The proposed circuit was requested by one of the avid readers of my blog. It is a low battery warning indicator circuit using op amp IC 741 and can be used for monitoring a particular low battery voltage threshold.
Circuit Operation
The circuit may be understood with the following points:
The entire configuration is wired around the IC 741 and it becomes the heart of the circuit.
Basically it is configured as a comparator with one of its inputs clamped to a fixed reference level while the other input used as the sensing terminal.
Here as can be seen in the diagram, the non inverting input is provided with a fixed reference voltage through a resistor zener network.
This input is fixed to about 5 volts.
The other inverting input pin #2 is wired via a preset to sense the input supply voltage from the source.
The preset is adjusted such that the voltage level at this input becomes lower than the fixed reference voltage at the other pin of the IC as soon as the source voltage becomes lower than the desired threshold level.
When this happens the output of the IC immediately becomes high, illuminating the connected LED.
The illuminated LED immediately provides the indication of a low voltage situation so that the required actions may be initiated.
Optionally, the output may be replaced by a piezo buzzer instead of the LED for getting an audible response of the above situation, eliminating the headache of monitoring the LED condition every now and then.
The above circuit can be modified by adding a relay stage for a latching facility, so that the relay latches and stops oscillations at the set low battery thresholds.
How to Setup this Low Battery Indicator Circuit
The above low battery indicator circuit can be even further improved in the following manner for controlling both lower and the upper charging thresholds:
Initially keep the 100K preset link disconnected.
Apply a 14.4V source from the "Battery" side and adjust the 10K preset such that the upper relay just activates, confirm the triggering by subsequently moving the preset to-and-fro.
Glue it once fixed.
The LED will respond by switching ON to the fixing of this preset.
Now reconnect the 100K preset feedback link, and reduce the input supply to about 11.2V. Make sure it has a high resistance initially.
Next, slowly adjust the 100K preset such that the relay just deactivates (at the 11.2V).
Confirm by flipping the preset as above. Ignore the lower relay as it will switch ON as soon as the input supply is switched ON, so its operation is obvious.
That's it, the low battery warning circuit is all set now and will accurately respond to the above settings or any different setting that may be preferred and implemented by the particular user.
IC 741 Low Battery Indicator Circuit with Relay Cut-off
The following circuit shows how the above low battery indicator can be enhanced with relays for achieving an automatic low charge and full charge cut for the connected battery, and also a cut off for the load during a low battery situation.
The upper relay becomes responsible for cutting off the battery during over charge and low discharge level, while the lower relay cuts of the load as soon as the battery reaches the unsafe low discharge zone and as soon as the upper relay reverts to charging mode
Questions & Answers
connect LEDs at the outputs of A1 and A2 and ground with 1k separate resistors.
Now apply 170V and adjust P2 such that A2 LED just lights up.
Next apply 245V and adjust P1 such tha A1 LED just lights up.
Your circuit is set now, the relay connected at the collector of the transistor will also respond correspondingly.
Swagatam, sir,
I think I've cracked it, Instead of the 10k preset, I used a voltage divider: a 12k in series with a 27K. From between these two, I was able to obtain a voltage of 3.86V at Pin 2 – Pin 3 had 4.49 VDC – and, which triggered the transistor via Pin 6 and tripped the relay, and the lights, connected to the normally open contacts, cut out.
Now to assemble all three circuits.
Stephen Gard
OK Stephan, that's great, no issues, if you have solved it you can go ahead with it…..
(Reposted without links)
Swagatam,
Thank your for your well-conducted and informative site. I have learned much from it. I am trying to adapt your 741 Low Battery Cut-Off circuit for use with a 6VDC lighting set-up, and I am having some difficulty. I hope you can advise me. Here is what I wish to do.
1. I have four high-power white LED spotlights. They each require 3.2VDC @ 34mA.The four lights, wired in parallel, draw 82mA at 3.2 VDC. So that is the load.
2. The supply is a single 6.5VDC SLA battery. I attach a 33 Ohm 1 Watt resistor in series with the negative supply to the 4 x LEDS, as a current-limiter. I wish to use ONE battery only for the entire circuit, no separate supplies. It will be installed in a friend's garden and not get much maintanence! So it must be fully 'automatic'.
