CHOKES - PAGE 2. Updated 2011.
This page 2 is about....

1. Filter chokes for "choke input" or LC filters in power supplies.

Fig 1, "Traditional" 2011 choke input + rectifier + following LC filter,
Fig 2, "Modern" 2011 choke input + rectifier + following RC filter,
Fig 3, Other 2007 choke input + rectifier + LC filter.
2. Design Method for Choke in LC PSU, steps (1) to (20),
Wire size table for sizes of available 200C temp rated polyester-imide magnet winding wire.
Fig 4, "Wasteless pattern" E&I lamination size relationships & details for L, H, T. 
Fig 5,
Table values list chosen values for T, S, Wire dia.

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 chokes used for dc supply to anodes, cathodes, go to
Chokes 3

1. Choke input power supplies.

Choke input power supplies also known as LC input supplies have been the best old fashioned
or traditional way to build a DC voltage rail supply, especially where the current draw will vary
by a factor of 10 or more as is the case with a power supply for a class AB, B or C amplifier.

A "choke input" basically has a HT winding with CT followed by two diodes, or a HT winding
with a four diode bridge with diodes connecting to a choke in series with a capacitance shunting
an RL load which changes value during normal use. The choke usually has more maximum
inductance and is larger than used in a CLC type of filter, but also has low winding resistance.
Rectifiers for high voltages were usually vacuum tube or mercury vapor diodes with limited
peak current abilities but with very adequate reverse voltage tolerance. With the advent of
silicon diodes which cost 1/100 of a tube rectifier, the choke input PSU has become fallen
from widespread use because of the cost savings possible by not using large heavy input choke
and vacuum diodes. Modern high value capacitors are now very cheap and reliable.

However, in a number of occasions I have found the choke input supply to provide a solution
where one wishes to use a HT winding of a given transformer where otherwise the the B+
voltage gained by Si diodes charging a capacitor directly would be much too high.

The schematic in Fig 1 below shows a traditional choke input PSU also with added LC filter.

Fig 1.

Fig 1 shows diodes charging capacitor C1 with continuous current through L1. L2 & C2 filter the B+
output which feeds the two OPT in a stereo amp with two 50W channels with 6550 or equivalents.

The L1 & C1 form a simple a second order filter with low F pole at 3.3Hz, which is also the Fo
for the 10H and 235uF. L1 is designed to have inductance = RL / 900 in Henrys where RL
= B+ Vdc / Idc minimum which in this case 44mA, So minimum RL = 11.4k so L = 12.6H.
As Idc increases to a possible maximum of 385mAdc, the L1 inductance should decrease to about 6H.
This choke L1 is known as a "swinging choke". At the idle current of 230mAdc, expect L1 = 10H.
L2 is a smaller choke to provide additional filtering at C2. The Fo of L2 & C2 = 6Hz. There is no
added resistance between L2 and C2 to provide damping of 6Hz resonance signals. The load at
the output of tubes plus bleeder resistance of 12k0 is not low enough to provide Fo damping.
Resonances at 3Hz from L1 & C1 will also appear at C2. Amplifier gain will be -3dB at 10Hz and
-12dB at 5Hz, so the amplifier will not excite much resonance at C2. The slight very low frequency
ripple in the B+ rail will not cause much IMD because the PP output stage has good CMR for signals
applied to the OPT CT.

The HT winding plus diodes establish an ac voltage of mainly 100Hz with even order harmonics
plus a Vdc content of 0.63 x Peak voltage of HT Vac. This is the *average* voltage present of the
100Hz V peak to peak which sits above the 0V rail as shown in the bottom left wave diagram.
The HT winding is a 583V - 0 - 583V winding with CT and each side of the CT primary delivers
60VA of power to make a total of 120VA delivered to L1 choke. +822Vpk is produced after the
diodes at L1 input. There is approximately 280Vrms of 100Hz signal applied to L1 input with about
15% of even order THD product. The L1,C1 and L2, C2 filters allows the LF content below 3Hz
and DC current to flow through to the amplifier load while 100Hz + harmonics are blocked by the
choke's high reactance at the AC frequencies, and while low reactance of C1, C2 shunt the signals
above 6Hz. The +500Vdc at 230mAdc at C2 will drop to about +485Vdc at Idc of 385mAdc if
we estimate the PSU resistance before the amplifier is a total of 100r. The choke input filter gives
fair regulation of the B+ output voltage where the current may vary from the idle condition to 2
times the idle current. Regulation depends on transformer and choke winding resistances and
diode resistances. The high peak current charging of the CLC filter do not occur. The diode charge
current flows continuously, so power dissipated in the HT winding is 30% less than in a typical
CLC supply providing the same VA. If no tubes are connected, the only  At

