I've got a question about power supplies

[myocardia]

So, what happens if I series the output from two 12v AC-DC converters, that have different output capabilities (amperages). I think I know that if I series two 12v @ 1A converters, I'll end up with 24v @ 1A, right? Or would it be 24v @ ½A?

Anyway, what I really need to know is what if I series one 12v converter that's 12v @ 1A with another that's 12v @ 1½A? Will I end up with 24v @ 1 or ½A, because it will be limited by the max output of the 1A converter, or will it be 24v @ the total of the two converters, divided by 2? Anyone know for sure?

 

[GuitarMachine]

well using the equation that 1 volt at 1 watt is 1 amp its pretty simple
so if you series two 12v @1A converters you have 24 watts right there and 24 volt since they are in series.

so its 1 amp at 24V

as for the next one 12v @ 1A = 12 W and 12V @ 1.5 A = 18W

so you have 24 V at 30W which is 1.25 amps


i hope im right

 

[Mark R]

You get the current rating or the lowest component; in a series circuit, the current is the same at all points.

The other catch is whether the outputs of the converters are floating or grounded. If the outputs are grounded (e.g. like in a PC PSU) then you can't connect them in series - because you'll simply short out one of the supplies. However, small supplies like those pregnant plug type converters are usually fully isolated, so there shouldn't be a problem.

The other problem with AC-DC converters in series, is that if one of them loses power input, you end up with reverse voltage across the other one, which will likely destroy it instantly.

 

[myocardia]

Originally posted by: Mark R
You get the current rating or the lowest component; in a series circuit, the current is the same at all points.

So that means that in the instance above, you'd end up with 24v @ 1A, or 24v @ ½A?

The other problem with AC-DC converters in series, is that if one of them loses power input, you end up with reverse voltage across the other one, which will likely destroy it instantly.

Okay, that's no problem in this instance, but thanks for the warning.