#143706

‘Using opamp cutoff’ I cannot understand uses of resistance and diodes(1N4148) properly. Is it possible for you to make me understand?
I have got your eMail but could not reply because the system failed to send it.

Author
#143709

OP amps are configured as comparators. The zener resistor provide the clamping volatge to the zeners, the presets or trimpots are used adjust the cut off threshold of the comparators.
For understanding comparators you can read the following article:
https://www.homemade-circuits.com/how-to-use-ic-741-as-comparator/
For opamp battery chargers you can read the following post:
https://www.homemade-circuits.com/opamp-low-high-battery-charger/

#106614

sir can you design online ups connection using with my module as mention in above comment

Author
#106619

I can design once the power supply option is correctly sorted out, as explained by me in the previous comment.

#106599

hi, I have make 12 v dc ups for my WIFI router using 3 li-ion battery in series with 3S 10A 12V 18650 Lithium Battery Charger Board Protection Module and output side i use XL6009 DC- DC Adjustable Step UP Boost Power Converter Module in this ups I am used 12 v 2.5 A input charger.
but I am facing issue when ups is in backup mode and discharge at 6-8 v approx. that time router can not start when input power will turn on what’s wrong with my ups please guide me.
thank you.

Author
#106607

Hi, your 12.6V li-ion battery should never be discharged below 9V, otherwise it might slowly damage the battery and also require high current initial charging.
Maybe your 2.5 amps is getting drained while charging the deeply discharged battery which is causing a severe voltage drop and preventing the router from starting.
Also make sure the boost converter is correctly set to produce 12.6V across the battrey terminals, and make sure the input current to the battery is at least 50% of the battery’s Ah rating

Author
#106608

…another point is that the 2.5 amp may not be sufficient enough to charge the battery and also simultaneously power the router. You might require a 4 amp or 5 amp input current.

#106613

ok sir thank you for suggestion I replace my charger with 5 Amp
one more question i nave one 12v 4 amp smps supply when i connect load on it it will give output in form of pulse i change dc output capacitor but still its blinking o/p what is problem. please give me suggestion

Author
#106618

Rahul, if your SMPS is blinking then that will not do. It means either your SMPS not able to accept any load or it is faulty.

Also, the boost charger must be removed, instead an SMPS can be purchased which has a preset setting which can be adjusted to get a 12.6V output.

Alternatively another approach would be to use a 0-12V transformer rated at 5 amp maybe. Then add a bridge rectifier and filter capacitor at its output so that its output increases to 17V.

Finally this 17V could be dropped to 12.6V through a transistor circuit for charging the battery and also drive the router.

#99861

Hi, thanks a lot for sharing your ideas.
The goal of my thruster project is to build a boat engine for my 4m catamaran. The system is based on an programed arduino board which controls an ESC. Furthermore the main power source for the thruster and electronics consists of a minimalistic internal 5S4P Li-ion battery pack. As a future optimization I would like to attach an additional external 5S70P Li-ion battery pack for longer distance journeys occasionally. For I do not want to add a mechanical switch just to control the change of power sources I’m thinking of integrate a kind of UPS module. The Idea is to plug the external source in and at the same time the internal source will be detached and vice versa. Maybe you have an idea to use on of your UPS templates? Specs: Rated Voltage of power sources: 18V. Current typical: 15A, 32A peak. Thanks a lot Swagatam for any reply.

Author
#99876

Hi Boris,

will a relay changeover work for you? It’s probably the simplest solution you can get, although it will consume some current for holding the contacts.

You can try the following simple configuration:

simplest relay chnageover circuit for DC UPS

The above relay is rated at just 10 amp only….so you have to replace the relay with a 50 amp type.

#99894

Hi Swagatam,

Your suggestion seems easy going. Thanks a lot. I suppose connecting the internal battery in order that the relay is not actively holding the contact is possible? And when I attach the external fat battery pack the relay is switching? Is there a specific order ref. for this 50 amp relay?

Author
#99972

Thank you Boris, you are right, you can connect the internal battery to “DC Source#2″ and the external battery to DC Source#1”, and that will do the job as desired.

Since the maximum current is not more than 15 amps, a 30 amp automotive relay should work satisfactorily, as shown below:

30 amp automobile relay compressed

#87314

A solution I found for delay/ reboot of router issue: add a dummy load at the input, with diode, it will decharge the capacitor of phone charger and tick relay much faster! Go for more than 200/300W and sees if it gets hot of course. Circuit with Tp4056 realised and it works fine.
Sorin’s idea.
Thx

#85557

Dear Mr Swagatam,

In recent days been through your website and found it really informative and interesting.
Am trying to build an uninterrupted power supply for my 12V 2A router.
Am using a 12V battery (7AH) and router adapter as 2 power sources connected via 2 relays (12v). The output of relay is connected to router. The relay is powered via the adapter.
Now the problem is because the output voltage of battery is around 13v, am using a voltage booster / stabilizer to output 12v from battery to router.
But when there is power off, the relay takes few milliseconds to change over and thus rebooting the router.
When power comes there is no loss and relay switches smoothly from battery to adapter current.
Please suggest how to handle this power loss during switching.

