The Cyber-Spy.Com Usenet Archive Feeds Directly
From The Open And Publicly Available Newsgroup
This Group And Thousands Of Others Are Available
On Most IS NNTP News Servers On Port 119.
Cyber-Spy.Com Is NOT Responsible For Any Topic,
Opinions Or Content Posted To This Or Any Other
Newsgroup. This Web Archive Of The Newsgroup And
Posts Are For Informational Purposes Only.
Subject: Re: PSU's: Exceeding transformer rated current
Date: Tue, 7 Jan 2003 23:55:08 -0000
NNTP-Posting-Host: pc2-heck1-4-cust111.hudd.cable.ntl.com (22.214.171.124)
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
"Richard" wrote in message
> "John Woodgate" wrote in message
> > You have a direct current in the secondary winding, which is the average
> > load current. This greatly increases the losses in the core, resulting
> > in the transformer heating up much more that you would expect from the
> > fact that the secondary current appears to be flowing for only half the
> > time.
> I don't doubt it, but I cannot see why. Rich.
Actually I think I see a part of the problem.
In my example there are two situations. The internal resistance of the primary is
Where the transformer supplies an ac current, ie say without any rectification, the
transformer rated at 10v rms @ 1A supplying it's max rating has 0.5 watts dissipated
in the internal resistance (I^2 x R; 1A x 0.5R) . Connect as half wave rectifier and
the available voltage for R load is 4v when dissipating 10 watts in R load and the
current is 2.4A. Dissipation in R internal is 2.4A ^2 x 0.5R = 2.88 watts. So, we
are supplying 10watts in both instances to a load, but greater copper losses where we
have used half-wave rectification. And that's assuming the 2.4A rms current, which
would be 6.7892504A peak would not melt the wire.
Go Back To The Cyber-Spy.Com
Usenet Web Archive Index Of
The sci.electronics.design Newsgroup