CHOKES - PAGE 3. Updated 2011.
This page 3 is about....


1. Chokes for DC anode feed,
General info about choke feed to anodes.

2. Choke with CT for balanced anode DC feed....
General info on chokes with CT for balanced feed of DC to anodes.
Fig 1, A balanced choke with ct is used for dc supply to a pair of EL84 in an LTP.
Fig 2, A balanced choke with ct is used as the biasing impedance for output tubes.
Design Method
Choke for AC only,
Steps (1) to (15).

3. Choke for anode DC feed with dc and ac currents.... 
Fig 3, Schematic for 45 watt SET amp with 2 x 845 or 211 tubes.
Design Method
Choke for AC and DC,
Steps (1) to (17), with notes about air gapping.

Other related pages :-

Basics about inductance and chokes, inductance test circuit,
Comparison of CRCRC filters with CLC filters,
Choke Design Method for
CLC filter, go to 
Chokes 1

For about
"choke input" or LC filters,
Choke Design Method for LC, steps (1) to (20), go to
Chokes 2

--------------------------------------------------------------------------------------------------------------------------

1. Chokes for DC feed - general info.
Chokes can be effectively used to feed dc current to a tube anode so that the anode Vdc
voltage remains stable because choke impedance at very low frequencies is little more than
the winding wire resistance. But as frequencies rise above 0.0Hz the choke impedance rises
at a rate of 6dB/octave until a peak is reached somewhere between 2kHz and 30kHz
depending on how much shunt capacitance exists within the choke and between the two
input terminals.
The low
winding resistance allows more DC current flow than using a resistor between the
B+ supply
rail.
The choke feed to an anode offers similar benefits to triode operation as
do active constant current sources to raise the total value of anode load ohms and allow
maximum triode voltage gain and minimum distortion.
With a choke, the B+ rail does not
need to be as high as with a resistance dc feed or CCS dc feed to anode because the
choke allows the anode signal voltage to swing higher than the B+ rail.

Because a choke
has high reactance much higher than any resistance DC feed at audio
F between 20Hz
and 20kHz, the signal voltage swing across the choke does not cause
high anode current change
so the choke does not consume much audio power and
there is more
available power to following capacitor coupled loads such as a volume
potentiometer or grid
bias resistors of output tubes. Cost and weight and chassis space
usually prohibit "choke loading" in all mass marketed amps designed by accountants.

If you know what you are doing, the choke works magic, and gives excellent sound.

There are some fanatics I know who do all they can to get rid of capacitors and resistances
throughout their audio circuits. I wish them well, but they have made their lives difficult,
and in many cases made the sound worse because so often they are not very well
educated about basic electronics.
For them, magnetic coupling through transformers is the only sure way to hi-fi nirvana.
Before reliable resistors and capacitors were made cheaply in the 1930s, transformer
coupling was commonly used between stages, but alas, the quality of them was awful
and sound was poor.

"Blocking caps" aka anode to grid coupling caps and resistors do not block fine music
flowing through an amplifier.
The main technical problem associated with choke loading or transformer loading of
intermediate amplifier stages is to obtain wide bandwidth and a flat response. It never
has been easy to make wide bandwidth interstage coupling transformers and in 1935
those available were very poor, causing high iron related distortion and bandwidth
limits. Using any tiny amount of global negative feedback around a 3 stage amp with
two interstage transformers is entirely prohibited because of the phase shifts created by
R&L and R&C time constants and cut off poles at frequencies close to 20Hz and 20kHz.
Such phase shifts will cause incurable oscillations. Even with the best modern design
and materials, transformer coupling between stages is a problem when NFB is applied.
So I have not bothered to ever make an amp with IST. However, one would be an idiot
to judge choke loading or choke feed as on judges the IST.

Therefore there is very little information here at this website about interstage transformers
which I have never found I have needed to use.


Usually the choke is used to supply dc to an anode of a triode with low Ra. Sometimes a
choke is used
between a cathode and 0V, where the effective Ra is many times less than
the data figures show for Ra because the cathode connection is an example of high local
NFB due to the "cathode follower" connection.  

High µ triodes with high Ra are usually unsuitable for choke feed because the reactance
of the choke must exceed Ra even at 20Hz so inductance must become extremely high.

But for good input triode for a power amp, I will parallel both triodes in one 6CG7, and use
the transistor CCS with 8mAdc, from B+ = +280Vdc, with Ea at 140Vdc. There is huge
voltage headroom, and much less THD than if I had dcRLa = 18k ohms. But I could also
use a choke of 100H plus a series resistance of 18k, and then loading is a minimum of 18k
at very low F, and rises to 25k at 29Hz, and to above 65k ohms by 100Hz, and 650k ohms
at 1kHz. The 18k in series with choke means that we always have dcRLa equal to 18k
minimum, with a vastly higher RL value for most of the AF band.
The benefits really start to show where you have EL84 or EL34 having to make a high
output voltage for 300B or 845 output triodes.

All iron wound choke feeds (and IST) create iron caused distortion. The Ra of the tube
is effectively connected in parallel with choke inductive reactance. The distortion caused by
iron cores is due to hysteresis and the equivalent circuit of a choke is that of a pure L
plus non linear resistance strapped across it which has minimum R at low  levels of signal
and at high levels of signal when saturation is approached. The lower the Ra and any other
R effectively in parallel with the choke, the lower the iron caused distortions. If you were to
use a pentode with very high Ra of perhaps 500k ohms, the distortion will be very bad,
so you never see pentodes in input or driver stages used with choke feed or to drive ISTs.
But a 1/2 6CG7 has Ra = 10k and the lower Ra reduces the effects of the non linearity of
inductance which is equivalent to a pure L strapped by a high value non linear resistance.
I will NEVER use a choke to
supply 0.7mAdc to 1/2 a 12AX7. The use of any pentode tube with choke supply is also
quite pointless due to distortions produced where Ra is high.

