The proposed circuit was requested by one of the avid readers of my blog. It is a low battery warning indicator circuit using opamp 741 and can be used for monitoring a particular low battery voltage threshold.
Circuit Description
The circuit may be understood with the following points:
1) The entire configuration is wired around the IC 741 and it becomes the heart of the circuit.
2) 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.
3) Here as can be seen in the diagram, the non inverting input is provided with a fixed reference voltage through a resistor zener network.
4) This input is fixed to about 5 volts.
5) The other inverting input pin #2 is wired via a preset to sense the input supply voltage from the source.
6) 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.
7) When this happens the output of the IC immediately becomes high, illuminating the connected LED.
8) The illuminated LED immediately provides the indication of a low voltage situation so that the required actions may be initiated.
9) 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 particular stage which may be relevant to the low battery cut of actions.
How to Setup this Low Battery Indicator Circuit
The above low battery indicator circuit can be even further improved in the following manner for controlling both lower and the upper charging thresholds:
Initially keep the 100K preset link disconnected.
Apply a 14.4V source from the "Battery" side 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.
Low Battery Indicator Circuit with Relay Cut-off
The following circuit shows how the above low battery indicator can be enhanced with relays for achieving an automatic low charge and full charge cut for the connected battery, and also a cut off for the load during a low battery situation.
The upper relay becomes responsible for cutting off the battery during over charge and low discharge level, while the lower relay cuts of the load as soon as the battery reaches the unsafe low discharge zone and as soon as the upper relay reverts to charging mode
muhammad ameer khan bin mustafah says
hi sir, i have some question regarding with 3rd circuit with relay
1. may i know the function of feedback loop of resistor 10K and 100K resistor at IC 741…why do we need to adjust it with 11.2 input being reduce as you mentioned.
2. based on the setup guide you explain, i’m confused where should 14.4V source should be applied at the circuit…
is it need to be supplied at battery input or 12V dc input
3. during the setup, should the battery and 12v dc input stay connected in the circuit when moving the preset to-and-fro
4. Lastly, can the 12dc input replace with charging input to recharge the battery.
thank you sir.
Swag says
Hi Muhammad, here are the answers
1) it is to enforce a small latching on the IC so that as soon as full charge is reached, the output gets latched. We must adjust this feedback resistor such that the latching effect breaks at the desired lower threshold which may be 11V for a 12V battery.
2) 14.4V must be applied from the battery side, so that this setup supply input imitates the battery voltage
3) No battery would be required because the input itself would be from the battery side.
4) Once the setup procedure is complete, the left side input must be connected with a battery to be charged, and the relay side must connected with the charging input voltage.
muhammad ameer khan says
Thank for the reply earlier sir, but i get something to ask
i try to run simulation in multisim but can’t get an result….during setup, when the dc souce applied at battery input is 14.4v and preset 10k ohm is adjust to turn on the op amp it activate the relay. But when the 14.4v is reduce to 11.2v as you mentionesd, the op amp automatically turn off because voltage is reduce causes voltage drop at 10k preset drop, hence this also deactivate relay… so at when i need to adjust the 100k preset, because the relay already turn off when voltage is reduce to 11.2v
Swag says
Hi Muhammad, I don’t recommend simulators for judging a circuit because simulator can be misleading if not optimized correctly, it is better to build and test it practically.
for the moment you can try removing the feedback completely and try again, and then try different fixed resistors for the feedback to test the difference in the low level triggering and which value gives the most appropriate results…you can also try other ICs LM321 and LM358 etc
muhammad ameer khan bin mustafah says
hello sir, im doing a project to charge up a 12V battery using 18v solar panel. So when the battery voltage drop at 5v the solar will charge up the battery and stop when the battery is full automatically. So can this circuit be used by adjusting the 1k pot or need other circuit??…please help me sir…thank you
Swag says
Hello Muhammad, it is possible but a 12V battery can be discharged only upto 10.5V , below this level it is strictly not recommended.
muhammad ameer khan bin mustafah says
thanks for the reply sir, but is there any other way to detect the 5v battery condition?
