A very simple low battery cut-off and overload protection circuit has been explained here.
The figure shows a very simple circuit set up which performs the function of an overload sensor and also as an under voltage detector.
In both the cases the circuit trips the relay for protecting the output under the above conditions.
How it Works
Transistor T1 is wired as a current sensor, where the resistor R1 forms the current to voltage converter.
The battery voltage has to pass through R1 before reaching the load at the output and therefore the current passing through it is proportionately transformed into voltage across it.
This voltage when crosses the 0.6V mark, triggers T1 into conduction.
The conduction of T1 grounds the base of T2 which gets immediately switched Off. The relay is also consequently switched OFF and so is the load.
T1 thus takes care of the over load and short circuit conditions.
Transistor T2 has been introduced for responding to T1's actions and also for detecting low voltage conditions.
When the battery voltage falls beyond a certain low voltage threshold, the base current of T2 becomes sufficiently low such that it's no longer able to hold the relay into conduction and switches it OFF and also the load.
Parts List
R1 = 0.6/Trip Current
R2 = 100 Ohms,
R3 =10k
R4 = 100K,
P1 = 10K PRESET
C1 = 100uF/25V
T1, T2 = BC547,
Diodes = 1N4148
Relay = As per the specs of the requirement.
Have questions? We love answering questions, Please feel free to post them through comments and get guaranteed replies within 5 hours. (Please Share and Bookmark us)
Abioye says
sir i have a 3000watt inverter built from a 650watt inverter with your honourable advice and guidance but i want to make a overload cutoff for it my maximum drain current from the battery is 125amp when i divide 3000watt by the batteries voltage which is 24v and it give me the R1 value of 0.0048 how many 0.1resistor do i wire in parrallel to achieve this value
Swag says
You will need 20 resistors in parallel. Each must be rated at 5 watts
Godson Ugo says
Hello sir Swagatam, I asked for an overload protection circuit for an inverter with auto-power off feature in the comment session of one of your posts and you referred me to this post. Having read through the post and comments, I concluded that I may not be able to use the circuit for the intended purpose. You explained in the comments that the inverter should me made the load but I’m afraid that I won’t be able to get the relay that can handle the huge current drain from the battery. So what I need is circuit that will be connected to the output side of the inverter transfo and which can operate a small 12V relay such that I can use the relay contacts to power off the inverter driver when overload is sensed at the inverter output. I’ll be glad if you could help me with such circuit. Thanks a lot sir.
Swag says
In that case you can try the concept explained in the following article:
https://homemade-circuits.com/2014/06/simple-current-sensor-circuit-modules.html
use the NPN version, and connect the collector with the mosfet gates via separate diodes or simply connect it with the “Ct” of the IC,…. for SG3525 you can use the PNP version, and connect the collector with pin#10 via a 1N4148 diode
Godson Ugo says
Ok sir. Thank you very much. I’ll check it out.
moses kwagam says
Thank u Swagatam and every one participating in this blog. Pls how do I determine the trip current and/or the value of R1 if I want to make 1KVA inverter and I want the inverter to cut off when it senses 800 watts of load
Swagatam says
thanks moses, it can be calculated by the formula:
R = 0.6 / trip current
trip current = wattage/Voltage…in your case it will be 800 / battery voltage
Faith Jumbo says
in the case of inverter the load is were the ac is connected but in on one of the comments i read u said the entire circuit is connected to the dc side so how do we monitor the load since the load are the appliances since the appliances are connected to the ac side
Swagatam says
the load cannot be an AC load here, the circuit is designed for DC loads, if an inverter is connected as the load then the appliances consumption would be ultimately drawn from the battery, and therefore the circuit would sense it and trip if an overload from the appliance was detected.
i.c.v. young king says
Pls sir can u send me the circuit diagram of low battery cut through this mail ,youngkingicv@gmail.com pls sir l'll be waiting thanks.
Swagatam says
ICV, you can refer to the following article for making a low battery cut off circuit
https://homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html
dawar jan says
How does the circuit work.esplain step by step. How can i change it to only cut off at low voltage.reply plz
dawar jan says
What do u mean when the voltage cross 0.6v
Swagatam says
it is the triggering base voltage for T1
MORRIS CERULLO says
What should i do for this circuit to cut off at 8volt when using 12v 70amp-hour battery?
Usa Juana says
Hello sir . am building the bubba osilator inverter and am adding the overload and no load detection circuits , please I have these question on those circuits .1, I have finished the oscilators and I am getting 1.4 Volts ac , is that OK?.2, for the sensing resistor of the overload circuit RESISTOR X I calculated 0.00909 ohm , so 12 0.1ohm in parallel will solve it but what wattage of resistor will be OK will 1/2 be OK as it is many?. 3,the no load detection circuit have 0.05 ohm so 2 0.1 ohm resistor will solve it , now this concern me more,what watts do I need to use so that it can with stand the 800watts inverter at its full operational power?
Swagatam says
Hello Usa,
please comment separately under those articles, so that I can refer to those diagrams and address the issues accordingly.
the bubba oscillator output needs to be processed to SPWM first in order to achieve sine wave output.
Shuddhatam Jain says
pl tell me where i purchases lm196 or 78h12a
vhafuwi says
Hi Swagatam
I require your assistance, I have installed an off-grid solar system for student residents ,just for them to use light , laptops and charge their cell phones , it is becoming difficult to control them since they have other high wattage appliances such as electric iron , boilers in their rooms, i require a help in designing a device that will shape the amount of watts or amps a students is allowed to use per room plug , the unit must only disconnect just the student who plug a overload appliance , i will appriciate your assistance , my email vhafuwi@gmail.com
Swagatam says
Hi vhafuwi,
you can try the following circuit
https://homemade-circuits.com/2015/03/mains-over-load-protector-circuit-for.html
install one such unit in every room after the mains DPDT switch
adelusi oluwatosin says
Thank you so much sir. sir how can i select R1 for 1.5kva inverter? i.e the appropriate current sensor resistor for 1.5kva. Thanks.
