In the following article I have explained 3 useful DC to DC uninterruptible power supply circuits or DC UPS circuits for low DC to DC uninterruptible power applications
The first idea below presents a DC UPS circuit can be used for providing back up power to modems or routers during mains failures, so that the broadband/WiFi connection never gets interrupted. The idea was requested by Mr. Galive.
Technical Specifications
I need a circuit like,
I have two 12v dc adapter(600mA and 2A).
When input Mains is present, with the 600ma adapter i want to charge the battery(7.5AH) and with the 2A adapter i want to use my wifi router.
when the AC mains fails the battery will backup my wifi router without interruption.like UPS.
MY modem is rated as 12V 2.0A. That is why i want to use two 12v dc adapter.
The Design
Two adapters actually are not required for the proposed application. A single adapter, probably the one which is being used for charging the laptop battery may be used for charging the external battery also.
Looking at the given DC modem UPS circuit diagram we can see a simple yet interesting configuration involving a couple of diodes D1, D2, and resistor R1.
Normally a laptop charger is specified with 18V, so for charging a 12V battery this needs to be lowered to 14V. This is easily done using a transistor zener stage.
When mains is present, the voltage at D1 cathode is more positive than D2, which keeps D2 reverse biassed. This allows only D1 to conduct, supplying the voltage from the adapter to the modem.
D2 being switched OFF, the connected battery starts receiving the required charging voltage via R1 and begins getting charged in the process.
In an event AC mains fails, D1 gets switched OFF, and therefore allows D2 to conduct, enabling the battery voltage to instantly reach the modem without causing any interruptions to the network.
R1 must be selected depending upon the charging current rate of the attached battery.
A much better and improved version of the above is shown in the following diagram:
2) 6V to 220V Boost UPS Circuit
The second circuit explains a simple boost converter UPS circuit for supplying an uninterruptible power to satellite TV set top boxes so that the offline recording is never allowed to fail during power outages. The idea was requested by Mr. Aniruddha Mukherji.
Technical Specifications
I am an enthusiast electronic hobbyist person. Though I know only the basics, I am sure you must be getting 100's of emails daily and I am completely betting on my luck if this one gets to your "eyes"
My requirement:
16 volt 1 amp DC backup for my apartment Tata sky centralized distribution panel.
Issue: My apartment maintenance people do not run backup (generator) during day time, I have a Tata sky DVR which fails to record since there is signal loss due to power failure.
Resolution:
I had thought of a small back up system,I had purchased a small 6 volt 11 watt CFL Ballast circuit thinking as cheap alternate solution, but the same failed to work.
Why I am looking for AC supply instead of DC?I do not want to tamper with their system and get penalized for whatsoever failures which may come to it due to natural course of operation.
Could you please help me with a very simple cost effective circuit that will give me 220 volt 20 watts power from 6 volt 5ah battery. To be precise 220 volts from 6 volt battery, as I have purchased a 6 volt 5 ah battery recently. The output wattage requirement is less than 20 watts, the
adapter ratings are :
Output - 16 volt 1 amp
Input - 240 volt .06 amp
I know you have lot of work, but if you could spare some time and help me with this it would be of great help. thank you
Thanks,
Aniruddha
The Design
Since today all electronic systems employ an SMPS power supply, the input does not necessarily need to be an AC for powering these equipment, rather an equivalent DC or pulsed DC also become useful and works as good.
Referring to the diagram above, a couple of sections can be seen, the IC1 configuration enables a 6V DC to be boosted to a much higher 220V pulsed DC through a boost converter topology using the IC 555 in its astable form. The extreme left side battery section ensures an changeover from mains to battery back up every time a power failure is sensed by the circuit.
The idea is pretty simple and does not require much of an elaboration.
How the Circuit Functions
IC1 is configured as an astable oscillator, which drives T1 and consequently L1 at the same frequency.
T1 induces the entire battery current across L1, causing a proportionately boosted voltage to appear across it during the OFF periods of the T1 (induced back EMF from L1).
L1 must be appropriately calculated such that it generates the required magnitude of voltage across the shown terminals.
The indicated 200 turns is tentatively figured out and might need much tweaking for achieving the intended 220V from the input 6V battery source.
T2 is introduced for regulating the output voltage to the desired safe levels, which is 220V here.