Swag says
for detection you can use the above transistor circuit, and set it for detecting 5V…
kabiru Ali says
sir do you have datasheets for these component：ic 7805, lm 317, lm 324？if yes pis help me with links
Swag says
datasheets can be easily found online, if you have any specific questions please ask me here I’ll clarify!
kabiru Ali says
sir ,it is 40Ah
Swag says
you can try the 12V 5 amp SMPS circuit which is included in this website…please search it online, the first result will be the one I have referred from this website
kabiru Ali says
yes sir that’s what i mean
Swag says
what is the Ah rating of the battery?
kabiru Ali says
thank .sir pls once again is there any circuit for transformerless ac to dc converter ,if yes pls send me the link
Swag says
transformerless is not possible, a minimum ferrite transformer has to be employed.
kabiru Ali says
thanks. Sir pls i need 12v battery charger circuit for lithium ion 18650 pack .
Swag says
Kabiru, you can modify the following article for charging your 12V battery
https://homemade-circuits.com/make-6v-4ah-automatic-battery-charger/
kabiru Ali says
sir ,good day .which pin-out of the relay conected to BC547
Swag says
the coil pinouts
olupot says
hello good morning thanks for the great work you do.
the circuit performed perfectly and effectively in turning off when the battery is low i set mine to cutoff at 11.6v
Swag says
Glad it worked, keep up the good work..
Godson says
Hello sir Swagatam,
Thank you so much for all your replies. I’m responding to the last reply that you gave as regards the second circuit.
Here are what I understand, and I need you to confirm them:
1. I’ll use the circuit as it is and remove the 100K feedback resistor.
2. I’ll use the relay contacts (N/O and COM) as the switch for the inverter driver.
3. In order to set the circuit, I’ll apply the lower voltage, say 10.5V for 12V battery, at the battery points in the circuit, and then turn the preset till the relay just deactivates, and then I’ll turn the preset back and fort to confirm the relay deactivation.
4. In order to use it for higher voltages, I don’t need to change any of the components, all I need to is to connect pin#7of the IC according to how it is in the link you gave above.
Please confirm those four points sir. I also have some questions:
1. How do I connect the LED so that it will light up only when the relay deactivates, i.e when low battery is detected.
2. I want to use a 7812 voltage regulator to power the inverter driver (if I’m using 24V and above), do I need to connect a capacitor across the input and ground and/or output and ground of the regulator? If yes, kindly suggest the value(s) of capacitor(s) to use. Thank you sir.
Swag says
Hello Godson, all your 4 points are correctly assumed, for above 24V application in addition to the pin#7 modification you must add a series resistor with the pin#3 preset, and increase the pin#2 resistor to 10K.
for LED indication you can simply connect an LED network having a series 1K resistor parallel to the relay coil.
for a 7812 IC you can connect a 10uF/50V at the input side and a 10uF/25V at the output, although this value is not critical.
Swag says
preset series resistor can be a 10K for supplies over 24V
Godson says
Ok sir. Thank you for the prompt reply. Its quite explanatory. But I want the LED to turn on only when the relay deactivates. If I connect it the way that you explained, it appears that it will remain turned on as long as the relay is activated.
Swag says
sorry, in that case you may connect the LED parallel to the collector emitter of the transistor or across pin#6 of the IC and the positive line…in both the cases use a 1K or upto a 4K7 resistor in series wit the LED
Godson says
Hello sir Swagatam,
Please I want to use the above schematic (second one) for low battery auto-cut off. I already have an automatic charger circuit which you recommended to me. I want to connect the circuit in such a way that the relay contacts will act as a switch to 7812 voltage regulator which will in-turn power the driver section of an inverter. My questions are:
1. How do I set it to activate/deactivate the relay only when battery is low, say about 10.5V, such that when the activation/deactivation occurs, the relay stays in that position until the inverter is switched off.
2. If I want to use it for higher battery voltages such as 24V, 36V, 48V and higher, how do I connect the opamp since it has a maximum supply voltage.
3. Do I need to change the values of the zener and feedback resistor for higher voltages?
Anticipating your usual prompt response. Thank you sir.
Swag says
Hello Godson,
The anseswers are as follows:
1) remove the 100K feedback resistor, it won’t be required. Through a separate SPDT switch connect the N/C of the relay with the base of the transistor, make sure to connect a series 10K resistor with this link. Initially keep this switch in the OFF position, feed the 10.5 or 10.7V to the circuit from the battery side and adjust that 10K preset such that the relay just clicks ON. In this position increase the input voltage to may be 12V or 14V, and switch ON the SPDT switch…now the circuit is all set, if the voltage drops below 10.5V the relay will switch OFF and latch up in that position until the SPDT switch is switched OFF and ON again, and if the voltage is found to be over 10.7V.