Swagatam says
adelusi, divide 1500 with the battery voltage, and then divide 0.6 with this result.
kamran abbas says
Hi How are you
Sir
How can I make digital numbers display for example I have inverter and i want show inverter output voltage on display and also battery voltage
Swagatam says
Hi Kamran, you can make and use the following circuit and apply it in your inverter for the results:
https://homemade-circuits.com/2013/05/make-this-simple-digital-voltmeter.html
alternatively you can buy a readymade digital voltmeter module and use it for the same.
rob wiseman says
Hi Swagman.
The P1 preset in your low voltage cutout says (10 preset) how many ohms is this component,,
regards Rob.
Swagatam says
Hi Rob, it's 10k, I have corrected the typo
Olaleye Kolade says
Thanks for this simple circuit.
Am having little problem about the circuit, I build just the circuit and set it, everything is working, but my question is where will I connect the terminal mark load? Is it directly from inverter transfo output
2. The terminal mark battery is it the battery for the inverter or another battery. Please explain
Thanks
Swagatam says
The terminals indicated as "load" is connected with the inverter input supply terminals which is normally supposed to be connected with the battery directly.
and the terminals marked as "battery" should be connected with the inverter battery….not anything external.
with the above set up the battery power now enters the inverter through the above circuit's relay so that the required cut-off actions can be executed
ataull hakim says
Sir, can i use it as a solar charge controller for LVD circuit?
Swagatam says
yes it can be used
Shuddhatam Jain says
Respectiv sir 3 problems come me for making circuit Low Battery Cut-off and Overload Protection Circuit
1. this circuit work for this battery 18650 li-ion 11.1v 6.6Ah battery ?
2. what is R1 = 0.6/Trip Current and how many watt?
3. what is setting of pre set is mark for this battery?
i use this battery pack in hero honda(super splender) biky for the self starting
pl help me
thankyou
Swagatam says
Shuddhatam, I have already answered to the same question in the other article….please check it
Angelous Chavez says
Sir, i interesting with this circuit. I am planning to make this circuit to add in my project. Can you give me s circuit for battery cut-off and overload protection for dc to ac inverters please if you have.
I am angelous from the philippines i am an electronics student.
I am hoping for your reply…
Thanks in advance.
Swagatam says
Angelous, I'll check and try to find one…
Shuddhatam Jain says
respective sir my name shuddhatam jain i likes your circuit digrame i need li-ion battery protection circuit i make Low Battery Cut-off and Overload Protection Circuit according to you in this circuit i don't understand what is meen by QUIZ = Explain the introduction of R4 and C1.
pl give mr reply
thankyou very much
Swagatam says
Hi Shuddhatam, thank you for appreciating my site.
the circuit required by you is already present in my blog, please use the search box above to find the preferred one of your choice.
R4 is for locking the relay when it clicks, and C4 is for preventing the chattering of the relay during the changeover action.
Sina Alimorady says
Hi Swagatam,
is this circuit suitable to use with https://homemade-circuits.com/2013/04/automatic-micro-ups-circuit.html in order to make the Micro UPS auto cut off the load from battery ??
if yes what changes i have to make to support 4 or 5AMP load like Micro UPS does ?
if not what do you suggest to use with Micro UPS ?
thank you again and again @
Swagatam says
Hi Sina, the above circuit is not quite efficient with its detection, and also it could make the entire design too lengthy, so it's better to use the last circuit from this article:
https://homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html
this will take care of the all the required safety conditions.
Ben says
How is the protection circuit connected to the inverter circuit?
Swagatam says
the "LOAD" is the inverter supply input….
Anonymous says
What is the purpose of R4 and C1?
Swagatam says
R4 is for making sure that the relay gets latched after activation when any of the situation is detected…C1 is for preventing the relay from chattering during the operating thresholds.
Anonymous says
The potentiometer is used to adjust the low voltage cut off correct?
Swagatam says
yes that's correct
Anonymous says
What range should the potentiometer be? How big? It says 10 above. 10 ohms? 10K ohms?
Swagatam says
10k ohms is the right value
Anonymous says
hi
Anonymous says
hello sir,
I tried above circuit but my load(LED Lamp) is getting supply when battery v/g is 16V.
Anonymous says
Hi Swagatam,
I submitted the comment above.
I'm now using two 12 volt, 15 amp-hour AGM deep cycle batteries in series. The load will be a 350 watt DC motor.
Shall I still use a 2N2222 for T2? The automobile relay should be rated up to 24V correct?
You said the wattage for R1 should be 10 watts. Should all the resistors be rated at 10 watts?
Once I have the circuit set up do I just use the potentiometer to adjust the cut of voltage?
Swagatam says
Hi anonymous,
an 30 amp automobile relay coil might require heavier current to operate therefore 2N2222 should be used since it is rated to handle upto 1 amp.
yes the relay coil should be rated at 24V
a 24V 350 watt motor will require around 14 amp current, therefore the sensing resistor should be
R = 0.6/14 = 0.042 ohms
wattage = 0.6 x 14 = 8.4w or 10 watts for max safety.
at 14 amp your 15 ah battery will become flat within 1/2 an hour or even less.
Anonymous says
Hi Swagatam,
I want to use two 11.1 volt, 12 Amp-Hour LiPo batteries in series. I'm guessing the cutoff voltage will be at 18 volts. The load will be a 500 watt DC motor. What values of the circuit diagram should I change to do this? What trip current will I use to calculate my R1?