Z1 should be therefore a 220V zener, which conducts only when this limit is exceeded, which forces T2 to conduct and ground pin5 of the IC, stalling the frequency at pin3 to a zero voltage.
The above process continuously readjusts itself rapidly ensuring a constant 220V at the output.
The adapter which can be seen at the extreme left is employed for two reasons, first to ensure that IC1 works continuously and produces the required 220V for the connected load regardless of the mains presence (just as we have in online UPS systems), and also to ensure a charging current for the battery when mains voltage is present.
The associated TIP122 transistor is positioned to generate a regulated 7V DC for the battery and also to restrict over charging of the battery .
Using Op Amp Cut OFF
If you want a precise circuit which will accurately monitor the DC UPS battery and implement the required over charge and low discharge cut OFFs, the following design may prove useful.
3) Redundant DC UPS Circuit
In this third concept below I have explained a couple of straightforward redundant UPS circuits for providing a secured uninterruptible power to crucial gadgets such as computer ATX or modems etc. The idea was requested by Mr. Shayan Firoozi.
Circuit Objectives and Requirements
- There are many products which has 2 input for different power supply,for example one for normal mains,one for generator or other mains,like servers,routers,and some critical equipment,we call it redundant power supplies
- I have an equipment which consumes 3 ampere in 12 volt dc,if I use 2 transfer with 12 volt,3 amp output which one take responsibility and which one is waiting for first loss?? Both are same on voltage and amperage,I don,t want them to work together,
- I want second power supply to be standby
- Just a simple question: What would happens if I replace battery with another 12 volt power supply ?? Will it work as a redundant or standby power supply ??
- Thanks for your answer in advanced And if it's possible tell us about model of diode and other components for 12 volt 3 ampere
The Design
As per the request, the circuit discussed in the above link can be modified to work with another DC power supply by eliminating the battery and associated stages as shown in the following form of redundant UPS circuit:
Using Two Power Supply Inputs
As we can see, the circuit is intended to work with a couple of power supplies having identical specs, such that whenever the primary power supply fails, the relay instantly changes over to the secondary power supply source ensuring an uninterruptible power supply to the connected load.
The diode D1 makes sure that while the primary power source is active and the relay in the deactivated position, it connects in series with D3 creating a greater forward drop than the primary supply diode D4...thus allowing the primary voltage to be in command and powering the load.
However as soon as the primary source goes through an outage, D4 is disabled, and for that split second D1 and D4 takes over powering the load, until the relay has changed over bypassing D1 and enabling the full rated power to the load.
The next diagram shows a method which allows a battery to be included within the proposed redundant UPS circuit, and the primary power source replaced with a solar panel, making the system a 3 way protected UPS circuit
Using Power Supply with Battery
Referring to the diagram, as long as the solar energy is available, the relay stays activated keeping the mains derived 14v supply cut off from the system.
The solar power in the meantime charges the battery and also the connected load via D1.
The battery power being slightly subdued than the solar panel power keeps D2 deactivated such that only D1 is allowed to carry the solar energy to the attached load at the output.
Using TIP122 for CV Battery Charging
The TIP122 ensures a regulated and safe over charging protected supply for the battery which charges solely through the panel voltage during day time.
As night sets in, the relay deactivates at some of time when the solar supply gets too weak to hold the relay activated.
The above changeover instantly switches the mains operated 14V into the system enabling the load to switch to the mains derived voltage without an interruption.
The battery power makes sure that while the relay is transferring over from the solar to the mains adapter supply, it compensates the split second changeover lapse in power by supplying its own power to the load, and inhibiting even a microsecond break of supply for the load.
The battery also forms the third "line of defense" in case both the primary and the secondary power happens to fail together, and is always positioned in the standby mode for the recommended redundant uninterruptible power supply circuit operation.
The first redundant UPS circuit incorporating two power sources can be better modified in the manner shown below, here the relay N/C can be seen directly connected with the load, thus enabling zero drop in the supply line:
Modem UPS using TP4056 Li-IOn Charger
If you are interested to make a 5 V DC UPS for your router using high end chargers such as TP4056 and boost converter modules, the following design could help:
The above design could be also built without a relay as given below:
Dear Sir
I have a 12v Battery 8Ah ups battery. I already have full charge auto cutt off and over discharge protection module.