The choke input supply ripple voltage will be higher L1 & C1 for the same amount of total L and C
used in a C-L-C filter. Despite its limitations, choke input supplies have admirers because the
continuous current flows do not produce switching pulses around ground paths and transformer
windings remain silent. The available HT supply voltage at the power transformer suits a wanted
lower Vdc to be produced. The regulation possible with tube diodes is acceptable and the tubes
do not exceed their limited current ratings. In this example I expect to get 500Vdc from a 583Vrms
transformer winding which is a Vac to Vdc conversion factor of 0.86. In practice, due to winding,
I would be lucky to see exactly what I have shown above.

The function of the bleeder resistors of 2 x 6k0 ensure Idc = approx 44mA without the amplifier
connected. The B+ voltage may rise to about +540Vdc when Idc = 44mA. But as Idc is reduced
further the B+ voltage will rise to the peak HT Vac which is +822 when Idc = 0.0mA. In fact
without a bleed resistor, there would be 3.6mAdc through the 220k across each 470uF capacitor.
All choke input filters will produce a "soaring B+" when the Idc is very low. Now the input plus
driver tubes plus any other R across caps etc will draw 75mA for both channels and this is enough
current to provide a bleeder current to stop the B+ soaring even if there are no output tubes
connected. Therefore in this case the bleeder R is optional, but we would want to make sure
all parts of the circuit will withstand +822Vdc applied until the tubes warm up and begin to
conduct their idle Idc. To stop soaring B+ even for the short period after turn on and to avoid
wasting 42mAdc of flow in the bleeder R there could be separate PT for the tube heaters and
bias supply and this is turned on by the on switch. 30seconds later the PT for HT is turned
by a delay circuit controlling a relay in the 240V winding. This would be the most traditional
approach without using the kind of immediately active solid state bleeder circuit I show in Fig 2
below to better manage the On-Off behavior. Bias failure protection could be powered by
the twin mains transformer option by turning off the HT if Idc in any one tube remains above
twice the idle value for longer than 4 seconds.

Fig 2.
Fig 2
shows a slightly different choke input PSU with slightly more modern attributes.
The general working is the same as Fig 1. The L1 choke used is designed to have high inductance
under low current idle conditions, but when current is increased from 230mAdc to 385mAdc
the B+ voltage drop may only be 17Vdc.

There are some modern tricks used in Fig 2. There is a diode bridge, very easy with Si diodes,
so the the HT winding uses thicker wire and 1/2 the turns of a CT winding which is easier to do.
There is also a 0.25uF plus 68r across L1 which causes L1 to about double its impedance at
100Hz due to damped parallel resonance. The value of 0.25uF is chosen for the value of the
L1 inductance at the idle current condition. At less Idc than at idle, the L1 inductance will rise
and Fo will drop below 100Hz, and with Idc at 385mAdc, L1 may be 6H so Fo will move
above 100Hz, but this is no problem because in practice the Idc will not much change because
most audio power is created by class A action of the OP tubes. But the reduced 100Hz ripple
reduction is always welcome, and achieved at minimal cost.

There is no need for L2 and 50r resistance will give an attenuation factor of about 1/6 of V ripple.
The use of the 50r & C2 filter instead of LC filter will provide damping of resonance in L1 & C1.
The Vripple will be higher than for LCLC type of filter in Fig1, but Vripple will be low enough at C2.
The input driver stages will have another RC filter for their B+ rails so filtering will be OK.

Q1 and Q2 are a pair of HV rated darlington pair connected transistors. You could use two HV
mosfets for the same same thing. The SS devices are connected in series to distribute the high
B+ Vdc rail voltage so the maximum possible Vdc between collector and emitter is +275Vdc
if the B+ rail was +550V. The 2k5 x 50W rated resistors may allow a maximum of 100mAdc
to flow through transistors with B+ = +503Vdc and the transistors will turned on fully and have
very low voltage across them so will be cool while resistors are hot. If I = 50maMadc, the Pd in
each 2k5 and in each darlington pair transistor = 6.3W.

