Power redundancy idea

[Nepalien]

Hello all,

I am using 7805 regulator from 12V source to power a modem. Current drain is about 500 mA. I also want to use a AC adapter (5V 1 Ampere model) at the same time. In other words when there is a electricity modem gets its voltage from AC adapter and during outage it gets from 12V battery via 7805 regulator.

So I am thinking of simply hooking the AC adapter in parallel with 7805 output. Would that be a wise move? Thanks.

-Nepalien

 

[Nigel Goodwin]

No, it would be a really bad move - you need two diodes to join them together (an OR gate).

 

[MikeMl]

Why dont you get a 9-12V AC adapter, and do the diode ORing on the input side of the 7805?

 

[Nepalien]

No, it would be a really bad move - you need two diodes to join them together (an OR gate).

I would appreciate if you can explain in terms of how I connect the diodes and where. Is this required to protect the battery from getting voltage from adapter? Thanks.

 

[MikeMl]

I would appreciate if you can explain in terms of how I connect the diodes and where. Is this required to protect the battery from getting voltage from adapter? Thanks.

Here is what I was talking about.

 

[Hero999]

Except V2 needs to be greater than V1 otherwise it will always run off the battery, so use a 15V wall wart.

I think it would be a better idea to use a 6V battery, a low drop-out regulator and a 9V wall wart.

 

[MikeMl]

Except V2 needs to be greater than V1 otherwise it will always run off the battery, so use a 15V wall wart...

Whoops, good catch.

 

[JimB]

Another idea would be to put a suitable resistor in parallelwith D2 in MikeMLs circuit and trickle charge the battery to ensure it is always fully charged.

JimB

 

[Hero999]

I forgot to say that, if a 6V battery is used, the diode in series with the battery would need to be a Schottky, otherwise the voltage drop will be far too great.

Here's a datasheet for a 5 regulator with a dropout voltage of about 240mV@500mA:
https://www.electro-tech-online.com/custompdfs/2010/04/LP8340.pdf

Here's a Schottky which drops 375mV at 500mA
https://www.electro-tech-online.com/custompdfs/2010/04/MB2FMBRS340.pdf

This means it'll run until the battery voltage is just over 5.6V before the output voltage drops below 5V.