Now I need an ups circuit for my Router and Onu both are 12V
Could you please help me sir
Hi Jobayer, you can try the following concept:
https://www.homemade-circuits.com/wp-content/uploads/2024/07/simple-DC-UPS.jpg
Will this circuit damage my battery ???
‘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.
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/
sir can you design online ups connection using with my module as mention in above comment
I can design once the power supply option is correctly sorted out, as explained by me in the previous comment.
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.
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
…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.
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
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.
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.
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:
The above relay is rated at just 10 amp only….so you have to replace the relay with a 50 amp type.
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?
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:
https://www.homemade-circuits.com/wp-content/uploads/2021/05/30-amp-automobile-relay-compressed.jpg
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
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
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
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
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.
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
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.
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…
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.
hi.
Which one can I use instead of 1n5402 diode?
thank you
Hi, you can use 1N5408
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.
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.
Thanks a lot. I will try this, if any problem I will contact with you.
Sure, no problem!
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.
Which circuit are your referring to exactly?
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.
Hi, the first circuit will be good for you requirement. You can try implementing the first one.
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
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
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.
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.
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.
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
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.
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.
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.
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)
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.
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.
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.
Hi, you can try the 6th design from this article;
https://www.homemade-circuits.com/how-to-make-efficient-led-emergency/
Sir I need your help I made 12v ups for router I used 5v 2A charger for input and connected to tp4056 and used 3*18650 batteries 3.7v 5000mAh in parallel combination then connected to dc booster xl6009 (and set the required voltages 12v)
But I connected the ups to my router it can’t run properly the power light of router starts blinking.
without connecting the ups and connected to the adapter its working fine without the power light blinking
My specifications of router are
Huawei 12V 1A
Waiting for reply
THANKS
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.
Dear Swagatam,
Thanks for publishing the circuits. First, by reading your article I see the power supply is assumed to be a laptop power supply which provides 18-19 V. Looking at the first and second circuits, the drawing indicates 12V sourced from what looks like “wire tap” from a laptop ps. This confused me a while as I was looking into using this circuit with a modem/router power supply that is rated at 12V 2A. This would not work, I believe since the 12V would not charge the battery, at least 14V should be required. Is that the case? Can you confirm my interpretation that the source should be 18-19 volt (or anything above 16V I guess). If that is indeed the case it may make sense to modify the drawing and state 18 – 19 volt source to avoid confusion.
Second, I wanted to go with the second circuit using LM317 until after discussing with a supplier who pointed out the second design will overcharge the battery, unlike the first design. How can I turn off the charging so it won’t be damaging the 12 V 7AH lead acid battery? Maybe add a simple circuit (found a couple online) that will switch off charging to the battery when it reaches 14.4 volt or so?
Thanks in advance for your help!
Dear Lord, yes you are right 12 V will not charge a 12 V battery. I have only tried to present a rough concept regarding the dC UPS and might have missed a few basic points in the design,and i am assuming the readers would take care of these by themselves or may consult me, as you have done.
Actually you will need 15 to 16V in the first circuit since an emitter follower transistor is used. To implement this you can make a transformer based adapter for the 12 V adapter. For this you will need a 0-12V transformer, a bridge rectifier, and a 1000uf/25v capacitor.
the second circuit will not overcharge the battery if the input supply is regulated at precise 14.1V. If I include an auto cut off that may unnecessarily complicate the simple design, which may not be required. To make a customized volatge regulator that will control the supply to exactly 14.1V you can apply a LM317 based design or a transistorized based regulator design.
Hello Sir…
I need to a circuit, DC UPS With 150A Battery charger I using Solar Power and power supply smps with 12V 150A Battery
Hello Ruwan,
you can try the following design
Replace the diodes with 20 amp diodes on heatsink
Hi sir, i am planning to run a 12v 20w bldc fan even during power cuts. So, i have used 12v 2a smps power supply to power the load directly. Using 12v 1a battery charger with auto cut off feature to charge the 12v 7ah battery. Using relay to switch between regulator power supply to battery power supply while power cuts. Also using over discharge protector bought on aliexpress at the output of battery to prevent over discharging. Everything seems OK. But, my doubt is 12v 1a charger stops charging when it reaches certain voltage ex: 14v and starts charging instantly when voltage falls below 14v during normal discharge of battery. This continuous charging will affect the battery or not.
