Air gap on SE output transformers etc.
Patrick Turner
when trying to build an OPT for a single ended amp with one
output device or a number of them in parallel
is the determination of the air gap as well as other considerations.
I'll try to make my logic flow about the issue as brief and easy
for anyone to follow. It is applicable for any type of SE amp from
something making 1 watt
with a 12AU7 or 60 watts with a pair of 845, or 100 watts with a pile
of mosfets.
The air gap name is a misnomer. It should really be called a plastic
gap, or paper gap
or non magnetic material gap in the magnetic circuit loop around the
core material.
But convention calls the gap an air gap, but of course
we use some plastic or paper which works like air, but which will
withstand the force of the iron trying to come closer together
under the magnetizing effect of the direct current flow in the
output tranny turns.
Let me outline the experience I have just had with my second
completed 60W SET monoblok amp using a pair of 845 in parallel.
The conditions of the tubes have Ea = +1,050V, Ia dc = 150 mA.
Anode load is 6k, and PO max = 60W with 4 ohms at the sec.
After several design attempts for the OPT details I came up with the
following recipe.
Core is a 72mm stack of 51mm tongue GOSS material with a maximum fully
interleaved µ = 17,000,
but with close butted Es and Is, and no real gap, µ = approximately
1,000.
The primary has 2,700 turns in 20 layers of 0.45mm Cu dia wire giving
135 turns per layer.
The 20 layers are divided into 4 primary sections of 5 layers each.
The secondary is 5 layers of 0.9mm Cu dia wire with each layer having 72
turns.
The 5 layers are interleaved with the 4 P sections
to give a pattern = S P S P S P S P S, or 5S x 4P.
The last S layer wound on is a secondary layer and it is divided into 4
sub sections of 18 turns each.
There is 0.05 polyester insulation between primary to primary layers and
0.5mm between
primary to secondary layers.
The OPT can be arranged for a wide range of output tubes and supply
voltages
varying from a quad of KT88/KT90/EL34 in PP or SE,
and for a pair of 845 or 211 in SE.
Other uses could be found if the S sections were further divided
to give a higher maximum turn ratio.
There is enough primary inductance available to allow PP operation of
the KT88 etc.,
without having to unpot the transformer and alter the air gap size.
The resulting turn ratio and impedance ratio options for 2 x 845 or 211
SE parallel
are as follows...
OPTION A,
2,700 : 72 turns with 5 x 72 turn S sections in parallel = 37.5:1.
Z ratio = 1,406:1.
The lowest anode load value that can be used without exceeding 10%
winding losses = 2.8k.
So the range of Z matches are
2.8k:2, 4.2k:3, 5.6k:4, 8.4k:6, 11.2k:8, 16.8k:12, and 22.4k:16.
Option B,
2,700 : 90 turns with 4 x 90 turn S sections each having 72 turns PLUS
18 turns in series, = 30.0:1.
Z ratio = 900:1.
The lowest anode load allowable will still be 2.8k, giving a range of Z
matches of
2.8k:3.1, 4.2k:4.7, 5.6k:6.2, 8.4k:9.4, 11.2k:12.4, and 16.8k:18.8.
Option A is the one I have for the OPT because my customer's speakers
are a
nominal 4 ohms, and with the added winding resistance the anode load is
near enough to 6k.
I get 60 watts at the output, and there is 600Vrms across the primary.
Its important to KNOW what the design anode signal voltage will be
because it affects
the iron's capability to operate without saturation at extreme low
frequencies.
If the E&I laminations were fully interleaved, and there was no NET dc
in one direction only,
as in the case of a PP amp, or "parafeed" SE arrangement, then the
maximum
iron permeability, µ will be 17,000, and the P inductance maximum =
2,047 Henrys
using GOSS E&I material I have.
With close butted Es and Is, but not interleaved, the µ is called the
effective µe,
and would be about 1,000, a huge reduction, because the act of merely
butting the iron
Es and Is is the equivalent of a quasi air gap and hence the iron path
is lengthened
and its the same as altering the iron permability to a lower and more
constant value
according to frequency and voltage levels.
We will see why will never want µe to exceed 1,000 in an SE amp.
There are TWO operating factors that cause the magnetization of the iron
to vary, the ac, and the dc.
The ac magnetisation in Telsa, Bac, is calculated for above 10Hz as
follows..
B = 22.6 x primary Vrms x 10,000 / ( stack height mm x tongue mm x
Frequency x primary turns )
22.6 and 10,000 are constants needed for all equations to work with the
units nominated.
In this case we wish to know what is happening to the iron at the lowest
frequency of interest, 20Hz,
and at the design Vac maximum or 600Vrms.