Thank you
Swagatam says
Hi Anonymous, if two are connected in series the cut off voltage should be 24V or somewhere near to it… how can it be 18V?
for current limiting R1 will be = 0.6/12 = 0.05 ohms using two 0.1 ohm resistors in parallel,wattage will 0.6 x 12 = 7.2 watts or simply 10 watts for better protection.
the relay will need to be replaced with a heavy duty automobile relay and T2 will need to be replaced with a 2N2222
Swagatam says
…ok got it, you meant to say the lower cut off voltage, yes that's right, 18V is correct.
Anonymous says
Hello Swagatam,
I want to use two 11.1 Volt, 12 Amp-hour LiPo batteries in series. I'm guessing the voltage cut-off will be at 18 volts. The load will be a 500 watt DC motor. What values in the circuit diagram should I change? What trip current would I use to calculate R1?
Thank you
Shankar ks says
hi sir, can i connect load cell directly to arduino. if no, please give the interface circuit to connect load cell to arduino(supply voltage should not more than 6v)
Swagatam says
Hi shankar, please provide the info in more detail regarding the application so that I can understand and design it better.
Yel Bordallo says
Hello sir can i used overload protection only in this circuit…what parts of this circuit i will remove…thank you hope for your quick response…
Swagatam says
Hello Yel, you'll have to use the entire circuit as shown even if you want only the overload to work….
Naveenkumar T N says
what is the maximum load it can handle
Swagatam says
will depend on the relay contact…..can be upgraded to any limits.
HuynhNguyet says
Hello Sir,
Intermediate relay only endure current 10A, how do we use for current 30A?
Thank you very much.
Swagatam says
hello Huyn, what's intermediate relay? relays are available in all shapes, sizes and ratings, 30amp are also available.
HuynhNguyet says
Hello Sir,
Where do load and battery connect?
HuynhNguyet says
Hello Sir
I'm sorry if my comment made you discomfort.
I change BC547 by C1815, that is ok.
Thank you very much for your design.
Swagatam says
OK that's great!
HuynhNguyet says
Hello Sir,
I don't understand: When the battery is 13V, there is not power to B pin of BC547 (T2), T2 can not active, so relay can not active.
Would you like to explain that? Thank you very much.
Swagatam says
hello Huyn, if you have selected 13V as the triggering threshold then you must adjust the preset until the relay gets activated at 13V…..
HuynhNguyet says
I'm sorry sir
I can not find any error. T2, T1, relay … are ok. I don't know why.
Swagatam says
there's no way a relay won't activate if the transistor is good, connected rightly, base getting 0.6V and the power supply current sufficient…
remove the emitter diode and check, may be it's faulty.
HuynhNguyet says
Hello Sir,
I made new board with new components, the relay does not active. When I check the load voltage is 1.3V at 13V battery.
Thank you very much for your support!
Swagatam says
Hello Huynh, try the following basic relay driver stage first, and check the response:
https://homemade-circuits.com/2012/01/how-to-make-relay-driver-stage-in.html
HuynhNguyet says
Hello Sir,
Does R1 relate this problem?
Swagatam says
No, R1 has no connection with this problem.
HuynhNguyet says
Hello Sir,
Does N/C pin of relay connect to R4 and NO pin connect to R3?
Thank you for your support!
Swagatam says
hello Hunnh, yes that's correct.
HuynhNguyet says
Hello Sir,
If I only use low bettery cut-off function, What component's I remove?
Thank you very much.
HuynhNguyet says
Hello Sir,
If I choice C1 = 100uF/50V, is it ok?
Thank you for your support.
Swagatam says
yes C1 = 100/50 will also do.
Swagatam says
you can try the following design which is without overload cut off
https://homemade-circuits.com/2011/12/simplest-smf-automotive-battery-charger.html
HuynhNguyet says
Hello Sir,
My equipment use 12V battery. I want to isolate between the equipment and battery when the battery voltage is 11.5V in order to protect battery. When battery voltage is 12V, it will supply power for the equipment again.
Thank you for your help.
Swagatam says
Hello Huyn, a 12V batt must be charged until it reaches 14V, therefore once your battery gets discharged to 11.5V, it should be charged to 14V before using…
the above linked design will do the above mentioned operations appropriately
HuynhNguyet says
Hello Sir,
I assembled this circuit. However, the relay do not active when the bettery is 13V (0V in the load). When I remove T2, I connect direct from relay->diode-> mass, the relay actived (the battery is 13V, 13V in the load). I don't known why.
Swagatam says
hello Huynh, either your transistors are not good, or the preset is not correctly set or T1 is interfering due to some incorrect connections, pls check everything again.
HuynhNguyet says
Hello Sir,
Thank you very much for your reply.
I will check T1 and some connection again. When I use Proteus software to sumilate, the result is the same my assembled circuit (the relay does not active when battery is 13V).
Swagatam says
check the base voltage of T2 at 13V, it should be 0.6V to activate the relay…the relay will not activate if its coil resistance is very low in that case R3 will need to be reduced for triggering the relay….1K could be tried for R3
HuynhNguyet says
Hello sir,
Thank you very much for your support.
I already check the base voltage of T2 at 13V, it is 0.6 V. I change R3 by 1k resistor, then I set P1 (P1=10k). The relay is not active (I am sure the relay is ok)
Swagatam says
Hello Huyn, connect a LED with a series 1k resistor parallel with the relay coil and check the response…if the LED glows/shuts off in response to the preset adjustments would mean the relay has problems or is not correctly matched.
HuynhNguyet says
Hello Sir,
Thank you very much for your support.
I already connect a Led with series 1k resistor parallel with the relay coil. The Led light. That it's mean the relay has problem?
Swagatam says
hello Huynh, short the collector/emitter of T2 manually, if the relay clicks would indicate a good relay but a faulty transistor…
HuynhNguyet says
Hello Sir,
Thank you very much for your support.
When I shorted the collector/emitter of T2 manually, the relay clicked. That it's mean T1 or T2 faulty?