Hi Mohammad, the ON/OFF charging and cut off at the full charge threshold will not harm the battery in anyway.
Thank you for this nice tutorial. I was looking for a laptop-style charging circuit. This helped me understand it quite well.
I wanted to create a Raspberry Pi Cluster containing max 8 Raspberry Pi’s. Each Raspberry Pi has 4 USB2 ports which can be used to connect to 4 external HDD’s leading to a max of 32 HDD’s. 32 external drives might be too much so I would limit this to 16. These HDD’s would need to be powered externally as the board itself does not provide sufficient power to each HDD. I believe the power required for each drive is 5V, 900mA. I also want to have a power module that consists of a battery that will be charged when external power is available. The power to the drives and Raspberry Pi’s should be provided from this battery when external power is not available. The switchover from A/C power to battery power should be seamless and should not cause the devices to power off.
I was thinking that using a laptop-style circuit for charging a battery while also providing power to the cluster would be required in this case. I also assumed that I would need splitter USB cables to pass data from the Pi board to the HDD and power from the power circuit.
What would you suggest as the circuit to power this cluster and what is the battery capacity would I need to support the power required for this cluster? Can you guide me on the circuit and the parts needed to build this power supply?
A/C -> Power supply ( rechargeable battery + circuit + board with USB ports for power to HDD’s and Pi’s)
Thanks, Glad you found the right article for your application. The first circuit can be ideally applied for powering the mentioned units.
For 16 units, you would require a supply of 5 V, 15 amp. This could be acquired from a ready made SMPS adapter. The battery could be rather large if it is a lead acid battery at 5 V, (15 x 10 = 150 Ah). This would provide a back up of around 9 hours for all the connected HDDs.
For this you can replace the adapter with a 5 V 15 amp SMPS input in the first diagram. Replace the transistor with a Darlington pair made using TIP35/2N2222
For the diodes you can use any 15 amp diode (on heatsink)
The zener diode must be experimented to at an output of around 6.5 V to 6.9V
R1 can be eliminated since 15 amp from the adapter is the right charging voltage for a 150 Ah battery.
I hope this will help you to achieve the intended purpose!
Hi and tanks for share all your knowledge.
I was looking at all the schematic and I think the second picture show what maybe will work with my system.
I have different network device connected all together to a single power supply.
The total load is around 7-8Amps at 12V but sometime I have spikes of 16A
I am looking to an UPS circuit that I can connect the entire system and charge a Lithium pack with max 14V 2A (This already have a balance charger installed inside).
I have a 15V power supply 18A but I am not sure what type of diode and resistance value to use for my environment.
Can you help me please
Hi, if you are interested in the second circuit, you can use it by replacing the LM317/R1 stage with your balance charger and the battery, and then terminate the battery positive through D2 for supplying the system.
I want limit the charger current to max 2A. What resistor value should I use? Also for 20A should I use 4x6A4 diodes in parallel ?
Thanks
You will need 1.25 / 2 = 0.625 Ohm resistor wattage will be 1.25 x 2 = 2.5 watts or 3 watt. For 20A load you will need a 25 amp diode
Thanks for the extra info.
For the diode, any schottky diode rated to 25 or + Amps should be fine?
Thanks a lot
Yes will do, but it can be very costly..
Dear Swagatam,
I was looking for ideas to make a good 12v ups for my modem and also a DVR. Your circuit ideas are very nice and informative. Now I will go ahead and build a unit.
Thank you for sharing these ideas and explaining them so well.
Best Regards,
Hari
I am glad you found them useful Harinath. I wish you all the best!
Hi Swagatam,
I don’t know much about circuits but am trying to adapt yours from the first image to run a 3.5A load backed-up by a 35AH battery. I have a 100W 13.5-18V adjustable power supply. Would my circuit at: https://www.lucidchart.com/publicSegments/view/4aeab8a0-f9b7-4575-be14-df8708da7af0/image.png work?
I’ve used 3x 2.2Ω resistors to give me a 12/3.3 = 3.6A charge. I can’t source a 14V 1W zener diode locally but can get a 14V 5W one … would this work? If I wanted to add a hold-up cap to the output (I have 1 10k uF 16V cap lying around) how would I wire it in?
