Re: low-cost 1800-amp heating source

In article <1186051537.860371.213280@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Winfield <winfieldhill@xxxxxxxxx> wrote:
Tony Williams wrote:

I was interested in Ken's idea, so sketched a
possible circuit, as below.

+ Pri - Load current
230V high +------////////---------->-----+
T1 ======== |
DC-------////////-----+ |(
+ Sec - | |( L(leak)
| |(
.----------------' | Win's Load.
| \
| - Sec + /Rload
+---////////---------DC \
T2 ======== |
230V low +-------////////----------------+
- Pri +


For quick estimates, could we guess that the DC current
would be 3 to 5 times higher than the peak AC current at
the VA rating of transformer?

I don't think it is even as high as that Win.

If you were to leave the Pri open circuit, connect a
voltage source to the Sec, and turn it up until the
current waveform had nice wide flat tops then the
measured Ipeak could be a reasonable estimate of the
polarising DC amps needed to saturate the core.

3 to 5 times the secondary magnetising current peak
is probably much lower than 3 to 5 times the peak
secondary load current.

The control power would then be the DC amps times
the winding resistances.... but see below on what
could buggerate that low control power.

Given that the DC winding resistance of the transformers is quite
low, the DC- control requirement doesn't look too bad. A fair
amount of DC control current, sure, but not too much voltage...

There could be a little ambush in the two-transformer
saturable reactor. Those two series'd secondaries
may not cancel completely, so there could be a
residual AC voltage at their terminals. I suspect
that the DC current source must have a high enough
compliance/supply voltage such that no AC current
flows in the secondaries.

The same (or worse) AC current flow could occur
with the circulating currents that would arise
from the paralleled windings in your previous post.

This is getting interesting.......

--
Tony Williams.
.

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