Seamlessly switch between two DC power supplies - please help

Thread Starter

GlynHanmer

Joined Feb 14, 2014
7
Hi,

I have a device that runs on a supply as low as 11.5 V and up to 17 V, which draws up to 17 A, but usually around 2 A.

I have a battery pack which is rated 16.5 V when fully charged and a DC wall adapter that provides 13.5 V.

I would like to be able to plug in the wall adapter while the battery is attached, have this trigger the battery supply to be "switched off" and power the device straight from the wall supply.

I have tried to find solutions online but most of the stuff I've found solves a similar problem but when the auxilliary supply, wall, is at a higher voltage than the primary, battery. These solutions included or'ing diodes and a couple of ICs.

The space that I have to build the circuit in is quite limited so I would like a solution as small as possible. As the device is a primarily a portable battery-powered device, I would like the extra circuitry to not drain too much power and have a little drop over diodes as possible etc. The device start-up time is also about 2 minutes so I would love to be able to plug in and switch the battery, then unplug without the device losing power. But these are things that we would want for most circuits anyway :)

I like to think that I have a fair bit of electronics knowledge but I'm guessing that compared to people here on the AAC it's very minimal. I am very willing to learn though.

If anyone has either a solution straight away or could point me in the right direction to get started, I'd very much appreciate it.

Thanks for any help in advance,
Glyn.
 
Last edited by a moderator:

studiot

Joined Nov 9, 2007
4,998
Since you have to manually turn on the mains adapter anyway, why can you not simply switch the other off (ie disconnect the battery) when the mains adapter is live?

You have enough headroom voltage to put a power diode in series with each supply to prevent backfeed from the other, if you ran them in parallel.

Obviously such diodes would need to be rated for the max current (plus a margin so you need 20A diodes).

Will changing the supply form 16.5 volts to 13.5 volts, less the diode drops, disrupt the action of your load?
 

alfacliff

Joined Dec 13, 2013
2,458
a 20 to 35 amp bridge rectifier is whats used for this. just hook the two supplies to the "ac" terminals, and take the dc from the + terminal. whichever has the most voltage will supply the circuit. the bridges are easier to mount and heatsink, they need no insulator from the heatsink.
 

studiot

Joined Nov 9, 2007
4,998
a 20 to 35 amp bridge rectifier is whats used for this
Perhaps you could suggest some bridges with an insertion Vdrop of less than 1.8 volts at 20 amps?

If you hook both supplies to the same (AC) terminals, why will they not fight?
 

MrChips

Joined Oct 2, 2009
30,706
He meant connect the +ve of the two supplies on opposite AC terminals. In effect he is using only two of the four rectifiers.
 

mcgyvr

Joined Oct 15, 2009
5,394
Just use one of those DC power jacks that can mechanically disconnect another source upon insertion of the wall adapter plug..
 

MrChips

Joined Oct 2, 2009
30,706
Just use one of those DC power jacks that can mechanically disconnect another source upon insertion of the wall adapter plug..
...assuming that it does not disrupt the supply.

Can the switch handle 20A?

What kind of jack does the OP's wall adapter use?

Another thought came to mind - What does the wall adapter look like? What is its current rating?
 

bountyhunter

Joined Sep 7, 2009
2,512
Since you have to manually turn on the mains adapter anyway, why can you not simply switch the other off (ie disconnect the battery) when the mains adapter is live?

You have enough headroom voltage to put a power diode in series with each supply to prevent backfeed from the other, if you ran them in parallel.

Obviously such diodes would need to be rated for the max current (plus a margin so you need 20A diodes).
And a pretty decent heatsink for the diodes.
 

bountyhunter

Joined Sep 7, 2009
2,512
The simplest solution is to make the wall adapter DC voltage greater than that of the battery.
That would work OK.

In most designs, they use a series FET for the battery pack and some sensing circuitry to open the FET when the other supply is present. That is so if there is a problem with the battery, you can still run off the wall wart.
 

studiot

Joined Nov 9, 2007
4,998
In most designs, they use a series FET for the battery pack and some sensing circuitry to open the FET when the other supply is present. That is so if there is a problem with the battery, you can still run off the wall wart.
You would still need protection steering diodes. Switching the source of 20 amps, without interruption, is no picinic
 

bountyhunter

Joined Sep 7, 2009
2,512
You would still need protection steering diodes. Switching the source of 20 amps, without interruption, is no picinic
True. back in the industry, we called this "hot swap" and it was indeed a nasty problem to solve without glitching the power line or damaging one of the units.
 

Thread Starter

GlynHanmer

Joined Feb 14, 2014
7
Thanks to everyone for all your help and suggestions. I've been away and busy with work since I posted the original post. I have come back to be overwhelmed by your willingness to help!

Some replies to your suggestions follow:

Since you have to manually turn on the mains adapter anyway, why can you not simply switch the other off (ie disconnect the battery) when the mains adapter is live?
If it were me just me who would be using the device, that would be absolutely fine. Unfortunately, it's going to be other people who I can't rely on to do this.

Will changing the supply form 16.5 volts to 13.5 volts, less the diode drops, disrupt the action of your load?
Since the 16.5 volt supply is a battery, that would mean not fully charging the battery, giving a much shorter battery life.

a 20 to 35 amp bridge rectifier is whats used for this. just hook the two supplies to the "ac" terminals, and take the dc from the + terminal. whichever has the most voltage will supply the circuit. the bridges are easier to mount and heatsink, they need no insulator from the heatsink.
If the wall supply had a greater voltage than the battery supply, this would be a suitable solution. Unfortunately, the supply is standardised and will only provide the 13.5 volts.

Another thought came to mind - What does the wall adapter look like? What is its current rating?
The wall adapter outputs its power through a four pin XLR cable. Different supplies are rated at different currents but I have been told that under certain circumstances the device can draw a lot of current.

Just use one of those DC power jacks that can mechanically disconnect another source upon insertion of the wall adapter plug.
This would be great and we'd already considered it. Unfortunately, the space we have to work with can't fit in the extra bit of the component that is used for the switching in one of those power jacks.

In most designs, they use a series FET for the battery pack and some sensing circuitry to open the FET when the other supply is present. That is so if there is a problem with the battery, you can still run off the wall wart.
This sounds like something I should look into. Would you be able to give me some more guidance as to where I should look to learn how to build this circuit please? Or a simple diagram to get me started?
Originally, I was looking into MOSFETs, but I was finding it very hard to find depletion mosfets that would "switch off" when the wall supply is present. I believe that JFETs are depletion mode as standard. I had a go at playing with a few JFETs in LTSpice but couldn't really get them to work how I thought they would.

Again, thanks to everyone for all your help.
 
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