Low Battery Cut-off and Overload Protection Circuit.

A very simple low battery cut-off and overload protection circuit has been explained here.

The figure shows a very simple circuit set up which performs the function of an overload sensor and also as an under voltage detector.
In both the cases the circuit trips the relay for protecting the output under the above conditions.

Transistor T1 is wired as a current sensor, where the resistor R1 forms the current to voltage converter.

The battery voltage has to pass through R1 before reaching the load at the output and therefore the current passing through it is proportionately transformed into voltage across it.

This voltage when crosses the 0.6V mark, triggers T1 into conduction.

The conduction of T1 grounds the base of T2 which gets immediately switched Off. The relay is also consequently switched OFF and so is the load.

T1 thus takes care of the over load and short circuit conditions.

Transistor T2 has been introduced for responding to T1's actions and also for detecting low voltage conditions.

When the battery voltage falls beyond a certain low voltage threshold, the base current of T2 becomes sufficiently low such that it's no longer able to hold the relay into conduction and switches it OFF and also the load.

QUIZ = Explain the introduction of R4 and C1.

Parts List

R1 = 0.6/Trip Current

R2 = 100 Ohms,

R3 =10k

R4 = 100K,

P1 = 10K PRESET

C1 = 100uF/25V

T1, T2 = BC547,

Diodes = 1N4148

Relay = As per the specs of the requirement.

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237 comments

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PETER
May 28, 2012 at 7:37 PM delete

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

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May 29, 2012 at 2:25 PM delete

Hi Peter,

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

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

Regards.

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PETER
May 30, 2012 at 9:29 PM delete

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

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May 31, 2012 at 6:12 PM delete

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

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

Regards.

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Anonymous
June 4, 2012 at 8:33 PM delete

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

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June 5, 2012 at 2:32 PM delete

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.

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Anonymous
June 5, 2012 at 8:30 PM delete

what will be the value of the preset

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July 12, 2012 at 12:02 AM delete

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

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July 14, 2012 at 7:13 PM delete

sorry, I did not understand your point.

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Anonymous
July 19, 2012 at 10:32 PM delete

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!

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July 20, 2012 at 1:13 PM delete

Namaste Ji,

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

You may find them through the search box given above.

Dhanyawaad!!

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July 23, 2012 at 12:02 PM delete

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

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July 23, 2012 at 6:23 PM delete

Hi Sandy,

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

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Anonymous
August 30, 2012 at 10:20 AM delete

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

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August 30, 2012 at 10:35 AM delete

Thankss!

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

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Anonymous
August 30, 2012 at 10:51 AM delete

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) ?

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August 30, 2012 at 3:02 PM delete

I think your calculations are absolutely correct...

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September 28, 2012 at 12:49 PM delete

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??

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September 28, 2012 at 1:36 PM delete

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

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

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michael.o
December 1, 2012 at 4:14 AM delete

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

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michael.o
December 1, 2012 at 4:31 AM delete

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

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December 1, 2012 at 5:43 PM delete

Yes, just use RL1 with appropriate contact current rating.

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December 1, 2012 at 5:45 PM delete

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,

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December 1, 2012 at 6:47 PM delete

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?

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December 1, 2012 at 7:47 PM delete

Hi Vinod,

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

Put two 0.1E in pararallel

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Michael. O
December 3, 2012 at 9:21 PM delete

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.

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toludan
December 4, 2012 at 11:51 AM delete

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

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December 4, 2012 at 1:35 PM delete

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.

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December 4, 2012 at 1:59 PM delete

You are welcome Toludan,

The entire circuit should be connected at the DC side

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toludan
December 5, 2012 at 1:02 AM delete

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

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December 5, 2012 at 11:54 AM delete

You can try this circuit:

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

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Yosua
December 7, 2012 at 9:59 AM delete

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!

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December 7, 2012 at 3:58 PM delete

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.....

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Michael. O
December 7, 2012 at 9:39 PM delete

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.

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December 8, 2012 at 9:26 AM delete

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.

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December 9, 2012 at 9:33 AM delete

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.

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Michael. O
December 12, 2012 at 12:40 AM delete

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

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December 12, 2012 at 9:04 AM delete

Please provide full specifications....

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Michael. O
December 12, 2012 at 3:01 PM delete

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 ?

