This page 1 is about :-

1. What is a choke? General information about inductance and filter chokes.

2. Measuring choke properties.

Fig1, Simple inductance tester.

3. Quick measurement of L.

4. Another quick measure of L.

5. Test CLC filter chokes with DC currents.

Fig 2, Testing choke in CLC filter with DC flow, setting the air gap.

6. Resonance in CLC filters.

Fig 3, curve of LC response.

7. Damping LC resonance.

8. Choke distortions.

9. Chokes for CLC in PSUs.

10. Design method for choke in CLC.

11. Winding method notes.

12. Permeability.

13. Formulas.

14. Other issues CLC.

15. Comparison of CRCRC filters with CLC filters.

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Other related pages :-

For "choke input" power supplies with LC filters and for dc supply to tubes

where a high impedance anode load is desired with high level Vac signals exist

across a choke, go to

Chokes 2

For chokes used for dc supply to anodes, cathodes, go to

Chokes 3

1. What is a "Choke"?

I do not know where the word "choke" came from but very soon after electricity was put

to use in hundreds of ways by 1900, people were winding coils of insulated copper wire

around iron to make an inductance to inhibit the flow of ac currents, or to "choke the flow"

of ac, alternating current.

At some point of electronic evolution, an inductance was found to be a good filter device

for reducing noise current flow between point A and B in a circuit. And indeed the right

quantity of inductance and wire resistance and low amount of stray capacitance may make

an excellent filter element. Whether we use a choke or not depends on the application,

size, weight, cost and perhaps and several other factors.

The unit of inductance is the Henry, named after Joseph Henry, a great American scientist

who died in 1879.

I don't want to get bogged down into the deep mathematics surrounding all inductors.

Definitions of Inductance are well covered elsewhere on the Internet, and readers need

to read MUCH MORE than I am saying here.

But basically, if there is change in current through a coil of insulated wire there will be a

change of magnetic field produced. This magnetic field change causes production of an

"electro motive force", EMF, which opposes the change in the flow of the current.

This phenomenon is difficult to understand. But if a length of anything which conducts

electrons is brought to be within a changing magnetic field, a voltage across the conductor

is produced which signifies current will flow if the conductor is part of a circuit.

The effect is larger when the conductor is a length of copper wire which has been wound

into a coil. Now if there is no changing outside magnetic field and someone applies a

changing or alternating AC voltage across the coil, there is a changing current flow ie,

alternating current, AC, and a changing magnetic field is immediately created, and this

field tries to prevent the flow of AC current flow causing the field.

It is as if someone unseen is applying some external magnetic field to prevent the current

flow by applying that field to oppose the initial current flow.

There is no outside Interfering Person, but there is magnetism.

The Sun and the Earth have giant magnetic fields and these have great effects on flows of

energetic particles. Magnetism cannot be seen, but we can determine where the

"lines of force" go and what strength of magnetic field may exist. So while not understanding

magnetism at the highest level of university education we can learn how to use it.

A steady unchanging magnetic field may very easily be produced by a coil of wire with

a steady unchanging DC flow of current. This has 101 applications in electro-magnets

used for solenoids and many other applications. In early loudspeakers the anode dc

supply current was fed through a field coil to generate a strong magnetic field within

which a voice coil could work as in an electric motor.

While there is an unchanging DC flow there is no magnetic field change to oppose

the current change in the coil producing the magnetic field. Therefore the inductor

seems to have a simple resistance equal to the measured resistance of the wire.

But as soon as we change the current, the coil reacts to the change in current to

oppose the flow. So coils of wire have Reactance. This reactance may be quantified

in ohms after calculating Vac applied across coil / Iac flow within coil but the reactance

calculated is only valid for the frequency of the AC. Reactance is proportional

to the frequency of the current flow. This reactance is the "frequency dependent

resistance" and is called the reactance of the inductance, or XL. It should always

be considered as being in series with the wire resistance which must never be

ignored unless it is a very low value. Sometimes the word "impedance" is also

used for an inductor because all "reactances" such as inductors OR capacitors

resistances consist of their pure reactance value dependent on frequency but

in series with their winding wire and lead resistance.

Inductor Reactance, XL, is always expressed as a number of ohms at a certain

frequency.

XL = F x 2 x pye x L, where X is in ohms, 2 x pye = 2 x 22/7 and L = Henrys.

The inductor impedance, ZL, includes more than just pure L, and includes winding

resistance Rw.

ZL = Square root of ( Rw squared + XL squared ) ohms.

If you had a 1.0 Henry inductor of 100 ohms winding resistance, then its ZL at 10Hz

= sq.rt. ( [100 x 100] + [ 62.8 x 62.8 ] ) ohms = 118 ohms.

At 100Hz, the ZL = 635 ohms.

XL = 628 ohms. So you can seen that the winding resistance of 100 ohms does

not just add to the XL of 628 ohms. There is a reason why this is so but I don't have

space to explain it here. But you need to remember this phenomenon. So at high

frequencies the winding resistance makes very little difference to the choke impedance

and for where XL > 10 x winding resistance plus any connected circuit resistance

the winding resistance may be neglected for most circuit calculations.

Suppose we have a simple circuit with AC voltage generator in series with a choke

with a coil of wire wound around an iron core in series with resistance. An equal

value of current will flow through amp, choke, and resistance and 0V ground path.

The magnitude of AC current flow depends on :-

A, The inductance value in Henrys, which can change depending on the properties

of iron.

B, The resistance in Ohms,

C, The magnitude of the voltage source amplitude in Volts.

D, The AC frequency.

E, The presence of steady DC current flow in addition to AC current. The DC current

tends to reduce the inductance value.

2. Measuring choke properties simply :-

Fig 1

The Fig1 test set up is shown with an inductance being tested to find its inductance value.

The same test set up could also be used for measuring capacitance values.

This simple test set up requires tools :-

A good signal generator, able to make 2Hz to 200kHz, sine wave, with accurate F indication.

Possible frequency meter if signal gene is poor

Dual trace analog oscilloscope with bandwidth up to at least 1MHz, with two probes, one for

channel A and B.

Audio amp, able to amplify 7Hz to 25kHz to 12Vrms, so not critical,

AC volt meter with multi ranges and bandwidth of 2Hz to 200kHz and giving Vrms.

Bread board, ie, sheet of ply say 200mm x 200mm with wood screws to which you can solder.

RCA leads, voltmeter leads and amplifier leads.

Solder, patience, persistence, and the will to understand and to never assume anything,

Note book and pencil.

PC and keyboard is NOT NEEDED.

Range of Resistors, 10W wire wound :- 10 ohms, 100ohms, 5W wire wound :- 1k0, 5k0, 10k0.

The most difficult basic thing here is the set up of oscilloscope for balanced mode, aka

differential mode. Learn how to read the oscilloscope manual and how to use what you have,

and if there is no manual, and/or your oscilloscope is single trace, then you have to do some

googling and research because I don't have the time to tell you how to use all the tools.

The signal generator should allow a constant level output voltage to be adjustable for each

of its range of frequencies, and its output resistance should be ideally no higher than 600

ohms so that connection to amplifier inputs does not have any effect on output signal purity

or level. Many "Signal generators" or "Function generators" have "decade" ranges say

2Hz to 20Hz, 20Hz to 200Hz and so on up to 200kHz which an audio tech uses to test audio

gear. The better and more expensive but fragile function generators have a top frequency

of 2MHz, with dc offset, sine wave, square wave, triangular wave, maybe saw tooth wave,

FM and AM modulation etc. But for this test you only need sine waves.

The amp raises the level of generator signal and converts the high output resistance of the

signal generator to a source of signal with a low resistance less than 1 ohm, so that anything

we might connect to the amp will not affect the voltage level at the amp. The resistors used

for testing are placed between amp and whatever L ( or capacitor ) is being measured.

The points of interest in the above circuit are the measured Vac voltages across the choke

or TP2 to TP3, and across the R or between TP1 to TP2.

The amplifier voltage is measured between TP1 and TP3.

TP3 is always regarded as 0Vac, and a reference voltage.

Suppose we wish to measure inductance of a filter choke, or speaker crossover choke.