2) for higher voltage you can modify the supply pin#7 as shown in the following figure, and add series resistor with the preset.
https://homemade-circuits.com/2012/08/make-this-48v-automatic-battery-charger.html
3) No need of changing anything else for higher voltages, except the above.
Swag says
sorry please ignore the first idea, it is incorrect and won’t work.
I’ll draw the design and update soon.
Godson says
Ok sir. Thank you very much for the prompt response. Actually I wanted use the “Battery Voltage Status Monitor Circuit” for this purpose by attaching a relay to the last LED output. But on a second thought, I decided to use a separate low battery indicator circuit and that’s why I’m asking questions on this post. And also, i don’t want to tamper with the pin#10 of the driver IC, that’s why i want to work with relays. I’ll wait for the updated design. Thank you sir.
Swag says
Godson, actually you can use the 2nd circuit exactly as given without changing anything, instead of supplying the charging voltage at the relay contacts you can configure it with your inverter driver circuit. as long as the voltage is above 10.5V, the relay will be in the activated mode powering the inverter driver stage, and the moment battery supply goes below 10.5V, the relay will switch OFF to the N/C point….
the 100K feedback resistor can be removed, it’s not relevant for this application
Swag says
the N/O point of the relay will need to be wired with the inverter driver circuit, and not the N/C….
Anonymous says
How can i modify the circuit to use 220-240 volt instead of 12V DC.
Thank you
Swagatam says
apply the preferred lower threshold mains AC value to the transformer based power supply which is used for powering the opamp circuit, then adjust the preset until the LED just illuminates
Nikhil Sonar says
hello sir,
good morning ,
i need solution for battery voltage indicator ,
there are six 12volt 7.2Ah batteries in series adding equal to 72 volt dc
i want to monitor online battery voltage continuously while indication for low battery indication,
i have seen many circuit ,all are up to 24 volt .
in my case it is 72 volt dc so how to monitor such high volt.
do you have any similar circuit .
thanks
Nikhil.
Swagatam says
hello nikhil,
you can make the following circuit, as many as required, and connect them all in parallel with your battery,.
https://homemade-circuits.com/2013/05/low-battery-indicator-circuit-using-two.html
after this you just have to set their presets at the respective low level thresholds, for getting the relevant battery level indications.
aizaz khan says
Sir i hav tried the following circuit but not working as it cant output the voltage and current i have tried my level best but unfortunatlly fail plz help i am so worried iTs my 20time i draw a circuit just hear the trip trip sound of relay but cant get output voltage whn i check by multimetr
Swagatam says
you may be connecting the relay contacts incorrectly, please show me the image of the relay, you can show me an online image link of that relay I'll identify the pinout details of the relayfor you.
aizaz khan says
Sir i am using exactly this one but colour is black 🙂
http://elgammalelectronics.com/Products/Details/7eb6a416-fe94-4359-a51f-fcb128443ba7
Swagatam says
this relay is already explained in the link which I provided you in the previous comment…I think you should first learn all the basics of electronics then start building circuits….
aizaz khan says
sir i have made the given third circuit but i cant get voltage on ic when i check on pin 7 and 4 and i also didnt get the relay connection as i think i have mistakenly wire the relay conection so which is nc and no in the diagram plz explain brefly
Swagatam says
Aizaz, Please check out the following article to learn exactly where the N/O and N/C pins are located in standard relays
https://homemade-circuits.com/2012/01/how-to-understand-and-use-relay-in.html
aizaz khan says
sir i have made the above 3rd diagram circuit and i have used everything as u mentioned the question i want to ask should i give it direct 10amp and 18v or should i make a adjustable as u have mentioned that first apply 11.2 volt etc so from where i apply this as i have transformer of 10amp and 18v and lm317 only drawn 1.5amp current so how i do that i am new to electronics so plzzzz support me i will very thankfull to u
Swagatam says
aizaz, for setting up the circuit as per the article, you will need an adjustable power supply circuit, if you don't have one you can build a temporary one by using an LM317 regulator IC.
while setting up the circuit the current input is not important, only the voltage must be set appropriately as mentioned.
please note that the 100K feed back preset was not calculated by me, rather chosen arbitrarily…so if you find the relay triping back at above 11V or 11.5V then you may have to use a higher rated preset, may be a 220K or 470K etc.
remember the current from the transformer while charging the battery must be 10 times lower than your battery AH rating.
aizaz khan says
Answer plz
aizaz khan says
And a 12 v relay are usually upto 25amp m i right sir?