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December 12, 2012 at 6:20 PM delete

Dear Michael,

booster circuit for what application?

What should be the voltage of it?

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

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

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Michael. O
December 14, 2012 at 2:45 AM delete

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

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December 14, 2012 at 10:01 AM delete

Do You want to boost a low watt inverter power to 1000 watts??

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Michael. O
December 15, 2012 at 12:57 AM delete

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)

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December 15, 2012 at 11:05 AM delete

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.

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Olayemi. F
December 18, 2012 at 1:57 AM delete

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?

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Michael. O
December 18, 2012 at 2:56 AM delete

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

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December 18, 2012 at 9:31 AM delete

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

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December 18, 2012 at 9:57 AM delete

Hello Olayemi,

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

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Olayemi. F
December 20, 2012 at 12:48 AM delete

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.

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December 20, 2012 at 9:24 AM delete

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.

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Michael. O
December 20, 2012 at 1:29 PM delete

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.

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Olayemi. F
December 20, 2012 at 4:21 PM delete

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?

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December 20, 2012 at 8:47 PM delete

It will work with an inverter also.

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December 20, 2012 at 10:04 PM delete

By Feedback circuit do you mean a regulated output inverter circuit??

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Michael. O
December 21, 2012 at 3:11 AM delete

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)

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December 21, 2012 at 3:56 PM delete

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....

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Olayemi. F
December 22, 2012 at 3:50 AM delete

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.

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December 22, 2012 at 2:00 PM delete

Yes it will work if R1 is selected correctly.

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February 1, 2013 at 4:59 PM delete

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

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February 1, 2013 at 9:33 PM delete

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.

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Anonymous
February 5, 2013 at 4:48 PM delete

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

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February 5, 2013 at 8:31 PM delete

Nothing needs to be changed, just adjust P1 for a 5.7V cut off

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Anonymous
February 6, 2013 at 10:51 AM delete

thanks a lot Mr. majumdar

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February 6, 2013 at 11:56 AM delete

P1 is a variable resistor for adjusting the low voltage cut off limit.

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Anonymous
February 9, 2013 at 12:18 AM delete

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.

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Michal
February 9, 2013 at 8:00 AM delete

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).

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February 9, 2013 at 12:50 PM delete

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

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February 9, 2013 at 1:30 PM delete

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)

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February 20, 2013 at 10:44 AM delete

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

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February 20, 2013 at 2:22 PM delete

Hi Prem,

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

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Anonymous
February 25, 2013 at 9:05 PM delete

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

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February 28, 2013 at 2:15 PM delete

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.

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February 28, 2013 at 2:29 PM delete

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

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February 28, 2013 at 2:39 PM delete

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.

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February 28, 2013 at 2:41 PM delete

Dear Vijay,

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

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February 28, 2013 at 2:58 PM delete

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?

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February 28, 2013 at 4:03 PM delete

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

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February 28, 2013 at 6:50 PM delete

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

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

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

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March 2, 2013 at 3:51 PM delete

Dear Sir
Any circuit available for 24
VDC battery low?

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March 2, 2013 at 9:17 PM delete

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

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March 4, 2013 at 5:08 PM delete

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

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March 4, 2013 at 8:46 PM delete

yes sir,

You are the best.
Thanks a lot.

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Anonymous
March 5, 2013 at 8:22 PM delete

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

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March 6, 2013 at 7:09 PM delete

yes you can connect it without problems.

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March 25, 2013 at 2:55 PM delete

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.

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March 25, 2013 at 3:19 PM delete

connect a 47uF/25V capacitor across the relay coil

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March 25, 2013 at 5:47 PM delete

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

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March 25, 2013 at 7:42 PM delete

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

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April 3, 2013 at 10:14 AM delete

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)

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April 3, 2013 at 11:58 AM delete

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.

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April 4, 2013 at 4:19 PM delete

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

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April 4, 2013 at 6:31 PM delete

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.

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April 14, 2013 at 5:47 PM delete

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.

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April 16, 2013 at 4:59 PM delete

what is the load wattage at the output??