Thank you so much for the great work & help.
Hi Danie, yes your circuit looks OK and will work, just make sure the battery gets the minimum 14V for charging. A 14V zener might not produce 14V at the output due to the TIP122 internal drop. So you may have a to add a couple of 1N4007 diodes in series with the 14V zener to ensure that the voltage at the emitter of TIP122 gets to 14V. The 1N4007 diode polarity will be opposite to the zener polarity.
If you want to include a capacitor at the output you can do it by joining the positive of the cap with the common cathode end of the 6A4 diodes, and negative terminal to the negative line.
hi dear swagatam
does the circuit in the following link work for 12 amps?
https://drive.google.com/file/d/1Npz2anCUy_pp1CLUHpua7mt7FOt-NX8T/view?usp=sharing
Hi Ali, it will work, but the diodes will need to be rated at 20 amps and on heatsink, the LM338 will also need a large heatsink
Dear Swagatam
to flow of 12 Amp, can I use SBL2040CT diode or two parallel 6A4 diodes ?
please guide me
Dear Ali, you can use two 6A4 in parallel by clamping them together on an aluminum heatsink
Thank you for your prompt response.
What is the role of 1k resistor in the first diagram?
without some kind of load the emitter side will show an incorrect reading on the meter while setting it up, connecting the 1K creates a dummy load for the transistor and allows a correct reading to appear on the meter.
1N4148 will not do, you must use the ones specified in the article, that is 1N5408 or 6A4
In the first circuit if you set the maximum output at the emitter of T1 to little below the specified full charge level of the battery, then extra no cut-off circuit would be required, for a 12V battery this could be set at 13.9V.
The negative line which is connected with the battery negative should eb connected with the load’s negative.
Hi Swagatam. I have a 9V 600mA router and I was wondering if I could use this circuit with 18650 cells in 3S2P or 3P formation with a TP4056 charging circuit? Using a DC-DC step down converter to drop the input charging voltage to 5V and then using another DC-DC boost converter to increasing it back to 9v at battery output. Is this convoluted and if yes then is there a better way?
Thanks.
Hi Nish,
You can use this circuit for the mentioned application, however you can keep things simple by eliminating the step down converter, and instead using the cells in series to match the 9V output, because using two converters together can waste some power in the process and also make the system unnecessarily bulky..
Hi Swagatam,
I have 1 switch [12V, 2A] & 1 router [12V, .5A], both of them needs a backup from 12V 7AH battery. So there will be 2 o/p from battery. I have following queries..
1) Can I use D1 & D2 as 1N4007, so that it can limit to 12V, 1Amp output of the circuit and can be input for switch?
2) How can I restrict the current to .5A [12V .5A] in another output, as an input of the router?
-Thanks
Sujay
Hi Sujay,
1N4007 will burn at 1 amp unless it is on some form of heatsink…you will have to use a higher rated diode such as 1N5408 etc. for allowing 1 amp current through it.
for restricting current you can use a simple LM317 IC and wire it as per the following concept
https://www.homemade-circuits.com/2013/06/universal-high-watt-led-current-limiter.html
Thanks Swag.
I need some more clarifications about the same. As router needs “.5A” current, I can use LM317 to restrict the current for that device input. For switch, hope I can directly connect through 1N5408 diode. Or in both scenario I should use LM317?
In precision current limiter, to calculate R1, Vref used as 1.25V. But here my Battery is of 12V & device input voltage also 12V, so Vref is practically 0. So how should I define Vref to calculate R1?
-Thanks
Sujay
Hi Sujay,
actually current control is not required for any of the devices because if the voltage specs for the devices match with the battery voltage, current would be immaterial. You can safely connect the devices directly with the battery.
as for the 1.25V ref, I don’t think it is dependent on the input supply, this reference will be available as long as the input supply is well over the 1.25V value
HI Sir, my router requiere 12v 0.6A I want to plug it straight to car cigarette socket, do I need any protection or just an adequat fuse. thank you.