Values could be anywhere between 0.1mH and 70H. For this test we need not worry about

DC flow and air gaps because I am trying to get readers to understand basic properties of

inductance.

Suppose we connect 1k0 resistance between TP1 and TP2 and have an unknown value of

inductance between TP2 and TP3.

For this example, suppose we have a choke with core of 25mm stack x 25mm tongue E&I

laminations, and has Es butted to Is with a slight but unknown gap between E&I, known as

the "air gap."

The wire size looks to be approximately 0.35mm dia. Let us say we measure the winding

resistance to be 50 ohms which is the same as if the wire was a straight length of unwound

wire. All inductors have some "dc resistance", ie, plain old resistance, and it must be

measured with a dc flow flow only, and R = V / I, which is Ohm's Law.

So all inductances are not perfect reactances and all are equivalent to a resistance-less

inductor plus a series R equal to the measured R. So 50 ohms is effectively in series with

the amount of choke inductance, and in the above circuit is in series with the 1,000 ohms

series R above. Because 50 ohms is 1/20 of the 1,000 ohms its effects on what we want

to measure can be neglected.

A choke may also have some capacitance between layers of windings which sum

to form the "self capacitance" of the winding when measured from one end of the

winding to the other. There is a small amount of capacitance between adjacent turns of

wire and it all adds up among all those turns to be significant. This C will become resonant

at some frequency determined by the formula :-

Fo = 5,035 / Sq.Rt ( L x C ) where Fo is in Hz, 5,035 is constant for all equations

to work, L is in milli-henrys, C is in uF.

If one were to apply the formula to a power supply filter choke of say 2Henrys with

a typical C = 250pF, then Fo = 7,120 Hz. XC = XL at Fo, so XL = XC = 89k ohms at

7,120Hz. The choke impedance at Fo will be higher than XL or XC could each be at

Fo because the impedance of a parallel tuned circuit is higher than either XL or XC.

The Q of the resonance peak is damped by resistances connected each side of the

choke and by the winding resistance. In general, the choke self capacitance may be

largely ignored for power supply applications but not for where the choke is used as

a load to an anode circuit in an audio amp stage. Nevertheless, the 250pF of typical

choke self capacitance is enough to convey HF switching noise of diodes to the following

audio circuit and if the ESR of the following electrolytic caps is high. The HF diode noise

which appears as little short duration bursts of noise at a 100Hz rate will find its way

to an amplifier output where it may be audible, and I often encounter such noise in my

servicing work. Therefore electrolytic capacitors should have plastic bypass capacitors

across them, 1uF adequately rated is usually enough.

Suppose we set the amp output level to 10.0Vrms, TP1 to TP3.

Suppose we set the frequency to 100 Hz, because we want to know what L the choke

has at about this F.

Suppose that in Fig 1 above we measure 1.59Vrms across Resistance R, TP1 to RP2.

And suppose we see it is a sine wave on the CRO, (cathode ray oscilloscope), and

distortion is less than 10%.

We can say there is a flow of current, IL, = 1.59V / 1,000 ohms = 0.00159Arms

= 1.59mArms. This flows from amp through L, through R and back to amp via the 0V

rail which consists of ordinary 5 amp rated hook up wire and has resistance less than

0.1 ohms. We also know that there is 10Vrms at the amp output. If we measured the

voltage across the choke we may find it is close to 10Vrms which is measured

between amp output and 0V. We might measure the choke voltage = 9.5Vrms

approximately.

The reactance XL = Voltage across L divided by current flow through L;

impedance follows Ohm's Law.

So XL = 9.5V / 1.59mA = 5,975 ohms.

XL = L x F x 2 x pye = L x F x 6.28, so L = XL / ( F x 6.28 ),

in this case L = 5,975 / ( 100 x 6.28 ) = 9.5 Henrys.

If we adjust the frequency downwards, leaving the applied amp voltage constant,

we will see that the voltage across choke TP2 to TP3 slowly reduces, and at some

point the measured voltage across 1k0 = measured voltage across choke, and both

will be equal to about 7.0Vac. I can predict that you may find the frequency is about

17Hz. Current in 1k0 = 0.007Amps, and in choke it is the same. Voltage across choke

= 7.0V. The reactance may be found in the same way as before, XL = VL / IL.

7V / 0.007A = 1,000 ohms, and inductance = 1,000 / ( 17Hz x 6.28 ) = 9.5H.

We may find the inductance has become higher at 30Hz than it is at 100Hz.

This is quite common and just because an iron cored inductance measures 9.5H

at one F, it does not mean it will be 9.5H at other F. The iron's permeability ( µ )

changes with frequency as well as with voltage amplitude. An air cored inductor

has a very constant inductance but once there is iron present the inductance

will be much increased over a wide range of LOW frequencies but not in a linear

manner. The non-linearities in a power supply circuit with an iron core may be

ignored, but not where the coke is used for loading an anode circuit.

Usually most E&I laminations have maximum µ at some F below 100Hz, often

around 30Hz. We can measure the choke reactance at any F and/or voltage level

and work out if the choke will suit our purpose. The measurements indicate that

as F is lowered, current flow increases, because XL reduces. Air cored inductors

have a constant inductance all F and all voltages for audio applications but it is

difficult to make large amounts of inductance without using iron cores. An air cored

10H filter choke without iron core and with 50 ohms wire resistance might be

50 times larger than the iron core choke and would be quite an improper and

expensive waste of copper. The choke, and only iron wound chokes have

any uses for audio gear as :-

1, Filter chokes for "capacitor input" CLC filters in power supplies covered in

this page,

2, Filter chokes for "choke input" or LC filters in power supplies, see Chokes 2.

3, DC supplies or sinks for anode or cathode, known as "choke loading",

or "choke feed", or "para-feed", see Chokes 3.

For filter inductances for speakers, attenuator networks in amplifiers, etc, air cored

chokes are often used because the amount of inductance required is usually much

less than 10H, and air cored chokes become easier and cheaper to make and have

a constant L value and do not generate any distortions. Some modern chokes used

as bass speaker filters to obstruct HF signals have pre-moulded iron dust cores or

a bar made of stacked laminations in a solenoid design to increase L and keep

winding R low. Usually the bar cored solenoid choke raises L by a factor of 4 and

because inductance is about proportional to the number of turns squared, a bar

cored choke can have 1/2 the turns as the air cored inductor of the same size

with the same wire. Chokes in series with midrange drivers or to shunt the LF

signals in HF tweeter crossover filters are invariably air cored.

For filter chokes in power supplies, we really only need to know what the inductance

will be at 100Hz because 100Hz is the frequency of the ripple voltage output from a

rectifier and is twice the mains frequency in UK, Europe, Australia, and elsewhere.

Any power supplies built in "50Hz" countries will work fine where 60Hz mains exists,

such as the USA, although iron wound components designed for 60Hz operation

may not work well when transported to 50Hz countries. I am Australian, and all my

formulas in this site are based around 50Hz mains and metric measurements unless

stated otherwise.

The above test circuit may be used for measuring a whole range of inductance types

for the audio band. The R value may be varied as desired. Practice with measuring and

calculating should train you to work precisely, like the dedicated technician you need

to be to avoid making amplifiers riddled with noise or with a bad smoking habit.

As I said above, Reactance is a name for both L and C circuit elements at a defined

frequency. The pure L or the C reactance actually has NO resistance and therefore

current flow through such reactances does not dissipate any heat, and no work is

done in the L or C. But where the current in reactance also flows in a pure resistance

or tube connected to a reactance, work is done and heating occurs in the resistance

or tube and Power = Vrms x Irms in Watts. It is also equal to

I squared x R, in Watts, or as Vrms across R squared / R, also in Watts.

Electrical energy is transferred by the L or C and can be temporarily stored but not lost

if the L or C is perfect, ie, has low resistance wire or leads. The energy in a choke

becomes stored in its magnetic field; increasing current increases the magnetic field,

and when the current declines the magnetic energy is released. Please search Google

for more extensive explanations about inductance and capacitance properties because

most people remain ignorant of such basic ideas and thus cannot ever learn to properly

design their amplifiers. Once a "hands on" understanding of how L, C and R

behave with AC signals, then other concepts become clear.

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3. Quick measuring of inductance, using Fig 1 schematic.

