connecting two 12V power supplies together

Power sources are not usually called "signals." You will be okay by diode ORing them...

It's unlikely that both 28V supplies will be=20 precisely the same voltage (due to design differences, component tolerances, etc.), =20 thus you will end up with an undesirable=20 current loop between the supplies.

What is the purpose of having two 28V supplies? Just to add current capacity?

I think that's what was bugging me... that the two power supplies aren't going to be clean regulated 28V sources... so I'm a little rusty on my electronics theory but there is something that leaves me uncomfortable about doing that.... just connecting them in parallel like that. I'm thinking a Diode OR type solution would be the way.... basically I'm trying to pass the DC power on to somewhere else... but the DC power may come from two different sources and either source (or both) may be plugged in at any time, but I just need to pass that 28V on through regardless of which source is connected.

much thanks

One way is to "diode or" the two 28V supplies. Connect the diode anodes to each of the supply outputs. Tie the cathodes together and to the input of the 5V regulator. Whichever supply has the highest output voltage will supply the load. This works best when the 28V supplies have current limiting. This ensures that when the dominant supply reaches its current limit, the other supply will take over.

yeah, if you're not concerned about sharing the load equally so as to maximize the output current available, just OR them and whoever's got the most voltage will carry it; being off constituting a very low output voltage.

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DIODe Or Solution sounds best... the only con about using that is the voltage drop across the diode, if at all possible I'd like to get the whole voltage across...

much appreciate the responses

Yeah, the diode drop can be a problem. You can minimize it by using Shottky diodes.

You can get MOSFET drivers that will implement the diode function with milliohm switch MOSFETs. see

Also, it's possible (if the only concern is disconnect) to use a single diode instead of two. The diode-connected supply will idle in normal operation, but will power the circuit if/when the direct-connect supply is removed. The efficiency loss is only for the duration of supply disconnect, and only for ONE of the two supplies.

To get the whole voltage across, see the diagram. To see why that is not necessary, read the last paragraph.

SupplyA ---+--->|---+ | | o | \\ | o------+----> to device | o | | | SupplyB ---+--->|---+

Use a switch (or relay) to designate one of the supplies as "primary". When it is removed, the device receives power from the "secondary" supply with a diode drop, until the switch is thrown (or relay transferred). That way, you avoid the diode drop most of the time.

But avoiding the diode drop is not helpful. You said the supplies are feeding a 5 volt regulator, so it will be dropping 7 volts anyway, and it doesn't matter whether the 7 volts is all dropped in the regulator, or ~ 6.3 is dropped in the regulator and ~.7 is dropped in the diode. So unless there is some other reason to avoid the diode drop, just use the diodes without the switch.

Ed

re

yeah, this is getting confusing. the title says two 12v supplies, the text says 2 28v supplies to be regulated down to 5v, now we're trying to avoid any voltage drop. ?

re

Thanks, for the relay idea.... a switch wouldn't work cause it has to be automatic.... and the power does more than just go to a regulator so getting the entire 12V passed is a concern, that's why I'm looking for something other than the Diode-OR.

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Ahhh.... sorry about the 12V/28V confusion..... that part is a bit arbitrary to me, there will be situations where I'm passing 12 and other situations where I'm passing 28. And the 5V regulator.... it's in the circuit as well but the power does more than just provide the input for the regulator.

Thanks

And what will you do for us if we tell you, other than disappear?

Ok. Whether a switch or a relay is used, there are implied assumptions in the circuit, because so far it is conceptual only, with no fixed details. (implied assumptions include supplies regulated within a diode drop of each other, a single known supply voltage, not the arbitrary input voltage you just mentioned in your reply to z, built in backfeed protection, current handling capability etc)

Assumptions can get you into trouble. :-(

Ed

You could use some electronics and big power FETs.

View in a fixed-width font such as Courier.

. . . . . VPC . .-----+-----+-----------------------------------. . | | | | . | | | | . | |B1 |B2 | . | --- --- L . | - - O . | | | N-CH2 A . | | | VN1 D . | | '-----+---D S ---. | . | | | G | | . | | | | | | . | | \\ ------- | VNC | . '-----|-------------|ARBITER|--+----------------' . | / ------- | . | | | | . | VN2 | G | . '-----------+---D S----' . . N-CH1 . .

Depending on the application, the arbiter could take any one of several forms, and more than a few of those require detecting a low batt threshold, and that is complicated when the battery is relatively arbitrary.

Except that your circuit doesn't work. The devices need to be "back-to-back" in series, as in this previously posted circuit...

(Body diodes _must_ be considered as sneak paths.)

...Jim Thompson

Hi Jim,

Fred's using N-ch mosfets with the drains on the negative inputs. The body diodes can't contribute to reverse current flow.

The only thing he loses (by not having back-to-back fets) is input overvoltage protection and the ability to "turn off" an input.

A similar arrangement would work for P-ch fets on the positive inputs with a controller such as the LTC4416. See, for example, figure 5 in this datasheet:

Regards, Allan

The circuit is a standard diode-OR when the FETs are off. Start from there...