For example, Ra of 1/2 12AX7 = 65k ohms, and if two 1/2 12AX7 are paralleled then
Ra = 33k. If there is a following cap coupled Rg = 220k, then Ra // RL total = 29k ohms,
so for the -3dB point to be 20Hz the XL reactance value at 20Hz should be 29k, so
choke L = XL / ( F x 6.28 ) = 29,000 / ( 20 x 6.28 ) = 230H, which is a lot, and this means
using many thousands of turns of wire about 0.15mm dia on a large core. Thinner wire
could be used, but it becomes a very fragile choke.
Iron caused distortion is high even if the core is GOSS and air gapped.  For such chokes,
it may be better to use 50% nickel cores, or mu-metal, and then you have terrible trouble
sourcing it. If you want to have a high value load to a 1/2 12AX7, then use a µ-follower
type of circuit with 2 x 1/2 12AX7 triodes in series, as in my 10 tube preamp.
Just as good is using both 1/2 12AX7 paralleled, and using a single transistor CCS
giving say 2mAdc, and this works better than any choke and without any iron caused
distortion and with wide bandwidth. If a pentode such as EF80/6BX6 were used instead
of 12AX7, Ra = 400k ohms and total load = maybe 180k and choke value would be
1,200 H, and response would still not be flat unless a loop of NFB was used, and then
gain is reduced, and you still won't get as good a result as with a triode without NFB.

I have never used a choke to supply dc to any 6CG7/6SN7 or 12AU7  power amp input
triode or preamp triode stage. I believe the use of a transistor Constant Current Source
is far easier, cheaper, and sounds better, and the headroom needed for the voltage
across the CCS is always available easily, and there is no audio power lost in a CCS
feed to an anode and the CCS has no LF or HF impedance reduction at extremes of the
AF band due to inductance and stray C. So the CCS produces negligible phase shift
and may be used where NFB is intended.

But I have found plenty of sonic benefit for choke loading feeding DC to anodes...

2. Choke with CT for balanced anode DC feed....
All my standard PP amps such as the 8585 have local cathode feedback windings in the
output stage, and up to around 80Vrms needs to be applied in two phases to the two
grid inputs of the output stage of the PP output circuit. When using 845 triodes without
any local cathode feedback the grid drive voltage needed is up to around 120Vrms per
grid. In about 2003 I realized the best way to drive any large octal output tube or any
other large tube was with a suitable smaller output pentode strapped in triode, typically
a 6BQ5/EL84, or perhaps an EL34 in triode. Such tubes have good voltage gain and
low Ra in triode mode and low THD at high voltage swing in either SE driver circuits
or balanced differential amps. Where resistance DC feed is considered the value of
R must be 10 x Ra at least. To get the wanted high Idc flow and high Ea across the
tube at idle, a very high B+ rail must be created. Such arrangements are for a wide
Va swing with little THD. For those who like DH triodes, type 45 or 2A3 make
excellent sounding PP driver stages, and Ming-Da amps use a pair of 300B to drive
PP 845.
( Ming-Da don't have choke feed as I describe, and what a pity! )

A PP or balanced triode driver stage should be capable of only 0.1% THD
( mainly 3H ) at 80Vrms from each anode so that the driver stage does contribute to
total distortion where there may be only 10dB of global NFB. In an SE amp, driver
THD should be less than 1% at 80Vrms, ( mainly 2H ). Such low THD is only
possible where the total load on the triodes is greater than 10 x Ra, or more.

The biasing resistance of most large tubes in output stages needs to be less than
100k for each output tube because after a couple of years use the grids tend to
become slightly positive with respect to the applied grid bias voltage, and sometimes
dangerously so. The gradual rise of positive grid current at idle in aging tubes can
raise tube working temperatures and cause the effect to get worse in an insidious
positive DC FB effect which may only be countered by use of low resistance grid
bias R. Quad-II amps have 680k use to bias KT66 grids and many of these amps
have come to me with red hot anodes and Eg at several +Vdc above their normal
0V bias levels. Rg bias should never exceed 150k, and I like to use 120k max.

Where a driver stage drives more than one output tube in parallel, it needs to be set
up so it can easily drive the grid bias R as well as the dc supply RLdc to the driver
tube. This RLdc should be 1/3 of the value of the following output stage bias R value.
Suppose you have an EL84 driver for two output tubes. If Rg bias was 120k for each,
total Rg bias = 60k. RLdc would be 22k, and if I have Idc = 16mA, then 352Vdc will
be across the 22k. If Ea = +280Vdc, and cathode bias = +15Vdc, B+ rail must be
+662Vdc. The load on the EL34 = 60k//22k = 16.1k, and this is not even above
10Ra if as Ra for EL84 in triode at 16mA = 2k2.

In my 8585 to ensure the RL seen by the driver triodes is as high a value as
possible for lowest distortions, I supply DC to anodes via a choke with CT with
fully interleaved E&I laminations with 25mm tongue size. There was little room for
the chokes so I used Stack = 10mm stack. N = 8,000 turns of 0.15mm Cu dia wire.
This makes an inductive reactance which has a maximum of 720H which is hundreds
of k-ohms at 1kHz, and at 20Hz = 90k, and 9k at very low Vo levels where the low
reactance will not cause distortions. This L has negligible loading on the triodes
compared to the capacitor coupled Rg bias of following stage. The choke is a
'balanced choke' and acts like a primary of a PP transformer which has no load.
The choke does not need to be air gapped if there is good dc balance of each
EL84 used, so iron µ remains high giving a large amount of inductance.