Swagatam says
No changes would be required for a 2N2222, a 12V relay may have a huge range, not limited to just 25Amp, it will depend on the user to select their preference.
however 25A will be enough for a 100 AH battery and 10A charging
aizaz khan says
2ndly i want to confrm should i make any changes if i use 2n222 transistor?
Swagatam says
click "load more" button below
aizaz khan says
Thank u sir let me try this circuit i want also is this circuit is fully automatic i mean it will automatically stop charging when battery is full
Swagatam says
yes it is fully automatic.
aizaz khan says
Sir you are superb blowing
Sir i need help i have few question will u confrm
Is this above circuit will work on 100ah battery 2ndly i want to confrm i dont understand how much we will apply voltage and current needed?
I am talking about 3rd ckt .
Sir what should be connected at the load and where is the red and green led whixh shows that battery is charging and full
I want to id so that i take help
Swagatam says
Thank you Aizaz,
you can use upto max 18V, and current can be upto any level that you prefer depending on the battery size. If your battery is 100AH in that case you can apply a 10 amp or 12 amp input….but make sure the relay contacts are rated to handle this much current….and if the relay is rated at some higher power then the transistor will also need to e upgraded, so better to go with a 2N2222 instead of the shown BC547
Swagatam says
the load can be an inverter or lamp anything that you intend to operate when the battery is fully charged.
for connecting the LEDs you can refer to the following circuit
https://homemade-circuits.com/2012/07/make-6v-4ah-automatic-battery-charger.html
Rupesh Raghatate says
hello sir
can i use this circuit for lithium ion battery whose max charge is 3.8- 4.2V.
i am asking this question bcoz my device is single battery operated and if i use above ckt, as my battery is drained my reference voltage will also vary since their is no other source of power ,
suppose i use zener of 3.3V for getting regulated reference and battery voltage drops below 3.3v ,
in this condition will this circuit work or not
Swagatam says
hello rupesh, the above circuits will not work with 3.3V supply, you can try the following design instead
https://homemade-circuits.com/2013/05/low-battery-indicator-circuit-using-two.html
Ainsworth Lynch says
Do you have any schematic with an adjustable power supply with atleast 50amp output
Swagatam says
you can try the design that's explained below:
http://makingcircuits.com/blog/2016/09/high-current-lm317-variable-power-supply-circuit.html
Anonymous says
Hello sir..please how can I make a battery charger with the rating 15v 5amp that can charge a car battery.. Please give me a step by step explanations.. I would realy appreciate it..thanks
Swagatam says
what is the battery AH rating? and also specify whether you want it to be automatic or manual OFF
olukunle ojewale says
dear sir i real enjoy your circuit design and analysis. can this 12volt be modified to 24volt. if yes what are the component to add
Mustafa says
Can I use this circuit for 14V DC or 16V DC?
dimoune says
Hello,
I have a 28v dc power supply and I am looking for a circuit to monitor voltage drop below 10v in the load. I want to have led light up when load voltage is above 10v and another led to light up if voltage drops below 10v. I also want the led to draw power from the power supply instead of the load. How can I achieve this please? Thank you.
Swagatam says
you can use the circuit, and connect one LED from positive to pin6 and another LED from pin6 to ground….both these LED must have a 10K resistor in series.
Use LM321 IC instead of 741 since LM321 can work with upto 32V supplies
Swagatam says
…and use 10k for the zener resistor
Ainsworth Lynch says
If the second diagram has both features then what is the extra feature of the 3rd diagram with the second relay
Swagatam says
the 3rd diagram allows the load to access both the SMPS voltage as well as the battery voltage depending on the upper relay switching, and the battery condition.
victory says
Pls sir i have made the first circuit nd it's working fine…my question is(1)is the second diagram used for cutting the relay when voltage increases to 14.4v cos it does not seem like a low battery circuit since preset is connected to pin 3 and zener to pin 2 and i want to build this second circuit for battery full cutoff at 14.4v .(2)i want to add the first circuit to my sg3524 inverter without relay driver stage so i'm planning on connecting pin6 of the op-amp to pin10 of sg3524 so that it will shutdown my inverter at the low voltage level will it work like that?(3)what will be the effect of adding a 2.2uf 400v fan capacitor in series to the load at the output of my inverter?