3. I have a solar-panel (5W 17 volt) that charges the battery and a darkness detector. Now I wish place your 741 Low Charge Cut-Off circuit (with relay) in between these two circuits, to prevent the 6.VDC SLA from discharging too far. The solar charger supplies only a small amount of current, and there will be no-sun days, of course.
If I understand the operation of your circuit, when the battery voltage drops, the 741, via Pin 6, allows current to flow through Q1 (the BC547), tripping the relay coil. My 4 x LED load is wired to the NORMALLY-CLOSED contacts of the relay (NEC EA2-4.5VDC), and when the coil is energised via Q2, they disengage the load by switching over to the NORMALLY-OPEN contact. So no current flows, and the 4 x LEDS go off.
That is what I wish to do, but I cannot make your circuit do this. The only change I made to your circuit was to the 10K pre-set pot. My pre-set over-heated. I measured the R, and replaced it with a 12k ohm 1 watt fixed resistor
b. This arrangement delivers 5.65V to pin 2 of the 741. Pin 3 is set at 4.52 V, from the junction between the 4.7V Zener and the 1K resistor connected to the +ve terminal of the battery.
c. To test, I connected one side of a diferent white LED load to the NC pins of the relay, and the other side to the battery ground. This load draws 2.8VDC, at 22mA.
d. I connect a tired old dry-cell battery to this circuit, and on my DMM, I watch the supply voltage drop.
4V99. 4V96. If I insert the DMM at Pin 2 and try to take a reading, the voltage goes down slowly.
e. So I connect a 2MegOhm pot in between the 12K and Pin 2, and bring the voltage down.
But now I notice that the voltage on Pin 3 is droppng as well. For example, from 4.351 to 4.348V.
It seems that the two pins will never achieve the diffential that will trip the relay.
When I connected the fixed resistor, I did not take it to ground, as the original circuit did, after the 10K preset. If I connect this fixed resistor to ground, Pin 2 goes low and the relay trips, but I don't think that is correct.
Can you please advise what I am doing wrong here?
Sorry for such a long letter!
Stephen Gard
Swagatam,
I am still testing to get exact values for my requirements. I am also drawing up the whole device in TinyCAD, and can send you a png of the completed three-part circuit, with test-points and voltages marked, if you wish.
Stephen Gard
Yes, why not Stephan, I'll be extremely happy to publish your efforts in my blog so that other can also benefit from it.
I appreciate your positive thoughts…kindly do send them.
And whenever you do something new in electronics, please make sure you share them here for us.
We all will love to see them.
Let me know if you need any further assistance?
Thanks very much.
Swagatam,
I added this circuit to my solar lighting set-up, but there is a problem. The relay must be activated to cut off the current to the LEDs. But this drains the battery even further. Finally, the battery is drained so much that it cannot activate the relay, which then returns to the NC condition, powering the LEDs, and draining the battery even further. After one dull day, when the solar panel could not recharge the battery, the lights do not work at all, because the battery was drained so low. Perhaps a timer would be better? I am running the whole set-up from just one 6VDC SLA battery, and don't want to add any other power sources. What do you suggest?
Stephen Gard
Stephen, the solution is to simply swap the pins of the IC, meaning connect the preset pin to pin#3 and the zener to pin#2.
In the above situation the load must be connected via the N/O contacts, so the LEDs stay illuminated as long as the relay is activated (voltage normal condition), as soon as the voltage drops below the lower threshold the relay and the LEDs all shut off without causing any further discharging issues.
Thank you, Swagatam, I carried out this mod, and now it all seems to working as I require, The set-up is currently undergoing a long-term (1 week or so) outdoor test, and after this, I will construct the final (weather sealed) installation and send you a schematic and some photos, as promised. I note that the circuits (there are three) still draw 12mA, even when the LED lights are off, but perhaps the solar-panel will keep ahead of this. The weather for the next week includes sunny and cloudy periods, so my solar-charged set-up should have a good test period,
Stephen Gard
That's great Stephen! I will certainly love to see the pics of the finished design, and also post it here for the viewers.
The quiescent consumption should not be more than 5mA according to me, i think something may be leaking through the output of the opamp, a 3V zener diode in series with the transistor could possibly fix this issue too.
