If you can get inside the power supplies and find the reference diode, use
just one of the 3 and join the rest to it, taking the others out of circuit.
This way most of the differential drift (ie drift between each one) will be
avoided and they will track better. Each psu will also have a pair of
resistors (or a short chain) across the output: think of a way to co-locate
these so they share the same temperature; even better, find a way of them
sharing the same pair - needs a bit of thought - I haven't got that far, but
exchanging them for high stability types would help and only cost a few $.
David
G3UNA
>I would put a diode rated for the current in series with each supply on one
> side and paralleled on the other. Just be sure the supplies are all set
> for
> the same voltage.
>
>
>> You might want to look up the connecting wire resistance per foot. Some
>> length will provide the small voltage drop mentioned by Bill.
>>
>> 73,
>> Gerald K5GW
>>
>>
>>
>> ORIGINAL MESSAGE:
>>
>> On Mon, 14 Dec 2009 09:51:35 -0600, chas <chasm@texas.net> wrote:
>>
>>> How do I connect three 30A power supplies together so as to obtain 90A
>>> ?
>>
>> REPLY:
>>
>> The usual way is to insert a small value resistor in the + lead if each
>> and then
>> parallel them. The value of the resistor should create a drop of about
>> .5
>> V or
>> so under full load. Be sure the output V of each supply is adjusted to
>> match the
>> others as close as possible.
>>
>> 73, Bill W6WRT
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