we need to change their value if yes then what will be value of components that to be sustain upto 24V level?

      i want to show the indication in different leds like green for live battery,yellow for medium and red for dead.for this is their need to change the circuit ?If yes then what??

      please do reply...

      Delete
  38. Hi. Do you know how costly in terms of power consumption this probe is? I would like to spend at most 10 mA for such functionality.

    I will use it in a solar energy system. I do not need a LED, but will use a relay to inform the controller about low voltage condition.

    ReplyDelete
    Replies
    1. With the relay switched ON, the power can be no less than 40mA (maximum)

      Delete
  39. Ecellent!!!
    So you are connecting the relays in series and both have to be charged in order for the load to function. Am I right?

    Does this also mean that in case of low battery level, both relay coils are not energized (i.e for minimum consumption in case the battery is very discharged)?

    ReplyDelete
    Replies
    1. Thanks!
      The relay contacts are in series, and the lower relay is wired in such a way that the load gets power from the charger DC source as long as it's present, and reverts to battery power during the absence of the source voltage.
      However the battery power is available to the lower relay only if its fully charged otherwise the load gets no power.

      The charging source can be a solar panel voltage or some other renewable option.

      Delete
  40. I see...

    I don't really mind the upper threshold thing but the rest of the circuit is perfect! So would you be kind enough to upload the previous version of the third circuit again please?

    I also wonder if there can be a hysteresis implemented in this circuit, preferably adjustable.

    Lastly, is there a particular reason you are using an opamp instead of a voltage comparator (i.e why not use LM393 instead of LM324)? I know it is not wrong, but I trust you might have a reason for your selection.

    Thanks again,
    Pete

    ReplyDelete
    Replies
    1. The present circuit can be also used for just monitoring the lower threshold, simply eliminate the second relay and the associated connections and wire the N/O contact of the upper relay with the load, I am assuming that you want to disconnect the load from the battery when the voltage reaches the lower threshold.

      Apply a sample lower threshold voltage to the circuit and adjust the 10k preset such that the relay just deactivates at this voltage, while doing this keep the 100k preset link disconnected, once the setting is done you may connect it back.

      The 100k preset link will provide the required hysteresis which can be adjusted by adjusting its value.

      LM324 is rated at 30V which allows it to be used with 24V supplies, that's why I preferred this IC.

      Delete
  41. So, if I need only the lower threshold version, I can use the second diagram and just a 10K fixed and 100K variable between pins 3 and 6 for the hysteresis. Right?

    Thanks again,
    Pete

    ReplyDelete
  42. Also, assume that pins 2 and 3 are interchanged for use as a high voltage indicator. Can u still add hysteresis in this case which will be a little lower than the high threshold voltage?

    This blog does amazing work mate, congratulations for running it!

    Pete.

    ReplyDelete
    Replies
    1. Interchanging the pins would make things complicated because we have the hysteresis loop involved....instead of thinking much, very simply the same circuit which is explained above can be used for the high voltage cut off also, just by pushing the preset setting at the required higher threshold, so when the battery voltage reaches say a set 14.3V mark, the relay actuates and cuts off the external power to the circuit and the battery...after this the circuit monitors the battery voltage and restores the relay back by switching it off once the battery voltage falls below the set lower threshold, determined by the hysteresis loop preset.

      You are most welcome....we all are enjoying your interesting feedbacks.

      Delete
  43. Hi I need to build an LED low battery indicator for a 7.2v battery. The low battery indication would be around 5 volts.
    How would I modify the circuit? would it work for my application.

    ReplyDelete
    Replies
    1. make the first circuit, don't modify anything, use it as is.

      Delete
  44. Just when I thought I had it all figured out, I am confused again!

    Let me put down all I need to do and see if it can be done:
    I need the circuit to keep the relay off (deenergized) for voltages bellow 22V and turn it back on at 24V.

    Do you think it can be done with this circuit (a little modified maybe)?

    ReplyDelete
    Replies
    1. yes you can achieve the results using the discussed circuit, 10k preset is for the high level (24V) and the 100k preset is for the lower level 22v.

      Delete
  45. Dear Swagatam,

    Your assistance during the past few days has been incredible, thanks to which I think I have managed to get close to the circuit I need.

    Having combined and the above info, a selector circuit I found in a different source and by adding my own ideas I came up with the following circuit:

    http://domoarchitecture.eu/scan.jpg

    I know there are issues which with my knowledge I cannot solve, but I think I am close (I will list these issues at the end of this post).