The resistance R may be used as shown.

The amplifier level is adjusted for 10.0Vrms.

The voltmeter is held across the R.

The frequency of the signal generator is adjusted until the voltmeter reading = 7.07Vrms.

The frequency where this occurs is recorded.

Now having achieved this, if you measure the voltage across the choke you will find that it

also is 7.07Vrms.

This might be very troubling because your mind will be telling you that if you have 10V

at the amp, and 7V across the choke, there should be 3V across the R, because 10 - 7 = 3.

But your meters are not telling a lie, and it is because of phase shift that the choke

voltage and R voltage when added can be greater than the amplifier input voltage.

If you were to connect TWO oscilloscope channels of a dual trace to view the amplifier

voltage TP1 to TP3, and the choke voltage between TP2 and TP3, then you would find

that the sine wave across L have peaks which are 1/4 of a wave ahead of the amp voltage

peaks. It seems as if the electricity has arrived at a destination before it left the amp.

But it is not so, and relative phase is merely being advanced. If F is reduced, phase

advance or "phase lead" as it is called increases towards a 90 degree maximum.

At some very low F, the XL falls to less than the winding resistance and the XL becomes

a negligible reactance, and the minimum voltage seen is about 0.47Vrms, because 1k0

and Rw of 50 ohms form a resistance divider. You won't be able to measure it because

it occurs at below 0.8Hz, and your gear reads unreliably at that F. Were you to reverse

the positions of L and R, you would see voltage across R fall above say 30Hz as the L

reactance increases. Relative phase between amp and 0V and R and 0V would show

the R voltage peaks behind the amp's, and this is "phase lag".

But wherever you have an R & L or an R & C, there will be a maximum of 90 degrees of

phase lead or lag after attenuation has begun to occur. Where R voltage = L or C voltage,

the phase shift = +/- 45 degrees. This sort of R, L and C behavior needs to be well observed,

and riveted firmly in your brain, if it ever manages to stay turned on properly.

When equal voltages exist across R and choke, then XL = R. From this inductance is

very easily calculated as L in Henrys = R / ( 6.28 x F ). R in ohms, F in Hz.

When equal voltages exist across R and capacitor, capacitance is easily calculated as

C in uF = 159,000 / ( R x F ). R in ohms and F in Hz.

For example, suppose the choke we were testing above had XL = 1,000 ohms at 14.5Hz,

then the L = 1,000 / ( 6.28 x 14.5 ) = 11H.

Suppose the frequency could not be lowered to enough to get VL = VR, then this

indicates XL is higher than 1,000 ohms. We will need to increase R to some arbitrary

value, say 10,000 ohms, and then if we found VL = VR at 14.5Hz,

L = 10,000 / ( 6.28 x 14.5 ) = 110H, which might be the case for the primary winding

of an output transformer.

We may wish to measure inductance at other frequencies of interest, such as

100Hz for a filter choke.

4. Another quick measuring of inductance with 25k pot,

instead of fixed R in Fig 1 schematic.

The amp is set up for 10.0Vrms output.

Frequency is set for the frequency of interest.

The potentiometer is adjusted until VR = VL = 7.07Vrms.

The value of R is measured with an ohm meter, or else a 100 ohm fixed R is placed between the

potentiometer and 0V so current in 100 ohms can be measured easily by reading the voltage

across 100 ohms and dividing by 100. The total of resistance of pot + 100 ohms

= ( V pot + 100 ) / current in 100 ohms. One way or another, we can find out the R between

choke and 0V. If measuring the pot resistance we need to disconnect the choke from amp to

prevent measuring the low winding resistance of the choke and to prevent the amp signal

interfering with the voltmeter.

Once we have the R value sorted, L is easily calculated as R / ( 6.28 x 100 ). Suppose we found

R = 5,966 ohms. Then L = 5,966 / ( 6.28 x 100 ) = 9.5H. This agrees with what we found earlier

when we measured the choke to be 9.5H using a fixed resistor of 1,000 ohms.

The frequency where VL = VR = 0.707 x source voltage is known as the LR network "pole" or

"cut off point" or -3dB point.

The iron cored inductance value changes for applied Vac levels, and during each cycle due to

"hysteresis". The hysteresis causes distortion currents. It is usually of no great importance in filter

chokes. Hysteresis is minimized by an air gap or use of grain oriented silicon steels. This

phenomena should be studied in elsewhere on the Net or in old books.

The saturation behavior of an iron cored filter choke is of very great interest to obtain the

maximum inductance with no saturation for for any choke or transformer winding. A filter choke

in a CLC filter will have considerable dc current flow and usually low AC flow, and the DC flow

will magnetize the iron at some point between zero and the maximum field strength possible in the

iron depending on its properties. The magnetic field intensity is measured in Tesla, and old iron E&I

lams may just manage 1Tesla, while GOSS lams may be magnetized to 1.6 Tesla.

5. Testing chokes for CLC filters with DC current, setting the air gap.

I have sometimes had to supply filter chokes or SE OPTs to people building tube amps and have

set the air gap prior to the sale but without the real amp circuit present. A simple test circuit may

be made with a low voltage dc current supply as in the Fig 2 schematic.

Fig 2.

The Fig2
circuit will allow safe measurement of a filter choke using a low voltage

secondary winding of
a 10VA transformer which is easy to find and very cheap.

Providing the DC current and capacitance values are the same as
in an amplifier with a

much higher B+, the ripple voltage calculations and filtering
action of the CLC will be

exactly the same. The dc current is the same as for a tube amp,
in this case 270mAdc.

The ripple voltages at 100Hz will be the same for the tube amp
PSU with higher voltages.

R1 is
adjusted so Idc = 270mAdc in this case.

Any surplus
transformer with winding over 12.6Vrms to 30Vrms may be used in
either bridge,

full wave, or doubler config to give Vdc at C1 not higher than
+40Vdc.

Setting the air gap in the choke
L1.

The choke
is shown as 3.3H in Fig 2. This value was used for the example
and we only have an

approximate idea of the inductance after winding the choke. The
air gap must be adjusted to

maximize the inductance and hence to achieve best filtering with
minimum possible ripple voltage

at C2.

DC current increases from zero to the wanted working value of µe will be reduced perhaps

50% because of the magnetizing effects of DC current..

To really
set the choke to get the best benefit, the choke MUST be tried
in Fig 1 test circuit or

in a real amplifier. The gap should be adjusted for size to get
the least ripple voltage at C2 of

the CLC filter with the wanted DC idle current flowing.

I place the
choke on a block of wood some distance away from the power
supply to gain

easy access to the gap between E and I or between halves of a
C-core. Once DC flow begins,

The Is tend to be drawn tightly against Es by magnetic forces
and there is no need for bolts

to be tight during adjustments.

I have my ac meter and work book set up close by to measure ac voltage across the choke.

I start
with no paper in the gap, measure ripple at C2, then place 1, 2,
3, 4, 5, 6 sheets of paper

right across the Es and Is in the gap increasingly, and record
the ripple at the second C.

A graph is then easily drawn, horizontal axis = sheets of paper,
vertical axis = ripple voltage.

It is important to measure the
gap, so that paper may be replaced with polyester

sheeting of the same total thickness as paper.

Paper
sheets from the exercise book may be used to establish the gap
size because they are a

convenient fine thickness of about 0.07mm. But nobody should ever assume
anything,

and the stack of say 120 sheets in note book needs to be
measured then divided by 120 to

find the thickness of each sheet. 7mm / 100 sheets = 0.07mm.

One may
find minimum Vripple occurs when say 2 sheets of paper
installed, with higher

Vripple each side of this null point. I add one sheet of paper
extra, and thus for me the gap

has been optimized in case Idc ever increases, as it may, during
class AB operation.

When this
is achieved, apply some adhesive to all papers, and tighten the
clamp bolts making

sure all Is are hard against Es and re-test to make sure Vripple
remains low.

If you have
no polyester sheeting, you may use paper sheets for permanent
air gapping

if you follow these steps.....