Hi Nooman, just do it through a 7812 IC, that will take care of everything
Having 3 cells would actually give me less than 12v (3.7V * 3 = 11.1v) that’s why I was thinking for 4. Will that be an issue? Also, how will the charging be limited else batteries will be ruined in no time on continuous charging
11.1V won’t be an issue for your modem…3 cells is the right number that you must use
as I mentioned in the my previous comment if the 4.2V per cell is prevented then an auto cut off won’t be required, because the cells will never be able to reach their full charge levels and therefore be safe.
Sorry, couldn’t follow this up earlier.
I will be using your second circuit with LM317, to be adjusted to 12V output. My input is 19.75V 3.5A using an old laptop charger. I am planning to use 3 cells 18650 in series for 1-3 mins backup
Can you please tell:
1. If 3 cells would be sufficient.
2. Can I use battery protection module in your circuit – just before the battery charge
3. Need help with R1 value with the values mentioned above.
3 cells would provide around 10V which won’t be sufficient for a 12V load.
yes you can use any battery protection module in between the LM317 and the battery
R1 = 1.25 / 0.5C, where C is the Ah value of the battery
Hi Swagat,
In an earlier post before last one, you asked for using 3 cells only
“11.1V won’t be an issue for your modem…3 cells is the right number that you must use
as I mentioned in the my previous comment if the 4.2V per cell is prevented then an auto cut off won’t be required, because the cells will never be able to reach their full charge levels and therefore be safe.”
I am fine to use 4 if that’s the case. Please clarify
Hi Drake, I said this with reference to the modem supply requirement, but as far charging is concerned 4 cells would required 16V and furthermore this level may be higher for the modem too, so it is better to use 3 cells, just check whether at 10V the modem still works satisfactorily or not, if yes then you an go ahead with 3 cells.
you can fix the charging level to 12V which will never allow the cells to get overcharged.
I am still confused. If you recall, I will be using chargeable batteries (I am planning to use 3.7 V 18650*4) in series as I need just 2-3 mins of backup . If I use LM7815 (7814 not available) in place of LM317, I will get 15V charging voltage and 1A fixed. How can I reduce the current here for batteries charge?
I think you must use 3 cells in series and not 4…so the full charge level becomes 4.2 x 3 = 12.6V.
even if you need 3,4 min backup the battery would be continuously charged by the system and therefore it would need some kind of limiter, the current actually won’t matter if the voltage is kept slightly below 12.6V… to implement this you can configure the LM317 as a voltage regulator instead of current limiter, and fine tune the output to may be around 12.3V, 7815 can be totally avoided
HI, Not an electrical student but can understand circuits to an extent. My Dlink router is 12V 1.5A listed. I have a power backup for my apartment but it takes 2-3 minutes to resume the power and in calls, this is very annoying to connect again. Can you recommend a circuit that can be used with AA rechargeable batteries? Like I mentioned, I would need backup only for max 5 mins. I would like to assemble myself so appreciate your detailed response.
yes I understood your requirement, you can still use the above concepts, for the battery you can use 8 rechargeable AAA cells in series….rest can be as is.
the first design looks more appropriate as it will not allow the batt to overcharge once the emitter side limit is correctly set.
the circuit (diodes) will automatically disconnect the battery as soon as the power resumes from your apartment backup.
Thanks for your prompt response. Are you recommending the one with TIP 122 circuit? Will the ratings for diode and resistor remains what you have recommended elsewhere?
yes the first design using TIP122, the base resistor can be increased to 1K that's not very crucial, but the zener value should be selected such that the voltage at emitter is slightly higher than the battery voltage that is around 13.5V, rest can be as given in the design
R1 can be removed and replaced with a short link since the 13.5V setting will never allow to battery to overcharge
I have found an old laptop charger which I would like to use. Its O/p is 19.75V 3.5A. Can I use your second circuit using LM317?
1. I see the direct o/p in your diagram from 12v to 12 v but that will not be case for me. How do I step it down? Can I use LM7812 there for reducing the o/p from 19.5v to 12v? What will be my current in that case?
2. While going through LM317 circuit, what should be value of R1? Can I use LM 7814 for charging the battery?
My router is rated at 12v, 1.5A.
1) yes you can use a 7812, the current will be limited to 1 amp.
2) you can use 7814 in conjunction with the LM317 for charging the battery.
3) R1 will be = 1.25 / charging current limit
Thanks for 1.5A, I was thinking of LM7912. Will that be ok as it outputs 1.5A?