Fig 3.
Shematic for LC+LC filter psu
Fig 3 shows a combination of choke input plus following LC filter in the classic LCLC arrangement
used in many older designs for where tubes operate in class AB with low bias currents.

However, perfectionists who like quiet rail voltages can use it for their class AB hi-fi amp with high
bias currents
and where Ia does not vary much in practice.

Fig 3 shows the input choke L1 and following choke L2 in the ground rail rather than in the B+ rail.
I have never found the placement of chokes in the B+ rail to work any better than being in the ground rail.
Such choke placement means the choke windings are at a slight negative dc voltage potential and thus
there is less voltage across insulation between windings earthy core iron.
The actual 0V connection
can ONLY be made where I have indicated, or else the benefits are entirely negated.
The wave form
after the diodes and at the input to the L1 choke is indicated, and is an inverted wave form compared
when the L1 is placed in the B+ rail.

The ripple voltage at bottom of C2 is less than 0.2Vrms, and if L2 was omitted, and C2 paralleled with
C3, then ripple
at B+ would be less than 0.1V, and quite OK for PP amps.

However, for SE amps 100mV of ripple may be too high, and the two LC sections are useful although
an RC section after L1&C2 where R = 50r as indicated in Fig 2 usually reduces Vripple enough plus
damps unwanted LF resonance. With SE amps PSU regulation is not critical because the Idc from PSU
to OPT remains constant.


It is difficult to make a choke for LC input which will be mechanically quiet. LC input chokes
tend to hum with vibration because of the high amplitude voltage of twice the mains frequency
plus harmonics which is applied across the choke. So the choke must not saturate because of the
combined dc current magnetization and the applied ac current magnetization. This type of choke
should be potted after being very well varnish impregnated to stop winding or core movements.
Potting will stop most stray radiated magnetic fields and quieten them. The varnishing is very
important to prevent windings moving because of the high AC voltage applied.

In 2004, I re-engineered a customer's Phase Linear 700W transistor amplifier. This amp had
+/- 87Vdc rail voltages which were too high for reliability. The existing PT was noisy, and heat
sinks seemed far too small. The customer had no need for 700W capability, but insisted on high
current ability.
I wound two chokes of about 0.35H for each + rail and -rail. These were well
varnished and potted with roof pitch in 1mm sheet steel cube shaped boxes with a side length
of 90mm.
The power transformer had a CT sec winding with 62V- 0 - 62V and there was
a 35A rated diode bridge. I obtained +/- 55Vdc which gave plenty of power. The PT noise
became negligible. 
The resulting "Phase Turner" amp has remained a very fine sub-woofer amp
for the last 7 years so far. Recordings of the Space Shuttle taking off rattled the windows.

If you insist you want a choke input filter, there are basic rules.

Minimum critical choke value, Henrys = RL / 900.

Where mains = 50Hz,
where RL = Vdc output / dc current at the minimum current.

The choke value is called the critical choke value. Choke input PSUs were often used for class AB
tube amps where the anode input current flowing into the output stage may have varied enormously.

The current flowing into the C1 resevoir cap through the choke L1 is continuous in each 1/2 wave
cycle. There is no heavy peak charge current over a small fraction of the wave cycle at the peak of
the wave as there is with a capacitor input CLC PSU. The LC input allows vacuum tube rectifiers
to be used and in high power and high voltage applications, use of mercury vapor vacuum tube diodes
such as 866 were good with a following class AB amp of kilowatt capability. The minimum and
maximum dc current flows must be known, and for some old fashioned amps the minimum current
is taken to be 10% of the maximum which will be calculated and become known. With LC inputs
The minimum must be allowed to flow even when output tubes are biased right off, as in the case
of a real class B or class C RF amp.

Unless there is some minimum current flow which is called the "bleed current", the B+ Vdc can
soar to the full value of 1.41 x the HT winding Vrms voltage, possibly stressing other components or
insulation layers especially if the DC voltage exceeds 1,000V when corona effects can start arcing
in amplifiers.