Swagatam says
OK now remove the collector of T1 from R3/P1 and check again, this time if the relay functions normally through T2 would indicate a faulty T1….otherwise it's T2 that may be doubtful..
HuynhNguyet says
Hello Sir,
I already remove the collector of T1 from R3/P1, the relay does not active at 13V. May be I will change T2 tomorow.
Thank you very much.
Swagatam says
OK!
HuynhNguyet says
Hello Sir,
I assembled this circuit. However, there is not the voltage in the load with any level battery voltage (11V – 13.5V).
Mouhammad Alnajjar says
well , I've made the circuit and put a red led to detect the passing of the current , and R1 = 3 resistors 2watt : 2x(1 ohm) and 2.7 ohm
I've tested it with 9v and 12v battery
when I change the preset the circuit doesn't deactivate !
although I've connected each point of the relay to the true point
can you help me please ?
thanks
youtu.be/AshLT5g010M
Swagatam says
for setting up the low voltage cut off, you will have to connect the 9V source to the "load" marked terminals and then adjust the preset until the relay just trips, after this the source may be removed from the "load" terminals and the actual 12V battery connected to the "battery" marked terminals for normal operations
R1 has not related to the above setting. It determines the overload threshold trip point for the relay.
Mouhammad Alnajjar says
hello sir
if the circuit current is 1.5A and i wanna cut-off at 10.5 volts which relay I have to use ?
if the cut-off volt is changed what is the rule to choose the suitable relay?
Swagatam says
hello Mouhammad,
the relay has no relation to amps and cut off voltage of the design….you just have to choose a relay whose coil voltage is matching with the supply voltage of the circuit and adjust the preset such that relay just cuts off at the specified lower threshold….the adjustment will need to be made by supplying this lower threshold through an external power supply while setting up.
Mouhammad Alnajjar says
thanks so much sir
there is something else :
is this circuit re-activate automatically after it is deactivated ?
because if it cuts-off at (e.g 11 volts) , the voltage will be more than 11 volts with after the loading ends, so the circuit will be activated then deactivated many times
Swagatam says
No that won't happen because of the transistors hysteresis properties…
Mouhammad Alnajjar says
ok that's good
but 'm facing a problem with relay connections
this is the diagram :
imagizer.imageshack.us/v2/280x200q90/537/zaHcTY.jpg
with the meter I've found the points (2,1) in the place of (3,5) and (3,5) in the place of (2,1) I don't know how or why
because I've found connection between the points 4 with 2 not 4 with 5 !
how can treat this please ?
Swagatam says
If you can show me the relay image I can try to help, without seeing the relay image it would be difficult to locate.
Mouhammad Alnajjar says
this is a video I've recorded :
youtu.be/bi4bgS_6YEU
in the website of the store I've bought from , there is the diagram of points connections the same I told you
www.matni.com/Arabic/Relay/RELAY.htm
the relay model is T-73 and in the website I think it's the same JQC-3F(T-73)
Swagatam says
the connections for this relay have been elaborately explained in this article, please check it out:
https://homemade-circuits.com/2012/01/how-to-understand-and-use-relay-in.html
Vishnu Menon says
How to make resistors of such low value 6miliohms etc here I am stuck do u have any idea pls let me know
Swagatam says
use a couple of inches long non-plated iron wire, or a meter long copper wire wound on a former, tweak and adjust the lengths by verifying the ohms through a suitable multimeter.
Vishnu Menon says
Sir ,
I didn't understand the preset part correctly can u explain its working in the given circuit
Swagatam says
Vishnu. apply the lower battery cut off voltage to the circuit through an adjustable power supply and adjust P1 until the relay just deactivates, once this is setting is done the circuit would automatically switch OFF whenever the battery voltage reaches this level
adelusi oluwatosin says
Thanks for time spending on this site. Sir can this circuit work for 3kva inverter? If yes kindly do the adjustment more so I need to know total calculation of R1. Thanks.
Swagatam says
Yes it will work with 3kva inverter, just modify the relay contact accordingly and use 8050 for T2.
Shahirah MNY says
hello sir, can this kind of cct can be modified to trip low standby current? thank you 🙂
Swagatam says
hello shahirah,yes it can be done by setting the preset appropriately.
sskopparthy says
sir, you told that "https://drive.google.com/file/d/0BytEbOgq6mqeQU5mdzU4Yl9kUDA/edit?usp=sharing" is the circuit low battery cut off(to walkabout in above comments). i would like you to help me to modify this circuit to suit my needs. i am not a genius and just a beginner. so, i dont know whether my idea is correct and it can be done or not. i wanted to tell that when load is applied to circuit, and when its voltage decreases to preset level, it cuts off the relay. but when load is disconnected, i observed that the battery regains its voltage to some extent. so, then the circuit may connect the load again……so, i request you to modify this circuit such that, once it cuts the load, it should stay in cutoff position only until a reset button is pressed….can this be done to this circuit, sir? im just a beginner, so please help me ………
Swagatam says
ss, there are two faults in the shown link, first the N/C should not be connected to ground, second, R3 other end must be connected to ground, rest everything looks OK.
it's highly unlikely that the relay would oscillate, due to the presence of the transistor hysteresis, so probably no latching feature would be required.
sskopparthy says
Sir, I connected the circuit in the correct way only as you said……before making this comment. But I failed to check the circuit diagram before giving the link(I erased n/c and connected r3 other end to ground but didn't save it but I thought I did). And thank you very much sir…now the circuit seems to be good working…..
Swagatam says
OK, no problem, that's great!
Shankar ks says
sir i need circuit for cutoff voltage with 2millivolt.
Swagatam says
use any of the following circuits, replace 741 with LM311
https://homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html
Shankar ks says
will that circuit need any other changes for cutoff voltage as 2 millivolt.