I have also tried the idea on my 300 watt amplifiers with 12 x 6550 per channel.
The 300 Watt amps have chokes with 32mm stack x 25 tongue with 5,000 turns of
0.2mm Cu dia wire. Such chokes have a loading effect of many more k-ohms than
the 16k I mentioned above.
The only problem of using just the CT choke is bandwidth.
To avoid the shunt C and shunt L effects of the choke used in a differential amp,
or Long Tail Pair, LTP, I place a resistance between the ends of the choke winding
and each anode, usually at least 2 x Ra. Fig 1 below shows the set up with 8k2.
Thus the triodes' lowest load value possible is 8k2 which is 3.7 x Ra for 1 x EL84
in triode. Phase shift caused by shunt C and shunt L is virtually eliminated,
and normal FB may be applied.


Fig 1.

schema-driver-ltp-choke-anode-aug07.gif
The Fig 1 above shows the choke with center tapped choke in the anode circuit.
The inductance with 5,000 turns with µe = 2,000 should reach 300H.

Doubling the stack height will double the inductance. GOSS can have µ up to
17,000 with fully interleaved lams, but if there is a small Ia imbalance of only a few
mA dc in the pair of EL84, then core can become dc magnetized and possibly saturate,
causing high distortion.

Partial air gapping will lessen the worst effects of possible DC imbalance in EL84.
the partial air gapping should reduce maximum µ to a µe of 2,000. To achieve such
a reduction of µ would be impossible with normal air gapping with all Es in one block
held tight against all Is, even if there was no actual real gap. Such close full gapping
will usually give µe between 500 and 1,200 for low µ or high µ laminations, but we
want about 2,000. To get this value, the Es and Is are assembled into the bobbin
in piles of Es and piles of Is in alternate directions. So instead of having each
consecutive E facing in opposite directions, you may have 5mm high piles of Es
facing one way, and a following 5mm pile facing the other way. For a stack height
total of 25mm, there could be 5 x 5mm piles of Es with 3 piles facing east, and 3
piles facing west, each with a 5mm pile of Is to close the end of the Es.
 
The exact number of laminations per pile MUST be carefully adjusted so each
pile has an equal number of laminations, but final number of piles gives the wanted
µe of 2,000. Most ppl don't bother doing this, but then I can only say what's best,
and if anyone wants to take the short cut and not use partial air gapping, then they
should not complain to me about the distortion.

The other method to stop dc magnetization is to use separate R&C cathode bias
networks for each EL84. 1k5 + 1,000uF would be OK, with a 2.2 uF cathode to cathode
to bypass electros at HF.

If you don't want to use a CCS with a transistor then take a resistor from bottom
of R&C networks to a -100Vdc rail, about 3k3 rated for 10W would be OK. THEN, use
a balanced drive to each EL84 grid from another LTP which in this case could be a
6CG7 or 6DJ8 with a transistor CCS. With balanced drive, the EL84 LTP requires
about +/- 4.5Vrms it each grid to give the +/- 80Vrms anode output. The 2 anode
outputs from EL84s should then remain very well balanced within 1% of each other.
 
8k2 would be rated for 5W. The typical shunt capacitance in L might be about 300pF,
so each EL84 anode "sees" 8k2 in series with 150pF, and with each EL84 Ra = 2k2.
As F rises above the audio band the 300pF shunts the choke with a pole at 25kHz.
The load "seen" by EL84 becomes 8k2, and gain reduces above 25kHz from 18 at
1kHz to about 15, or 2dB, and maximum undistorted output voltage is reduced. The
ultimate HF pole at EL84 anodes is caused by output stage Miller C and stray C of
perhaps 200pF will be due to 8k2 in parallel with Ra of 2k2, ie, 1.k7. So HF pole
will be at around 460kHz. Such excellent driver amplifier performance is seldom
seen in many commercially made amps and is not seen from many famous designs
such as the Williamson amplifier.
At the low F end of band below 20Hz, the choke reactance falls to zero at DC, but
the EL84 have 8k2 minimum load, So the HF AND LF will be well extended, and
the 8k2 will isolate L and C from having much effect on gain, and the slight reduction
of gain at below 40Hz and above 30kHz is a shelved response and entirely benign.
High signal voltages are NEVER needed below 20Hz or above 20kHz, and the amp
will be stable with even a high amount global NFB. 

Fig 2.

schema-driver-ltp-choke-biasing-aug07.GIF
In this Fig 2 schematic, the choke with CT and 2x 8k2 are used for biasing the output
grids to obtain a very low biasing resistance at DC. The 2 x EL84 can be set up
without much regard for Idc imbalance causing choke saturation because 2 x 18k
are used for Idc to each EL84 anode, and there is no Idc in the choke because of the
coupling caps.
The loading of the tubes is similar to Fig1. The coupling cap values
have to be high because the C + L form a HPF with ultimate -12dB/octave attenuation
and if the L has a minimum L of say 100H at very low signal level then the LF pole
needs to be very low, in this case, I expect at 2.2Hz, when open loop gain elsewhere
is very low, so stability with NFB should be OK. Just using 3.3uF would give a pole
at 12Hz, and too near the audio band. But with 100uF, at 2.2Hz the XC = 722 ohms.
The addition of a series 8k2 damps the attenuation and gives a shelving effect and
100uF + 8k2 give a pole at 0.193Hz, so I doubt there will be severe phase shift  to
worry about.
Maybe C = 22uF is OK, but this has to be trialled and with NFB applied
to ensure LF stability. Such things can give unpredictable behavior.

2. Design Method, - Choke for ac only.
Design a choke for L1
in the above Fig 1.

(1) Choose series R between anodes and ends of winding to be at least
from 3 x Ra to 5 x Ra of tube.

Tube = EL84 triode, Ra at Ia = 14mA = 2k2.
Series R minimum = 3 x 2k2 = 6k6.
Fig 1 shows 8k2, to suit available B+ and wanted Ea.

(2) List the maximum operating voltage and current conditions.
Vac. Maximum possible undistorted signal Va-a across choke winding from
tubes. 
From Fig 1, Max Va-a = 2 x 140Vrms = 280Vrms.
Iac. Maximum nominal Iac = max Vac / Reactance at 20Hz.
Reactance 20Hz > 20 x Ra where Ra is dynamic anode resistance of one tube.