After adjusting the gap correctly the choke yoke bolts are drawn
up tight to prevent easy

movement or gap increase. Any looseness between bobbin and core
should be wedged

with some scrap cardboard, and a layer of cardboard pushed
between winding and core

to prevent arcing. The completed choke is then soaked in a vat
of polyurethane air cure

furniture varnish for several hours, then excess varnish drained
off and the choke allowed

to air dry. The varnish will impregnate the paper and cardboard
and then slowly harden

and resist moisture ingress to stop rusting of the iron in the
gap which will expand the gap.

Setting the
gap to suit the current to get maximal filtering takes is a much
faster way to get

the best from a given choke than by tedious calculations
required to do it any other way.

I know of no simulation software available to predict the best
gap size.

After the
gap size has been established, if the Idc temporarily increases
in the case of a

class AB amp, the choke L will reduce slightly and the core
might even saturate, but the

slightly increased Vripple will not affect the sound. In well
made class AB amps, the Idc

flow rarely increases hugely because the 470uF or more from OPT
CT to 0V contains a

huge reserve of energy, like a battery, and transients in music
pull all the extra AB energy

from this capacitance. In all the cases of 30 watt AB amps,
rarely does the B+ move more

than 1V even with loud passages.

If Idc
falls below the rated Idc, Inductance slightly increases, and
filtering is improved.

accurately because the mains voltage level is constantly changing amplitude because other

people who have many types of equipment powered off the mains will be switching on

and off thus causing switching transients. It is not unusual to find that the signal at C2

jitters up and down with low frequencies with a peak to peak voltage = +/- 0.05V.

Most of the mains LF noise is below 15Hz because the capacitors have become high

value reactances which do not shunt LF signals and choke reactance has become low which

allows LF signals to pass. The LF mains noise does not affect the sound. But to measure

ripple voltage when it is below 15Hz I have shown the RCRCR filter in Fig 2 for reducing

artifacts below 100Hz. The two CR filter sections with C = 0.047uF and R = 100k each

give a pole at 33Hz, and give about 1dB attenuation at 100Hz, but about 20dB at 10Hz

and 40dB at 2Hz.

An oscilloscope with
calibrated mV setting for amplitude is the best way to look at
the 100Hz

wave form and determine its amplitude even with some trace
movements at lower frequencies.

The wave form of 100Hz ripple should appear without much
distortion.

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6.Resonance in CLC filters.

In Fig 2, the voltages present include the Vdc rail voltage plus LF noise voltages with F below 100Hz,

and there is the angular wave of 100Hz ripple voltage where there are many harmonics of 100Hz.

In the example shown the ratio of noise and ripple voltage to B+ supply rail is a small fraction and many

manufacturers of tube amps do not bother to filter the B+ rail any further before applying the voltage to

the CT of an OPT in an amplifier. In the past, quite low values of C1 were chosen to suit tube rectifiers

such as in Quad-II where C1 was 16uF and Vripple was about 17Vrms with Ia of 150mAdc. In the

example of Fig 2 or in the 5050 power supply schematic , C1 = 235uF, and Vripple is still significant

at 2.5Vrms for 270mAdc. The higher Idc, the greater becomes Vripple.

For complete prevention of generation of intermodulation artifacts the B+ rail at OPT connection should

be well filtered, especially in the case of SE triode amps. The LC filter is a simple good filtering solution

but the filtering effect reduces as frequency is lowered, and the effects of series resonance should be

prevented.

Fig 3.

Fig 3 shows an equivalent
model circuit of a typical CLC filter used to smooth the B+ rail

of an amplifier.

Two
possible responses are shown in the graph for 3.3H plus 470uF.
Both give about

-56dB attenuation at 100Hz. The L and C form a series resonant
circuit at 4Hz.

The reactance of L and C are both each 85 ohms at 4Hz, but
together they make

a load which is much lower value at 4Hz.

The voltage
generator in Fig 3 shows the mains
supply with perhaps only 5 ohms

output resistance at the wall socket.

There is a
considerable amount of LF noise and amplitude change in mains
supplies.

This means that the LC filter will pass much of the LF mains
noise below the LC

cut off frequency to the B+ rail.

Suppose
that Source R1 in parallel with reactance of C1 may be a low
value less

than say 30 ohms which can be possible in many amp PSUs. Suppose
the choke

Rw has very low resistance and R2 is not used. Suppose the amp
R3 load is very

high. Suppose the amp is my 5050 where B+ = +515V and Idc draw =
270mA.

This may seem to be load of 1.9k. However, the supply is for 4 x
6550 in UL

mode where the Ra anode resistance for each tube at very low
frequencies and with

fixed bias is about 1.3k per per 6550, or about 325r for the 4
output tubes. This

dynamic resistance value is the load R3 on the output of the B+
rail supply. If the

4 tubes were in beam tetrode mode with regulated screen supply
each tube's

Ra = 32k, so the total load would be 8k.

Where the
R3 "output terminating load resistance" is well above XL1or
XC2 at Fo,

and where R1 "input terminating resistance" is well below XL1 or
XC2, and where

Rw and R2 is low then any input at Fo will generate high current
through the low

impedance undamped resonant circuit formed by L1 and C2, and the
resonant peak

shown in the graph will appear in the response. It is possible
that the peak in the

response at the filter output at top of C2 may be a higher
voltage than that applied

to the input of L1 & C2. The reason for this is that the LC
circuit has a high Q at Fo,

ie, any R in series with LC or effectively shunting either L or
C is not present,

so there is little "damping."

Fig 3 shows
the response of the filter with low R1 and R2 with high R3 for
3.3H

plus 470uF plotted as the "undamped response". If we used an
audio amp with

low Rout below 1 ohm with ability to amplify sine waves between
1Hz and 1kHz,

and we had a 3.3H plus 470uF connected in series across the amp
output, and

we applied constant level of sine waves between 1Hz and 1kHz, we
would see

curve like the undamped response. At 4Hz, there may only be
1Vrms at the input

to LC, but across the C we might measure 5Vrms. The LC filter
can produce a

"bouncing B+" rail at Fo, even though it filters the 100Hz
ripple so well. To avoid

such effects, we choose low values of C and high values of L to
get Fo to be

well below the audio F band.

Consider
the situation where a L1 = 5H plus C2 = 16uF for a power amp
supply

made in a typical amp in 1955. At 100Hz, XC = 100r, and XL =
3,140r. 100Hz

attenuation is good at 0.032 = -30dB. Fo = 17.8Hz, and much too
close to the

audio band. XL = XC = 558r. The pair of 6L6 in beam tetrode used
for output

tubes gives a high R load well above XL & XC so the B+ rail
would appear to

bounce around considerably at 17.8Hz. Some of this bouncing B+
would appear

at the OPT secondary, and especially if the tubes were
configured as SE triodes,

where the triode Ra is low, RL is high, and the B+ rail bounce
voltage mainly

exists across the OPT primary load.

In many
amplifier PSUs with CLC filter, the R1 source resistance may
quite low

and XC1 may also be low if made a high enough C value. But often
we do not

want to make C1 a very high value because we do not want high
charge currents

through diodes which help power transformers to be noisy and nor
do we want

high inrush currents at turn on. We mainly depend on L1 and C2
for

filtering the B+ rail.

7. Critical Damping of LC
filters.

The cut off
frequency of the LC filter may be made to be the same F as the
resonant

Fo for the LC. First we may assume the R1 resistance with Si
diodes plus parallel

XC1 will be a lot less than XL2 or XC2 at their Fo.

There are
two places in the Fig 3 circuit where added series resistance is
permissible

to "damp" the LC filter response. ( The added resistance acts
like shock absorbers

on the suspension of a car to prevent the wheels bouncing at
their resonant frequency

with highly undesirable effects. ) L or C Reactance acts like a
spring, and just the

right added R reduces resonant behavior so that the peak in the
undamped

response disappears leaving the damped response we wish so that
low frequency

input signal below Fo from mains are not boosted at Fo at the B+
rail.

The series
LC may be damped with R connected across C, or the same amount

of R connected in series with LC.

The added R need not be any higher than 1.41 x XC2 at the Fo. We have already

established that in most tube amps the tube load will not offer sufficiently low R

close to XC2 at Fo. Therefore there is only one other place, at R2. We will always

wish to keep Rw low to avoid choke heating.