Swagatam says
victory, the second circuit has both the features, it will switch ON the relay when the battery reaches the full charge value…and will switch OFF the relay when the battery reaches the low charge level…the low charge cut-of setting is determined by the value of the feedback resistor connected across pin6 and pin3 of the IC.
yes it will work in tat way with SG3524 shut down pin….
capacitor should be added in parallel to the load, not in series…it will help to modify the square wave to a sine wave.
Nelio Abreu says
Hi Swagatam,
I'm back again…. 🙂 …. I'm working on a digital thermometer using ICL7106 and an LCD display. The LCD has an icon of BAT for low battery indication. The main power supply is a 9V battery, reduced in 5V by a 7805 regulator. I need to have the icon BAT light on (which is terminal 40 of the LCD and was not included in the original project) whenever the battery goes below the minimum of the regulator, which is 4V. The LCD is common Anode. How can I interface this circuit with the digital thermometer and what changes do I have to do?
Best Regards.
Nélio
Swagatam says
Hi Nelio,
you'll need to first confirm regarding what logic level pin40 is assigned for (zero or high?) After that you can use an opamp based low voltage detector circuit for activating this pin with the relevant logic level when the low voltage is detected….for converting it into a digital thermometer you'll need to use a temperature to volatge converter stage and integrate this with the voltage sensor input of the IC 7106
Nelio Abreu says
Hi Swagatam,
The part of the thermometer I have. All I need is to trigger the BAT icon in the LCD when the battery goes low. How to I know what is the logic level of pin 40 of the LCD?
Best Regards,
Nélio
Swagatam says
Hi Nelio,
you can manually inject a 5V supply through a 100k resistor into this pin, and/or do the same by connecting the pin to ground via a 100k resistor….and check the response on the low batt icon…this will let you know the logic assignment for this pin.
Shayan Firoozi says
Thank you ,
it's really good circuit , but i can not understand , in wanna use 2nd circuit with your great Auto Micro UPS circuit , but what is charging voltage input ?? is this circuit has built in charger ?!! i'm really confused !!
i wanna Auto Micro UPS to cut off the load when 12volt battery is going down and reaches its critical voltage
who to use this circuit with Auto Micro UPS ?
thanks in advanced
Swagatam says
This diagram is correct:
https://drive.google.com/file/d/0B2VgP4YQwP4ETzczQ2Y3VExKX00/view?usp=sharing
you can use the second circuit from the above article and use it as indicated in your diagram.
Shayan Firoozi says
Great circuit , i wanna use it with Automatic Micro UPS , but what AMP does it support ? i need to use with 5 AMP Micro UPS
Swagatam says
thanks, it will support any desired amp, 5 amp can be easily used with the shown circuits
Dr Ankit says
Dear Sir, The circuit is working as intended (using PNP so that relay is activated when voltage is low). However on borderline voltage, the relay chatters. Can something be done to avoid this eg adding delay with IC 555 or some other way. Thanks
Swagatam says
Dear Ankit, just connect a 100uF/25V cap in parallel with the relay coil, this will take care of the issue
Dr Ankit says
Dear Sir, I connected 100uF/25V cap in parallel with relay. Though this works fine when I test it at my work bench with a dc adapter, but as soon as I use the circuit in car (12V 70 Ah battery), the relay again starts chattering at borderline voltage.
I maintained the polarity of caps (+ to + of relay)
The slight modification that has been done to your original circuit (in addition to using PNP) is following
1. I am using 2 relays in parallel (as I need 2 switches and double pole relay not available) (diode put across both relays)
2. The 100uF 25 V caps have been connected to each of the relays separately
3. I have connected a buzzer in parallel to the LED (buzzer spec 3V to 30 V)
Kindly advise to prevent relay chattering
Thanks
Swagatam says
Dear Ankit, do you have the feedback resistor in place? (between pin6 and pin3)
anyway, try increasing the value of the cap to 1000uF
or try placing the 100uF across the base and emitter of the BC557
Dr Ankit says
Yes, the resistor between pin 6 and 3 is there.
