Subject: Re: PSU's: Exceeding transformer rated current
Date: Sat, 11 Jan 2003 17:39:12 -0000
NNTP-Posting-Host: pc1-heck1-4-cust126.hudd.cable.ntl.com (18.104.22.168)
X-Newsreader: Microsoft Outlook Express 6.00.2800.1106
"John Popelish" wrote in message
> Richard wrote:
> > "Don't know whether this forum is appropriate, but here goes.
> > In a half-wave rectifier circuit, a purely resistive load, no filtering, could
> > double the rated secondary current? The transformer will not get any hotter than
> > being loaded at rated current output, but would the secondary windings be prone
> > fail for overcurrent? TIA. Rich."
> Double the current produces 4 times the resistive heating. If that
> current occurs only half the time (half wave operation) then the
> effective heat would be half of that 4 times. If the half wave
> current was 1.4 times the rated current, this would double the IR
> heating, but the 50% duty cycle would reduce the average heat back to
> rated. All this assumes that you somehow compensate for the DC
> component (caused by IR drop reducing only one half of the applied
> voltage) added to the primary voltage that will soon saturate the
As you know, doubling the current produces 4 times the resistive heating, if the
resistance stays the same. Because then the voltage has to double to produce the
double in current.
With HW operation, the effective (rms) voltage reduces by half, 10v rms becomes 5 v
rms to the load. So, if rms current were to be the same, changing from FW to HW, we
would need to half the load resistance. Power dissipated then would be half the FW
heating position. So, you then have to half the load resistance again to bring it up
to the original resistive heating that we had with FW.
All this as you say is ignoring the complications introduced from parameters
associated with the supplying source, such as source internal resistance, core