Regards,
Sunil Kumar

Author
#85567

Dear Sunil, you can use the first or the second circuit for your application which will ensure that the power is transferred without any interruption.

To add a regulated 12V for the router, you can add the following circuit, between the D1/D2 cathodes and the router.

Low-Dropout 5V, 12V Regulator Circuits using Transistors

#82568

I tried the last option with the TP4056+ StepUp + Relay. But it is not working as intended, because the relay is not switching anough fast. So when Power fail the router is rebooting because relay is not anough fast 🙁
And I tried also the option with the relay, directly TP4056+MT3906, but the green led of the TP4056 are never getting on. Why? because the TP4056 IC is behaving in a way, when a certain Amp flow through it, below 1/10 of the charge current the charge is completed and the led turn on. So it is never getting on because, always aroun 700mA is flowing through the IC. I hoocked a ampmeter and I descovered this. Because the router is always sucking, dicharge/recharge at the same time I guess…
That is why, I first went for the option with the relay, I even added capacitor 2200 even parallel, it is still rebooting. I want the battery to be charge lonely by the TP4056 and light on when it is completed, and kick fast when the power fail…
Is it possible to use BC547 as a switching relay instead of a magnetic relay maybe, for it to switch faster??
Thx

Author
#82575

The diode version is perfect and has no problems. Your router is consuming all the current because your supply source is not rated with enough current to charge the battery and the router together. You can either use a higher rated input source or add a 5 ohm or 10 ohm 1 watt resistor in series with the router and check the response.

The low current at the input source is also the reason why the relay is not operating quickly enough.

#82586

the issue is not in the turning ON of the relay, it is more in its turning OFF.
When the Relay is on NO position (no power applied to the coils pin + Router Power by the Backup) and I simulated Power Back (Let say Main electricity came back) so Coils of Relay are again powered, and it switch from NO to NC: here there is no problem, it is working fine. The rooter do not reboot. So enough currrent is coming to the Relay to do the task required.
The issue is in the opposite, when the relay need to OPEN, when it need to go NO! In this small milli second of time, the Router is rebooting. So the relay take to long to open, and to switch to battery supply.
PS: My Router is only 450mA and I changed the Rprog of the TP4056 to charge the battery at a slower current rated. it is not helping. I am using 12v – 1A adatper, and still the same.
So maybe better to use Solid State Relay go get a faster switch between DC Battery Backup for the Router?? What do you think??
Thanx

Author
#82591

try putting a 1N5408 diode before the 2200uF capacitor, and possibly put another 2200uF capacitor in parallel to increase the back up power. This should should take care of the slight millisecond delay.

For the diode version a 1 amp transformer might not help at all, because the battery will itself require a current equal to its mAh rating. Try a 2 amp or a 3 amp adapter and check the results.

#82592

Thx you for your help, I did not find any 1N5408 diode, but I got some 1N5401 and also 1N5399.
So maybe try with 1N5401. I dont lose much trying…
About capacitor, I paralleled 2 2200uF yesterday and still delay >>Router Reboot 🙁
For Power Supply, I am currently testing directly with Lab Power Supply as main Power Source, that can deliver till 10A if required, so current part is covered, no stay the delay part of the relay…

Author
#82608

Yes a diode will make a lot of difference, since it will make sure the capacitor charge goes entirely to the router only. The 10 amp power supply should more than enough to keep the charging operation sustained while also operating the router.

#82333

hi.
Which one can I use instead of 1n5402 diode?
thank you

Author
#82336

Hi, you can use 1N5408

#80720

Hi sir, I have a problem of power cuts, so I want a mini ups.I have a onu device which specification is 12v, 0.5A and a router which specification is 12v,0.5A. So I want a power backup system (2 hours power backup minimums) for those devices. In the market I found mini ups for the router but its very expensive. So I want to make the safest home made mini ups, which is not damage the equipment. Tell me how to make this. Thank you.

Author
#80724

Hi Krishendu, you can build the first circuit which is the simplest and the safest. You will required a 12 V, 4 Ah battery for a 2 hour back up. Make sure to adjust the output from the TIP122 to exactly 14.1 V.

#80733

Please tell me about Intelligent electrical safety protection circuits with battery management systems in this project. Which type of battery and module or component I used? And please tell the input adapter specification.

Author
#80744

Which circuit are your referring to exactly?

#80732

Thanks a lot. I will try this, if any problem I will contact with you.

Author
#80736

Sure, no problem!