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April 16, 2013 at 6:11 PM delete

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

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Ola2
April 18, 2013 at 2:32 AM delete

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

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April 18, 2013 at 3:08 PM delete

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

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June 19, 2013 at 5:52 AM delete

swagatam,
Im trying to implement your circuit into a battery powered project i have to use it only as a low voltage disconnect. I have a smart charger to control over voltage so Im searching for a small simple circuit i can build to put into my projects to stop the batteries discharging to far. Your circuit seems perfect for this.

i have built the following circuit in circuit lab does it look ok for a simple LVD? this is my first circuit design so any help will be appreciated.

https://www.circuitlab.com/circuit/227r5y/screenshot/1024x768/

Also i am wandering if i can implement a set reconnect voltage to stop the circuit reconnecting immediately when the load is removed hence increasing voltage which would continually occur i think this is called hysteresis but its a bit above my level?

Also i believe the bottom outlet is the protected (switched) outlet is this correct if so where does the top outlet go to ground?

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June 19, 2013 at 3:39 PM delete

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.

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June 19, 2013 at 3:42 PM delete

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.

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Anonymous
August 13, 2013 at 11:53 AM delete

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

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August 13, 2013 at 8:14 PM delete

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.

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Allnight
August 19, 2013 at 10:05 AM delete

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.

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August 19, 2013 at 7:49 PM delete

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.

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September 2, 2013 at 6:21 PM delete

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.

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Anonymous
September 10, 2013 at 2:18 PM delete

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.

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September 10, 2013 at 6:51 PM delete

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

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November 26, 2013 at 8:43 PM delete

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?

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November 27, 2013 at 10:22 AM delete

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?

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November 29, 2013 at 11:06 AM delete

Continuing the discussion of just having low voltage cutoff circuit. You mention(I'm quoting you) "Keep only D1, D2, T2 and the relay, eliminate everything else.

However R3 now gets replaced by a preset, the center goes to the base of T2, the other two terminals go to the positive and the ground respectively.

For safety add a 1K resistor with the base of T2 which then can be connected to the center tag of the preset."

ok following your instructions and basing from the updated diagram, lead me to redraw it to fit for a low voltage cutoff circuit http://imageshack.com/a/img443/5532/6s0f.png.

is that the schematic for a adjustable low voltage cutoff circuit?

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November 29, 2013 at 2:05 PM delete

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

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November 29, 2013 at 4:18 PM delete

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

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November 30, 2013 at 10:23 AM delete

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.

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January 17, 2014 at 3:37 PM delete

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

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January 18, 2014 at 11:13 AM delete

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

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February 20, 2014 at 12:42 AM delete

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..

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February 20, 2014 at 11:27 AM delete

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

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March 27, 2014 at 1:28 AM delete

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.

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March 27, 2014 at 5:28 PM delete

Hello Oluwaseyi,

if you are using a battery then the settings should be easier to implement, if the input is through a power supply then it would need to be regulated.

If you find it difficult then it could go for an IC based design.

You can try the last charger circuit for your 100 ah battery from the following link, just replace the shown LM338 with LM396 or LM196

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

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April 1, 2014 at 5:37 PM delete

Thanks for this circuit.

How can I modify this to use solid state relay?

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April 2, 2014 at 11:33 AM delete

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.

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April 16, 2014 at 10:43 AM delete

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

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April 16, 2014 at 1:58 PM delete

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

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May 5, 2014 at 12:15 PM delete

sir i need circuit for cutoff voltage with 2millivolt.

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May 6, 2014 at 7:26 PM delete

use any of the following circuits, replace 741 with LM311

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

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May 8, 2014 at 12:20 PM delete

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 .........

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May 8, 2014 at 6:42 PM delete

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.

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May 9, 2014 at 9:00 PM delete

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.....

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May 13, 2014 at 10:51 AM delete

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

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May 13, 2014 at 8:12 PM delete

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

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May 26, 2014 at 3:17 PM delete

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

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May 27, 2014 at 9:19 AM delete

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

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July 27, 2014 at 6:50 AM delete

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.

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July 27, 2014 at 6:08 PM delete

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

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August 4, 2014 at 9:14 AM delete

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

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August 5, 2014 at 9:12 AM delete

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

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August 23, 2014 at 12:19 PM delete

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

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August 23, 2014 at 9:20 PM delete

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.

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September 24, 2014 at 2:54 AM delete

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?

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September 24, 2014 at 10:29 AM delete

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.

