I know this question has been asked quite a few times on this site. The closest one i could find is: make-9v-ups-for-router-with-tp4056-and-mt3608.

While there are UPS systems designed for routers, in my country, they literally cost more than the router. I am just trying to see if I can learn anything from trying to put together my own UPS system.

Problem: I have a 12V 1.5A Nokia router. During our daily power cuts (my apartment backup kicks-in in about 5s to at most 1min) my router decides to give up on life and doom me to the internet black hole.

I just need to have a UPS system to ensure uninterrupted internet during the switch over from mains to apartment backup.

This is what I have so far (with my amateur reasoning of it):

Possible UPS Circuit Design for 12v 1.5A Router

My router comes with a 12V 1.5A DC adapter that I am planning to replace with a 12V 2A DC adapter. Connected to this would be a 2S1P Li-Ion 18650 batteries (Samsung make) that will be charged through a 2S BMS board (TP5100). The output from this backup has to go through a XL6009E1 to meet the 12V supply required by my router.

I plan to setup the Li-Ion batteries in series using Nickel strips and with glass fuses rated at 2A each. I wanted the backup to kick in only once the mains went out and so I have got a IRF9520 P-MOSFET along with 1N5408 diodes (the only ones I could get my hands on) for switching the power supply and for reverse voltage protection.

Along with these, I got a couple of pull down resistors (100K), capacitors (4.7μF 25V Electrolytic capacitors) and a heat sink for a TO 220 package.

Will this circuit work to charge my batteries and provide power to the router and then successfully switch to backup for those precious few seconds when the mains cut out?

I just want to make sure I don't burn the house down or something.

  • 1
    \$\begingroup\$ Why is the input to the XL6009E1 taken from P+P- instead of B+B- ? What is the voltage at P+P- when there is no 12V supply from the DC adaptor ? Can you add a link to 2SBMS board data sheet ? \$\endgroup\$
    – AJN
    Jul 17, 2020 at 16:53
  • \$\begingroup\$ For a couple reasons, I would move the DC-DC converter to the very end of the circuit. First, you have two diode drops between the input and output; so your output will be closer to 11V. Also, that converter may be designed for isolation between the input and output; you lose that by shorting the two negative terminals, which may have other affects as well. \$\endgroup\$
    – mbedded
    Jul 17, 2020 at 20:30
  • \$\begingroup\$ @AJN I took the input from P+P- because I thought the TP5100 is capable of limiting discharge of the Li-Ion batteries and I wanted to use that capability. From my limited understanding the voltage at this point will be around 7.4V (assuming nominal voltage of 3.7V per battery). TP5100 has poor documentation apparently and the best I could find was TP5100 EN Documentation. I don't really get most of the technical details. \$\endgroup\$
    – Jithin
    Jul 18, 2020 at 6:06
  • \$\begingroup\$ @mbedded, To be honest I think I know why the D4 diode is used here. If i figured out the reason for the D2 diode then i don't remember it anymore. As you would have figured out the diagram is adopted from a different setup that I found on forums. I can't remember where or find it again. The output from the DC-DC converter would have to go through the MOSFET switch right? Can you please elaborate on where you would place it? Thanks \$\endgroup\$
    – Jithin
    Jul 18, 2020 at 6:10
  • \$\begingroup\$ @AJN, Here is a youtube video where he takes a look at this board TP5100 \$\endgroup\$
    – Jithin
    Jul 18, 2020 at 6:14

2 Answers 2


1 - Nothing else than the batteries shall be connected to B+/B- ports of the 2S BMS module. Its purpose is to protect the batteries for over charging, over discharging and over current. User access to charge and discharge is done using P+/P- ports only. You're right on this point with your first schematic.

2 - Connections must be short (low resistance) between the batteries and the 2S BMS module. So fuses are definitively not welcomed in these lines and could render the things worse regarding the voltage control. (Better put one in the P+ line if you want)

3 - You have to control the charge of the batteries (the BMS module doesn't do that). A module based on a TP5100 could do that. But you've to remove or change the limiting current, sets by default to 2A - way too much - by removing one R100 (0.1 ohm SMD): it will drop the current to 1A, but replacing both of them by a unique 0.2 ohm will provide 0.5A to the batteries, which seems much adequate in your case.

For different value, R(ohm) = 0.1 / I(A)

Don't forget to put the "SET" jumper in place for two Li-ion cells.

Input from your 12V/2A power supply, output to the P+ of the BMS module. Common GND.

4 - As suggested, the buck-boost XL6009 (sepic?) converter must be at the very end of the chain to precisely control the voltage delivered to the switch/router.

5 - You may use the external switching circuit proposed in the TP5100 datasheet, but you need to use the pin 6 of the TP5100 chip, which isn't available easily on the Chinese module. But if you're skilled enough in delicate soldering, it could be possible.

Keeping it simple, I think in your case that a double schottky diode in TO220 package could do the trick: one anode to the positive of the 12V/2A power supply, the other anode to the P+ of the BMS and the common cathode to the positive input of the buck/boost XL6009 converter. Double diode, could be replaced by any single diode, but select the schottky diode with the lowest Vf available according to the expected current to limit looses in them.

6 - One latest recommendation: the readily available TP5100 Chinese module doesn't provide thermal control of the batteries while charging (even if the chip itself is capable of doing that), so keeping the charging current low is the safest way.

Hope this could help you.


Below are my recommended changes; I believe this circuit is easier to understand, ultimately making it safer for a beginner.


There are two basic changes here:

  • DC-DC converter now is directly driving the output, giving you better control over that output
  • Power source is controlled by a simple "wired OR" using the two diodes

The FET switch was more efficient than the diodes, but I saw its complexity as a liability for you. Keep it simple for starters, and maybe upgrade if your battery life is insufficient.

The "diode OR" will supply power from the higher voltage input, so it uses your power supply when operational and the battery otherwise.

I have not fully reviewed your components; make sure all the power ratings are compatible, and especially make sure your battery cells are suitable for the charger and discharge. Power circuits can be dangerous; use at your own risk!

One final note. Make sure the supply is powerful enough to both power the router and charge the battery - which it will need to do when line power is first restored; you may need to reduce the charge current (your module seems to be adjustable) to meet the supply's limit.

  • \$\begingroup\$ This may or may not work with his boost converter on a 12 Volt input while trying to output 12 Volts. Boost converters expect an input voltage lower than the output voltage, but, depending on the design, may or may not pass input current onto the output when the input voltage reaches the output voltage. Of course, he could always replace the mains powered supply with a 9 Volt unit.... \$\endgroup\$
    – Hitek
    Oct 1, 2020 at 17:47
  • \$\begingroup\$ You may be right; the documentation for that module is pretty scant, but the controller's boost example did not allow Vin=Vout. That's part of the reason I included the comment about reviewing components (and removed part details from my block diagram). I was only addressing the high level design. \$\endgroup\$
    – mbedded
    Oct 1, 2020 at 18:26

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