EL84 triode Ra = 2k2.
20 x 2k2 = 44k0. Ia max = 280V / 44,000 ohms = 6.35mA rms.

(3) Are there two balanced Idc flows anywhere that are to be balanced?

Yes, there are two equal Idc flows in opposite directions for each 1/2 winding.

Therefore assembled core laminations will have µe max < 2,000, and core
will have partial air gap.

NOTE. If you answered No, it means the choke has NO Idc flows at all, then
µe may be as high as maximum possible and core will have no air gap.

(4)  Define Idc flows at the zero signal condition.

L1 has 28mAdc supplied to CT. Wire each side of CT has 14mAdc.


(5)  Define worst fault condition possible which could increase Idle Idc
above value stated in (4).

Worst case fault will be where cathode is at 0V and B+ is at +425Vdc,
and anode is shorted to cathode. If the winding has Rw = 0.0 ohms, Max Idc
= B+ / series resistor = 425V / 8k2 = 52mAdc.

(6) Apply safety factor of 2.0 to Idc value calculated in (5), to get Idc wire rating.
2.0 x 52mA dc = 104mAdc.

(7) Calculate theoretical wire size able to survive a tube fault condition with
3A/sq.mm max.
Wire Cu dia = 0.65 x square root Idc,
Where wire Cu dia in mm,  0.65 is a constant for all equations, Idc in amps dc.

Theoretical wire size = 0.65 x sq.rt 0.104 = 0.21 mm dia.

(8) Select nearest wire size available from table......

Table 1.
table-wire-sizes.GIF
Table says 0.212mm nearest, but is not available.

Use Cu dia = 0.200mm, = 0.245mm o/a dia including enamel.


(8) Calculate wanted maximum inductance of choke.

Calculate wanted maximum coil reactance at 20Hz.

NOTE. From step (2), each tube should have "see"inductive reactance = 10 x Ra.
With two tubes, the two Ra are effectively in series with each other to give
Ra-a = 4k4, so XL = 10 x 4k4 = 44k0.
 
Calculate minimum inductance required at 20Hz, and at high Va operation
with Bmax > 0.2 Tesla. Henrys = XL / ( 2 pye x F in Hz )
= 44,000 / (6.28 x 20) = 350H.

(9) Choose Wasteless E&I core T x S dimensions from Table 2 for 4 different
T sizes of core with 3 different wire sizes from Table 2 :-
Table 2.
1
2
3
4
5
6
Wasteless E&I
Square Afe,
T mm x S mm
Core window size
L mm x H mm
Bobbin winding area
L mm x H mm
Turns 0.15mm Cu
dia,  oad 0.188mm
Inductance, Henrys
µe = 2,000
Turns 0.20mm Cu dia,
oad 0.245mm
Henrys
µe = 2,000
Turns 0.25mm Cu dia
oad 0.301mm
Henrys
µe = 2,000
25mm x 25mm
37.5 x 12.5
33 x 9.5 = 313sq.mm
8,000t
720H
5,000t
281H
3,000t
101H
28mm x 28mm
42 x 14
38 x 10 =  380sq.mm
10,000t
1,260H
6,000t
453H
4,000t
201H
32mm x 32mm
48 x 16
44 x 12 = 528sq.mm
14,000t
2,822H
8,000t
921H
5,200t
389H
38mm x 38mm
57 x 19
52 x 15 = 780sq.mm
20,000
6,840H
11,500t
2,261H
8,000t
1,094H
Examine column No 5 for wire size from step (7), wire Cu size = 0.2mm.
Search for nearest Inductance below value calculated in step (8)

Choose T x S = 25mm x 25mm, core window 37.5mm x 12.5mm, N = 5,000 turns, L = 281H.

NOTE. Inductance and XL is directly proportional to S.

Increase stack height S for selected core to increase L to wanted value in step (8)
Revised S = 25mm x wanted L / L in table 2,
= 25mm x 350 / 281 =  31.14 mm.

Use bobbin made for T = 25mm, and S = 32mm.


NOTE. With Wasteless E&I lams, the core window area = 0.75 x T squared. If C-cores are used,
the core window area is often much more than 0.75 x T squared, especially if just one C-core
is used. For example, one might have a C-core with window 39mm x 13mm, and strip build up
or T = 12mm. If the strip width is 25mm, then Afe = 12 x 25. The window is similar to wasteless
material with T 25mm, but C-core inductance will be 140H with same 5,000 turns.
One might source a C-core with strip width of 50mm to get the 281H.
C-cores have maximum µ of about 12,000 with well polished cuts in the wound GOSS strip used.
The fact a cut is used and 1/2 cores are tight together well does mean there is a gap, because if
the core strip was continually wound with no gap, then µ can be 40,000 for GOSS. The only way
to adjust the µ down to a wanted lower µe is to add some plastic sheeting between both C-core
cuts and I suggest 0.01mm to 0.03mm may be fine. It is a lot easier to adjust air gaps in C-cores.
To replace 25T E&I lams, C-cores with window 38mm x 13mm are fine and with build up of 12mm.
Then use two C-cores side by side to make a 00 pattern and then select the strip with to alter S.
There are C-cores with Lx H = 38 x 13, but build up of only 8mm, giving T = 16 for double cores
so strip width needs to be increased to 40mm or you stack 4 x 20mm strip width C-cores.
Whatever is used for a core, you can't compromise the design. Using exotic materials such as
nickel laminations, mu-metal, amorphous etc may seem just fine, but all parameters must be
complied with including saturation behavior with DC offset and / or low frequency saturation.

NOTE. The Inductance calculated so far has been based on performance at maximum Va-a
and 20Hz. But µe will become a lot less at very low signal voltage levels, hence the XL will
become lower at low signal levels.