At Fo = 4Hz, with 470uF, XC2 = 85r. 1.41 x XC2 = 120r. Let us say we measure

R1 source to be 25r, and choke Rw is at most 40r to keep the choke from heating

up with the DC flow of 0.27Adc. Let us also say the tube load was say 400r.

Damping resistance required = 1.41 x XC.

Added resistance R2 = [ ( 1.41 x XC ) - Rw - R1] in parallel with R3.

= [ 120r - 40 - 25 ] // 400 = 48 ohms.

We would use 47r, 10W rated. The Vdc drop is 12.7V with 0.27Adc, and Pd = 3.42W.

Whenever R is added between the B+ rail and tubes the regulation of B+ voltage

is reduced. But in most tube amps this is of no concern because with music

signals the audio power is mostly generated by class A action where the Idc

flow remains nearly constant. The added 47r will help to prevent instability at LF.

After connection of the damping resistance into an LC filter, the B+ rail circuit

should be examined with a CRO to measure the reduction of LF noise.

The estimated R2 may need slight increase, or decrease, or to be left out

altogether if there is extremely low CRO trace bounce without the R2.

With a preamp, the tube load is usually a very high ohm load, but Idc is very

low, and if a choke is used the damping R may be increased significantly to

give a response which is overdamped, and a slower rate of increase of

attenuation to the ultimate 12dB/octave possible. In preamps with sufficiently

high HT windings there may be +360Vdc at C1 and only +320Vdc needed

for the anode supplies, and at say 40mAdc flow. Chokes are not needed if

one uses a CRCRC with 3 x 470uF caps and each R = 330r. Each CR

section has a LF pole at 1Hz, and gives attenuation of 100Hz of 40dB.

Two such RC section reduce 100Hz C1 ripple of 0.19Vrms to 0.02mV at C3.

The filter choke in a CLC supply may be made to be resonant at 100Hz ripple

frequency. This is done by adding a capacitor & series R across the choke.

The R is usually 2 x Rw, and the cap is calculated when the choke inductance

is known. This will have the effect of further reducing 100Hz VR at C2 by a

maximum of approximately -12dB, ie, 1/4 of VR without the added resonant C.

The attenuation of 100Hz harmonics should remain the same but these will be

at a very low level compared to the choke without a 100Hz resonance.

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

Fo for any LC =
5,035

sq.root ( L x C )

Where Fo is frequency of resonance in Hz, L in millihenrys, C in
uF.

Suppose the
choke was 1Henry, and C = 470uF,

Fo =
5,035

sq.root (
1,000 x 470 )

= 7.3 Hz.

This F is a
little high. But to halve Fo, the C must be quadrupled, OR L
quadrupled. Or the C

doubled AND L doubled. After seeing how much effort went into
the choke and how large the

choke would need to be if we wanted 4 times the L with 270mA, it
may be easier, cheaper,

and just as effective to add three more 470uF caps in parallel
with C2, so the C2 = 1,880uF.

However, then the inrush current will be quite high as C1
charges with low wire resistance to C2.

The law of diminishing returns is in operation, because reducing
ripple voltage from say

10mV to 2.5mV is usually inaudible. Having 1,880uF to anchor the
B+ end of the OPT to 0V

does not do anything better than would a 470uF. Hum in B+ rails
when ripple is low could be

from other sources other than an imperfectly filtered B+ supply
rail. However, the hum at

the output of my SE 55 Watt monobloc amps with parallel 845
measured only 0.25mV.
------------------------------------------------------------------------------------------------------------------------

The iron cored choke is not easily modeled by an equivalent circuit arrangement. Choke

reactance varies with frequency applied AC voltage amplitude and permeability. During each

wave the choke appear to have less reactance either side of the zero crossing point for the wave.

The value of dynamic inductance will rise until until the core material becomes fully magnetized

and cannot be made to increase its magnetic field. This point is called core saturation. Once the

inductor voltage increases further the reactance becomes zero and the choke acts as a slightly

non linear resistance equal to the winding wire resistance.

This sudden collapse of inductor reactance during wave cycles may be a very undesirable

behavior in most applications wherever we might want to use a choke or transformer winding

and may cause very serious distortions to wave forms or damage to tubes. When the applied

voltage across the choke is reduced during a sine wave cycle the saturation ceases and magnetic

field re-establishes itself and until the voltage amplitude swings to the same value but for the

opposite wave crest so that distortion created is mainly 3H, 5H, 7H etc. The resulting

distortions are not unlike the dreaded crossover distortions in underbiased class AB amps

when the Idc in each PP tube is too low.

Where a choke is used to supply dc to an anode which is CR coupled to a following amp

stage the distortion may be easily reduced to negligible levels by use of resistance in series

between the choke and anode. The tube should always be a low Ra triode and never ever

be a high Ra pentode unless triode connected. Whenever a very high driving source resistance

is effectively an AC constant current driver source the voltage across a choke or unloaded

transformer primary winding will become distorted by the non linear magnetizing currents

generated in the iron wound coil. The "generator" of such distortion currents within the choke

has a finite impedance value and it is usually lower than the Ra of a pentode but higher than

Ra of a triode. The low Ra of a triode is able to shunt the "distortion generator" within the

choke, so the lower the Ra is, the less iron caused distortion is generated.

Iron wound components when well designed contribute far less distortions to signals than

do the vacuum tubes themselves unless the tubes are beam tetrodes or pentodes without any

NFB applied. The iron caused distortions without NFB can become considerable, as in the

case of the thousands of radio sets made with a 6V6 to power an OPT with no NFB used

anywhere. It is common for many guitar amps with PP output beam tetrodes or pentodes

tubes to have no NFB anywhere thus resulting in high amounts of THD and IMD. The guitar

amp distortions are well tolerated but the awful audio quality from many radio sets was horrid

and permitted only low levels of operation. The quality was made all the worse by the

accountants who ensured quality was minimized in favor of lower production costs and

higher company profits.

For a gapped choke.

The air gapped iron cored choke exhibits a much greater ability to withstand a high AC

voltage being applied without saturation during the wave cycle compared to an un-gapped

choke with the same amount of inductance. This is because the air gapped choke will have

many more turns and a larger Afe to achieve the same inductance of the non gapped choke.

The Bac max does not depend upon permeability, but upon V, Afe, F and No of turns of wire.

So the gapped choke of say 1H will have the same inductance reactance as the un-gapped

choke of 1H, but the gapped choke will sustain a higher AC voltage or sustain the same

voltage but down to a lower frequency without increasing Bmax to saturation. The gapped

choke reactance value is held more constant for the whole wave cycle and for different

frequencies. The air gap reduces the maximum possible inductance to being perhaps 1/20

of the maximum value without a gap because the inductance is proportional to permeability,

and the air gap effectively reduces permeability by artificially increasing the magnetic path

length. So an air gapped choke will need to have much more iron and wire turns to give

the same inductance as an ungapped choke.

In essence, the air gapped choke tends to behave as constant value of inductive reactance

when the frequency is low rather than an higher inductance suffering saturation during

portions of sine wave cycles. In most cases for an air gapped choke or SE transformer

primary there is a steady DC current flow. The air gap much reduces permeability and

therefore reduces DC core saturation, because the core Bdc max is proportional to µ as

well as other things. The presence of the DC Bdc field intensity lowers the maximum

possible AC voltage which may be applied. But the choke or SE OPT allows some

useful function to be achieved with iron and wire when all factors are balanced with

good design and calculations. With DC current present in a choke or OPT primary, any

AC voltage acts to further increase core magnetization or decrease it during each wave cycle.

here is no zero of the ungapped choke with no DC current. Therefore at low AC levels

the choke distortion is low, but generally the gapped choke or SE primary winding

generates even numbered H, mainly 2H.

The saturation effects are due to voltages applied and independent of load currents.

9. Chokes for CLC filtering in power supplies.

A choke in a CLC filter is an effective way to greatly reduce power supply rectifier noise and

mainly 100Hz hum from entering the B+ voltage rail and audio signal path. The aim of using

a choke is to attenuate 100Hz hum by a large factor which justifies the resources spent on

the choke which are greater than if we used an CRCRC filter. A typical use for a choke might

be in a two channel amp where each channel needs 125mA and it should not have more

than 5mV of ripple voltage at the connections to OPT especially if the amp is a Single Ended

Triode variety with little global NFB and which are far more prone to B+ rail hum than PP

amps which rely on the common mode rejection of noise. SE amps are also all pure class A

so Ia does not change, so good regulation of the B+ supply voltage naturally exists, and the

direct voltage drop across any resistance in the hum filter does not matter if we allow for it

in the choice of power transformer and design the resistance to cope with any heat produced.