    My knowledge is limited (I am neither an expert nor a novice in electronics), so I would like to hear your opinion and recommendations.

    To my understanding, what I expect it to do is as follows:

    LM324 will monitor the voltage and turn off the relay at the bottom of four scales: 18.5-20V, 20-22V, 22-24V, 24-28V adjusted by the various 10K presets.

    The circuit starts (battery full) with the relay armed (load connected) and the 24-28V LED on (left hand side set of LEDs). The selector circuit (top part) assures that only the selected OPAMP arms the relay although all of the OPAMP outputs will be high at the same time (this I need for a later function, see end of next paragraph).

    During discharge, when 24V are reached the equivalent output (1) will go low and the relay will disarm, disconnecting the load. At the same time BC337 (NPN) will conduct and light up the three right hand side LEDs, to indicate scale options that are available.

    Once a button on the selector is pressed for a lower scale (i.e 22-24V), the relay at the lower circuit will arm by the second OPAMP (output 7) and the load will be connected. Same goes for the equivalent left hand side LED. The same applies for the 20-22V scale, but only 2 LEDs will light from the right hand side set).

    Now, if the last scale (18-20V) is selected, when 18V are reached the 4th OPAMP of the LM324 will disconnect the load once again but at the same time the SPDT relay at the top left of my picture which is connected as a latching relay will disarm and take power completely off of the circuit to minimize consumption as the battery will be very deeply discharged now. In order to start the system again, manual reset will be required via the push button near the latching relay, once the battery is charged again.

    The purpose of this circuit is for a solar system used to power an emergency communications station. This is the reason it cannot be set in a completely automatic way. Although deep discharging of the batteries is not recommended, in such stations the operator will have to judge whether he must deep discharge the battery or not, in case the type of emergency makes it inevitable. Under normal situations, the power will cut off at 24V automatically protecting the batteries, but it is important that the operator will have the option to continue drowning the batteries if necessary.

    So here are my issues:
    1. I have a fear that in the way the circuit works, the outputs go high, not low when thresholds are met. If it is so the circuit will not operate in the intended way, especially the latch relay function.

    2. I have a feeling that there will be a lot of small and big flaws in my additions to the design. I am confident in building it when it is all designed, but designing of circuits is a skill I only have to certain extend.

    3. I wonder if I can substitute the relays of the top part of the circuit with some kind of low power MOSFET. If so, please give me a clue as to how to connect them.

    4. I am also thinking that the disconnection of the load will cause the system to do some chasing effect, so I wonder if some small hysteresis can be added.

    Please get back to me with you thoughts; I believe this can be a useful circuit if ever completed for many people (maybe by adjusting the thresholds to higher values, not everybody needs to drain their batteries so much).

    My gratitude once again for the tones of information you have helped me to collect during the past week.

    Regards,
    Pete

    ReplyDelete
    Replies
    1. Hi Pete,

      Please give me some time to think, because the circuit is quite elaborate and a bit puzzling too, I'll revert once I grasp all the details about its functioning.

      Regards.

      Delete
  46. Sure my friend, I hope my long description gives you a good clue. Please get back to me if you need more info...

    Good day,
    Pete

    ReplyDelete
    Replies
    1. Hi Pete,

      Please refer to the following article, I have presented a simplified version of your circuit at the end of the page, I think it will do the job, but not entirely sure:

      http://homemadecircuitsandschematics.blogspot.in/2013/07/selectable-4-step-low-voltage-battery.html

      Delete
    2. I tried the circuit by using LM741. It is working Perfectly thanks Swagatham

      Delete
  47. hello i had constructed your second circuit http://homemadecircuitsandschematics.blogspot.in/2011/12/highly-accurate-mains-high-and-low.htm
    there doesnot occur any cutoff either in higher range or in lower range ...what problem might had occur......would you plz solve my problem
    mine lower voltage level is 170 and higher is 245

    ReplyDelete
    Replies
    1. connect LEDs at the outputs of A1 and A2 and ground with 1k separate resistors.

      Now apply 170V and adjust P2 such that A2 LED just lights up.
      Next apply 245V and adjust P1 such tha A1 LED just lights up.

      Your circuit is set now, the relay connected at the collector of the transistor will also respond correspondingly.

      Delete
  48. Swagatam, sir,

    I think I've cracked it, Instead of the 10k preset, I used a voltage divider: a 12k in series with a 27K. From between these two, I was able to obtain a voltage of 3.86V at Pin 2 - Pin 3 had 4.49 VDC - and, which triggered the transistor via Pin 6 and tripped the relay, and the lights, connected to the normally open contacts, cut out.