Fig 3. An additional diagram of wasteless lamination sizes :-
Wasteless2-E&I-lam-dimensions.GIF
-------------------------------------------------------------------------------------------------------------------------------
(10) Check Fsat, Bac, Possible Bdc, and operation of the design so far.


Frequency of core saturation, Fsat, with maximum Va-a, and maximum Bac of
1.5Tesla and with no DC magnetization
in one direction, ie, Bdc = 0.0 Tesla,
Fsat = 22.6 x V x 10,000 / ( Afe x 1.5 x N ) in Hz.

This example, 
Fsat = 22.6 x 280 x 10,000 / ( 32 x 25 x 1.5 x 5,000 ) = 10.54Hz.

Is Fsat < 20Hz ? if Yes, design is so far OK.

Possible Bdc, ie, DC core magnetization can be due to unbalanced Ia in the two tubes.
Assume maximum possible Idc imbalance = 33% of wanted Ia in one tube of pair.
Wanted Ia in one tube = 14mAdc. Net Ia dc in one direction = 33% of 14mA = 4.62mAdc.

Bdc = 12.6 x Idc x N x µe / ( iron ML x 10,000 ) Tesla.

This case,
Bdc = 12.6 x 0.00462 x 5,000 x 2,000 / ( 140 x 10,000 ) = 0.4158Tesla

is Bdc < 0.75Tesla? if Yes, design is OK so far.

NOTE. The choke designed so far has ability to swing Va-a = 280Vrms, much higher
than the anticipated maximum of 80Vrms needed in amps I build with CFB in output stage.
The only tubes I know which would need a high Va-a approaching 280Vrms will be
the 845, often with grid biased at -190Vdc, and thus able to accept drive = 134Vrms at
each grid for class AB1 operation with Ea = +1,200V, and Ia at idle = 44mA dc.
However, in the McIntosh type of schematic the amount of OPT CFB = 50%, and there
may be 140Vrms at both anode and cathode of a 6550 in two phases, and the Va-k
required may be 30Vrms, so grid input needed could be 170Vrms. The McIntosh achieves
such high grid drive voltage by bootstrapping the driver tube anode RL to the OPT.
But this is a sample of mild positive feedback, and to avoid this the CT choke might be used
instead with some slight increase in EL84 Ea to produce a wanted 340Va-a, or better
still to use a pair of EL34 in triode with 20mAdc each and Ea = +400Vdc, and Va-a = 450Vrms
can be expected.
----------------------------------------------------------------------------------------------------------------------------

3. Design Method, - Choke for ac PLUS dc.

Fig 4.
graph-GOSS-core-permeability.gif
Fig 4 above is a reproduction of the curves for GOSS made in 1955 and as shown
in the Radiotron Designer's Handbook, 4th Ed, 1955, Page 244. The curves A to F give
permeability for GOSS laminations at various amounts of DC magnetization depending
on turns and Idc flow between zero Idc for curve A, to heavy Idc for curve F.
The graph was probably drawn by a keen manufacturer's employee, and it appears the
GOSS core used has a plan shape and dimensions in the top left insert picture
The iron magnetic path mL = 5.6 inches = 142mm. The core window is 37.5mm x 12.5mm,
and actual coil length is about 33mm winding width across the bobbin.

There are basic issues here which confuse nearly everyone I know! The study of
electro magnetic phenomena and descriptions of magnetic coil properties and behavior
involve magnetic study language which has evolved over about the last 200 years as
knowledge increased, and concepts were defined and described mathematically.
Letters of the alphabet are used to nominate the concepts and mathematical
quantities.
 
First mathematical quantity to think about is magneto motive force also known as F,
(nothing to do with F for frequency). F = N x I where N = turns of a coil and I is in Amps.

Second is the magnetic field, H, which exists within the coil inner area for a given
length of magnetic path. The magnetic field per length is sometimes stated as
H = N x I / Metre. Or it is stated in a unit called the Oersted, after a Mr Oersted.
Most units of magnetism are named after fellows who were famous for discoveries,
such as Ampere, Volt, Oersted, Maxwell, Lorentz, Gilbert, Tesla, and I suggest you
search on each via google because the history of discovery is a path to understanding.

One Oersted, Oe, is where you have a 1 metre long solenoid coil of wire with 79.6 turns
and where current = 1Amp. The definition is based on an air cored winding which has
an infinite length but we only concerned with 1 metre of that length and for a variety
of reasons which I won't mention now, 1.0 Oe = 79.6Amp-turns per metre.
Or you could have 1.0 Oe = 2 Amp-turns per inch.
If you have a toroidal winding with wires all around the toroid "circle" average circumference,
then you have a magnetic path equal to the circumference, and magnetic field inside the coil
will be similar to an infinitely long toroid. If you put turns around an iron core on a bobbin
then Magnetic length path mL is all around the iron, because the iron core creates an equal
magnetic field all around the iron, even if there are many layers of turns along only about
1/3 of the mL. The addition of an air gap can change H value because the mL is increased
with the air gap. One might assume H for a bar core or no core at all will be N x I / coil length,
some 33mm in the above core sample above, and NOT N x I / 142mm which would be a
much smaller H measurement because mL along the coil and outside the coil is a much
longer distance measurement.

Third. Permeability, µ, is 1.0 for an air cored coil.
µ = B / H where B is the magnetic field density in Gauss lines of magnetic force or Tesla
over a given sectional area across the magnetic field. I Tesla = 10,000 Gauss.
So B = µ x H, and for an air core coil B = H, because µ = 1.0.
But when a GOSS core is used, µ can be up to 40,000 depending on the material and
how it is assembled. But nevertheless,  B = µ x H = µ x N x I per a given length which varies
according to any air gap.
Depending on how the GOSS material is assembled, µ in a GOSS core is rarely ever the
maximum possible unless the core is a toroidal type and no DC "flux" is present.
So the term µ should not be used. The permeability that should be used is effectively
less than maximum known as µe.
µe is a result of an air gap or equivalent of air gap placed in the iron path length. There
is more about air gaps below, but for now....