Which option will be better increasing value of each cap to 1000 or placing cap between E and B
Swagatam says
first try the E,B, option if it doesn't then you can try the 1000uF method
Dr Ankit says
Dear Sir, I made the circuit exactly according to second schematic and it is working fine. Just one query, In this Initially at normal voltage the relay is activated and when the voltage drops below the low level then the relay gets deactivated. CAN this function the OTHER way ie at normal voltage the relay remains deactivated but when voltage drops it activates.
Further what change may be required if low voltage cut is to be set at around 12V (currently it is 5V probably)
Thanks
Swagatam says
Dear Ankit, what you are asking is not recommended because if you allow the opposite to happen that would mean at the lower critical level the relay coil will be energized and this would lead to even further consumption of the battery which will in turn discharge the battery even faster…….. this won't be a favorable situation for the circuit and the battery.
by the way you can do it by simply changing the NPN transistor with a PNP…and connect the relay diode stage across its collector and ground.
Dr Ankit says
Thanks Sir. As per your suggestion of using PNP, will the ground be emitter or the base.
Swagatam says
emitter will go to the positive….
Dr Ankit says
Dear Sir, I did all the changes with pnp BC 557. But now the circuit is not working. Kindly advise
Swagatam says
Dear Ankit, you might have done something incorrect with the connections for sure…otherwise there's no reason why it wouldn't respond….relay should be connected between collector and ground….emitter to positive supply….base to opamp resistor/LED……relay diode cathode must go to transistor collector…anode to ground
the LED polarity needs to be reversed.
Dr Ankit says
Dear Sir, Thanks for the reply. After reversing the LED polarity, the modified circuit is working fine. As a final piece of advise regarding this circuit, kindly let me know the following-
1. What is the appox consumption of the circuit (on 12V battery) (when the relay is in deactivated stage), and
2. Is it OK to keep the circuit connected 24X7 or should it be switched on only intermittently (ie will continuous operation reduce the life of circuit)
Thanks
Swagatam says
Dear Ankit,
increase the value of the perset to 100k, and same for the zener resistor, this will allow just a few mA current to be consumed by the circuit (only as long as the relay is not active)
the circuit can be kept powered ON infinitely without any harm
Dr Ankit says
Dear Sir
Kindly advise if 4.7V zener diode is not available then can 3.3v or 7.5v zener diode be used.
Swagatam says
Dear Ankit, yes any zener below 8V will be fine…
Dr Ankit says
Dear Sir
Thanks for your prompt reply to my other query and referring this circuit for my purpose. Kindly help me with the following queries regarding this circuit
1. To use a relay can the LED and its resistance (as in first diagram) be simply replaced by the relay OR is it mandatory to do the other changes of 2nd diagram like addition of transistor.
If I want to use <12v as the voltage to activate the relay will the required zener diode be of 12v OR as shown in diag 4v is needed.
3. What will be time required by the relay to act. (will the relay be immediately energized as soon as voltage falls below set 12v Or will some time be required and Vice Versa when the voltage again resumed to normal).
Thanks
Swagatam says
Dear Ankit,
the IC will not be able to handle a relay directly, therefore a transistor stage will be required as shown in the second diagram.
The zener diodes should have a much lesser value than that of the supply voltage level, so for a 12V supply any value between 3v and 6V would be OK for the zener diodes.
the relay will act almost immediately as soon as the voltage crosses the set threshold limit.
Swagatam says
….if the 100K feedback resistor is used then the restoration will not be immediate rather as per the 100K response…lower values will produce relatively quicker restoration….as the resistor value is increased the restoration point will move backwards (will be delayed proportionately)
ogunyemi gbenga says
hello
i used the above circuit to shutdown my inverter when battery voltage is 9.5v but inverter goes on and off severally at the set voltage.
Swagatam says
that can never happen….use the second circuit, remove the 100k resistor and adjust the 10k preset such that the relay just cuts off at 9.5V and activates at around 9.7V….once this is done the circuit will implement the same accurately each time the battery voltage falls to this level.
vijeesh kumar says
hai sir,
I made this circuit it is working perfect with sugar cube relay but it cant pull the coil of a 6 amp 200 ohm 12v relay,I am making a 800 watt relay,since the current has to pass through the relay I have to use a high amp rating relay,how to solve it.