In about 2005 I completely re-engineered a very poorly designed CR Audio Developments
"Woodham" 5050 stereo amp with a pair of PP channels with KT88. The original amp had HT
transformer voltage of 487Vrms and the cap input with Si diode bridge gave up to +650Vdc.
This powered a very poor imitation of the McIntosh style of output stage. One could only get 40
Watts of audio for a short time before smoke arose from this horrid amplifier. The anode to anode
loading was found to be about 1/3 of the correct value for such a high B+ voltage.

I found I had a choke which could be suitable. So after a few tests I potted it and got B+ of  +430Vdc
at the at the reservoir cap after the choke at full current draw. I converted the amp to 50% UL
because the toroidal OPTs had 4 primary windings which could be re-arranged. I got 45 reliable
Watts at about 1/5 of the THD/IMD and better damping factor with revised loading for the OPT etc.
I made a solid state shunt regulated supply for input tubes which ensures 30mA is
drawn right after switch on so B+ does not rise above +450Vdc, and as the output stage current
begins to flow to a normal level of 250mA, the shunt regulator reduces its bleed current "wastage"
of energy, and B+ remains steady at +420V despite the huge change in current.

In this amp, when Idc = 30mA, RL = 450V / 0.03A = 15,000 ohms. To ensure the choke would
work with low current to keep +Vdc less than +450V, the choke should have had a maximum L
value = 16.7H. The choke only had 9H at 30mA. At 250mA, when the amp load reduced to 1,680
ohms, the choke inductance fell to about 4Henrys due to the heavy DC flow. The result was that
the "knee" of the regulation "curve"
wasn't sharp, and at low Idc the voltage at the PSU tended to soar above +500V. With a choke in
a CLC supply, the Bdc can be allowed to be quite high at the highest Idc, because the Bac is
quite negligible. But in the choke for LC input, the Bac is MUCH greater, and for most amps we
should make sure Bdc = Bac = 0.6Tesla maximum, for maximum Idc. Thus the choke for LC
input is always going to be larger than the choke for CLC.

There is a simple remedy for where there is only 1/2 the wanted maximum critical inductance.
An R+C Zobel network is connected across the choke so that the L and the C value have a
resonance at the ripple frequency where the L value is slightly above where Idc is minimum.

I placed a 0.33uF cap plus 100 ohms for a Zobel network across the choke when its value = 8H,
so that the cap and the L were resonant at near 100Hz. The effect was quite remarkable,
and this choke plus Zobel acted almost identically to a choke of 16H, or twice what I had,
and B+ soaring was prevented, and Vdc better regulated, and a low bleed current was required.
So if one wants 250mA maximum, bleed current should be at least 25mA.
For best natural B+ regulation, the power transformer and choke should have low winding
resistances. If the choke was 50ohms, the change in current of 200mAdc would cause an
unavoidable drop of 10Vdc. This isn't bad, but never use a choke of 500 ohms! It will
smoke, and regulate poorly.

2. Design Method - Choke for LC PSU.
Now let us consider the design of a choke L1 in Fig 2 where there will be +435Vdc at C1, and
300mAdc maximum continuous with LC input filter. Let us design the choke without reliance
on a Zobel resonant RC network across the choke to improve 100Hz filtering at low Idc levels.
The Zobel may be added to the circuit upon completion to not only improve 100Hz noise rejection,
but to provide a resistance load at switching frequencies to damp diode switching transient emf
which could apply excessively high voltages to the power transformer insulation. If the maximum
L is available is thus sufficient, and the Bdc at full current still permits the choke to work as a pure
inductance without ac saturation, the choke is well designed. The dc current density should not
exceed 2A/ at maximum Idc.
To calculate wire Cu dia, d :-
For 1A/, d = 1.13 x sq.rt I,
For 2A/, d = 0.8 x sq.rt I,
For 3A/, d = 0.65 x sq.rt I, I in Amps, d in mm.
(1) What is the wanted range of DC currents at the nominal B+ Vdc rail?
What is the range of load values supplied by power from the power supply?
Idc min = 30mAdc, Wanted B+ at C1 = +435V, Maximum RL = Vdc / Idc = 14,166 ohms.

Idc max = 300mAdc, Expect Vdc = +425V, Minimum RL = Vdc / Idc = 1,417 ohms.

Maximum allowable Rw = Minimum RL / 30.
Rw = RL minimum / 30  = 1,417 / 30 = 47 ohms.