I once found this formula which works for me with iron cored chokes and SE OPTs :-

Bdc = 12.6 x N x µe x Idc 
             10,000 x mL     
Where Bdc in Tesla,  N is turns, µe is effective permeability, Idc is dc current in Amps,
mL in mm is the iron path length not including any air gap, and 12.6 = 4 x pye and a constant,
and 10,000 is a constant to convert Gauss to Tesla.

Now core magnetization is due to DC current plus AC current. The effects of DC give
Bdc, ie, field density due to DC, and effects of AC give Bac.

Look along curve F for H = 10.0 Oe, and for 1,000 turns with 25mAdc.
We see that if the Bac = 1.25 Tesla, or 12,500 Gauss, then µe = 580, about 1/10 of the
µ = 5,200 maximum for the core shown with no Idc.

According to my formula above, Bdc = 12.6 x 1,000 x 580 x 0.025 / (10,000 x 142 )
= 0.129Tesla.
Totals of Bdc and Bac = 0.129T + 1.25T = 1.379Tesla, and getting close to saturation.

The calculations of Bdc and Bac and air gap is what becomes critical for
designing chokes
with both AC + DC present.

The point I make is that the higher the DC current the lower the µe must become
for a choke to avoid core saturation from DC while allowing enough "headroom"
for audio signal voltages to be applied across the choke without causing saturation
at low frequencies.
---------------------------------------------------------------------------------------------------------------------------------

Example of a choke with AC and DC :-

Here is the driver stage for driving a pair of SE 845 in parallel to make 55 watts of pure class A.
Fig 5.

845set-jul08-1-schem-input-driver.GIF
Fig 5 needs explaining.
V1 has dc feed to anode through an MJE350 which acts to give effective collector
resistance of over 5 Meg ohms at all F with a small amount of shunt capacitance under
100pF. It acts like a constant current source, CCS. So the anode load experienced by
V1 is R11, 180k which is OK to bias the 3 following EL84. The paralleled 6CG7 anode
Ra = 5k0 approx and so total anode load = 36 x Ra, so THD is negligible.


There is nothing to be gained by using a choke to feed Idc to V1 so I don't have a design
recipe here for one but anyone is free to design a choke using general principles below.
I can offer a wild guess, use RLa = 39k, 5W, and a choke with L = 120H.


V2, 3, 4, are 3 paralleled EL84 in triode giving Ra = 700 ohms. The Choke L1 has air
gapped wasteless E&I GOSS laminations with 5,000 turns of 0.2mm dia wire on a core with
T = 25mm, S = 32mm, but could be 38mm or 50mm. The gap was set to give approximately
for 60H at 36mA.
60H at 20Hz has reactive XL = 7k5. The added series R = 8k0, so R&L resultant load = 11k.
This 11k is in parallel with capacitor coupled bias resistance of 23k0 for 2 x 845, so final
total load becomes approximately 7k7. Thus total RLa = 10 > 10 x Ra, so THD is low
even at 107Vrms. The THD is nearly all 2H which cancels the 2H produced in the 845 SE
output stage, so at low levels the amp produces similar low levels of THD compared to a
good PP amp of the same power. The maximum drive voltage needed for class A1 SE 845
is 107Vrms, but the driver stage should be able to produce 1.5 times more to help keep
THD minimal at much lower levels. At 100Hz the XL increases to 37k, and load on EL84
triodes rises from 7k7 at 20Hz to above 15k, and RL = > 20 x Ra, so THD is low.
The increasing XL with higher F means that for most audio F the loading is close to 21k.

Unlike the filter chokes mentioned elsewhere for power supplies, the choke with a large signal
voltage across it resulting in low AC current while also having low Idc it does not
need to have
low dc resistance. The DC current density should not exceed 3Amps/sq.mm even in if Idc
is doubled. The ac current can be neglected because signal voltage variations just increase
and decrease total idle DC current so average current always remains equal to DC current.
The winding resistance
acts in series with L, along with any other added series "damping"
resistance, seen here as R23, 8k0.


The circuit MUST be designed so the choke can survive without fusing open even if a short
circuit occurs between between anode and cathode in one or more tubes.
In the Fig 5 schematic, if the anode connection went to 0V, there is +624Vdc across
about 8k6 including R23 plus choke Rw, so Idc = 73mA, so about 2 x idle DC of 36mAdc.
At idle, R23 generates 10.5 Watts of heat, so R23 needs to be rated for at least 20Watts.
But with Idc = 72mA during a worst case fault condition heat generated in the R23 8k0
= 41 Watts, and R23 might fuse open, but only cost $5 to replace, if 4 x ceramic wire
wound R are used. There is nothing to stop anyone using aluminium clad resistors
screwed to the chassis and with a 50W rating. The 72mA in the choke with Rw = 500
ohms generates 2.6Watts of heat, and the choke will survive quite OK.


Design steps for L1 in Fig 5 above :-
(1)  Choose series R between anodes and ends of winding to be from
5 x Ra to 12 x Ra of tube.

Tubes = 3 x EL84 triode. Ra of each = 2k1 approx with Ia = 12mA in each.
Ra of 3 in parallel = 700 ohms.
Series R could range between 3k5 to 8k4, depending on available B+.
B+ for a driver stage with choke should be at least 1.5 x Ea.
Fig 5 shows B+ = +624Vdc, about 2 x Ea, OK.

Fig 5 shows R23 = 8k0, to suit available B+ and wanted Ea.

(2) List the maximum operating voltage and current conditions.
Calculate wanted inductance L1.
Vac Maximum possible undistorted signal Va across choke winding from
3 x EL84 parallel tubes. 
In Fig 5, Max Va = 107Vrms at clipping, but want maximum possible at 1.5 x
clipping level = 160Vrms.