Swagatam says
no changes would be required…it will respond even to the minutest changes between its inputs.
Rolando Avecilla says
Thanks for this circuit.
How can I modify this to use solid state relay?
Swagatam says
It may be done wit the following mods:
Remove the relay and R4 entirely.
replace the relay coil connection points with the battery poles.
T2 may be upgraded as per the load amp specs.
Lester Johnson says
Sir . Will you low battery cut off circuit, turn on the inverter when the battery is recharge?
Swagatam says
it may be done by incorporating another relay parallel to the existing one and by wiring its contacts appropriately for the intended chageover
Oluwaseyi Onasami says
Hello Swagatam,
I have tried working with the Overload, Short Circuit and Load Battery Cut-Off Circuit but having challenges setting the cut-off voltage (10.5V) and getting the right value for R1. I want to use it for 2KVA inverter.
Secondly, can you help with a 12V, 100AH charger circuit with constant current charging and overcharge protection at 13.5V.
Thanks in this regard.
Swagatam says
Hello Oluwaseyi,
if you are using a battery then the settings should be easier to implement, if the input is through a power supply then it would need to be regulated.
If you find it difficult then it could go for an IC based design.
You can try the last charger circuit for your 100 ah battery from the following link, just replace the shown LM338 with LM396 or LM196
https://homemade-circuits.com/2012/07/making-simple-smart-automatic-battery.html
Muhammad Junaid says
Hello Sir I want to make an overload protection circuit which can sustain upto 600Watts or as per requirement by use of Potentiometer. Please help me out how to modify this circuit for this purpose..
Swagatam says
Hello Muhammad, you can do it by replacing R2 with a 1k pot. the ends of the pot will go to transistor base and R1, while the center will touch the ground
Debashees Sikder says
Sir,
Can I use this circuit with sg3525 without relay?I am thinking that I can use a PNP transistor , am I right?
Swagatam says
The above circuit is suitable only with a relay, for 3524 IC only an opamp circuit would be suitable
Anonymous says
Hi!
I found your circuit very interesting.
Is there problem to make it work at 24V?To protect 2 SLA.
Exept relays what else do i have to change?
Thanks in advance.
Joca.
Swagatam says
Make R3 = 22k, and R2 = 1k, that's all, other components will remain as is except the relay which should be 24V rated.
tervkal says
Hi, thank you for your great cutoff circuit. I suppose the cutoff voltage is changing by temperature quite large range in your circuit. A base-emitter voltage of the bipolar transistor has a temperature coefficient at least -2mV/C. That is not a problem because a battery to be charged only in specified temperature range.
Swagatam says
you are right, thanks!
Allnight says
New question about this: Could two of these circuits be hooked up to the same relay?
I have an RV with two sets of batteries: house and chassis. When the engine is running (the chassis battery is over 13V), I want to connect the two sets of batteries. When the house is plugged in (house battery over 13V), connect the two sets of batteries.
Also, if the voltage of either side goes over 16V, disconnect them (that part may be harder, and not an original requirement, it just occurred to me).
I was hoping to build two of these circuits, one hooked up to the chassis battery and the other hooked up to the house battery, and have a single relay connect the two batteries. I want to limit the current that crosses the relay to about 30amps (the max output of the house battery charger, as the engine alternator can put out 190amps).
I'm currently hunting for a relay that will strongly move to open when the coil isn't energized, as I don't want bumps in the road to accidentally flip the relay to the connected state.
Swagatam says
Yes that would be possible, you just have to make the collectors of T2 common from the two circuits, and join with the relay coil.
For 30 amps you would need a powerful relay which would obviously have a tough electromagnetic connection, so hopefully it won't be rattled by the "bumps"
T2s will need to be appropriately dimensioned as per the relay coil ratings.
Swagatam says
No this circuit cannot be used in any manner for feeding a 3525 input, because the IC requires definite logic level which cannot be achieved from a transistor, you will have to incorporate a 741 IC for it.
Anonymous says
dear swagtam
can i use bc 557 to get +ve signal at emmiter to give pin no 10 of sg 3525 to stop action when voltage goes below the cuttoff level.and what will be the changes in the circuit.
thanks
Chris Edwards says
swagatam,
Im trying to implement your circuit into a battery powered project i have to use it only as a low voltage disconnect. I have a smart charger to control over voltage so Im searching for a small simple circuit i can build to put into my projects to stop the batteries discharging to far. Your circuit seems perfect for this.
i have built the following circuit in circuit lab does it look ok for a simple LVD? this is my first circuit design so any help will be appreciated.
https://www.circuitlab.com/circuit/227r5y/screenshot/1024×768/
Also i am wandering if i can implement a set reconnect voltage to stop the circuit reconnecting immediately when the load is removed hence increasing voltage which would continually occur i think this is called hysteresis but its a bit above my level?
Also i believe the bottom outlet is the protected (switched) outlet is this correct if so where does the top outlet go to ground?
Swagatam says
Hi Chris,
The diagram looks OK to me although the setting up procedure for such crude circuits is always time consuming.
the circuit which you have shown already consists enough hysteresis so you won't have to add anything more, it would work just fine the way it's shown.
Swagatam says
correction: the relay contact shown toward the arrow mark shouldn't be connected to ground, this will short circuit the battery and instantly damage everything…..you can leave it open.
Robert Random says
Hi Swagatam,
Would there be a drain of the battery through P1, D1, R1, T1 when the circuit is off due to battery depletion? Thanks. ~Robert
https://www.circuitlab.com/circuit/227r5y/screenshot/1024×768/
Swagatam says
Hi Robert, it's possible only through P1, that is through the end terminals of the preset
Ola2 says
hello sir, how many 0.1 ohms resistor i will connect in paralle to get 0.09 ohms? Thank u sir.
Swagatam says
there's hardly any difference, both are approximately the same.