The best
way to design a filter choke for CLC use is to work from first
principles with

Hanna's method clearly spelled out in the Radiotron Designer's
Handbook, 4th Ed, 1955,

pages 247 to 250. But I have never found time to convert Hanna's
Method to metric units

which were introduced to Australia in the 1960s, so anyone using
Hanna's Method will

have to work with inch dimensions.

There is another good little book, 'Coil Design and Construction
Manual' B.B. Babani,

first printed 1960, except that it does not delve into air gap
setting very well. When I

use Hanna or Babani to check my design for a choke, I simply go
back to using Imperial

measures of inches which I still know well.

But not all
of you will beg borrow or steal a copy of RDH4, and since most
men never

read books any more, I will prod your lazy brains with an
alternative and include a typical

example most everyone will find useful at some time if they
build any tube amp supply

which needs a supply from 50mAdc to 270mAdc, and for any B+
voltage from +250Vdc

to +550Vdc.

Suppose we
want a choke able to take 270mAdc flow, and which will perhaps
fit onto a

chassis bolted underneath, and not dissipate more than 4 Watts.
Now 4W = I squared x R,

so, R = Pd / I squared =
4W / ( 0.27 x 0.27 ) = 55 ohms wire resistance.

For all
copper windings the maximum current density in the wire should
not exceed 3A/sq.mm.

You will always find that if you have more than this current
density the wound iron cored

article tends to get too hot. In summer, a tube amp chassis gets
quite warm so let us not have

hot little chokes, or worse, hot big chokes.

As the core
size is increased, the turns and their length increase, so Rw
increases, but the

larger size allows a greater area for heat radiation and
convection cooling so as long as the

current density is maintained low enough the larger choke with
higher Rw won't get much

hotter than a smaller one.

We would
need useful amount of inductance, at least above say 2
Henrys. This is a small

filter choke L value compared to the usual 5H to 10H much
favored in 1955. But in 1960,

PSU filter capacitors were very low values of between 4uF and
32uF and everyone put up

with much more noise in amplifiers. Now we would use between
100uF and 470uF without

any hesitation because the modern electrolytic capacitors now
made are fabulously reliable

with higher ripple current ratings and designed for continual
hard labor in switch mode

power supplies in 1,001 locations including your PC PSU, where
50Hz mains is directly

rectified into a large value C and a 470uF cap rated for 450Vdc
costs less than a hamburger.

And the size of the 470uF cap is now not much larger than many
32uF caps of 1955.

So plenty of C can always be found to complement a 2H choke.

10. Design Method - Choke for
CLC PSU.

Nearly all
my choke designs start with a choice of GOSS E&I laminated
core with 25mm

tongue with 25mm stack. This size of choke can dissipate about 4
Watts of heat without

rising in temperature too much. It will be air gapped, and with
an air gap the maximum

µ of the laminations without an air gap need not be high
because the gap effectively

reduces the µ to a similar value even if the iron material
is GOSS EI lams, GOSS C-cores,

or plain cheap quality high loss non oriented transformer iron
with µ max = 2,500 which

often may be found in an old transformer core. So if we know the
dc flow, we can

tailor the winding resistance, Rw to suit, and get the best
available amount of inductance

filtering. In heavy dc current situations, the Rw needs to be
low, eg, for an anode supply

to output tubes, and in low dc current circuits the Rw does not
need to be low. In preamps

there is seldom any need for a choke, but a high resistance type
with high L value may

be used if available.

So let us
accept the simple premise of a 25mm x 25mm core area using
wasteless pattern

E&I, and 4 Watts heat dissipation. This will have a winding
area available in the bobbin

= 33mm x 9 mm = 300 sq.mm. Turn length, Lt, = 140mm , iron
magnetic path length,

ML = 140mm.

A. First question :- "what is the
wire size required for the proposed dc current?"

It is
mathematically difficult to relate the wanted dc wire
resistance, wire copper dia, and

bobbin winding area, and turn number all in the one equation,
and it is because of the

enamel thickness involved, which varies in thickness for
different wire size, and has a

greater fraction of wire dia when dia is small.

So I have
prepared a useful table for everyone to read off the number of
turns

required for a given current. The inductance isn't given,
because gapping and other

considerations vary widely.

Winding
window area = 300sq.mm, average turn length = 140mm, iron
magnetic

path length = 140mm.

B. This table allows a simple
choice to find wire size for a winding

around a 25mm stack of 25mm tongue E&I.

Cu wire dia, mm |
turns, max. |
Resistance, ohms |
DC current max |
L, Henrys µe = 200 |

0.20 |
4,800 |
380 |
89mA |
25.0 |

0.25 |
3,100 |
157 |
138mA |
10.8 |

0.30 |
2,200 |
77 |
197mA |
5.4 |

0.35 |
1,600 |
42 |
267mA |
2.88 |

0.40 |
1,300 |
26 |
339mA |
1.9 |

0.45 |
1,050 |
17 |
420mA |
1.24 |

0.50 |
870 |
11 |
522mA |
0.85 |

0.60 |
600 |
5.3 |
752mA |
0.40 |

0.75 |
390 |
2.2 |
1.16A |
0.17 |

1.00 |
230 |
0.73 |
2.02A |
0.06 |

1.40 |
120 |
0.19 |
4.0A |
0.016 |

2.00 |
60 |
0.05 |
7.7A |
0.004 |

The wire current density is all slightly under 3A per sq.mm for all wire sizes.

If current = 270mA dc, simply choose the wire size for the current rating just above

270mA, ie, 339mA and the wire Copper dia = 0.40mm, and you should get 1,300

turns onto the bobbin. Or put another way, if we had a spool of 0.40 wire, we

know we could have 1,300turns and with 330mA.

The table does not show that if the stack of laminations is increased, the inductance

increases in direct proportion, so that if stack was 50mm, then L would double

with the same current. Winding turn length and Rw also increases so heat losses

also increase, but because the surface area increases, the temperature should

not increase.

C. I did come up with some useful formulas....

For where N and Lt and d is known...

Winding resistance, Rw = N x Lt x 0.0000226 for all core sizes.d squared

where Rw is in ohms,

N is turns,

Lt is turn length in mm,

0.0000226 is a constant for all equations and because the resistance in ohms of

1 metre of 1mm dia wire = 0.0226 ohms, and d is the copper dia of the wire we are

using in mm.

From the last equation, and for a 25mm stack of 25mm tongue size,

Rw = 1300 x 140 x 0.0000226 / 0.4 x 0.4 = 25.7 ohms.

D. Here is a formula for a core with equal stack and tongue size T to be

found for a given wire size and wanted winding resistance...

T = 25 x cube root of Rw x cube root of ( d squared x od squared ),

where 25 is a constant,

Rw = ohms,

d = Cu dia of wire,

od = overall wire dia including enamel.

So for 26 ohms and d = 0.4mm, and od = 0.47mm,

T = 25 x 2.92 x 0.328 = 23.94, and we would choose 25mm.

If the resistance was to be 50 ohms, with the same wire size,

T could be 25 x 3.68 x 0.328 = 30.2, so we could choose T = 32mm.

Then actual number of turns can easily be worked out once a core window size

becomes known from the T dimension and a wire size has been chosen has been

chosen. The winding window area size = length of window x height of window

= 1.5T x 0.5T = 0.75T squared for wasteless pattern E&I lams. For T = 25mm,

Area window = 468 sq.mm. But allowances for the bobbin cheeks and bobbin base

and for final winding clearance from iron all equal to approximately 1.7 mm all

around reduce winding area to 34.1mm x 9.1mm = 310sq.mm, or 0.5 x T squared.

E. Calculating turns N.

The copper wire dia may 0.4mm, but the overall diameter of class 2 high temp

rated polyester-imide enameled magnet winding wire will be 0.47mm, and thus

possible turns are :-

N = Bobbin Window area / oa wire dia squared = 0.5 x ( T/ oa wire dia ).

