Re: DC magnetisation of transfromers and chokes.

Patrick Turner wrote

The formula for woking out turns on an output transformer is

N = 22.5 x V x 10,000 / ( Afe x B x F )

where N is turns,
V is the voltage in Vrms across the winding,
10,000 is a constant for all equations,
Afe is the core area in sq. mm,
B is the flux density in Tesla,
F is the frequency.

I try to have the ac Bmax = 1.2 Tesla at 14Hz, so the bass response of
any amp is very good.

However, where you have an OPT for an SE amp there is DC in the core and
one has to limit ac Bmax to 0.6T
approximately to allow for 0.6T of dc magnetization of the core
in a transformer carrying DC through its primary.

Some interpretation here may help answer your question, because this paragraph
assumes that AC and DC effects are comparable. That is, you choose 0.6 because
it is half of 1.2, regardless of the fact that one is DC and the other AC.

OTOH, it may appear you are assuming a linear relationship: that half the
current will get you half way to your allowable Bmax. But also you realise, at
least in the case of AC, that µ (well done for finding the right character all
on your own this time!) is not constant and that therefore the relationship
between current and B is not linear.

I suggest part of this apparent anomoly can be explained as follows.

The BH curve for the material in question has no significant frequency
component. If you take the section of the curve between H(0) and H(your Bmax),
you will have an "S"-shaped curve. If you assume that the "S" is symmetrical,
then half of H, and hence half of current, will get you half way to Bmax.

It does not by any means follow that 1/10th of the current will get you 1/10th
of the way to Bmax, even though it will get you to 1/10th of H(Bmax).

Because H is proportional to current, you can say that 6ma of unbalanced DC
through a PP transformer will subtract 6ma from the unbalanced Imax in that
direction, where Imax is the current required to produce Bmax. The peak-to-peak
value of unbalanced Imax will be reduced by 12ma. This is not a very useful
indication, however, *because it does not take into account secondary current*.
For in-bandwidth AC, the secondary current cancels most of the field.


The formulas I have give the DC flux in Tesla = 12.6 x µe x N x Idc / (
FeMl x 10,000 ),

where 12.6 is a constant, µe is the effective permeability when a gap is
installed, and is often a value
below 500, and calculated with another formula taking into account the u
max for interleaved material
and the air gap,
N is the primary turns,
Idc is in amps dc,
FeMl = iron core magnetic path length in millimetres,
10,000 is a constant for all equations.

Er...eh? What is it? What are it's units?


For example, a 100mm stack of 50mm tongue E&I material with 600mA of
Idc, and 1,480 turns and µe = 184
and FeMl = 280mm has a dc field strength

= 12.6 x 184 x 1,480 x 0.6 / ( 280 x 10,000 ) = 0.73 Tesla,
so if total ac and dc flux cannot exceed 1.2T, then
I could have 0.46T max for ac Bmax.


However, if there was not a gap in the transformer, how does one
calculate the DC field strength?
Is µe just µ with ungapped material?
in my case µ is 17,000 maximum for the lams that I have when testing
with ac.

There is a case where I want to build a push-pull amp with two 1,480
turn windings
in series with balanced dc flows in each 1/2 of the transformer, in the
standard fashion.
The PP has the two windings in series, with a CT taken to the B+, and a
fully interleaved stack of
GOSS, 100 mm stack of 50 mm tongue material.

If there is a DC imbalance to the two tubes working in PP, say 6mA, then
I could have a net
dc flow from one end of 2,960 turns to the other of 6mA, so the using
the above formula,

dc Bmax = 12.6 x 17,000 x 2,960 x 0.006 / ( 280 x 10,000 ) = 1.36 Tesla

Bmax makes no sense with DC, surely? You just mean B.

Are you sure you aren't mixing up units? Have you omitted a factor of 4pi? Or is
that where the 12.6 comes from?...

Aha...I see...12.6/10,000 = 4pi * 10^-4...that rings a bell.

So my points about the non-linearity of µ, above, don't seem to entirely explain
this particular anomoly without calling into question your method of calculating
Bmax for AC. The difference also seems too great to explain by non-linearity.

Consider this. µ is defined as the slope of the BH curve, rather than the static
ratio. i.e. dB/dH rather than B/H. Taking the S-shape of the curve, the slope of
the centre section of the "S" shape is considerably steeper than the slope of a
line drawn from the origin to your point of Bmax. If your value of µ is for the
slope of that linear, centre portion of the "S", then your formula will
underestimate the H, and hence the current, required to produce Bmax. But this
underestimate will be the same for AC and DC.

This seems a "silly result" because in many amps I have seen a dc
imbalance, or net dc flow in one direction
and seen no saturation problems.

What would you regard as saturation problems? What did you measure? You seem to
be confusing several issues. In the first case, when you calculate the the Bmax
of your gapped transformer, you consider voltage rather than current.
Effectively you use the voltage to derive the current through the primary
inductance, and hence the field strength. If you were to measure the primary
current, then for the same result you would need to do so with an open
secondary, otherwise you would be mixing up balanced, non-magnetising current
with unbalanced, magnetising current.

When you come to DC, you measure current rather than voltage, and this is
entirely magnetising current because it is outside the bandwidth of the
transformer so there is no opposition from the secondary.

When you talk of "saturation problems", you seem to be confused between the two
views. It is quite possible that a PP transformer could be close to saturation
with 6ma of unbalanced bias current, and yet still be capable of 100ma
(whatever) AC swing without saturation, as long as the AC is within the
bandwidth of the transformer. It is only the unbalanced, magnetising current
that counts.

So unbalanced DC will reduce the frequency at which saturation will occur at
full power, and decrease the output voltage at which it will occur. However, a
decent amp should in neither case be limited by saturation...there should be
some margin between saturation and the low-frequency roll-off due to primary
inductance, and between max power and in-band saturation voltage.

So "saturation problems" may not have been readily apparent.


Does anyone have a better idea how to **calculate** the dc flux density
in gapped or ungapped
laminated steel cores?

Use the BH curve for the material. Lamination doesn't matter, but alignment
does, so to be precise you need to divide the core into sections with different
curves and add them up like you do with the gap and the iron.

perhaps the µ for dc use varies non linearly with dc flow so that µ is
low at low dc and high with high dc,
similarly to what ac voltage produces.

Indeed. In which case you may be overestimating your iron requirement for AC,
depending on the conditions applying to your quoted value of µ.

cheers, Ian


.

Relevant Pages

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