Shankar ks says
sir when i connect the 25watts load through this circuit to 12volt 7.5AH battery the load is running well. but when i replace battery with 60watts solar pannel the relay is vibrating why? what can i do. value of R1 i used is 0.1ohm 2watts.
Swagatam says
what is the load wattage at the output??
Shankar ks says
40watts
Swagatam says
load is OK….voltage input to the circuit should not exceed above 14V, it should be near about the relay voltage, please check it.
Shankar ks says
sir i connected this circuit to 20ah 12volt battery. when i shorted the output
the spark is appeared in circuit and then the circuit is not working why.(but the circuit connection is correct)
Swagatam says
What is the value of R1 that you have selected? It is the main component which decides the functioning of the circuit.
Don't short the output to check….instead use a 2200uF/50V capacitor, add a 10K resistor across leads of this capacitor, then connect this assembly at the output for simulating a short for a second, use it as many times you like. This will not damage the battery.
Shankar ks says
value of R1 i used is 0.1ohm .but i check the value of resistor in multimeter it shows as 0.3ohm.
Swagatam says
with 0.1 ohms, the short circuiting should pass about 7 amps of current at the output only then the relay would conduct.
remove the base capacitor and then check again.
also connect around a 33uF/25V capacitor across the relay coil.
Shankar ks says
sir when short circuit is made at output the relay is not holded in normaly closed position but it is vibrating like buzzer what can i do. even if R4 is used.
Swagatam says
connect a 47uF/25V capacitor across the relay coil
Shankar ks says
when i connect 47uf capacitor the frequency vibration decreases and not completely eliminated(i also tried 470uf capacitor) plz tell solution.
Swagatam says
this is the only correct solution, if it's not still not working then there might be some other abnormal issue with your circuit.
Anonymous says
can i connect this circuit directly to 12volt 100AH battery,will there any damage to circuit in such high current.
Swagatam says
yes you can connect it without problems.
Vijay Basutkar says
Thank you sir
Vijay Basutkar says
Sir I used 324 and circuit working good, but 1k resistor getting hot.what can I do?
Swagatam says
Make it 10K, it won't get hot.
Vijay Basutkar says
yes sir,
You are the best.
Thanks a lot.
Vijay Basutkar says
Dear Sir,
I want a Only Battery Low voltage circuit,which is consume very low loss itself. Please hepl me.
Thanks & Regards
Swagatam says
Dear Vijay,
what is the battery voltage that you want to monitor??
Vijay Basutkar says
Thank you very much for your valuable time.
I want to monitor 12VDC battery
And second is 40 battery bank voltege, it is 480 VDC & i want 435 VDC cutoff, Is this posible?
Vijay Basutkar says
1) 12 VDC
2) 480 VDC (40 battery require cut off 435 vdc)
Swagatam says
For 12V, you may try the second circuit explained in the following link:
https://homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html
For 480V, I'll have to design a specific circuit, may take some time, I'll inform you when it's posted.
Vijay Basutkar says
Thank You Sir
Vijay Basutkar says
Dear Sir
Any circuit available for 24
VDC battery low?
Swagatam says
Dear Vijay,
you may try the following circuit, just replace 741opamp with IC324 opamp
Vijay Basutkar says
DEAR SIR
WHAT IS THE SELF CONSUMPTION OF THIS CIRCUITE AND CAN I TAKE ONE BATTERY VOLTAGE FROM 40 SERIES BATTERY BANK FOR BATTERY SENSE.
Swagatam says
The self consumption is around 30-40 mA if a 400 ohm relay is used.
If your batteries are connected in series, each might have a slightly different discharge rate…..I think that won't be advisable.
Anonymous says
sir what is the relay specification i need for 12 volt battery.
Swagatam says
12V/400 ohms/spdt
Premdeepak Antony Peedikayil says
Res Sir,
I've used for Low cut off circuit a preset 2nd leg connected to the base of BC547(No resistor in between), first and sec and third to +ve & -ve respectively, BC547 emmiter through a diode to the -ve, collector to Relay with Diode, now when I apply 11.5 and adjust preset to deactivate the relay, but later even If I give 14v also the relay does not activate in this scenario, how will the circuit work , please help sir, where am I gone wrong, should I add some components to make it work.
Prem
Swagatam says
Hi Prem,
I couldn't understand the connections properly, anyway you can add one more diode in series with D1 and check.
Anonymous says
dear sir , i am going to use 150 Ah battery and 150 watts D.C load the battery should cut off at 11.1 volts .please give me your opinion.
Swagatam says
apply 11.1 volts to the circuit and adjust the preset such that the relay just deactivates.
Faridul Islam says
sir
please help me ,what is P1
Swagatam says
P1 is a variable resistor for adjusting the low voltage cut off limit.
Anonymous says
Hello,
thanks for the circuit( low cut only ) now i want to know how to change the circuit for 6 volt battery.thanks in advance
Swagatam says
Nothing needs to be changed, just adjust P1 for a 5.7V cut off
Anonymous says
thanks a lot Mr. majumdar
Premdeepak Antony Peedikayil says
Res Sir,
Thank you for this circuit, Please explain to me how to set this circuit for low voltage protection when I am using a 12v battery 30 Ah battery and using a 30w 1 amp solar panel, what to modify and how to calculate R1 please reply to my Query
Swagatam says
Adjust the preset such that the relay just deactivates at 11.5V.
The circuit will work for your application, please read the article I have explained everything there.
Olayemi. F says
When an inverter is overload can this circuit above cut-off the inverter? Becuase you are said that it work with an inverter. Thank u sir.
Swagatam says
Yes it will work if R1 is selected correctly.
Michael. O says
Yes sir, i need feedback circuit diagram for an inverter to regulate output inverter. Pldase sir help me to send it to this E-mail (Olayemi.michael@yahoo.com)
Swagatam says
OK Michael, I'll try to design it, I'll post it in my blog after a few days because already there are quite a few circuits requested by other readers which are to be updated….