If using 7814 instead of LM317, the charging current will be 1A. Will that mean that I need R1 as 1.25 Ohm?
you can use 7912.
the charging current is supposed to be extremely low, may be around 1/20 of the battery AH if it is an SMF type of batt.
therefore R will need to be calculated as per this charging current rate
good day sir
i am using normal inverter at home. the modem also connected to the normal power supply at home.when a power failure ( mains ) happens the inverter take some micro secs to switch over to battery and viceversa . so in that period the modem restarts and will take time to function . i need a backup for modem to avoid this operation, modem using 9v dc supply.( i installed a 9v dc batery to overcome this in parallel with power supply but it didnt work out still modem restart during this small changeover time )
can u help me to sort it out . (i dont want to install a separate full ups only for modem ).
regards
john
good day john, a modem will not reset if the changeover is within microseconds…it will happen only if the delay lasts over 500ms approximately.
you can probably try solving the issue by connecting a high value filter capacitor across the supply terminals of the modem. it should be at least a 6800uF/25V rated capacitor or above.
Can 12V 2A (DC) Circuit be connected to 12V 9AH (DC) ?
yes it's possible
Hi there I was wondering if I'm charging the battery with an outside source charger I just need to add a diode to the positive line of the battery, so I don't get a feedback to the battery is that correct?
Hi, only D1 and D2 are required no other diode needs to b included no matter from where you are charging the battery
hi
i have a maintenance free 12v bike battery,and 12v input modem, could i use your above circuit .
yes you can use them without any issues…
Hello sir,
I need you help. This is what i m looking for. My router adapter rating is 9V 600mA. And i want to use 12V 7AH battery. which circuit will be batter in my case? How much backup i will get from this battery.
Sorry for my english
Hello Rahul,
sorry, none of the circuits will work because a 9V source can never charge a 12V battery….so you will have to procure a 14V adapter, or build a 14V adapter using a 0-12/1amp transformer along with a bridge rectifier/capacitor output network, for the application
I have a 16,5 VAC – 1,5 AMPS transformer. Can I use it with a full rectifier bridge (or 4x1n4007)? (I think I obtain 23.3V on the output)
23V will be high and dangerous for a 18V router, unless you step it down through a regulator IC
Not a electrical student. Have a wifi router. Adaptor says power output 18v/1a. Couldn't get a UPS for this router in market. Only 12v is available. Can you help me with this. Need to make my own ups.
you can buy a 0-12V transformer and connect a diode bridge rectifier to its output along with a filter capacitor, this will provide you an output of around 19V which can be used for driving your router.
Thanks sir for your kind reply.
Sir I got the idea of emitter follower cct for transistor but I forgot to mention in previous question that my battery is 12 volt.
One thing more I want to ask that, does the formula R= 1.25/charging current apply on the second circuit only and Ohm's law apply on the first curcuit for the the value of R1. If yes then for the 1st cct, is the value of R1=12/2 = 6 Ohms or 14/2= 7 Ohms for 2 Amps of charging current. And what will be the wattage rating of other componants like zener diod for this charging current.
You got everything right Alis!
the wattage can be calculated with the following formula
P = R x I^2 = 6 x 2×2 = 6 x 4 = 24 watts
By the way the current is supposed to be 1/10th of the battery AH…I am not sure what's 2 amp in your case.
Dear Swagatam.
I tried to make the circuit according to the diagram, but i got an issue.
The line going to the laptop has very low and fluctuating amperage although the voltage is fine. I am actually trying to power two routers and it works flawlessly on one router, but it only works with the mains connected in the second router. Once the mains is disconnected and battery is connected, the amps is too low and fluctuating. What could be the reason for this? Please give feedback
Dear Kailash,
check your adapter specifications or change with a higher current adapter, I think it may be low in current and therefore not able to charge the external battery and the laptop together.
Dear Swagatam,
Thank you for your prompt response.
I bought the adapter just for this project according to the specs you had given on one of the comments.
It is a 14 volts adapter with 2.37 amps current. I could not find the 3 amps one in the market so I got the 2.37 amps one.
I cannot afford to buy another adapter at this moment, so is there anything that could be done for instance by adding transistors or any other parts in the same configuration.
I would be very grateful if you could provide the solution.
