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.
The"LOAD" terminals in the above diagram is supposed to be connected with the inverter +/- supply terminals.
This implies that the battery current from the right side has to pass through R1 before reaching the inverter, enabling the sensing circuit around R1 to sense a possible over current or overload situation.
CORRECTION:
The above shown circuit will not initiate unless the relay is actuated manually through a push switch as shown below:
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.
Inverter Overload Cut-OFF using Opamp
In the above paragraphs I have explained a very simple concept of inverter overload cut-off using only transistors.
However a cut off system using only transistors cannot be very accurate and sharp.
In order to get a precision inverter overload and short circuit cut off circuit the use of an opamp based design becomes imperative.
The following diagram shows a simple battery overload controller circuit using a single opamp 741 and a relay driver stage.
How it Works
The opamp is configured as a simple comparator circuit. he inverting input of the opamp is clamped at a fixed 0.6 V using a 1N4148 diode.
The non-inverting input of the op amp is connected with the negative line of the circuit through a over-current sensor resistor Rx.
Due to inverter overload or short circuit or over current conditions, a voltage drop develops across the resistor Rx which can exceed the 0.6V as per the calculated value of the RX, and cause the non-inverting input of the opamp potential to go higher then its inverter 0.6V potential.
This causes the op amp output to turn high activating the transistors and tripping the relay.
When power is first switched ON, and assuming the inverter is working normally without an overload, the voltage developed across RX is minimal, which keeps the pin3 potential of the opamp the opamp lower than the pin2 potential.
This allows the output of the opamp to be low ensuring that the transistor is switched OFF, and relay contacts stays at the N/C point.
Due to this the 12V is able to reach the inverter and operate it normally.
However, as soon as an overload or over current happens at the inverter side, a large amount of current passes through the RX resistor, causing a voltage drop to develop across pin3 of the IC.
When this voltage drop exceeds the 0.6V reference level of the pin2 of the IC, the output of the op amp goes high, causing the transistor to switch ON and trigger the relay.
The relay contacts now shift from N/C to N/O switching of power to the inverter and thereby averting the short circuit or overload conditions.
The N/O contact can be seen attached with the base of the relay driver transistor, which ensures that as soon as the an overload is detected the relay contact quickly latches the transistor, switching the power permanently off for the inverter.
The power can be restored only by disconnecting the 12 V battery input, but before that it must be ensured that the short circuit or the over load condition is appropriately removed from the inverter side.
What do u mean when the voltage cross 0.6v
it is the triggering base voltage for T1
Hello sir I have challenge on how to build h_bridg inverter
What should i do for this circuit to cut off at 8volt when using 12v 70amp-hour battery?
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?
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.
pl tell me where i purchases lm196 or 78h12a
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
Hi vhafuwi,
you can try the following circuit
https://www.homemade-circuits.com/2015/03/mains-over-load-protector-circuit-for.html
install one such unit in every room after the mains DPDT switch
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.
adelusi, divide 1500 with the battery voltage, and then divide 0.6 with this result.
Hi Robert, it's possible only through P1, that is through the end terminals of the preset
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/
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
Hi Kamran, you can make and use the following circuit and apply it in your inverter for the results:
https://www.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.
Hi Swagman.
The P1 preset in your low voltage cutout says (10 preset) how many ohms is this component,,
regards Rob.
Hi Rob, it's 10k, I have corrected the typo
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
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
Sir, can i use it as a solar charge controller for LVD circuit?
yes it can be used
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
Shuddhatam, I have already answered to the same question in the other article….please check it
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.
Angelous, I'll check and try to find one…
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
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.
Hi Swagatam,
is this circuit suitable to use with https://www.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 @
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://www.homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html
this will take care of the all the required safety conditions.
How is the protection circuit connected to the inverter circuit?
the "LOAD" is the inverter supply input….
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.
yes that's correct
10k ohms is the right value
…ok got it, you meant to say the lower cut off voltage, yes that's right, 18V is correct.
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)
Hi shankar, please provide the info in more detail regarding the application so that I can understand and design it better.
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…
Hello Yel, you'll have to use the entire circuit as shown even if you want only the overload to work….
what is the maximum load it can handle
will depend on the relay contact…..can be upgraded to any limits.
Hello Sir,
Intermediate relay only endure current 10A, how do we use for current 30A?
Thank you very much.
hello Huyn, what's intermediate relay? relays are available in all shapes, sizes and ratings, 30amp are also available.
Hello Sir,
Where do load and battery connect?
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.
OK that's great!