Olayemi. F says
What u are saying is that it will not work with an inverter? Which circuit i can use for overload cut off of an inverter?
Swagatam says
It will work with an inverter also.
Michael. O says
Please sir, i need feedback circuit diagram for an inverter and also sin wave inverter circuit diagram. Please sir, help me to send it to this E-mail (olayemi.michael@yahoo.com) . Thank sir.
Swagatam says
By Feedback circuit do you mean a regulated output inverter circuit??
Olayemi. F says
This circuit, is it work with an inverter for overload cut-off? If i connecte two of 0.1 ohm's resistor in paralle for R1 in order to cut -off the load at exacly 160watt. Is it this two of 0.1 ohm's afect the seting of low battery cut-off. Thank u sir.
Swagatam says
According to me this circuit will work for all DC applications which requires overload sensing and tripping.
R1 will not affect the low voltage cut off setting.
Michael. O says
Why are u afraid about booster circuit. It can not work with inverter or what?
Swagatam says
It's not possible to increase watts of an inverter by using external circuits.
Olayemi. F says
Hello sir, i build this circuit above, i used 10k variable resistor for R3 and also i connect two of 0.1 ohm's in paralle for R1, by the time am turning variable resistor to set low voltage cut-off that 10k was cash fire. What can i do to set low voltage cut-off at 10.5v?
Swagatam says
Hello Olayemi,
I have made some changes in the diagram please do it according to it….
Michael. O says
Yes sir, want booster circuit diagram to boost low watt inverter to 1000watt. Please sir, help me to send it to this E-mail (Olayemi.michael@yahoo.com)
Swagatam says
I am afraid that would be only possible by upgrading the transformer of the existing inverter to 1000 watts and by using a battery capable of supplying 1000 watts of sustained power, no external circuit can work for implementing this.
Michael. O says
Thank u sir, i need 1000watt booster circuit diagram for inverter at 220v ac- 240v ac(olayemi.michael@yahoo.com).
Swagatam says
Do You want to boost a low watt inverter power to 1000 watts??
Michael. O says
Thank u sir, for your response to me. Sir, i need 1ooowatt booster circuit daigram. My E-maill (Olayemi.michael@yahoo.com) and also i build battery charger for lead acid battery, the battery is 100AH,12V, I whant battery to cut-off at 13.6v what can i do ?
Swagatam says
Dear Michael,
booster circuit for what application?
What should be the voltage of it?
You can try the last battery charger circuit given in this article:
https://homemade-circuits.com/2012/07/making-simple-smart-automatic-battery.html
Michael. O says
Thank u for your response to me. Please i need booster circuit. E.g 1000watt to 2000watt (olayemi.michael@yahoo.com)
Swagatam says
Please provide full specifications….
joynal says
Hi . i need low battery procted output with current limet output. Circuit . anyone help me. Email : joynal_307@yahoo.com. send me this emil address.
Michael. O says
Thank u for response to me. How many 0.1ohm's resistor, i will connect in paralle to get 0.007 ohm's. (2) i build 1kva inverter, i add 260 load, the output voltage inverter droup from 250v to 165v. I used SG3524 with 4N35 optocoupler, and also i used this mosft IRFP26N60K each of them is 22A,600V,300Watt, 4+4 in each saide 12v 100AH battery dry cell, what can i do to make output voltage to being permanet at 250v ? (2) i need booster circuit. E.g 1000watt to 2000watt. E-mail (olayemi.michael@yahoo.com) . Thank u.
Swagatam says
Connect 14 of them in parallel.
I am afraid you cannot modify your inverter so simply.
Without studying the fault correctl it won't be easy to rectify the problem.
Yosua says
Hi,
It's not written in the text, but it seams that R4 and C1 are here to create a gap between the cutoff voltage and switch-on voltage (to avoid ON – OFF oscillations), now I want to use the circuit to automatically shut down the supply from a car accessories outlet when the car engine is off and automatically reconnect when the engine is started again. If I adjust R3 for a cutoff voltage at 13 volt, what would be the value of R4 for the circuit to switch on again at 13.8 volt when the alternator is charging? Thank you!
Swagatam says
Hi,
Yes C1 is placed to create a gap before final switch on, however R4 is placed for latching the system.
For your application you won't require this circuit. Simply derive a DC by rectifying the alternator voltage and apply it to a suitable relay coil directly, and wire up the contacts of the relay with the battery and the outlet…..
toludan says
pls i wanna ask if it is possible to make timer that will on for 24hour and off for six hours, to control a charger, thanks so much
Swagatam says
You can try this circuit:
https://homemade-circuits.com/2012/04/how-to-make-incubator-timer-egg.html
toludan says
thanks so much for this circuit, i am a fan of ur circuit. please i have a question. for tripping overload with R1, should it be connected to ac load side or dc side that is battery, and what is the arrow of the overload and low battery output of R1 connected to. thanks
Swagatam says
You are welcome Toludan,
The entire circuit should be connected at the DC side
Michael. O says
I =w/v =1000/12 =83.3A therefore R1 =0.6/83.3 = 0.007 ohm's how can i get this value of resistor. (2) i build 1kva with optocoupler 4N35, i used SG3524 as an oscillator, when i add the load of 220watt the output voltage droup from 240v(ac) to 199v(ac) what can i do to it.
Swagatam says
You may have to use many 0.1 Ohm in parallel for achieving that value, or simply a calculated iron wire piece will do the job.
If the voltage is dropping means either your battery is not sufficiently rated or the the battery and the transformer together are not sufficiently rated.
Michael. O says
Thank you
Vinod KC says
Hi Swagat,
My inverter's cut-off limit is 120w and my input voltage is 10. So the trip current is 12A. As per your formula,0.6/12 = 0.05 Ohm. Where to find such a low value resistor?