So B = 22.6 x 600 x 10,000 / ( 72 x 51 x 20 x 2,700 )
= 0.683 Tesla.
Let's say 0.7 Tesla.
At this point, it isn't necessary to know if the air gap variation
affects the ac Bmax at all or not.
The µe isn't in the above equation, so apparently the Bac isn't affected
by the µe.
The iron can be magnetised up to around 1.6 Tesla before the rapid onset
of
full magnetization and magnetic saturation. Beyound 1.6T, the iron
molecules cannot
be magnetized any further and there is no increasing magnetic field with
increasing voltage applied
to oppose the flow of current as a result of the applied Vac, and for
part of the wave form the
OPT primary inductance falls towards zero Henrys, and the OPT has
horridly low
shunting load about equal to the low winding resistance of the primary.
So you cannot and must not magnetize the OPT core beyond its capability
which is about
1.6 T for GOSS and up to 2.4T for cobalt alloy cores which cost about as
much as NASA's budget
per anum if you could get hold of some. Some ppl would use
50% of the laminations in the form of nickel, and 50% GOSS, and
that's what the original Ongaku OPTs had originally for a lone but
splendid 211.
I do not know what the max B for the Ni-Fe core is, but perhaps the
Ongaku designer
who died a year ago had many reasons; I know he sure had a lot of
claims,
but perhaps the Ni-Fe has a smidgin less distortion than just plain
GOSS.
GOSS varies now depending on where its made and the heat treatments
involved,
and in 1955 the highest µ GOSS went about 5,000, but now its 17,000
from my sources, so perhaps I don't need any damn expensive nickel.
But for brevity, let us stay with 1.6 Tesla.
If we know the ac will magnetize the iron to 0.7T, then the dc could be
allowed
to magnetize the iron a further 0.9T, before saturation commenced.
In fact though, this sails too close to the wind, and we would try to
have the Bac < 0.8T, and the Bdc < 0.8T.
The Bac and Bdc don't need to be equal, but could be 2:1 ratio, but
generally the 1:1 ratio is the best because you will have made efficient
use of the wire and iron,
and you won't need a forklift to lift the darn OPT.
And you won't need a zillion P turns, which might ruin the HF
performance.
How to choose the core size and calculate the turns and wire sizes is
all dealt with at my website.
Now we want to know what the Bdc will be.
We also want to know how much primary inductance is required
because this Lp is in parallel with the anode load on the triodes.
As the F goes lower, the reactance in ohms of the Lp goes lower.
So to make 60 watts into the nominal anode load of 6k, and be able
to get good LF extension of the amp, we wouldn't want XL, the inductive
shunt reactance
of the primary inductance to be less than 6,000 ohms.
When XL = RL, the load seen by the anodes is 0.707 x 6,000 ohms, and
there will be a 45 degree
phase shift, and the load line will become an elipse, and the full 60
Watts just isn't quite available like it is at say
1 kHz, the reference mid frequency power, and cut off distortion will
occur dur to the
eliptical load line beginning to traverse the Ia = 0.0mA line on the
load line graphs.
So we will want XL = 6,000 ohms at no higher F than 20Hz, preferably
below 20Hz.
XL = L Henrys x Freq x 2 x pye = L x F x 6.28. ( 6.28 x F is otherwise
known as theta.F, a very common quantity
used in reactance figurings. Theta is a greek funny type of "w". )
So L = XL / ( F x 6.28 ), nice and simple eh.
Here Lp minimum required = 6,000 / ( 20 x 6.28 ) = 47.77 Henrys.
Let us aim for about 50H.
The formula for inductance with an iron core is
L Henrys = 1.26 x Np squared x stack mm x tongue mm x µe / ( ML of iron
in mm x 1,000,000,000).
1.26 and 1,000,000,000 are constants to make the equation work in metric
millimetres.
I only use metric, so don't ever get confused and try to work in inches
and other old fashioned things like feet, yards, gallons, walnuts,
cigars etc.
So from the above equation, we could work out what µe we would want
if we know all the other things in the equation, and we do, so,
µe needed to get wanted Lp = Lp x ML x 1,000,000,00 / ( 1.26 x Np
squared x S x T )
The iron magnetic length for wasteless pattern 51 tongue E&I material =
280mm.
All wasteless pattern E&I material has ML = 5.5 x tongue size, S, in mm.
So the required µe = 50 x 280 x 1,000,000,000 / ( 1.26 x 2,700 x 2,700
x 51 x 72 )
= 415.
The µe, effective permeability is just a number, with no units, OK.
Its actually the number of times the air cored inductance is increased
by the presence of the iron.
Now what value is the Bdc?
Bdc IS affected by gap in the core, and the µe becomes very important.
