[Ian Iveson]

PS I haven't distinguished between PP and SE, and obviously
for the latter you need some bias anyway. The important
point is that the steepness of the sigma depends on where
you are on the curve *and* on the signal amplitude. Small
signals will have flatter loops the closer you get to zero
bias, but a larger signal will result in a steeper loop at
that same low bias.

What I particularly want to attack is the notion that, for
PP, zero inductance at the crossing point leads to a kind of
crossover distortion. ********. The sigmoid is always
steepest at the crossing point. The problem for an unbiased
PP transformer is that the sigmoid is flatter for smaller
signals, and very flat indeed for infinitesimal signals, so
bass extension may be lost *for small signals only*.

Finally, back to your question. For highest max power you
should bias the transformer half way between zero and max
current, rather as you might a power valve for SE. That's
also likely to be the point for lowest distortion, happily.
Because of the gap, the BH curve is shallow and so B is more
linear wrt H, and where you are on the curve doesn't matter
so much.

Is it not an assumption of your example specs that bias will
be half the saturation current?

Ian

"Ian Iveson" wrote in
message ...
So my question is, how does the inductance vary with (dc)
current. If it measures higher with lower dc current then
if I use a 10W instead of a 5W but with a lower dc
current, do I get even higher inductance because of the
lower current?

This isn't a simple question, IIRC from longago posts. I
venture the following expecting a tongue-lashing.

The inductance varies with the slope of the BH curve. If
you look at one common, DC depiction of that curve, it
appears as a sigmoid, shallow at the zero crossing because
the magnetic domains exhibit what you might call stiction,
becoming steeper, and then shallow again as the majority
of domains reach their elastic limit, and some reach the
yield point.

The yielding dissipates power, resulting in hysteresis.
Consequently, for AC, the sigmoid becomes a loop, steepest
at the zero crossing, and shallow at both extremes of its
variation. What's more, any AC signal will generate a
sigmoid, with the magnetic bias affecting how flat or
steep the sigma-shaped loop is, according to where its
centre is on to the DC BH curve.

Break for ridicule.

Now, it seems from the DC curve that inductance should be
zero as B crosses the H axis. Much nonsense arises from
this interpretation. In fact, for pure AC the curve is
steepest there, both on the upward and downward journey.

However the sigmoid is flattish for small AC signals, so
the mean inductance is lowish. Then it gets steeper as the
signal increases, so the mean inductance rises, and then
begins to flatten out at its extremes and the inductance
falls again, as the core saturation approaches.

Still with me?

Add DC and small signals see a higher inductance, but the
average inductance for larger signals doesn't increase
very much, especially if the signal results in the core
getting to the sharper curve approaching saturation.

I would have thought that the great benefit of using a
transformer rated for high DC current, assuming the same
rated inductance, would be that it gives you greater
separation between roll-off and saturation. That means you
can use more feedback to extend its bass range at a given
max signal. Alternatively, some designs bias the SE
transformer and deliberately oversize it, to avoid the
bass disappearing at small signal levels. So they say, I
think.

So, the answer is: yes and no.

Ian


"Ian Bell" wrote in message
...
Just wanted to add a related question. In looking up SE
transformers on the net many that quote inductance do so
at a specified dc current. So the difference between a 5W
and a 10W transformer might typically be the 5W has a 10H
primary inductance measure at a dc current of 48mA
whereas a 10W type would have an inductance of 15H at a
dc current of 64mA.


Cheers

Ian