Thank you once again.
Dear Kailash, please try the new adapter and see how it performs.
for a better response you can try connecting a 2200uF/25V cap across the D1/D2 junction and ground.
Dear Swagatam,
I think that there has been some misunderstanding regarding my situation.
The circiut works fine when the adapter is connected to the main line of 220 volts. Once I switch the main line off and the circuit is running on battery, It can power one of the routers, but the other fluctuates. I do not think that this is the issue with the adapter. I might have done some mistake with circuit itself. I am a bit confused with the final part of the circuit. From the d2 and d1 junction it goes to the +ve of the modem. What about the -ve of the modem?I have connected the negative of the adapter to the negative of the battery using a wire from which I have cut two areas which connect to the -ve of the modem and to the ground. Is this correct? I think that this could be the issue. Please shed some light on this.
Dear Kailash. yes I misunderstood and thought that the issue was when the mains was ON.
The negative line shown with the earthing sign should be connected in common to all the negatives of the respective devices….
If the voltage is not dropping then the current fluctuations can be ignored…please measure the voltages across the specific routers and confirm whether it's constant or not.
Dear Swagatam,
Thank you for your prompt response.
Could you please post picture of the capacitor connection. I am new to all this stuffs. I do not think that I could do it without a picture.
Dear Kailash,
It's easy, solder the long lead of the capacitor with the D1/D2 junction, and the smaller lead with the bottom negative line or the earth line
Dear Swagatam.
I tried to make the above circuit, but I have come accross a certain issue.
The amperage on one of the connection is very low and fluctuating.
I am new to electronics, so could you please shed some light on this matter.
The line connected to the laptop has the problem. The one going to the router is just fine.
Thank you for your time
Thanks a lot for nice and great information.
Sir, I just want to know that what phenomena in the first circuit is to protect overcharging the battery, that is less current, voltage or anything else.
Secondly if I want to charge my 9Ah AGM battery with 2 Amps for quicker charging what changes should I made in first cct, what will be the value of R1 and across it what will be the value of resister in series of Led to indicate charging.
Thanks again.
Thanks Alis,
in the first circuit, the transistor is wired in the emitter follower mode in which the transistor is simply not allowed to conduct or is "choked" as soon as the emitter voltage reaches the base voltage level or at a difference of 0.6V (close to it.)
therefore assuming if the base zener voltage is 14.6V, then as soon the battery voltage reaches 14V, the transistor is unable to conduct any further, unless the battery drops to below 13.8V
for 9V battery, you can use a 11V zener diode
R1 can be calculated by using ohm's law
R = V/I
9/0.5 = 18 ohms
watt = 18 x I^2 = 4.5 watts or 5 watts
if an LEd is connected in parallel to R1, then a 100 ohm resistor will be enough to safeguard it
Can u please tell me…. Is this circuit protects battery from over charging
the first one will do it..
Also, please kindly list the components that i should use to complete the circuit.
click the diagram to enlarge, and write down the components as it's shown, your retailer will understand and give you the components accordingly…
Thank you for your quick reply. But, you didn't mention which circuit should i use? And, would the 12 V adapter safe?
Rusan, you did not ask me about the circuit selection therefore I did not suggest about it.
both the circuits are correct and will work.
you can use any of those as per your convenience..
12V adapter will not charge 12V battery…so it must be set to 14.4V adapter
Hello, Swagatam.
I want to power my router which requires 9 volt 1 amp but i have two 6 volt 4.5 Ah Lead Acid Battery. So, i want to use two 6 V batteries to power my router. Is it possible? I would be very thankful if you could help me.
Hi Rusan,
yes you can do that, just be sure to add a 7809 regulator across the output, that is across the cathode junction of D1/D2 and ground
Sir can I use a Transformer with bridge connected instead of 12v adapter and can I also just skip voltage regulator and directly connect through R1/D1 & which R1 resistor should I choose if I need 2.5~3.0ampere current for 12v battery. Thanks for your help & It is humble request to you please reply this message.
Adnan, yes you can do it in that way, but please note the battery charging section is crude therefore it's better to use a reduced charging rate.
you can use 1/20th of the battery AH as the charging rate.
Use Ohms law for R1, that is R = V/I, where I is the charging rate as described above, V is the supply voltage