Iac Maximum = max wanted Vac / wanted RL total at 20Hz.
Reactance at 20Hz = at least 10 x Ra where Ra is dynamic anode resistance of
one tube more paralleled.

3 x parallel EL84 triode Ra = 700 ohms, wanted XL = 7k0
L1 = XL / 6.28 x F = 7,000 / 6.28 x 20 = 55.7H.

Calculate Total load at 20Hz :-
XL at 20Hz in series with added R = 7k0 + 8k0 =
11k0 approx,
and this in parallel with cap coupled load of 23k0 = 7k3

Total RLa = 7k3.

Iac max = wanted max Vac / RL total = 160Vrms / 7k3 = 22mA

Peak Ia change = 1.414 x Ia rms = +/- 31mA peak.

(3) Calculate Idc at idle.
Max +/- Idc from step (2) = +/-31peak.
Allow idle Idc = +10% more than peak Ia change, but not less, say 36mAdc.


B+ = +642Vdc, and Ea = +300Vdc + Ek of +10Vdc = +310Vdc.
Assume Rw = 500 ohms. R23 = 8k0.
Iadc at idle = B+ less (Ea + Ek) / ( Rw + R series ) = 624V - 310V / ( 500 + 8,000 )
= 314V / 8k5 = 36.9mAdc, near enough.


(4)  Define normal B+ and Ea and Ia conditions and worst case fault condition
possible which could increase Idle Idc above value stated in (3).

Worst case fault will be where cathode is at 0V and B+ is at +624Vdc,
and anode is shorted to cathode. If the winding has Rw = 500ohms, Max Idc
= B+ / ( Rw + series resistor ) = 624V / ( 500 + 8,000 ) = 73mAdc.

(6) Safety factor need not be applied because the fault Idc level is only twice
the idle current. Wire must be rated for at least 73mAdc or slightly more at 3A/sq.mm.
Rate wire at say 100mAdc.

(7) Calculate theoretical wire size able to survive a tube fault condition with
3A/sq.mm max.
Wire Cu dia = 0.65 x square root Idc,
Where wire Cu dia in mm,  0.65 is a constant for all equations, Idc in amps dc.

Theoretical wire size = 0.65 x sq.rt 0.10 = 0.2055 mm Cu dia.

(8) Select nearest wire size available from table......
table-wire-sizes.GIF
Select wire size = 0.20mm Cu dia = 0.245mm oa dia.

(9) Go to step (9) above ( for PP circuit ) and read Table 2 showing wasteless
E&I laminated core dimensions and wire sizes and turns possible.
See column 5 for 0.20 Cu dia wire. Try use of core T25mm x S25mm, and turns = 5,000.

(10)
Calculate core stack required for maximum Bac of 0.6Tesla at 14Hz.

NOTE. The use of GOSS allows for the total of Bdc + Bac = 1.6Tesla, with Bac and Bdc
each = 0.8Tesla, but for design purposes let us limit each to 0.6 Tesla.

Bac = 22.6 x V x 10,000 / ( S x T x F x N )
where Bac = maximum ac magnetic field density in Tesla,
22.6 and 10,000 are constants for all equations, V is applied Vrms across choke.
S x T = Stack x Tongue in mm and is sectional area Afe in sq.mm of central leg of core,
F is the frequency of AC signal, N is number of turns.

Therefore S = 22.6 x V x 10,000 / ( Bac x T x F x N )

Signal Voltage maximum likely = 160Vrms, Bmax = 0.6Tesla at 14Hz

Theoretical S = 22.6 x 160 x 10,000 / ( 0.6 x 25 x 14 x 5,000 ) =
34mm.

Is this calculated S between 0.75T and 2T?

NOTE. If S required for a given T is more than 2T, consider choosing a core with larger T.
If S is calculated at less than 0.75T, consider choice of smaller T, or larger wire diameter.

NOTE. Many people wanting to make a choke will already have some core material
taken from some old transformer. They will make a bobbin easily, and only need to
purchase the right size of wire. The Stack height may be calculated for a wide range of T,
and too small a stack may satisfy Bac < 0.6Tesla, and Bdc < 0.6Tesla but give too little
inductance which will require stack to be increased. The winding resistance will rise but
is of no concern in this choke application so the wire size will not need alteration.

Calculated S = 34mm, and is between 0.75T and 2T, S is OK.


Choose nearest convenient bobbin size above calculated S size.

Use bobbin for T = 25mm, S = 38mm, 1" x 1.5".

NOTE.  If a new choice of T is made, re-calculate for new selection of core material.
 

(11) Calculate µe, effective core permeability, µe, and calculate air gap required
and with maximum Bdc = 0.6Tesla.

µe = Bdc x 10,000 x Iron ML / ( 12.6 x N x Idc )
where µe = effective permeability with air gap and presence of Idc, Bdc in Tesla,
10,000 and 12.6 are constants for all equations, ML in mm and is the iron core
magnetic path length without air gap, N is the number of turns, Idc is in Amps DC
at wanted design idle value, (not Idc for fault condition.)

Wasteless pattern mL for T = 25mm = 5.6 x T = 142mm.

µe = 0.6 x 10,000 x 142 / ( 12.6 x 5,000 x 0.036 ) = 375.

(12) Calculate Inductance for calculated µe.
 
L  =  1.26 x Nsquared x Afe x µe  / ( 1,000,000,000 x mL )
where L in Henrys, N is turns, Afe is Smm x Tmm, µe is effective permeability
with air gap and Idc, 1.26 and 1,000,000,000 are constants, mL is Iron magnetic
path length, all dimensions in mm.

µe = 375, S = 38mm, T = 25, mL = 142mm, N = 5,000 turns,

L = 1.26 x 5,000 x 5,000 x 25 x 38 x 375 / ( 1,000,000,000 x 142 ) = 79H.

(13) Is calculated inductance OK, equal or more than specified in step (2) ?

