Well, first of all, the linear type in that range aren't inexpensive . . .
Switchers are cheaper. That's why the use them so much. All that would
happen is that your voltage would drop out of regulation because the ripple
would cause the input voltage to be too low. I'd be not the least bit
afraid to try it. It probably won't break anything except maybe the fuse on
the supply.
-----Original Message-----
From: Aaron Christopher Finney <af-list_at_lafleur.wfi-inc.com>
To: Discussion re-collecting of classic computers
<classiccmp_at_u.washington.edu>
Date: Thursday, July 01, 1999 4:09 PM
Subject: Re: DC Power supply question...duh...
>Thanks, everyone who answered. I think I'm going to spring for another PS
>that I know is rated for the full 12 amps. Just out of curiosity, what
>could conceivably happen if I ran it at 12 amps and it couldn't handle the
>load? By the way, it is indeed an inexpensive, linear type...
>
>Thanks again,
>
>Aaron
>
>On Thu, 1 Jul 1999, Tony Duell wrote:
>
>> >
>> > I am risking the effects of that famous quote, "Better to be thought a
>> > fool..."
>> >
>> > I have a 15V DC powersupply that has no other rating on its sticker
than
>> > 180 watts. Now, with my [extremely] limited knowledge, I did the simple
>> > math and worked out that I could safely pull 12 amps out of this
>> > PS...am I right? Does it really work this way?
>>
>> In theory, yes.
>>
>> DC power = I.V. So 180W at 15V is, indeed, 12A.
>>
>> However, that is taking the '180W' to be the output power of the supply.
>> In many cases, the 180W is the maximum power it will consume from the
>> mains. And as power supplies are not 100% efficient, you don't get out
>> all the power you put in. In fact, the efficiency might only be 50%
>> (particularly if it's a linear PSU), so you could only draw 6A from it.
>>
>> I'd probably not take more than that from it, unless I _knew_ it was
>> designed to stand it.
>>
>> -tony
>>
>
Received on Thu Jul 01 1999 - 17:12:57 BST