The following article explaining a dual battery changeover relay circuit was requested by Mr.Raja so that it could become possible to switch between his old and new inverter batteries automatically, eliminating manual interventions. Let's read it in details.
Technical Specifications
"Dear Swagatam sir,
Sir, i bought a new 12v 110ah lead acid battery for the dc home lighting system.
I had another 12v 110ah battery which is about 8year old. ( which is connected earlier in the same lighting system in my house itself) . But the old one has approximately 25ah capacity as i calculated.
But it is not sufficient to glow light for about 5 hours of night.(i.e. From 6pm to 11pm) So i want to use old and new Battery. But i can't join them in parallel, as old one may takes charge from new battery, which decrease the life of new one (as i think)
Therefore at present i am using a'Two-way' switch to switch on and off between old and new battery. Whenever lighting system controller shows red light, i.e. At about 11.5v, i manually operate the two way switch to switch on the new battery.
Now , please give me the circuit to switch on and off between two batteries in such a way that, initially lighting system operates with old bty and as the voltage of old bty decreases (below 11.5v) then only switch on to new one.. Thanking you"
The Circuit Design
The designed idea of the proposed dual battery changeover circuit or rather old to new battery changeover circuit may be understood with the following points:
Referring to the circuit diagram we see a IC 4017 which functions as a sequence toggler or switcher.
The IC will shift its output from pin#3 to pin#2 and then to pin#4 in response to every positive pulse at its input pin#14.
pin#4 of the IC has been connected to the reset pin#15 of the IC, means the moment the logic sequence reaches pin#4, the sequence gets reset back to pin#3 so that the cycle can repeat.
Here the input pulse is derived from the low battery warning indicator from the existing inverter system.
The LED light from the low battery warning has been integrated to an LDR using a light proof tubing.
Initially when the power is switched ON, the 1uF capacitor resets the IC so that the logic sequence initiates from pin#3.
The relay at this point remains connected with its N/C pole connecting the old battery positive with the inverter.
The inverter starts operating, draining the old battery
On reaching the low battery threshold, the inverter low battery LED illuminates which instantly lowers the enclosed LDR resistance supplying a positive pulse to the pin#14 of the IC.
The IC responds shifting the logic sequence from pin#3 to pin#2.
Since pin#2 of the IC is connected to the relay driver transistor, the relay immediately activates, switching the new battery into action through its N/O contacts.
The new battery being fully charged toggles OFF the low battery indicator light, the IC goes into a standby position holding the situation intact.... until, the new battery also reaches a low-battery condition switching ON the LED and resetting the IC into its initial position.
The cycle then repeats producing the required automatic dual battery changeover actions.
washington says
Dear Swagatam,
I enjoy and follow your well explained circuits which opens my brainstorming data bits.
I have 2x24v dc battery banks which I use to alternate my hybrid inverter supply when I experience prolonged power outages. I manually changeover when a low alarm sounds (24.5vdc) to either bank depending on the last bank discharged. This manual changeover affords me a bump-less power transfer thus you do not see any dips in power. The low battery gets charged immediately .I need help to design a circuit using dc/dc SSRs (SSR-40DD FOTEK) for the changeover switch achieving the same or better and avoiding damage to the SSRs. The max load will be less than 30amps.
Best Regards
Washington David Masunga
Harare . Zimbabwe
Swag says
Thanks Washington, appreciate your thoughts,
Here’s one SSR concept which perhaps you can try for your mentioned application
https://homemade-circuits.com/12v-dc-solid-state-relay-ssr-100-amps/
let me know if it seems suitable..
washington says
Thanks Swag
I greatly appreciate your quick response.I shall try the circuit during the week and send you feedback.
Regards
Washington
Swag says
Thanks Washington, please keep up the good work…
Unknown says
Sir what is it's applications..
sadique temitayo says
Pls i tried this circuit out and notin happens my 4017 was jss static on output of pin3 pls i nid reply quick
Swagatam says
the LED needs to illuminate sharply or suddenly, if it illuminates slowly then the circuit might fail to respond
Sham says
Hi Swagatam,
I tried the ckt with 7 amps relay. But as the load increased, inverter shutdown. Looks like 7 amps was not sufficient to power the inverter. It was before reading your reply.
My UPS, iam using it as inverter since iam not charging battery through Mains, is 600VA capacity.
1. So to power this system my relay should be 600/12 = 50 or 60 amps right?
2. Or as you said i have a battery of 125 ah, is it 125 ah relay? if yes, Then i think it will be expensive switching..
Swagatam says
Hi Sham,
125 AH would be the maximum safe limit, which would ensure that even if the battery was discharged at the full AH rate, the relay would still hopefully not burn or fuse.
However if you intend to use the battery at a normal 30 to 40amp rate in that case a 60 amp relay would be fine.
Sham says
Hi Swagatam,
Nice article. I am having the exact situation as explained above by Mr. Raja. I am trying this idea using MC. When voltage drops to 11.0v on load, i want to switch to new battery. my question is 1. What current rating of relay do i need to use? is it the sugar cube relay[ 7amp] or more?
2. Do i need to connect the -ve of both relays together?
3. Instead of relay can i use mosfet?( for efficiency and more battery time).
4. if yes, which one? I am planning to connect the MC and switching ckt directly to battery terminal for powering the system, which may drain the battery eventually over a period of time.
5. Like wise, i think i can use the same logic to charge the old battery first and next new one. So at what level should i stop charging the old battery before switching over to new one?
6. Or how long should i wait (after reaching battery voltage 13.6v) before switching to next battery?
Kindly help me in this. Awaiting your replay.
Regards,
Sham.
Swagatam says
Thank you Sham,
Are your questions with reference to the above circuit? because I cannot see two relays in the diagram.
Relay contact rating (amps) will need to be selected as per the battery AH rating, it can be same as the AH value of the battery.
The battery charging can be stopped and toggled once it reaches the 14.3v mark.
Swagatam says
according to me the relay option is better and much safer than mosfets
Anonymous says
Sir,
would you pls kindly let me know if any free simulator software on web ?
Regards
Maung Pru
Email : mpru69@yahoo.com
Swagatam says
You can try this one:
http://www.falstad.com/circuit/
Anonymous says
Hi Swagatam,
i am now testing the circuit… i do not know what is wrong co'z it is not working.
the numbers? is it pin or output?
the capacitor? is it electrolitic or ceramic?
Regards,
Swagatam says
The outputs are the pin numbers. Capacitor is not critical can be any type but polarity must be observed if it's an electrolytic.
Initially do not connect the LDR, generate pulse at pin#14 manually by touching the positive to pin14…. the output should sequence from 3 to 2 then back to 3.
Make sure the 1M resistor is connected to pin#14.
Swagatam says
Pin#13, pin#8 should be connected to the negative of the supply….
Anonymous says
can i test this to a simulator?
Swagatam says
yes you can do it….
Rajagilse says
Thank you sir, thank you once again for your lightening fast reply i.e. Solution circuit
Swagatam says
you are welcome!!