Swagatam says
Hi Vijeesh,
according to me a BC547 transistor should be able to pull a 200 ohm relay easily, try reducing the base resistor to 4k7 and check the response, or you can also try a 8050 or a 2N2222 in place of the BC547 for the same….
Sharanabsava says
Swagatam Majumdar …sr….
My ups is 40W ..
Battery is fully charged is there ..
But my ups is showing battery low why sr…
Swagatam says
sharanabsava, check the battery voltage while the inverter is operating and without any load….if the battery volatge drops then that would indicate an incorrectly charged battery or a faulty battery…also check the current with an ammeter in series with the batt positive….if it indicates high consumption without any load then the fault could be in the inverter, not the battery
RON VAN DKN says
can i use this circuit for 24v gel battery with 180 ah??
Swagatam says
yes can be used..use a 324 IC for the opamp.
anthony mercier says
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.
Swagatam says
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.
Carl Corbeau says
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
Swagatam says
Hi Carl,
The lower relay is not required for your application.
The circuit is a battery over charge/discharge controller.
Ashok Dhenge says
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.
Swagatam says
Connect the pinouts as shown in the third diagram.
all types of 741 will work for this circuit
beelal saeed says
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
Swagatam says
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
Ashok Dhenge says
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?
Swagatam says
you can use it…there's only one battery supply that connects with the load
shifa nadeem says
HI
Can you please explain what is the purpose of using a bipolar transistor in the 3rd circuit?
why are we using it?
Swagatam says
for driving the relay.
Dharma Raj says
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..
Swagatam says
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.
MidstLifeCrisis says
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
Swagatam says
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.
MidstLifeCrisis says
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
Swagatam says
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.
MidstLifeCrisis says
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
Swagatam says
That's great Stephen, I congratulate you on that. Best wishes to you.
MidstLifeCrisis says
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
Swagatam says
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.
MidstLifeCrisis says
(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
Stephen Gard says
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
Swagatam says
OK Stephan, that's great, no issues, if you have solved it you can go ahead with it…..
Anonymous says
hello i had constructed your second circuit https://homemade-circuits.com/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
Swagatam says
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.
Anonymous says
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
Swagatam says
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:
https://homemade-circuits.com/2013/07/selectable-4-step-low-voltage-battery.html
Guru Sivakumar says
I tried the circuit by using LM741. It is working Perfectly thanks Swagatham
Swagatam says
You are welcome Guru!
Anonymous says
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
Swagatam says
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.
Anonymous says
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)?
Swagatam says
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.
Anonymous says
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.
Swagatam says
make the first circuit, don't modify anything, use it as is.
Anonymous says
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.
Swagatam says
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.
Anonymous says
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
Anonymous says
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
Swagatam says
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.
Anonymous says
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)?
Swagatam says
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.
Anonymous says
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.
Swagatam says
With the relay switched ON, the power can be no less than 40mA (maximum)
Anonymous says
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?
Swagatam says
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)
Anonymous says
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…
Lall Kumaran says
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
Lall Kumaran says
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
Swagatam says
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.
Lall Kumaran says
Thank u……i'll try this and give a feedback
sultan says
plz tell how to modify this circuit and turn it into a over voltage cutoff ,
basically the opposite of low voltage cutoff
Swagatam says
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.
Anonymous says
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…
Swagatam says
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.
bashir ahmad says
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
Swagatam says
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
Premdeepak Antony Peedikayil says
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.
Swagatam says
yes your connection is correct.
Premdeepak Antony Peedikayil says
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.
Swagatam says
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.
Premdeepak Antony Peedikayil says
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
Swagatam says
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.
Mike says
Instead of the LED or piezo can this be used to switch a 12V relay off so the battery wont fully discharge?
Swagatam says
Hi Mike, see the second circuit, it has a relay.
Mike says
What is that red thing on pin 6 of the 741? Is that a resistor and LED?
Swagatam says
yes it's a resistor and a LED, 10K resistor and a red 5mm led.
Mike says
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?
Swagatam says
everything is very flexible with such comparators, you may adjust any of the two.
Anonymous says
can you tell me the name of the box after the pin#6
Swagatam says
10K, 1/4 watt resistor
Premdeepak Antony Peedikayil says
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
Swagatam says
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.