#80383

Hi,i have a project which must supply power to the router and the radio (for a home wifi),while charging the backup battery when the a.c mains available and if the mains fail the battery must take over and supply power to the wifi system.i just wanted to know how to design it.

Author
#80394

Hi, the first circuit will be good for you requirement. You can try implementing the first one.

#79666

hello.
I want to provide uninterrupted power for my arduino project.have 12v-7ah lead battery for backup.I want to use 12v dc adapter.Would you recommend me a circuit diagram to make a backup power supply?Please also write the materials I will use.thank you

Author
#79697

Hello, the first circuit is easiest and reasonably safe design which you can implement. But 12V adapter will not do, it must be a 15 V minimum, and the current rating must be 10% of the battery Ah value, meaning it should be 15 V/ 0.7 amps or 1 amp

#79596

Hi,
thank you for sharing this circuits. I want to use your first circuit to power a router and a modem ( both at 12V /1.3A each ). I have a 15.5V (laptop) adaptor and I want to use it for power a circuit that will enable to charge a 12V SMF type battery of 7Ah ( with charging around 14V/0.35A ) and to supply the necessary 12V for both devices. The maximum load estimated current is around 3A ( router + modem ), so 1N5402 it’s a good option for D1 and D2. For charging the battery the maximum current is ratted at 2A ( that value is written on the battery as initial current, even if I measured only 0.35A charging current ). So my 90W adaptor is sufficient. What I need is a steady 12V at output when AC is present or not. I don’t want to supply 13.5-15V to the 12V input modem , because I’m not sure that this will be ok for the modem.
Please tell me how can I step down from 15.5V to around 12V at output ? If a put a series of 3 or 4 1N5402 diodes instead of D1, this will drop the voltage to around 15.5-4×0.7 = 12.7V ( if the drop forward voltage is 0.7V on each diode ) but this value on last diode cathode in the series ( and D2 cathode ) will make the make the D2 to conduct. So this in not a good solution.
What simple voltage regulator to use at output to make sure that I will have around 12V when AC is present or not, even in the situation that his input is less then 15V ? ( when AC is missing and the load is present, the battery voltage will decrease in time from 13.7V to 12V, so this range must be the input voltage for the voltage regulator ).
Is there any other simple solution to power my 12V modem and router , without an voltage regulator at output ?
Thank you.

Author
#79612

Hi, You can use a 7812 IC for dropping the higher voltage input to 12V or slightly lower. Use it after the D1/D2 junction. Even at voltages below 12V, the 7812 will continue to conduct and allow the voltage to reach the load.

#79618

Hi, thank you for your support.
I see that 7812 IC have 1A , is that corect ? If so, that is not enough because the load is about 3A* ( 2.6 A, router 1.3A + modem 1.3A ). Is there any other version of regulator with a minimum of 3A ?
In the case of a missing AC, the battery voltage will drop significantly after an hour ( I will use a 7AH battery ) at about 12.5V or so. In this case, what will be the voltage after the regulator ? If it will be around 10-11v, this will be not so good. I am hoping to design a circuit that will be usefull for a few hours, not for 20 minutes.
* Or there is another possibility: because only one device ( modem ) requires 12V, I need only a 1.5A regulator. ( The voltage input for the router is 10-57 DV )
A few questions about the first diagram/circuit: my adaptor has 15.5V, so what is the R1 voltage/ R1 value for a battery of 7AH , and what will be the charging voltage on battery ? ( I know that the battery current will be between 0.35-0.7A ) The other components from the transistor + Zener stage remains the same in my case ? ( 14V Zener, two resistors of 220 ohm and 1kOhm )
Thank you.

Author
#79624

The TIP122 base zener diode for the charger section should be actually a 15V zener, so that the output can be 14v, this will need to be fixed with some experimentation.

#79626

Hi Swagatam. Ok, I’ll try that. So I will must have 15V at the base of TIP122 and the transistor will be off when the emitter voltage reach about 14.4V, right ? I have find your Customizing Zener Diode Value exemple here, and I will follow it. https://www.homemade-circuits.com/zener-diode-circuits-characteristics-calculations/

But I can’t understand the last part of “You can adjust R1 using Ohms law so that 0.7 amp is able to reach the battery. R1 = 12 / 0.7 = 17 ohms / 10 watts” . The voltage across R1 is not 12V, right ?
With Kirchhoff’s law: with Zener of 15V and with VBE of 0.6V, if we have 14V on battery ( charging voltage ) and around but max 14.4V at emitter (Zener 15V – 0.6 VBE = 14.4V ), we are left with only 0.4V over R1 ( in this case R1=0.4V/0.7A =0.57 ohm ). Please correct me if I’m wrong, which is probably the case.
( Or maybe I will charge the battery at a slightly less voltage, and this will set the R1 voltage at about 1V; Or maybe is better to custom a transistor Base Voltage to a slightly higher value, at 15.5V )

For the 12V output additional stage, an adjustable LM317 module like this is ok ?
I prefer an already made module instead of experimentation and tweaking of a custom made TIP122/resistor//zener stage and I’m not very experienced. And it’s cheap, it’s about 1.5 Euro in my country.