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.
hello Huyn, if you have selected 13V as the triggering threshold then you must adjust the preset until the relay gets activated at 13V…..
I'm sorry sir
I can not find any error. T2, T1, relay … are ok. I don't know why.
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.
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!
Hello Huynh, try the following basic relay driver stage first, and check the response:
https://www.homemade-circuits.com/2012/01/how-to-make-relay-driver-stage-in.html
Hello Sir,
Does R1 relate this problem?
No, R1 has no connection with this problem.
Hello Sir,
Does N/C pin of relay connect to R4 and NO pin connect to R3?
Thank you for your support!
hello Hunnh, yes that's correct.
Hello Sir,
If I only use low bettery cut-off function, What component's I remove?
Thank you very much.
Hello Sir,
If I choice C1 = 100uF/50V, is it ok?
Thank you for your support.
yes C1 = 100/50 will also do.
you can try the following design which is without overload cut off
https://www.homemade-circuits.com/2011/12/simplest-smf-automotive-battery-charger.html
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.
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
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.
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.
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).
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
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)
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.
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?
hello Huynh, short the collector/emitter of T2 manually, if the relay clicks would indicate a good relay but a faulty transistor…
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?
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..
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.
OK!
Hello Sir,
I assembled this circuit. However, there is not the voltage in the load with any level battery voltage (11V – 13.5V).
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
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.
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?
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.
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
No that won't happen because of the transistors hysteresis properties…
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 ?
If you can show me the relay image I can try to help, without seeing the relay image it would be difficult to locate.
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
http://www.matni.com/Arabic/Relay/RELAY.htm
the relay model is T-73 and in the website I think it's the same JQC-3F(T-73)
the connections for this relay have been elaborately explained in this article, please check it out:
https://www.homemade-circuits.com/2012/01/how-to-understand-and-use-relay-in.html
How to make resistors of such low value 6miliohms etc here I am stuck do u have any idea pls let me know
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.
Sir ,
I didn't understand the preset part correctly can u explain its working in the given circuit
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
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.
Yes it will work with 3kva inverter, just modify the relay contact accordingly and use 8050 for T2.
hello sir, can this kind of cct can be modified to trip low standby current? thank you 🙂
hello shahirah,yes it can be done by setting the preset appropriately.
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 ………
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.
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…..
OK, no problem, that's great!
sir i need circuit for cutoff voltage with 2millivolt.
use any of the following circuits, replace 741 with LM311
https://www.homemade-circuits.com/2011/12/how-to-make-simple-low-battery-voltage.html
will that circuit need any other changes for cutoff voltage as 2 millivolt.
no changes would be required…it will respond even to the minutest changes between its inputs.
Thanks for this circuit.
How can I modify this to use solid state relay?
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.
Sir . Will you low battery cut off circuit, turn on the inverter when the battery is recharge?
it may be done by incorporating another relay parallel to the existing one and by wiring its contacts appropriately for the intended chageover
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.
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://www.homemade-circuits.com/2012/07/making-simple-smart-automatic-battery.html
can you share your overload design circuit
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..
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
Sir,
Can I use this circuit with sg3525 without relay?I am thinking that I can use a PNP transistor , am I right?
The above circuit is suitable only with a relay, for 3524 IC only an opamp circuit would be suitable
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.
Now i tried google drive, if you can view the image. https://drive.google.com/file/d/0BytEbOgq6mqeQU5mdzU4Yl9kUDA/edit?usp=sharing
the image link is not opening in my computer.
It should be OK if you have done as per the mentioned instructions.
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?
Hello sir, I understand everything else about this circuit except for (R1) please i do not know what am suppose to use, if it,s resistor, fuse or breaker as (0.6/Trip Current) because what i understand is that the negative taminal of the better is suppose go to the inverter negative source which means using a resistor is not going to be a good idea, please i need your help because i have already finished building this circuit with a (0.22 ohm) resistor in place of (R1) but i need to know exactly am suppose to use before testing it because am sure. Thanks
R1 is a resistor, its wattage should be = 0.6 x maximum trip current value.
You are supposed to do exactly what is shown in the diagram, the negatives have to connect through the resistor R1.
Here’s another similar design you can try which is thoroughly tested by me:
https://www.homemade-circuits.com/how-to-make-automotive-electronic-fuse/
OK, thanks 4 ur quick respond, i really do appreciate.
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?
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?
Make R3 = 22k, and R2 = 1k, that's all, other components will remain as is except the relay which should be 24V rated.
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.
you are right, thanks!
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.
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.
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.