Bdc field strength for an OPT ( or choke ) = 12.6 x µe x N x Idc / ( ML
x 10,000 )
So, is the µe = 415 OK with regard to Bdc?
Bdc in this case = 12.6 x 415 x 2,700 x 0.15 Amps dc / ( 280mm x 10,000
)
12.6 and 10,000 are constants for all equations.
Bdc = 0.756 Tesla.
This is very lucky now isn't it! the wanted Bdc < 0.8T, but note that
the Bdc depends on the µe, and if µe was only 200, then Lp would be too
low,
but Bdc would be nice and low, and if the µe was say 800,
the Lp would be unecessarily high, and Bdc too high.
How do we make sure the µe is really near what we calculated above?
The air gap distance in mm is set to give the wanted µe.
Now with Es and Is jammed together but not interleaved, µe is
approximately about 1,000.
µe = µe max of iron butted close without a real gap / ( 1 + [ µe max x
gap in mm / ML of iron in mm ] )
Solving for gap, g, = ML x ( [ µe mx / µe ] - 1 ) / µe mx.
In this case without a real gap, µe = 1,000 and wanted µe = 415, so
gap = 280 x ( [ 1,000/415 ] -1 ) / 1,000.
= 0.39mm.
Now this figure is the total air gap distance.
In butted E and I there is an air gap each side of each window in the
core for the wire,
so the actual gap becomes 1/2 the above calculated value, so about 0.2mm
might be OK.
If you were very lucky, and just fitted 0.20mm of material for the gap,
then all the above wanted conditions might be right, and you'd not have
to test the
OPT in the amp to check that indeed you have 50H of inductance,
and that indeed there isn't any saturation above 20Hz at full output
voltage.
Hmm, too many might be this, might be that, so we need to prove
ourselves right.
There is a second way to calculate the gap.
This way says the max µe is the value with all Es and Is maximally
interleaved,
in this case, its 17,000.
So the equation for gap g = 280 x ( [ 17,000 / 415 ] -1 ) / 17,000
= 0.65mm.
But the only time the maximum µ of the iron is at 17,000 is when Bac
= 0.6 Tesla, ie, when the anode signal voltage is
about 3/4 of the maximum.
At levels we are interested in is well below the PO max level, and
at say 20% of the VO max, or at 120Vrms, and at this lower Bac, the µ
will be
also and if you plotted the curve for µ versus Bac,
you'd find µ at about 8,000 at 120Vrms of signal with high µ material
like mine.
And at anode voltage of 10Vrms, µ will have fallen to maybe 1,000....
The µ without a gap varies widely with voltage across the coil and with
frequency,
and somewhat non-linearly, hence the inductance changes, and hence load
value
with voltage hence the iron causes some distortion in the signal.
The gap reduces this distortion because it makes the µ much more
constant
over a wide F range and voltage range. The gap has a dominant effect.
Geesuss! bloomin complex eh?
But it IS what all good designers will definately consider on the way
towards the
final adjustment of the gap size.
So lets try µ max = 4,000, a value YOUR GOSS steel sample
taken from some second hand OPT YOU may have.
gap, g = 280 x ( [ 4,000 / 415 ] -1 ) / 4,000.
= 0.6mm
From these calculations you should notice that once the absolute maximum
µ possible for the iron is above say 3,000, and below 20,000, then the
gap needed to give µe = 415
won't vary very much at all, or less than 10%.
I used the low u max value of 1,000, because it seems to me to work
better with GOSS.
With crummy quality non oriented transformer iron, but still with 3% Si
content,
absolute max µ possible is around 3,500, and at low levels we are
interested in its maybe only 1,000.
In my OPT I have just installed in my customer's amp,
I fitted 0.25mm plastic gapping material across the core so the gap =
0.5mm.
Then I thought some measurements and wave form inspections were
necessary to check what I had done.
To measure the primary inductance, one needs to
have no load connected, and to measure ac voltage across the OPT
primary,
and ac current flow through the primary.
I connected a 10 ohm 5 watt R between anodes and the OPT input.
Then I set up the amp, and turned on the power, and began measuring.
Now when I said I had no load, I really told a fib.
I did have 8.2 ohms plus a series 0.22 uF Zobel network across the
secondary output coil,
one end of which is grounded.
So there is the lumped total input primary capacitance in parallel with
Lp.
The Zobel is is necessary for unconditional HF stability with my 8 dB
global NFB connected.
But at between 7 Hz and 70Hz, the area we are interested in Lp, the
Zobel
is a high impedance load no lower than many megohms so it has no effect
on measurements.
I set the anode signal voltage at 600V and 50Hz, and there was no
serious distortions,
and I measured the IL, and got Lp = 54H, not bad.