Swagatam,
I got this all installed and working, but by that time, I had learned a lot more about solar lighting, and realised that my approach to this project was wrong: all I needed was a Joule Thief! However, the experience was very helpful, and your low-voltage cut-off circuit works fine. I plan to use it for a 6VDC solar-powered water pump… wilol report back when that's done.
Thanks for your help and advice,
Stephen Gard
That's great Stephen, I congratulate you on that. Best wishes to you.
hai swagatam!!! I had connected this as your circuit.. but when the circuit cut off the output..the battery voltage increases then the relay deactivates and the output works and it again activates and cut the output …this happens simultaneously ..how to cure it..
Hi Dharma,
Please connect a feed back resistor link from pin6 to pin3 as shown in the last diagram, this will prevent the circuit from oscillating at the threshold points.
HI
Can you please explain what is the purpose of using a bipolar transistor in the 3rd circuit?
why are we using it?
for driving the relay.
hey
can i used 3rd circuit diagram for 12 volt 7.2 to 45 AH battery? And in this circuit there is two battery supply for one is relay operate & another is for load?
you can use it…there's only one battery supply that connects with the load
Hello sir,
Plz I need clarification on ur last circuit dat has two relay. I built d circuit for my 12v inverter as charge control but d problem is i got d first conrol using 10k VR for 14.5v auto cuttoff but.for low batteery sensing I could not get what I want to deactivate at 10.5v. All I got was it deactivate at above 11v and even that I have to tuned d 100k VR toward zero ohm so what do i do to achieve low battary dectation at exactly 10.5v instead of above 11v
Hello Beelal,
try a lower value pot in place of 100k, may be a 33k pot and reduce the series 10k fixed resistor with 1k
Hello,
i see 2nd diagram 10 k preset connect @ 2 pin & in 3rd diagram 10K preset connect @ 3pin of IC. Another question is that I have UA 741C IC can i use it or there is difference IN UA 741 & LM741 please guide as i used it for 12 volt 7.2 AH battery.
Connect the pinouts as shown in the third diagram.
all types of 741 will work for this circuit
Hi Swagatam,
You have recommended this third version for my hi/low contacter.
You suggested that I don't require the second relay.
Which one do you mean and what was it's original function?
Thanks for this
Yours
Carl
Hi Carl,
The lower relay is not required for your application.
The circuit is a battery over charge/discharge controller.
Hi swagatam,
For the third circuit, I would like to know the exact values of the potentiometers, because i am only going to put resisters instead of potentiometers.my circuit will be using 12v input, 12v battery and the load current will be 3a.
Hi Anthony, the simplest way of implementing it is to first set the thresholds with pots and then measure the pot resistances to get the exact values of the fixed resistances.
The pot can be removed thereafter and replaced with the measured equivalent resistances.
can i use this circuit for 24v gel battery with 180 ah??
yes can be used..use a 324 IC for the opamp.
hai sir,
I made this circuit it is working perfect with sugar cube relay but it cant pull the coil of a 6 amp 200 ohm 12v relay,I am making a 800 watt relay,since the current has to pass through the relay I have to use a high amp rating relay,how to solve it.
Hi Vijeesh,
according to me a BC547 transistor should be able to pull a 200 ohm relay easily, try reducing the base resistor to 4k7 and check the response, or you can also try a 8050 or a 2N2222 in place of the BC547 for the same….
Swagatam Majumdar …sr….
My ups is 40W ..
Battery is fully charged is there ..
But my ups is showing battery low why sr…
sharanabsava, check the battery voltage while the inverter is operating and without any load….if the battery volatge drops then that would indicate an incorrectly charged battery or a faulty battery…also check the current with an ammeter in series with the batt positive….if it indicates high consumption without any load then the fault could be in the inverter, not the battery
hello
i used the above circuit to shutdown my inverter when battery voltage is 9.5v but inverter goes on and off severally at the set voltage.
that can never happen….use the second circuit, remove the 100k resistor and adjust the 10k preset such that the relay just cuts off at 9.5V and activates at around 9.7V….once this is done the circuit will implement the same accurately each time the battery voltage falls to this level.