(3) Maximum wanted L.
L max  = max RL / 900 = 14,166 / 900 = 15.7H.

(4) Minimum  wanted L.
L min = min RL / 900 = 1,417 / 900 = 1.57

Maximum Cu current density = 2A/, For 2A/, d = 0.8 x sq.rt I, d in mm, Idc in Amps.

Cu d = 0.8 x sq.rt 0.3 = 0.438mm.

(6) Select wire size from table......

Wire size table.

Wire will be 0.45Cu dia , o/a dia including enamel = 0.516mm.

(7) Select T dimension of E&I wasteless laminations.
Fig 4.

Try choosing T = 32mm.

(8) Read table for area for winding wire, WA = 12mm x 44mm = 528

(9) Calculate Number of turns N  =  WA / oa d squared, for coil without insulation layers.
N = 528 / ( 0.52 x 0.52 ) = 1,950 turns.

(10) Calculate Turn Length TL = Rw x 100,000 x d squared / ( N x 2.26 )
where current density = 2A per

= 47 x 100,000 x 0.45 x 0.45 / ( 1,950 x 2.26 ) = 216mm.

(11) Calculate Stack height S = 0.5 x ( TL - [3.6 x T] )
= 0.5 x ( 216 - [3.6 x 32] ) =  51mm.

Is calculated S between 0.75T and 2T?

yes, we may proceed, S = 51mm.

NOTE. If S required for a given T is more than 2T, consider choice of larger T.
If S is calculated at less than 0.75T, consider choice of smaller T, or larger wire diameter.
For a given wire size and maximum allowable Rw, there is an optimal Afe which gives the
most inductance.

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 over a wide range during some trial and
error calculations until a solution appears. Too small a stack will become obvious, and
for more L the stack needs to be increased. But if turn length increases to raise Rw too much
then a slightly larger wire size may be needed if Rw is to be kept low with highest possible
inductance and always without core saturation.

If a new choice of T is made, go back to step (7) and re-calculate.

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

Bac = 22.6 x 240 x 10,000 / ( 51 x 32 x 100 x 1,952 ) = 0.17 Tesla.

(13) State allowed total maximum Bac + Bdc, medium grade silicon steel E&I core, = 1.2 Tesla.

(14) Calculate maximum allowable Bdc at Idc maximum = Total ( Bac + Bdc ) - Bac
= 1.2 - 0.17 = 1.03, say 1.0Tesla.

(15) Calculate 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,
where Bdc in Tesla,
10,000 and 12.6 are constants for all equations,
Iron ML is the core magnetic path length without air gap,
N is the number of turns,
Idc is in Amps DC.

ML for T = 32mm for wasteless pattern = 5.6 x T = 179mm.

µe = 1.0 x 10,000  x 179 ( 12.6 x 1,952 x 0.3 ) = 242.

(16) Calculate Inductance, L  = ( 1.26 x Nsquared x Afe x µe ) / ( 1,000,000,000 x ML )
where L in Henrys,
N is turns,
Afe is S x T,
µe is effective permeability with air gap and Idc,
1.26 and 1,000,000,000 are constants,
ML is Iron magnetic path length in mm.

L =  ( 1.26 x 1,952 x 1,952 x 51 x 32 x 242 ) / ( 1,000,000,000 x 179 ) = 10.6H.

(17) Calculate air gap and gap material thickness.
Note, µe = µ / ( 1 + [ µ  x  gap / ML ] ),
Therefore gap = ML x ( µ - µe ) / ( µe x µ ),
Where µe is effective permeability with air gap, µ is the maximum possible permeability
with E&I lams maximally interleaved,
gap is the total "air gap" in mm consisting of a
gap or gaps filled with plastic sheeting within iron magnetic path,
ML is the iron
magnetic path length in mm.

Estimate or state maximum possible µ for medium grade E&I lam material, µ = 3,000.

Gap = 179 x ( 3,000 - 242 ) / ( 242 x 3000 ) = 0.67mm
State number of gaps in magnetic path = 2,
Gap material = calculated gap / 2 = 0.33mm.

(18) Calculate inductance with Idc = 1/10 of maximum.
L is proportional to µe; at low Idc µe will be 1.5 the µe at high Idc so L at low Idc is 1.5 times greater.
L at high Idc x increase in µe = 10.6H x 1.5 = 15.9H.