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September 24, 2014 at 7:32 PM delete

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

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September 25, 2014 at 9:54 PM delete

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

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September 26, 2014 at 9:04 PM delete

ok that's good

but 'm facing a problem with relay connections

this is the diagram :
http://imagizer.imageshack.us/v2/280x200q90/537/zaHcTY.jpg

with the meter I've found the points (2,1) in the place of (3,5) and (3,5) in the place of (2,1) I don't know how or why

because I've found connection between the points 4 with 2 not 4 with 5 !

how can treat this please ?

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September 27, 2014 at 11:40 AM delete

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

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September 27, 2014 at 10:59 PM delete

this is a video I've recorded :
http://youtu.be/bi4bgS_6YEU

in the website of the store I've bought from , there is the diagram of points connections the same I told you
http://www.matni.com/Arabic/Relay/RELAY.htm

the relay model is T-73 and in the website I think it's the same JQC-3F(T-73)

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September 28, 2014 at 7:28 PM delete

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

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

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October 1, 2014 at 4:22 AM delete

well , I've made the circuit and put a red led to detect the passing of the current , and R1 = 3 resistors 2watt : 2x(1 ohm) and 2.7 ohm
I've tested it with 9v and 12v battery
when I change the preset the circuit doesn't deactivate !
although I've connected each point of the relay to the true point
can you help me please ?
thanks

http://youtu.be/AshLT5g010M

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October 1, 2014 at 11:35 AM delete

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.

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October 9, 2014 at 2:00 PM delete

Hello Sir,

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

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October 9, 2014 at 3:22 PM delete

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.

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October 9, 2014 at 8:02 PM delete

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.

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October 10, 2014 at 6:27 AM delete

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).

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October 10, 2014 at 11:51 AM delete

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

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October 13, 2014 at 8:23 AM delete

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)

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October 13, 2014 at 10:42 AM delete

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.

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October 13, 2014 at 2:06 PM delete

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?

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October 13, 2014 at 5:47 PM delete

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

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October 14, 2014 at 6:44 AM delete

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?

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October 14, 2014 at 1:57 PM delete

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..

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October 14, 2014 at 3:23 PM delete

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.

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October 15, 2014 at 2:39 PM delete

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

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October 15, 2014 at 8:29 PM delete

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.

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October 16, 2014 at 2:45 PM delete

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!

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October 16, 2014 at 9:21 PM delete

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

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

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October 17, 2014 at 3:14 PM delete

Hello Sir,
Does R1 relate this problem?

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October 18, 2014 at 10:06 AM delete

No, R1 has no connection with this problem.

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October 22, 2014 at 9:04 AM delete

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

Thank you for your support!

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October 23, 2014 at 7:53 AM delete

Hello Sir,

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

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October 23, 2014 at 9:26 AM delete

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

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October 23, 2014 at 12:47 PM delete

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

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

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October 23, 2014 at 3:13 PM delete

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.

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October 23, 2014 at 7:41 PM delete

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

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October 24, 2014 at 12:48 PM delete

Hello Sir,
Where do load and battery connect?

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October 31, 2014 at 6:58 AM delete

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.

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November 3, 2014 at 9:41 AM delete

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.

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November 3, 2014 at 2:16 PM delete

hello Huyn, if you have selected 13V as the triggering threshold then you must adjust the preset until the relay gets activated at 13V.....

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November 5, 2014 at 7:28 AM delete

Hello Sir,
Intermediate relay only endure current 10A, how do we use for current 30A?
Thank you very much.

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November 5, 2014 at 8:19 PM delete

hello Huyn, what's intermediate relay? relays are available in all shapes, sizes and ratings, 30amp are also available.

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November 25, 2014 at 12:39 AM delete

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...

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November 25, 2014 at 2:43 PM delete

Hello Yel, you'll have to use the entire circuit as shown even if you want only the overload to work....

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December 14, 2014 at 7:27 PM delete

what is the maximum load it can handle

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December 15, 2014 at 3:11 PM delete

will depend on the relay contact.....can be upgraded to any limits.

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April 10, 2015 at 5:15 PM delete

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)

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April 10, 2015 at 9:06 PM delete

Hi shankar, please provide the info in more detail regarding the application so that I can understand and design it better.

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Anonymous
April 15, 2015 at 10:33 AM delete

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

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Anonymous
April 15, 2015 at 10:39 AM delete

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

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