Re: Low THD transformer circuit
- From: Eeyore <rabbitsfriendsandrelations@xxxxxxxxxxx>
- Date: Sun, 19 Nov 2006 12:12:33 +0000
Robert Baer wrote:
Eeyore wrote:
Robert Baer wrote:We were speaking of "ideal"; in practice the input is equal to the
Eeyore wrote:
martin griffith wrote:
Eeyore <rabbitsfriendsandrelations@xxxxxxxxxxx> wrote:
John Larkin wrote:
martin griffith<mart_in_medina@xxxxxxxxxxx> wrote:
Just found this
http://www.pat2pdf.org/patents/pat4614914.pdf
It's how Audio Percision let very low thd from transformers
Why not load the transformer secondary into a summing point
How about the real load ?
and drive
the primary through a resistor, or from a current source? That would
keep the core at nearly zero flux.
I'm not clear how you can have near zero flux though.
Graham
I've seen it done on Quad amps, apparently even with ferrite
transformers for the input stage. The Beeb had a go at it as well,
sorry no URLs
There's this little equation that says E = -N.dthi/dt though !
No flux means no voltage.
Graham
Yep!
And an (ideal) op-amp works with zero voltage difference at the
input, where zero input times infinite gain gives exactly what was
needed at the output.
That's a very silly reply.
Graham
output divided by the open-loop gain.
So an error voltage (input) of a microvolt, times a gain of a million
gives one volt drive for the feedback (and load).
Thus even a modest open-loop gainof a million, one gets reasonably
close to the ideal.
You won't find gains of 1/2 million at moderate audio frequencies.
More like a few thousand.
Graham
.
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