Sharanjit Singh says
Hi,
i want to make 54V low battery indicator, when the battery voltage comes down to 46 volts
how should i go about it?
Swagatam says
Hi, this circuit can be used for upto 24V application only, not above that…
Anonymous says
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.
Anonymous says
u mean the 4v7 zenar diode ?
Anonymous says
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.
Swagatam says
Connect a 1N4148 diode across pin3 and pin6 of the IC.
anode to pin3….cathode to pin6
Anonymous says
ok sir,i will do it and give you feedback. thanks.
Anonymous says
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.
Swagatam says
Replace the diode with a 10K resistor and check…
Anonymous says
sir, help me i want construct 500watt inverter without transformer, what do i do?
Swagatam says
you can try the following circuit:
https://homemade-circuits.com/2012/05/how-to-make-transformerless-inverter.html
Anonymous says
sir, how can i construct 12v inverter with at least 500watt without transformer.
Swagatam says
using a ferrite transformer
Anonymous says
ok sir i will try and tell u later . thanks
Anonymous says
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.
Swagatam says
for over 20 volts, change the 741 opamp, replace with one of the LM324 opamps.
LM324 has 4 opamps, use any one of them.
Anonymous says
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
Swagatam says
It will be difficult for me to check the fault without actually seeing the circuit board or the diagram.
Swagatam says
it should work actually, not sure what's wrong with your circuit.
Anonymous says
please explain sir,
Swagatam says
put pin#2 at zener, and pin#3 at 10K preset.
Anonymous says
dear sir , how can i modify as over voltage protection for battery
Swagatam says
just interchange the input pin numbers of the IC.
Anonymous says
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
Swagatam says
Hi,
Just replace the zener with two 1n4148 diodes in series, cathode toward ground. No further modifications would be required.
Babita Joshi says
how is indicator different from voltage regulator???
Swagatam says
An indicator will only indicate while a regulator will control the voltage as well…..
Anonymous says
Hi Swagatam,
for the relay version? what is the value of resistor connected to LED and what type of transistor do i use?
Regards,
Swagatam says
resistor can be a 10K 1/4 watt, and the transistor BC547…
Anonymous says
Hi,
can i integrate this circuit to this blog? https://homemade-circuits.com/2012/11/automatic-dual-battery-changeover-relay.html
Regards,
Swagatam says
Hi,
Yes you may do it.
Anonymous says
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!
Swagatam says
Well done. Thanks!
Anonymous says
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.
Swagatam says
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.
Anonymous says
I was looking into placing a voltage regulator at the IC's input (pin 7)…could this be used to reduce the input Vcc
Swagatam says
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.
Anonymous says
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?
Swagatam says
That's it…Thanks!
Anonymous says
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.
Swagatam says
That's all you would need, the battery to be monitored is connected at the +/- points.
Anonymous says
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…:(((
Swagatam says
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.
Anonymous says
what is a 10k preset?please post a list of equipment and what battery to use?(battery spec.)
Swagatam says
10k preset is a variable resistor, the circuit will monitor all types of battery voltages….
Anonymous says
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!
Swagatam says
The above circuit is the only way of using a 741 for low battery indications…
Anonymous says
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.
Swagatam says
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.
Ajay says
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.
Swagatam says
Hi Ajay,
This concept has been covered in the following article:
https://homemade-circuits.com/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.
Ajay says
Thanks for the info and the link Swagatam.
Ajay says
Hi Swagatam,
Can you please suggest a better circuit as you are saying that the above concept can't be very effective.
Swagatam says
Hi Ajay,
May be this circuit can be used in conjunction with a LM338 IC for complete cut off:
https://homemade-circuits.com/2011/12/high-current-10-to-20-amp-automatic.html
Sandip says
Dear Swagatam,
Can the above circuit be modified to place a piezoelectric buzzer after the LED?
Swagatam says
Dear Sandip,
Yes, just replace the led resistor with the buzzer….
be sure to connect a 3 v zener in series with them.
Sandip says
Thank you for your reply. I tried it works great.
Ajay says
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
Swagatam says
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.
Ajay says
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.
Ajay says
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.
Swagatam says
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.
Anonymous says
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
ojasvi says
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?
femmy says
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?
raj mohan says
hai.. can u help me.. i have created a blog comutronics.blogspot.com/ for Ece students.. how can i get it by google search