If not, try increasing stack height. Inductance is directly proportional to stack height.
There is no need to re-calculate Bac which becomes lower with higher stack height.
There is no need to increase stack height to achieve inductance above maximum
specified in step (12), but if the budget permits, and size allows, there is also no
to not increase stack height up to 2T.

Inductance will be 79H  and is above 55H in step (2), with S = 38T. Air gap may be
calculated, but can be adjusted to give L = 55H to 80H  so that µe = 375 approx.

Can Stack height be increased?

Yes, to 2T, so revised stack S = 2T = 50mm.

(14) What is revised Inductance with changed stack size?
Revised L = previously calculated L x new larger stack height / former stack height.

Revised L = 79H x 50 / 38 = 104H.

(15) Calculate Theoretical air gap and gap material thickness.

Air gap, ag = mL x ( µ - µe ) / ( µe x µ ),
Where air gap ag  is the total gap in mm consisting of a gap or summed gaps filled with
non magnetic sheeting within the iron magnetic path,
mL.
µe is effective permeability
with air gap,
µ is the maximum possible permeability with E&I lams maximally
interleaved,
mL is the iron magnetic path length in mm.

Let me assume the maximum possible µ for choke core E&I lam material = 3,000.

Air Gap, ag = 142 x ( 3,000 - 375 ) / ( 375 x 3000 ) = 0.33mm

State number of gaps in magnetic path = 2, see above plan view of E&I wasteless
laminations.
Therefore
Gap material = calculated gap / 2 = 0.165mm.

(16) Calculate TL, Turn Length = 2T + 2S + 3.142T / 2.

T = 25mm, S = 50mm,


TL = 50 + 100 + 39.3 = 189.3mm.

(17)  Calculate winding resistance.

Rw = TL x N x 2.26 / ( 100,000 x d squared ),
where Rw in ohms, TL in mm, 2.26 is a constant for a 100 metre length of
1.0mm Cu dia wire, 100,000 is a constant = 100,000mm in 100 metres and
d = Cu dia in mm of wire used.

Rw = 189 x 5,000 x 2.26 / ( 100,000 x 0.20 x 0.20 ) = 534
ohms.

(18) Calculate idle heat generated = Rw x Idc squared.

Heat = 534 x 0.036 x 0.036 = 0.69Watts, therefore T rise will be low.

NOTE. The calculated value of µe = 375 is quite high. The calculated µe may
not be able to be achieved even if the calculated air gap is set up in a given E&I
lam core, or even if there is no real air gap and all Is are butted hard against all Es.
This is because the join between E and I is imperfect between iron laminations
with crystal grain structures at a 90 degree angle, so the butted join is an
"equivalent" of an air gap which does not really exist. The highest µe achievable
might be 1/10 x the maximum possible µ if all lams had no DC flow magnetization
and all laminations were maximally interleaved. Hence if the max µ = 3,000, perhaps
max µe would be 300 even with close butting of Es against Is. Therefore maximum
achievable inductance would be 66H, not 100H as we have calculated.
Therefore using a core with S = 50mm is definitely a good idea.

To be 100% sure that there is enough inductance and that saturation will not occur
over 14Hz at the maximum possible Vac across the choke, the choke inductance
MUST be measured with the specified idle dc flow present and gap adjusted for
maximum L which occurs when the XL is highest. It is best to do this measurement
and adjustment with the choke in the circuit. Sheets of thin polyester say 0.05mm
can be used to make up the gap material but make sure the Es and Is are kept
together when measuring different stacks of gap material. In this case, 0.165mm
has been calculated for right across the length of Is, so you may find that 0.15mm
with 3 x 0.05 or 0.2mm with 4 x 0.05 to be best.
If you let the gap become too low,
then distortion at 160Vrms and at 20Hz will exceed 3%, even though inductance
is high. This will be due to µe becoming too high, and thus Bdc is too high, and
combined Bac + Bdc is too high, and core saturates.
If the gap is too large, µe becomes too low, and inductance becomes low, and
distortion is due to total RL being low. But this condition is better than having gap
to small, because the core does not saturate. 

If µe cannot be raised to wanted value even with all Es tight against all Is, it may
be raised enough by what is known as partial air gapping. Instead of having one
big stack of Es and another of Is, the full height stack is divided into a number of
equal sized piles of Es and Is. if each lamination has thickness of say 0.5mm for
NOSS laminations then for a 50mm stack there are 100 laminations. These can
be divided into 10 piles with each having 10 laminations. Each pile of 10 Es is
inserted to the core facing in alternate directions, and a pile of 10s placed to close
The choke is then tested with DC flow and XL measured to see what inductance
at a low signal level of say 10Vrms across the choke has been achieved. If the
inductance is higher than calculated, the µe is maybe too high, and Bdc will be
too high, so iron will saturate at high Va levels at low F. So perhaps the use of 5
piles or maybe only 3 piles are needed to get µe up just far enough.

Use of the best GOSS E&I lams may give µ max = 17,000, well above a sample of
NOSS which may only give 3,000. If the GOSS core choke has
T = 25mm and S = 50mm, and we want µe = 375,
then air gap =
142 x ( 17,000 - 375 ) / ( 375 x 17,000 ) = 0.37mm.
Notice that this gap is not much bigger than the 0.33mm
calculated for
material with µ = 3,000. But this all suggests that the µe could be higher than
375 if the GOSS was close butted, so therefore the partial interleaving would
not be needed nor be desirable as it would raise µe to a value too high.

You will find that any air gap added to a choke core or SE OPT core will reduce the
available inductance, but it also tends to make the choke or OPT have a more
constant amount of inductance for a wide range of F and voltage levels because
some of the constancy of the air cored winding becomes evident in the iron cored
coil.

But its no wonder that not one commercial amp maker would use a choke like
I have suggested, they all find it far to difficult to spend the time to get it right.

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