From: John Popelish
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Subject: Re: PSU's: Exceeding transformer rated current
Date: Wed, 08 Jan 2003 01:52:00 GMT
NNTP-Posting-Date: Tue, 07 Jan 2003 20:52:00 EST
> "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.
> You are saying that a transformer rated at 1A rms current when supplying power during
> both halves of the cycle cannot supply 1A rms if supplying power for only half the
> In both cases I^2 losses (I think) would at least be the same. But the peak
> currents would be different, In first case, the peak current is 1.414A, I'm not
> really sure what the peak current would have to be for 1A rms in the second case.
> Mind gone blank. Bedtime for me. Rich.
The limitation is not caused by the heating of the copper windings
during the half cycle that current flows, but during the other half
cycle. With a half wave output, the primary winding has IR drop only
during one half cycle, so there is a bit more voltage applied during
the unused half cycle. The core integrates the total volts over time,
so this total, instead of adding up to zero, accumulates flux in the
direction that the stronger half cycles drive it. In a few cycles,
the core saturates before the end of the unused half cycle and the
winding inductance essentially disappears, leaving the primary winding
as a low value resistor across the line for the remainder of that
half cycle and for the same fraction of all those future half cycles.
This big blast of primary current once a cycle is what accounts for
the rapid heating of a transformer loaded with a half wave rectified
load. In effect, the transformer is a magnetic amp that switches from
an inductor to a short circuit every cycle, with a switching delay
based on the DC passing through the secondary.