Dear Sir
Thanks for your prompt reply to my other query and referring this circuit for my purpose. Kindly help me with the following queries regarding this circuit
1. To use a relay can the LED and its resistance (as in first diagram) be simply replaced by the relay OR is it mandatory to do the other changes of 2nd diagram like addition of transistor.
If I want to use <12v as the voltage to activate the relay will the required zener diode be of 12v OR as shown in diag 4v is needed.
3. What will be time required by the relay to act. (will the relay be immediately energized as soon as voltage falls below set 12v Or will some time be required and Vice Versa when the voltage again resumed to normal).
Thanks
Dear Ankit,
the IC will not be able to handle a relay directly, therefore a transistor stage will be required as shown in the second diagram.
The zener diodes should have a much lesser value than that of the supply voltage level, so for a 12V supply any value between 3v and 6V would be OK for the zener diodes.
the relay will act almost immediately as soon as the voltage crosses the set threshold limit.
….if the 100K feedback resistor is used then the restoration will not be immediate rather as per the 100K response…lower values will produce relatively quicker restoration….as the resistor value is increased the restoration point will move backwards (will be delayed proportionately)
Dear Sir
Kindly advise if 4.7V zener diode is not available then can 3.3v or 7.5v zener diode be used.
Dear Ankit, yes any zener below 8V will be fine…
Dear Sir, I made the circuit exactly according to second schematic and it is working fine. Just one query, In this Initially at normal voltage the relay is activated and when the voltage drops below the low level then the relay gets deactivated. CAN this function the OTHER way ie at normal voltage the relay remains deactivated but when voltage drops it activates.
Further what change may be required if low voltage cut is to be set at around 12V (currently it is 5V probably)
Thanks
Dear Ankit, what you are asking is not recommended because if you allow the opposite to happen that would mean at the lower critical level the relay coil will be energized and this would lead to even further consumption of the battery which will in turn discharge the battery even faster…….. this won't be a favorable situation for the circuit and the battery.
by the way you can do it by simply changing the NPN transistor with a PNP…and connect the relay diode stage across its collector and ground.
Thanks Sir. As per your suggestion of using PNP, will the ground be emitter or the base.
emitter will go to the positive….
Dear Sir, I did all the changes with pnp BC 557. But now the circuit is not working. Kindly advise
Dear Ankit, you might have done something incorrect with the connections for sure…otherwise there's no reason why it wouldn't respond….relay should be connected between collector and ground….emitter to positive supply….base to opamp resistor/LED……relay diode cathode must go to transistor collector…anode to ground
the LED polarity needs to be reversed.
Dear Sir, Thanks for the reply. After reversing the LED polarity, the modified circuit is working fine. As a final piece of advise regarding this circuit, kindly let me know the following-
1. What is the appox consumption of the circuit (on 12V battery) (when the relay is in deactivated stage), and
2. Is it OK to keep the circuit connected 24X7 or should it be switched on only intermittently (ie will continuous operation reduce the life of circuit)
Thanks
Dear Ankit,
increase the value of the perset to 100k, and same for the zener resistor, this will allow just a few mA current to be consumed by the circuit (only as long as the relay is not active)
the circuit can be kept powered ON infinitely without any harm
Dear Sir, The circuit is working as intended (using PNP so that relay is activated when voltage is low). However on borderline voltage, the relay chatters. Can something be done to avoid this eg adding delay with IC 555 or some other way. Thanks
Dear Ankit, just connect a 100uF/25V cap in parallel with the relay coil, this will take care of the issue
Dear Sir, I connected 100uF/25V cap in parallel with relay. Though this works fine when I test it at my work bench with a dc adapter, but as soon as I use the circuit in car (12V 70 Ah battery), the relay again starts chattering at borderline voltage.
I maintained the polarity of caps (+ to + of relay)
The slight modification that has been done to your original circuit (in addition to using PNP) is following
1. I am using 2 relays in parallel (as I need 2 switches and double pole relay not available) (diode put across both relays)
2. The 100uF 25 V caps have been connected to each of the relays separately
3. I have connected a buzzer in parallel to the LED (buzzer spec 3V to 30 V)
Kindly advise to prevent relay chattering
Thanks