If we have wire = 0.4Cu dia, then oa dia = 0.47mm.

If we choose T = 25mm, then N = 0.5 x ( 25/0.47 )squared = 1,414 turns which

is close to what is conservatively estimated in the table above.

Should anyone wish to use a larger stack of 25mm laminations, then the turn

length will increase from the 140mm used to make the table, and winding resistance

will increase, but as I mentioned, this is of little concern if the current density

remains constant.

Let T = S = 32mm.

If wire is 0.47mm oa, N = 2,317 approx and inductance will be much increased,

all other things being equal. The same inductance may need a stack S of 75mm

with T = 25mm if N is only 1,400t. For most E&I cores, the stack should not exceed

3 x tongue width. The higher the stack, the more difficult it becomes to clamp all

Is tight against the Es, and a special clamp must be arranged.

11. Winding.

Neat layer winding methods are best where patience, skill, and machine lathes allow.

These days polyester-imide coated high temp grade 2 winding wire is much more

rugged than fragile enamels of the past and wire can be wound on without layering

neatly, but gradually letting turns pile up while slowly feeding on wire while *slowly*

traversing the wire across between cheeks to avoid wire crossings at a wide angle,

and and thus preventing a small wire to wire pressure area which could lead to a

short circuit between one point in the winding and another. Such shorts prevent

any inductance being maintained. The level of the wire should be kept free as

possible from forming large humps and troughs while traversing the bobbin

especially when you get close to nearly filling the bobbin. Skill and practice gets

it right. Tension in the wire is light, and varnishing can be done as you go with

Wattyl 7008 floor varnish daubed on after each 200 turns. This terrible smelling

and possibly toxic varnish has a pot life of hours before setting to become a

rugged hard polyurethane which promotes heat transfer from inner winding layers.

Once winding is complete, some insulation tape should be wound twice over the

top of windings for basic protection and the maximum height checked to ensure

it can later fit into the bobbin hole without touching the iron anywhere, thus allowing

another layer of 0.2mm polyester sheet to be slid between iron and winding to

ensure arcing cannot occur between iron and coil wire. Terminations on the bobbin

should be provided where wire is less than 0.75mm dia.

Some of you may have the book, Coil Design and Construction Manual B.B. Babani,

first printed 1960 and re-printed 14 times I know of. Mine is a 1991 copy, somewhat

grubby from from so much use.

Some people
will just try to read the tables in the book and wind something,

and maybe they end up with a sub optimum choke that lacks enough
inductance

which is all too easy to do especially in the case of a choke
for a choke input power

supply. Babani's book with its tables requires someone to have
lots of experience

and an IQ = 259 to be assured of not making a mistake. Neither
exists in the

minds of many DIYers.

People may
find some old and useless transformer with an open winding which

may yield a suitable amount of iron for a choke. Chokes in PSU
do not demand

that the iron be top grade GOSS. I have obtained material from
re-cycled cores

often. First you remove any bell ends, yokes and bolts if
possible for re-use.

To extract E&I laminations from a well varnished transformer
is difficult because

they are all glued together by varnish. If you simply place the
transformer into

a small wood fire for 20 minutes and heat iron until just cherry
red hot, the heat

vaporizes and burns off any plastics. Next day when the lams
have cooled, the

E&I lams will all just fall out loose, and the heat will
have annealed the iron and

won't affect the iron's magnetic qualities.

Unfortunately
many
people
won't
be
able to light a fire anywhere to fry old chokes

and transformers and the smoke is toxic, although the heath
hazard produced by

DIY people doing dirty filthy craftwork processes is
infinitesimal compared to what

mainstream industries are doing 24/7. But once the old iron is
cooked and had

cooled the copper turns will be free of insulation and bobbin
and may easily be

cut loose and removed to a re-cycling bin. Don't cool the hot
cores in water.

Don't try to clean the oxide layer off the laminations, it
insulates the iron laminations

from each other which prevents eddy current losses.

If the
material is not 25mm tongue size, the turns and current and wire
size will

have to be worked out to suit the window size.

12. Permeability.

At this point I must discuss the iron permeability known as
µ because for all

the chokes we make, it will be varied to suit the purpose of the
choke which

usually means the iron will all be placed into the bobbin in a
few basically different

ways apart from method A below, each method has the effect of
effectively

lengthening the magnetic circuit length, and thus reducing the
maximum possible

permeability, µ, to effective permeability, µe.

(A)
All
Es
and
Is
are
reversed
in
direction
as
they
are
stacked
up, and E&I are thus

maximally overlapped and interleaved. This gives the choke the
maximum inductance

it can ever attain with NO DC present because µ will be at
a maximum and for high

grade GOSS material it can be 17,000. This method is usually
only ever used where

no dc flow is present.

(B)
All
Es
are
piled
together
facing
the
same
direction
as
they
are
stacked
and a

pile of Is clamped to the pile of Es with polyester sheet
gapping material inserted when

we know what size to use. Even without any gap and Is are hard
against Es, the

maximum µ available will probably be approximately 1/10 of
the value it was when

maximal interleaving above was used in (A). The resulting
µe for anything other than

maximal interleaving is the effective µ, known as
µe. This is because the core has

imperfect mating surfaces, and the change of grain direction
between all E and all I

has the effect of an equivalent gap larger than what is actually
physically present.

So the maximum µe for a butted core without a gap can only
be measured if we

need to know what it is. It cannot be accurately calculated.

(C) Es for the whole
choke are divided into into say 5 piles of equal height. Is for
the

whole choke are divided into 5 piles of equal height also equal
to piles of Es. A pile

of Es is inserted to through the bobbin hole, and a pile of Is
are butted to the pile of Es.

Then another pile of Es is inserted in the opposite direction
and another pile of Is

placed to close this pile of Es. The process of alternately
directioned piles of Es and Is

continues until all Es and Is have been assembled. Tightening of
the finished stacks

may be done with some additional Es and Is placed on the top and
bottom piles.

When bolted partially tight with holding yokes the Es and Is may
be tapped up close.

This method allows the µe to be intermediate between being
maximal with maximal

interleaving and the µ/10 value or less achieved above in
(B). This method is rarely

ever used, but applied sometimes to PP OPT or in a balanced
choke with a CT on

the winding where one wants the core to resist being saturated
by unbalanced DC

currents in each 1/2 primary, and still be able to get a high
enough and more constant

amount of inductance, with freedom from saturation.

For all
amplifier power supply filter chokes, there will ALWAYS be a
full gap between

all Es and Is and ONLY method (B) is ever used.

When
E&I are stacked close, maximum inductance without a dc flow
can be determined

with the test shown on the top of this page where we apply a
signal at 100Hz at about

10Vrms and measure current in a sensing resistance, and work out
the ZL at 100Hz.

Then the inductance is calculated from L = ZL / 628.

13. Formulas, µe, air gap,
Bdc, Bac.

A. For all choke
inductances, L = 1.26 x Nsquared x Afe x
µe

1,000,000,000 x ML

1.26 and 1,000,000,000 are constant for all equations,

N is the turns,

Afe is the cross sectional area of the core in sq.mm,

and µe is the effective permeability,

and ML is the iron path length.

So µe = 1,000,000,000 x ML x L

1.26 x Nsquared x Afe.

Let us suppose the maximum L of the example inductance
with 1,300t on

Afe = 625s.mm with a close butted core = 3H, and with no dc
present.

µe
max = 1,000,000,000 x 140 x
3

1.26 x 1,300 x
1,300 x 625

= 315.

If the
µe was reduced by the presence of dc or by placing a gap
into the core

with sheets of paper so µe = 200, L would become 1.9H.

NOTE Adding a gap of 1 sheet of
0.07mm notebook paper for a gap

right across the core gives a
total REAL gap of 0.14mm ! ! ! !

This is
because in an E&I core there are TWO gaps in ONE magnetic
length,

one each side of the holes bounded by E legs and I.

Major
errors in gap length calculations and choke function can arise
if this

fact is not remembered carefully.

B. So adding a gap.

Add gap =
0.14mm will change µe.