    Now to assemble all three circuits.

    Stephen Gard

    ReplyDelete
    Replies
    1. OK Stephan, that's great, no issues, if you have solved it you can go ahead with it.....

      Delete
  49. (Reposted without links)

    Swagatam,

    Thank your for your well-conducted and informative site. I have learned much from it. I am trying to adapt your 741 Low Battery Cut-Off circuit for use with a 6VDC lighting set-up, and I am having some difficulty. I hope you can advise me. Here is what I wish to do.

    1. I have four high-power white LED spotlights. They each require 3.2VDC @ 34mA.The four lights, wired in parallel, draw 82mA at 3.2 VDC. So that is the load.

    2. The supply is a single 6.5VDC SLA battery. I attach a 33 Ohm 1 Watt resistor in series with the negative supply to the 4 x LEDS, as a current-limiter. I wish to use ONE battery only for the entire circuit, no separate supplies. It will be installed in a friend's garden and not get much maintanence! So it must be fully 'automatic'.

    3. I have a solar-panel (5W 17 volt) that charges the battery and a darkness detector. Now I wish place your 741 Low Charge Cut-Off circuit (with relay) in between these two circuits, to prevent the 6.VDC SLA from discharging too far. The solar charger supplies only a small amount of current, and there will be no-sun days, of course.

    If I understand the operation of your circuit, when the battery voltage drops, the 741, via Pin 6, allows current to flow through Q1 (the BC547), tripping the relay coil. My 4 x LED load is wired to the NORMALLY-CLOSED contacts of the relay (NEC EA2-4.5VDC), and when the coil is energised via Q2, they disengage the load by switching over to the NORMALLY-OPEN contact. So no current flows, and the 4 x LEDS go off.

    That is what I wish to do, but I cannot make your circuit do this. The only change I made to your circuit was to the 10K pre-set pot. My pre-set over-heated. I measured the R, and replaced it with a 12k ohm 1 watt fixed resistor


    b. This arrangement delivers 5.65V to pin 2 of the 741. Pin 3 is set at 4.52 V, from the junction between the 4.7V Zener and the 1K resistor connected to the +ve terminal of the battery.

    c. To test, I connected one side of a diferent white LED load to the NC pins of the relay, and the other side to the battery ground. This load draws 2.8VDC, at 22mA.

    d. I connect a tired old dry-cell battery to this circuit, and on my DMM, I watch the supply voltage drop.

    4V99. 4V96. If I insert the DMM at Pin 2 and try to take a reading, the voltage goes down slowly.

    e. So I connect a 2MegOhm pot in between the 12K and Pin 2, and bring the voltage down.

    But now I notice that the voltage on Pin 3 is droppng as well. For example, from 4.351 to 4.348V.

    It seems that the two pins will never achieve the diffential that will trip the relay.

    When I connected the fixed resistor, I did not take it to ground, as the original circuit did, after the 10K preset. If I connect this fixed resistor to ground, Pin 2 goes low and the relay trips, but I don't think that is correct.

    Can you please advise what I am doing wrong here?

    Sorry for such a long letter!

    Stephen Gard






    ReplyDelete
  50. Swagatam,

    I am still testing to get exact values for my requirements. I am also drawing up the whole device in TinyCAD, and can send you a png of the completed three-part circuit, with test-points and voltages marked, if you wish.

    Stephen Gard

    ReplyDelete
    Replies
    1. Yes, why not Stephan, I'll be extremely happy to publish your efforts in my blog so that other can also benefit from it.

      I appreciate your positive thoughts...kindly do send them.

      And whenever you do something new in electronics, please make sure you share them here for us.

      We all will love to see them.

      Let me know if you need any further assistance?

      Thanks very much.

      Delete
  51. Swagatam,

    I added this circuit to my solar lighting set-up, but there is a problem. The relay must be activated to cut off the current to the LEDs. But this drains the battery even further. Finally, the battery is drained so much that it cannot activate the relay, which then returns to the NC condition, powering the LEDs, and draining the battery even further. After one dull day, when the solar panel could not recharge the battery, the lights do not work at all, because the battery was drained so low. Perhaps a timer would be better? I am running the whole set-up from just one 6VDC SLA battery, and don't want to add any other power sources. What do you suggest?

