I am a learning newbie. I am trying to find online a circuit that is safe and effective for creating a battery back up for my modem which is 12 volts, 1.5 amps, my router which is 12 volt and 3 amps, and a WYZE camera, which is 5 volts, 1 amp. I would also like to add a charging port for a phone.
i have a 12 volt, 19 ah sealed battery for the project. I would like everything to run off mains power and the battery to float charge (but not overcharge) when power is available but switch to battery automatically when there is a power failure. I will have to figure out some sort of housing to put everything in to put beside tv in living room. Can anyone help me find a plan?

 

OK, an easy set of devices to power until you get to the camera and phone charging parts. IF you have an adjustable and regulated 12 volt mains powered DC supply then it can "just" be a matter of setting the power supply voltage to the battery recommended float charge voltage and connecting it to the battery. Other schemes will be a bit more complex, if that sort of supply is not what you have.

 

As MB2 suggested you could use a mains-powered 12V adjustable, ≥8A supply set to the battery float voltage (typically about 13.5-13.8V) which should be okay to power the router and modem directly, with a 12V to 5V buck converter to power the camera.
The charging port could be a car-type USB adapter that converts the battery voltage to 5V.

 

As MB2 suggested you could use a mains-powered 12V adjustable, ≥8A supply set to the battery float voltage (typically about 13.5-13.8V) which should be okay to power the router and modem directly, with a 12V to 5V buck converter to power the camera.
The charging port could be a car-type USB adapter that converts the battery voltage to 5V.

But won’t that reduce battery life by constantly pulling off the battery instead of the battery just be in the back up

 

But won’t that reduce battery life by constantly pulling off the battery instead of the battery just be in the back up

No.
The 13.5V supply is providing the modules' current along with also keeping the battery trickle charged.
Under those conditions, a small current is flowing into the battery, not out of it.

 

My caution with this suggestion, AGAIN, is that setting the supply voltage exactly right is rather critical. Really!

 

Here's an example of a 12V supply you could use.
It has a 10A max,, 8A continuous power rating, with the voltage adjustable ±15% which should go high enough for the battery trickle charge voltage.

It should be put in a vented enclosure to avoid overheating.

 

So connect the power supply to the battery. Adjust to 13.8 volts. Also connect to the battery the jacks to power the router and modem? Are any diodes used?

 

Also what would be a good diy housing?

 

So connect the power supply to the battery. Adjust to 13.8 volts. Also connect to the battery the jacks to power the router and modem? Are any diodes used?

May want to use a 10A Schottky diode in parallel with a resistor, with that in series with the battery connection to limit the battery charge current after the battery has been discharged.
The diode would carry the current in the forward direction with low voltage drop when powering the devices, but block current when charging, so the parallel resistor then limits the charge current.
A resistor value of about 1 ohm should work for that to give a maximum charge current of 0.1C to 0.2C, where C is the Ah rating of the battery (1.9 to 2.8 amps for your battery).

If you want to improve efficiency and minimize the diode forward voltage drop, you could use an "ideal diode" circuit, which has a P-MOSFET and an op amp to generate a forward-drop of just the load current times the MOSFET's on-resistance, which can be very low.

what would be a good diy housing?

Don't have any recommendations.
Look up "vented electronic project box" in Google or Amazon and you should find something appropriate.

 

Limiting the charge current is definitely important. Note that a 1 ohm resistor will need to handle significant power; at the suggested max of 2.8 amps, P = I*I*R = (almost) 8 watts. And that assumes the battery voltage is only 2.8V below the power supply voltage. But what if the battery runs all the way down? When the power comes back on, the battery might be 12V or so lower than the supply, so the resistor will dissipate P = V * V / R = 144 watts. Unless you have a very large resistor, it is going to let out the magic smoke.

A larger value resistor will dissipate less heat, but will also cause the battery to charge more slowly and you still have to get the float voltage exactly right to avoid damaging the battery. Even then, it won't be ideal, as lead acid batteries are ideally charged at a higher voltage then floated at a slightly lower voltage (to minimize sulfation without slowly boiling off the water in the electrolyte).

An alternative to using a resistor is to use a battery charger to charge and maintain the 19 AH battery. Look for one that can be left connected indefinitely. Something like the CTK56-865 would be appropriate. With only a 0.8A charging current, it would take a day to recharge the battery, but for a backup application that would probably be fast enough. You could use a higher-current charger, but the only benefit would be that the battery would be recharged and ready for another power outage sooner after the power comes back on.

Connect the battery to your modem and router through a diode that can handle 5A. To handle that much current, you could put multiple diodes in parallel (leave some space between them for airflow) or use use an "ideal diode" like the SZ-D40100 (although it is huge overkill). Note that circuits using just a P-channel MOSFET with no ideal diode controller can protect against reverse polarity, but not against reverse current flow, so they won't work here.

Also connect a power supply through another 5A diode to the modem and router. That will form a diode-OR which will supply current from whichever source has the higher voltage. Set the power supply voltage to be higher than the battery charging voltage (maybe 14.2V to 14.5V). It should be able to supply as much current as the modem and router typically use (measure it; probably less than 5A), but would not have to handle the peak current, as the battery can help out with occasional surges.

I realize that is more complicated and more expensive than you were probably planning on, but not smoking when the power comes back on after a long outage is worth something Also, a nice side effect of a good battery charger is that it can indicate the the state of the connected battery (for example, the CTEK will show the charging state within a few minutes and indicate some other problems after a few hours). FWIW, I'm not associated with CTEK; it is just the brand that I am familiar with.