(19) Is there sufficient inductance at low Idc?
State wanted max L from step (3) = 15.7H.

Calculated max L value in step (18) is more than stated in Step (3) so choke design is OK.

Here is a table of calculated parameters if 32mm Tongue size E&I laminations are used.

(20) If winding wire choice was revised for a larger size, calculate new winding resistance,
Rw =
Rw = N x TL x 2.26 / ( 100,000 x d x d ). This should be less than calculated
maximum allowed Rw.

Could a core with T = 25mm be used?
Steps 1 to 20....

Let us trial the method above....

(1) RL max = 425 / 0.03 = 14.2k, RL min = 425 / 0.3 = 1.42k.

(2) Rw max = 1.4k / 30 = 47r.

(3) L max = 14.2k / 900 =  15.7H.

(4) L min = 1.4k / 900 = 1.57H.

Cu d = 0.8 x sq.rt 0.3 = 0.438mm.

Wire from size table will be 0.45Cu dia , o/a dia including enamel = 0.516mm.

(7) Select T = 25mm.

(8) WA = 33.5 x 8.5 = 285

N = 285 / ( 0.52 x 0.52 ) = 1,053 turns.

(10) TL =
47 x 100,000 x 0.45 x 0.45 / ( 1,053 x 2.26 ) = 400mm.

(11) S =
0.5 x ( 400 - [3.6 x 25] ) =  155mm.

Is S between 0.75T and 2T?

No, S is much more than 2T.

Therefore return to step (7), try T = 38mm.

(8) WA = 53 x 15 =795

(9) N = 795 / ( 0.52 x 0.52 ) =  2,940 turns.

(10) TL = 47 x 100,000 x 0.45 x 0.45 / ( 2,940 x 2.26 ) = 143mm.

(11) S = 0.5 x ( 143 - [3.6 x 38] ) = 6.2mm

Is S between 0.75T and 2T

No, S is much below 0.75T.

Therefore we might try choosing T = 32mm, but also
could try increasing wire size.

Go back to step (6) and select wire size about 1.4 times previous diameter.

Wire could be 0.6mm Cu dia = 0.675mm oa.

(9) N = 795 / ( 0.675 x 0.675 ) = 1,745 turns,

(10) TL =
47 x 100,000 x 0.45 x 0.45 / ( 1,745 x 2.26 ) = 242mm.

S = 0.5 x ( 242 - [3.6 x 38] ) = 53mm.

Is S between 0.75T and 2T?

Yes, proceed on.....

Bac = 22.6 x 240 x 10,000 / ( 53 x 38 x 100 x 1,745 ) = 0.15 Tesla.

medium grade silicon steel E&I core, = 1.2 Tesla.

) allowable Bdc at Idc maximum = Total ( Bac + Bdc ) - Bac
= 1.2 - 0.15 = 1.05 Tesla.

µe = ( 1.05 x 10,000 x 213 ) / ( 12.6 x 1,745 x 0.3 ) = 339.

L =  ( 1.26 x 1,745 x 1,745 x 53 x 38 x 399 ) / ( 1,000,000,000 x 213 ) = 14.4H.

Air Gap = 213 x ( 3,000 - 399 ) / ( 399 x 3000 ) = 0.46mm.
Therefore use 0.23mm thick gapping material across 3T length of lamination.

L at high Idc x increase in µe = 14.4H x 1.5 = 21.6H.

(19) Is there sufficient inductance at low Idc?
State wanted max L from step (3) = 15.7H.

Calculated max L value in step (18) is more than stated in Step (3) so choke design is OK.

Rw = N x TL x 2.26 / ( 100,000 x d x d )
= 1,745 x 242 x 2.26 / ( 100,000 x 0.6 x 0.6 ) = 26.5 ohms.

This figure of 26.5 ohms is a delightfully low Rw for this PSU, but it is somewhat
unnecessary and probably more than half the weight of the power transformer for the
two channels for the example amp. We could consider trialling a wire size of 0.56mm
and re-calculating.

If the power supply becomes too heavy, consider making the PSU on a separate chassis
and the two 50W audio channels can be on another chassis which will be easier to move
around while ensuring there is less chance of noise from power supply reaching the
amplifier output. This is especially so with triode amps with little or no global NFB.

Fig 5.