Swagatam says
Hi Vinod,
If you are using it in the DC side then it's correct….
Put two 0.1E in pararallel
michael.o says
overload short and low battery protected , if i want inverter to be cut off at exacly 1kva what can i do?
Swagatam says
divide 1kva with input voltage, you'll get the trip current, now solve the following to set the trip at 1000 watts.
R1 = 0.6/Trip Current,
michael.o says
overload short and low battery protected is it work for 1kva inverter
Swagatam says
Yes, just use RL1 with appropriate contact current rating.
happygolucky says
hey.. i am building a 0-15v 1 amp DC power supply and need to add an overload/short-circuit protection circuit. how can i modify your circuit to remove the low battery cutoff feature?? ps- suppose the trip current is 2 amp, R1=0.6/2=0.3ohms… 0.3ohm= how is that possible??
Swagatam says
Please try the following circuit, it's much more effective and accurate:
https://homemade-circuits.com/2012/04/how-to-make-automotive-electronic-fuse.html
By the way 0.3 Ohm is correct, you have to employ a resistor of this value for controlling current above 2 amps.
Anonymous says
HI,
great and simple circuit, thanks for shearing with us.
can you explain how to choose R1. or more detail about how to make it.
regards
Swagatam says
Thankss!
R1 = 0.6 divided by the max specified current limit to the output
Anonymous says
Thanks for your quick respond,
If current will 10A, R1= 0.6/10 = 0.12oms
what about the wattage of R1 is it 100W(w=12/.12) ?
Swagatam says
I think your calculations are absolutely correct…
Michal says
Hi, I'm probably missing something, but I think the calculations are wrong. For maximum current 10A R1 should be 0.06 Ohm (not 0.12). And the maximum voltage on R1 is 0.6V when the current is 10A, so the resistor must be able to handle just 6W (not 100W).
Swagatam says
That's correct, may be i didn't check the previous calculations properly.
Watts = V x I = I.R x I = I^2 x R = 10x10x0.06 = 6 watts (correct)
sandy says
hi swagatham
plz design a simple low voltage cutoff with buzzer..which can be used in between a charger and inverter
Swagatam says
Hi Sandy,
I'll try to do it soon…will inform you if it's done.
Anonymous says
Namaste Swagatam ji,
I am using AC/DC autocharger 12V to charge Lead acid sealed battery 12V. Can I connect the autocharger parallel to the output ? Will it charge and cut off at high voltage said 14.0~15V and if the battery reach 10.6V will the low voltage cut off and prevent over discharge of the battery. What relay and resistor to use? Can I use variable resistor to adjust the high & low cut voltage ? What resistor to use? And how to adjust the desire voltage? Do I need to separate the high voltage cut off circuit portion (to between the auto charger and battery) and low voltage cut off circuit portion (to between the battery and 10W LED lighting output.
Thank you for you help in advance. Dhanyavaad!
savyasachi says
I think this what you said.[img src="i1243.photobucket.com/albums/gg546/vinodvinu2/th_lowbatteryandoverloadbatterycutoffcircuit.png"]
Anonymous says
what will be the value of the preset
Anonymous says
hi Swagatam, i found this so interesting almighty will keep strengthen you. pls i will like you to tell me what to do if what i need from the circuit is just low battery cut-off.
thanks
Walkabout says
I easily get confused, after I drew the schematic diagram. My questions are;
1st:There are 3 connections for R3 preset, the 3rd will not be connected, right?
2nd: Where will the negative polarity of the battery be connected?
3rd: T2 base is connected to the 1K, while the other two terminals go to the positive and the ground respectively. Is the emitter the ground?
Could help me redraw the schematic?
And for just overload cut-off circuit, what would I need in a circuit and what would the schematic look like?
Swagatam says
1) In the above updated diagram R3 is a fixed resistor P1 is the preset, yes, any two terminals of the preset can be selected, the center lead being the mandatory one.
2)the line which joins R1, T1 emiter, C1 negative and D1 cathode is the negative line.
3) please refer to the above diagram, I am not sure which diagram you are referring to?
Walkabout says
Continuing the discussion of just having low voltage cutoff circuit. You mention(I'm quoting you) "Keep only D1, D2, T2 and the relay, eliminate everything else.
However R3 now gets replaced by a preset, the center goes to the base of T2, the other two terminals go to the positive and the ground respectively.
For safety add a 1K resistor with the base of T2 which then can be connected to the center tag of the preset."
ok following your instructions and basing from the updated diagram, lead me to redraw it to fit for a low voltage cutoff circuit imageshack.com/a/img443/5532/6s0f.png.
is that the schematic for a adjustable low voltage cutoff circuit?
Swagatam says
the image link is not opening in my computer.
It should be OK if you have done as per the mentioned instructions.
Walkabout says
Now i tried google drive, if you can view the image. https://drive.google.com/file/d/0BytEbOgq6mqeQU5mdzU4Yl9kUDA/edit?usp=sharing
Swagatam says
R3 preset bottom free end should be connected to the negative line.
the indicated +/- lines are correct. The relay out is positive, the bottom rail is the negative.
PETER says
hi,
am very impress with the way you have been responding to my questions and i say very big thank you.
one more thing sir, in your reply to my questions above you said i should upgrade relay fine but what will be the upgrade for R1 and R2. thank you
PETER says
Thank you very much for your quick response. I have few questions to ask 1. can the over load section of this circuit handle high voltage for instance if the capacity of an inverter is 1kva can the overload protect section protect the inverter from been overloaded beyond 1kva (2) if I want to increase the threshold i.e. want to use it for inverter and want the inverter to shut-down when the battery is drain up to about 10-10.5v, if yes does it require modification if so what are the modification. Thanks a lot