Then I lowered the Va to 100Vrms, and measured between 7hz and 2 kHz.
Lp remained at an average of 50.0 H.
How's that for calculated guessing?
But while measuring above 300Hz, there was huge peak in the graph I drew
for Lp
versus frequency, and L seemed to go up to 133H at 680Hz, so why?
Well, there is that shunt capacitance lurking, and to get a resonant Fo
= 680Hz with
50H of Lp the C must be about 0.0015uF, which probably seems right
for a big OPT like mine with so much interleaving.
I placed a 0.01uF across the Lp and made C = 0.0115total, and the Fo
moved down to 200,
and with a similarly high Q.
So it IS the C reacting with the Lp.
It is not at all important because the speaker load when connected is
vastly lower than the
reactive impedances of Lp or Csh or their resonant Z.
One cannot avoid the effects of Lp and csh, but they can be minimised
to have zero sonic effects, and to affect measurements in a completely
negligible manner.
Just don't wind the P and S windings with real thin insulation between
them.
Csh will then become high, and at F above 5kHz, this will seriously load
the amp down
badly, and the Fo between Csh and the leakage inductance will be a lot
lower F than anyone wants,
maybe only 16kHz, and at this second Fo point at HF the leakage L and
the C are a series resonant
circuit and a very low Z at Fo, so the response will have a savage dip
with the amp under near short circuit conditions at Fo.
I had a good look with my CRO at the sine waves at 600V as the F was
lowered.
The onset of visible cut off distortion was 19Hz.
Most ppl wouldn't know the difference between the waves for eliptical
reactive R&L loadings
and the waves when saturation begins.
At the onset of the distortions due to L&R loads becoming lower than the
rated mid F load of 6k in my case,
there is a flat line that forms on the right hand of the sine wave
crests gradually.
The saturation is where you get great sudden chunks out of the wave as
suddenly the
load becomes a short circuit for part of the wave, ugly, but in my case
at F and at voltage levels which won't occur when the amp is used,
unless it was used as a
sub woofer amp, and there were teenagers at the volume controls.
My OPT passed all its tests, and shall be potted tomorrow.
The potting of transformers is somewhat a vexation for many ppl.
But I think I have found a good cheap effective way to do it.
I bought a 4 litre tin of vinyl liquid resin and hardener used for
making fibreglass yachts and 1,001 other things mankind is addicted to.
I seived some washed river sand I got from a local concrete supplier for
free.
The seive is just an old framed flyscreen with aluminium fly wire, and
placed over a wheelbarrow.
Sand from the supplier is poured over the seive, and moved around with a
hand so that
large lumps and garbage are removed.
I then heated the sand for some hours in an old frypan at max temp to
dry it.
The OPT is placed onto the bench upside down within its pot with its
holding bolts,
and dry COOL sand poured in around the OPT.
Then I whack the pot which is a sheet steel box I made with the handle
of a hammer
about 500 times from all different directions and gradually the sand
works its way
into all crevices and compacts tightly because of the vibration.
Finally when no more settlement occurs and when the level is still about
15mm down from the lip
of the pot, I cease the vibrating.
Then i spray the sand surface liberally with clear cheap enamel from a
can, and leave to harden for a day.
Next day the final 15mm of pot is filled with a mix of the resin and
sand in about equal volumes,
and this forms an extremely strong concrete that binds the can bottom to
the bottom of the
transformer.
The sand and resin is a self levelling mix for about 20 minutes before
it all begins to gel and cure.
Don't use excessive hardener, less than normal is best.
I have done the same with power trannies to lower the natural hum you
get.
With the PT fixed to the chassis of the PSU or amp, the PT can be fully
tested with load
BEFORE the potting is done. Its better to know your work is right before
you seal it away.
Potting is done while the PT is all mounted up, connected, and bolted to
the chassis which
is upside down. There is room to get the sand in through a funnel, and
the final concrete.
Some black paint over the concrete makes it look nicer.
In the event that a tranny fails, the 15mm of epoxy concrete can easily
be chiseled away,
and sand poured out, and the pot retained for re-use.
The errant transformer can be examined, and if the fault is in the lead
in wiring, you may find it
and cure a problem cheaply. But usually a shorted turn/s means the
tranny is useless,
so the bolts and angle lugs and yokes are all removed, and the tranny
placed in an open log
fire just hot enough to heat the tranny to dull red for 1/2 an hour
and next day you wil find the plastics are all vaporized, and wire can
be cut through easy with a
small angle grinder, and the laminations will fall apart easy and be
entirely re-usable.
The heat won't damage the magnetic properties, maybe even improve them.
I've done barrowloads of old open circuit OPTs, PTs, chokes et all.
Its a good way to re-cycle gear.
Patrick Turner.