LM317 buck converter

Btw, don’t you know an adjustable module that will serve the purpose of an auto cut-off for battery charging ? 😀

Thank you very much for your support.

#79643

Hi Swagatam,
I tried to find a cheap (10-20 Euro) DC to DC UPS readymade system, but unfortunately I didn’t find any with my desired requirements ( 3A at 12V). If you do know one, please tell me.

From your information, I will have 15-1.2 = 13.8 V on the emitter, but that is the voltage that is needed for battery charging, so I running out of voltage for R1.
Because my laptop AC adapter is one in steps, maybe is better to use 17DC output for my design. With a 16V Zener – 1.2V VBE – 1 Volt across R1 = 13.8V on battery and that is ok ( 13.7 more or less, depending on the battery loading current which is varying ). In this case R1 = 1V/0.7A = 1.4Ohm but I think 2 ohm will do the job. ( or maybe 3 ohm, I just measured a similar battery and at discharged level of 12.5V across battery, it uses 0.35A for charging ).
About the rest of the components: I think R of 1kohm is ok, but the R of 220ohm maybe must be replace with a smaller one, maybe 30- 50 ohms. Tell me please if all this values are correct or not and if this 17V DC design are feasible.

Thank you for the link about LM317 Variable Switch Mode Power Supply (SMPS) and for the information about the TIP122.

Author
#79649

Thank you Cristi, I too tried but couldn’t find any cheap readymade DC to DC UPS online, that’s indeed very strange.

I think the R1 voltage must not be considered in the calculation, because as the battery charges this drop will reduce until eventually it reaches a zero.

TIP122 being a Darlington has a very high gain, around 1000, so reducing the base resistor value might not matter too must. For analysis this could be calculated using the following formula:

R = (15V – Battery discharge V + 1.2) x 1000 / 0.7 amp

Although this formula is actually for a collector load, I think it can be also used for an emitter load since the current is through collector emitter.

Author
#79631

Thanks Cristi, your calculations are right. In a hurry I forgot to deduct the load voltage specification from the supply voltage. However it should be 12V that must be considered in the calculation for the battery voltage, so R1 will be 14 – 12 / 0.7 = 2.85 ohms, 1.4 watts…. a 2 ohm will also do.
Since TIP122 is a Darlington, the emitter voltage will be 1.2 V less, that is 15 – 1.2 = 13.8 V to be precise.

A buck converter is the best option but if a readymade thing is what you want to include then why not buy the whole system readymade ? You may easily find a DC to DC UPS with all the required facilities very cheaply 😀

The tip122 based regulator is perhaps one of the easiest designs, which requires very basic adjustments using a multimeter, and it is quite foolproof.

And it will also do the auto-cut off job at the threshold level.

The 317 buck can be also built at home using the following concept:

LM317 Variable Switch Mode Power Supply (SMPS)

Author
#79623

Hi, If 7812 does not suit your application, you can use an additional TIP122/resistor//zener emitter follower stage at the output side of the D1/D2 junction. The zener could be a customized 13 V zener which will allow the output to be around 12 V. This will also drop around 1 V which is the minimum that you can get, since all linear regulator will drop some voltage. The tip122 will require a large heatsink for delivering 3 amp.

You can adjust R1 using Ohms law so that 0.7 amp is able to reach the battery. R1 = 12 / 0.7 = 17 ohms / 10 watts

#78902

Hi.
I dont know English.excuse me.
I read the article from the translation.I am not very successful in electronics.my question is;
I want to run the system with an adapter.
when the power is cut off, I want the battery to be activated automatically.
battery is 12v power.please would you recommend a circuit?
thank you.

Author
#78911

Hi, you can try the 6th design from this article;

https://www.homemade-circuits.com/how-to-make-efficient-led-emergency/

#119697

Design a 5V, 9V and 12V output UPS that could be used for supplying power to router during power failure from battery. It will be charged by AC supply as well as Solar Panels.

Output voltages = 5V, 9V and 12V

Output current = 1A

Battery capacity = 20000 mAH

DC-DC Converter

Switching Relay

Battery charging circuit

Sockets type = USB, Pin type

Solar Panel = 5 – 10 W

Author
#78872

Hello Salman, it’s because your booster is not working correctly, and it is unable to supply the required amount of current to the router.
Instead of using such complicated circuits, you can use a LM317 based charger, adjust the output to 12.2 V, and current to 1 amp. And attach the 3 li-ion in series with its output and simultaneously connect the router also, all will work perfectly, without any issues.

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