    Stephen Gard

    ReplyDelete
    Replies
    1. Stephen, the solution is to simply swap the pins of the IC, meaning connect the preset pin to pin#3 and the zener to pin#2.
      In the above situation the load must be connected via the N/O contacts, so the LEDs stay illuminated as long as the relay is activated (voltage normal condition), as soon as the voltage drops below the lower threshold the relay and the LEDs all shut off without causing any further discharging issues.

      Delete
    2. Thank you, Swagatam, I carried out this mod, and now it all seems to working as I require, The set-up is currently undergoing a long-term (1 week or so) outdoor test, and after this, I will construct the final (weather sealed) installation and send you a schematic and some photos, as promised. I note that the circuits (there are three) still draw 12mA, even when the LED lights are off, but perhaps the solar-panel will keep ahead of this. The weather for the next week includes sunny and cloudy periods, so my solar-charged set-up should have a good test period,

      Stephen Gard

      Delete
    3. That's great Stephen! I will certainly love to see the pics of the finished design, and also post it here for the viewers.

      The quiescent consumption should not be more than 5mA according to me, i think something may be leaking through the output of the opamp, a 3V zener diode in series with the transistor could possibly fix this issue too.

      Delete
    4. Swagatam,

      I got this all installed and working, but by that time, I had learned a lot more about solar lighting, and realised that my approach to this project was wrong: all I needed was a Joule Thief! However, the experience was very helpful, and your low-voltage cut-off circuit works fine. I plan to use it for a 6VDC solar-powered water pump... wilol report back when that's done.

      Thanks for your help and advice,

      Stephen Gard

      Delete
    5. That's great Stephen, I congratulate you on that. Best wishes to you.

      Delete
  52. hai swagatam!!! I had connected this as your circuit.. but when the circuit cut off the output..the battery voltage increases then the relay deactivates and the output works and it again activates and cut the output ...this happens simultaneously ..how to cure it..

    ReplyDelete
    Replies
    1. Hi Dharma,

      Please connect a feed back resistor link from pin6 to pin3 as shown in the last diagram, this will prevent the circuit from oscillating at the threshold points.

      Delete
  53. HI
    Can you please explain what is the purpose of using a bipolar transistor in the 3rd circuit?
    why are we using it?

    ReplyDelete
  54. hey
    can i used 3rd circuit diagram for 12 volt 7.2 to 45 AH battery? And in this circuit there is two battery supply for one is relay operate & another is for load?

    ReplyDelete
    Replies
    1. you can use it...there's only one battery supply that connects with the load

      Delete
  55. Hello sir,
    Plz I need clarification on ur last circuit dat has two relay. I built d circuit for my 12v inverter as charge control but d problem is i got d first conrol using 10k VR for 14.5v auto cuttoff but.for low batteery sensing I could not get what I want to deactivate at 10.5v. All I got was it deactivate at above 11v and even that I have to tuned d 100k VR toward zero ohm so what do i do to achieve low battary dectation at exactly 10.5v instead of above 11v

    ReplyDelete
    Replies
    1. Hello Beelal,
      try a lower value pot in place of 100k, may be a 33k pot and reduce the series 10k fixed resistor with 1k

      Delete
  56. Hello,
    i see 2nd diagram 10 k preset connect @ 2 pin & in 3rd diagram 10K preset connect @ 3pin of IC. Another question is that I have UA 741C IC can i use it or there is difference IN UA 741 & LM741 please guide as i used it for 12 volt 7.2 AH battery.

    ReplyDelete
    Replies
    1. Connect the pinouts as shown in the third diagram.

      all types of 741 will work for this circuit

      Delete
  57. Hi Swagatam,

    You have recommended this third version for my hi/low contacter.
    You suggested that I don't require the second relay.
    Which one do you mean and what was it's original function?

    Thanks for this
    Yours
    Carl

    ReplyDelete
    Replies
    1. Hi Carl,

      The lower relay is not required for your application.

      The circuit is a battery over charge/discharge controller.

      Delete
  58. Hi swagatam,
    For the third circuit, I would like to know the exact values of the potentiometers, because i am only going to put resisters instead of potentiometers.my circuit will be using 12v input, 12v battery and the load current will be 3a.

    ReplyDelete
    Replies
    1. Hi Anthony, the simplest way of implementing it is to first set the thresholds with pots and then measure the pot resistances to get the exact values of the fixed resistances.
      The pot can be removed thereafter and replaced with the measured equivalent resistances.

      Delete
  59. can i use this circuit for 24v gel battery with 180 ah??

    ReplyDelete
    Replies
    1. yes can be used..use a 324 IC for the opamp.

      Delete

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