L Henrys,
at 0.27
L Henrys,
at 0.027

Fig 5 table values list chosen values for T, S, Wire dia. It is assumed maximum
Bac + Bdc allowed = 1.2 Tesla.
Values for N, TL, Rw, Bac, Bdc, µe, L minimum, L maximum may be calculated.

N for given Cu dia wire size = window winding area / oa dia of wire squared, ie,
N = ( [1.5T - 4] x [0.5T - 4] ) / ( oad x oad ).
TL = 2S + 3.6T, therefore S = 0.5TL - 1.8T. Numbers given are constants
valid for wasteless E&I only.
The following have numbers in equations which are all constants....
Rw = N x TL x 2.26 / ( 100,000 x d x d ),
therefore TL = ( Rw x 100,000 x d x d ) / ( N x 2.26 ).

Bac = ( 22.6 x Vrms x 10,000)  / ( T x S x F x N ).

Bdc allowable for given choke = Rated maximum total of (Bac + Bdc) - Bac calculated.

Rated maximum total of (Bac + Bdc) for very old low grade iron may be 0.9Tesla;
with some Si content but not grain oriented, 1.2Tesla;
and for GOSS perhaps 1.4Tesla.

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

L = ( 1.26 x N x N x T x S x µe ) / ( 1,000,000,000 x ML )

L Henrys at low Idc = 1.5 x L Henrys at high Idc. Approximate!

NOTE. Where the stack height becomes very low, the Afe becomes so low the Bac then
becomes very high. The Bdc is limited to the difference between maximum total B and Bac.
Bdc is proportional to µe so the gap must be increased reduce the dc magnetization. In the
above case where S = 6.3mm, µe has dropped to 17.6 by use of a 10mm air gap.
Regardless of stack height, if the Vac applied is high enough and frequency low enough, Bac
could be quite high with a high stack of laminations, and Bdc would still have to limited by
reducing µe by widening the air gap.

But consider the equation for an iron cored inductor in Fig 5 table above with T = 32, S = 51,
µe = 283, Iron ML = 179mm, and N = 1,950.
Inductance = 1.26 x 1,950 squared x 32 x 51 x 283 / ( 1,000,000,000 x 179 ) = 12.4H.
The air gap is 0.57, and we have assumed maximum possible µ = 3,000.

The equation could be reduced to L = 7.82 x µe / ML. = 7.82 x 283 / 179, So µe / ML = 1.58.

If the lams were interleaved with no gap, L = 7.82 x 3,000 / 179, and µ / ML = 16.7, so L = 131H.

The air gap effectively lengthens the iron path length from 179 to 1,891mm. The air gap is
only 0.57mm, and 0.57mm x 3,000 = 1,710mm, and if we add 179mm we get 1,889mm almost
exactly 1,891mm, which illustrates that the magnetic path length is effectively 179mm of iron plus
0.57mm of air which sums to being the same as 1,889mm of iron all with µ = 3,000.

If the Is were removed away from the Es of this sample choke the Iron ML would reduce to
138mm, and the "gap" would be the distance from centre of 3 legs of the E, roughly 32mm.
The ML total path L become [(32 x 3,000) + 138], and µ / ML = 3,000 / 96,138.
Or it can be expressed as µe / ML = 4.3 / 138, because we consider the wide gap has reduced
µ to µe = 4.3. Inductance of the choke would be 0.188H. In practice if you add a bar core
made of laminations the increase of inductance above the air cored coil is about 4 times. This can
be very useful where we wish to make a low loss bass speaker crossover coil with low Rw.
A given air core coil of 2.5mH may be raised to 10mH with a bar core of about the same length
as the winding. This can also be useful where a choke is wanted for CLC low voltage DC supply.
The E&I lams are better, but it is possible to use a bar core solenoid.
As far as I know, the equations I have used from text books to make up my above table become
unreliable once the air gap exceeds 1/5 of the width of the iron - due to fringing and path length shape.
The calculation of an air cored inductor involves Wheeler's Formula or something invented elsewhere.
There are at least several online calculators for air cored inductances. I have never ever seen an online
calculator for Hanna's method spelled out in RDH4, or any other program offered for filter choke design.

I think it would be quite easy to prepare step by step choke design program using Hanna's Method.
My own program seems to work OK though.

The inductance equation for iron cored L cannot be simply adapted for air cored coils or solenoid coils.

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