µe
= µe max of iron
butted close without a real gap

1
+
(
µ
max
x
gap
in
mm / ML of iron in mm )

Suppose in
this case µe =
315

1+
(
315
x
0.14
/
140 )

= 315 / 1.315

= 239

C. DC field strength Bdc.

What is the
magnetic field strength Bdc when we have 270mA,

core = 25 x 25, and µe = 239?

The dc field strength for a choke, Bdc = 12.6 x µe x
N x Idc

ML x 10,000

where Bdc is in Tesla, ue is effective permeability,

N is the turns,

Idc in Amps dc,

ML is the magnetic path length of the iron in mm,

and 12.6 and 10,000 are constants for all equations to work.

In this
case Bdc = 12.6 x 239
x 1,300 x 0.27

140 x 10,000

= 0.755Tesla

This means
that the iron is magnetized by the dc current to a large portion

of its full capability. If the iron is GOSS, it may saturate at
1.5Tesla,

and if it is old low grade iron maybe at 1Tesla.

Maximum Bdc should not exceed about
1Tesla for CLC chokes.

We also
need to keep in mind the AC field strength, Bac, and and keep
the total

of Bdc + Bac to below 1.2Tesla.

In the case of the CLC filter we have, the Vac across the choke
is only 1.2Vrms.

D. AC field strength, Bac.

The AC field
strength, Bac = 22.6 x Vrms x 10,000

Afe x N x F

where 22.6 and 10,000 = constants, Vrms = voltage across the
coil,

Afe = core section area, N = turns, F = frequency in Hz.

In this case, Bac = 22.6
x 1.2 x 10,000

625 x 1,300 x 100

= 0.0033 Tesla,

This Bac is quite insignificant, and may be ignored as it will
make little difference to the

sum of Bdc and Bac.

How well will this choke work with dc flow in a real circuit?

The CLC
choke performance depends heavily on the gap size, which is
adjusted as

I have said above.

At turn on in a tube amp with CLC supply, C1 and C2 require charging up to full working

voltage, say +420Vdc, and the total C could be 5 x 470uF = 2,350 uF. The flow of current

in the first few mains cycles at turn on during charge up is quite high. There is also

magnetization current of the mains transformer, and the current is only limited by the

mains wiring resistance, series resistances of windings, and diode resistance and

winding resistance of the choke. The high charge current can cause a mains fuse

or any secondary fuse we may have in the HT winding to blow all too easily.

And even when all is charged up, and the amp is working there will be quite high

peak charge currents of perhaps 1.3A peak flowing into C1. Where a voltage doubler

is used, the peak charge currents at the HT winding will be twice those at a full wave

winding, maybe 2.6A.

It is good practice to arrange a resistance in series with the mains of say 100 ohms

with 20 Watt rating which causes a delay of B+ rise to 2/3 its final working value after

4 seconds. After 4 seconds a relay shunts the 100 ohms and B+ continues to rise to a

peak value. This will much reduce and limit "inrush" charge currents and initial filament

currents. Further reductions in peak charge currents into C1 is possible using permanent

series R between Si diodes and C1, where R may be approximately 7 x XC, so that

if C1 was 470uF, then XC = 3.4 and R = or 22 ohms, 10W rated. The effect of the

22ohms increases the time it takes to charge C1, and the saw tooth Vripple wave form

will show charge time for caps lasting nearly as long as the discharge time when DC

flows to the amp. The Vripple measurement does not change with or without the

22 ohms. Ripple current = Vripple / XC, which always works out to 1.41 x Idc.

If Idc = 0.27A, then Ir = 0.38A rms, and with C1 470u, Vr = Ir x XC = .38 x 3.4 = 1.3Vrms.

But the wave is not a true sine wave, but result is close enough. Peak current

cannot be less than 1.41 x 2 x 0.27A, say 0.76A because during cap charging the diodes

must have to charge the cap plus supply the amp current. Its not unusual that maximum

peak charge currents when C1 becomes a huge value is 6 x Idc, or 1.62A, and only

limited by transformer and diode resistances. By a huge value, it is when the time

taken for caps to charge = about 1/6 of the total cycle time for charge and discharge.

Although the CRO sees the saw tooth wave as an innocent looking wave form the current

wave form is a very different picture. Basically, you simply want the cap charge time to

be just less than half the total charge and discharge time so to work out the series R,

inspect the wave form and add the R to give you the near equal times. This added R

usually reduces transformer noise and reduces heat in the HT winding. The heat in the

series R will be considerable, and more than if you try to calculate Heat = V squared / R.

If a CT winding is used, the 22 ohms can be placed between CT and 0V, or 22 ohms at

each end of the winding to diodes. If a doubler is used, one would have two 22 ohms

in parallel for 11 ohms in series with the one doubler winding.

If you measure the performance of the B+ power supply with added R to reduce peak

charge currents, you will see these resistors cause slight de-regulation of the B+, but

as I have said elsewhere it is not important because all your music is made while the

tubes are in class A and there is very little change in current from the power supply,

so very little B+ voltage rail change.

The +Vdc will also be slightly lower than the possible maximum at C1. If instead of +440Vdc

at 270mA, you get only +430Vdc with limiting resistors, its OK, because the power

transformer will run a little cooler, and instead of dissipating so much heat in the HT winding

with such high charge currents, the peak currents are lower, and heat is mainly in the limiting

resistors. If the fuses don't blow during a fault, or protection circuits fail, you want some

cheap resistors to fuse open rather than have a destroyed output transformer or power

transformer.

Peak charge current can be limited further by using a lower value for C1, retaining such

resistances, and placing more C after the choke.

Tube rectifiers of any kind will destroy themselves if the capacitance values exceed allowable

values given for the wanted Idc given in the data sheets. So for 270mA dc and 440Vdc, and a

full wave CT winding, one may have TWO GZ34, one on each phase of the winding.

C1 = 100uF would be the safe value, and ripple voltage will be much higher, at about 6Vrms.

The reactance of 100uF at 100Hz = 16 ohms, and if there is 6Vrms at 100uF, ripple current =

375mA, which is about 1.41 x 270mA. The natural "on" resistance of the GZ34 is the anode

resistance of the diode tube, and is high, and which limits peak charge, so added resistances

have to be high to achieve any peak I reductions. In fact with tube rectifiers, the Vac to Vdc

ratio is often a low 1.2:1 or lower loaded, but with Si diodes it about 1.35:1 loaded. Tube

rectifiers get hot because average voltage x average current = large waste of heat.

Sometimes I have rugged polyester motor start capacitors for C1, and sometimes because

I have a transformer that has a HT winding with a very high Vac and no taps for adjustment.

In 2009, I made two 60Watt SE monoblocs each needing +480Vdc at 330mAdc, so 660mA

total from one large power supply chassis. HT winding gave 420Vrms giving a maximum no

load Vdc = 590Vdc, and working Vdc of +560Vdc if C1 was a high value, so somehow

I had to drop from +560Vdc to +480Vdc which is 80Vdc at 0.66A, and power wasted

= 53 Watts, which I didn't want to waste. I used C1 = 30uF using 2 x 60uF in series,

each rated for 400Vdc and Vripple was very high, but charge currents low, and no heat

is liberated because of AC in a reactive C or L component, so I got the +480Vdc I wanted.

C1 was followed by a 4Kg choke of about 10H, and then by 4 x 470uF series /parallel for 470uF

total and Vripple was less than 35mV at C2. I had an RC filter following that with R = 50ohms

and C = 705 uF and Vripple < 2mV. The added second RC filter damped the resonance in first

CLC filter at about 3Hz.

If you have a pair of tube rectifiers to feed 100uF and then have choke plus an enormous

following C value, during charge up the tube rectifier really struggles, and can experience

excessive peak currents causing internal arcing inside the tube. Too many arcing occasions

hasten the death of the tube rectifier. Such GZ34 might be OK in a Quad 22 with pathetically

low value C and almost non existent B+ rail filtering before the CT of OPT. Tube rectifiers just

do not belong in modern amps with large amounts of C, and I never use them.

The filter choke in a CLC supply may be made to be resonant at 100Hz ripple frequency.

This is done by adding a 400V rated polyester capacitor plus series R across the choke.

The R is usually 2 x Rw, and the cap is calculated when the choke inductance is known.

This will have the effect of further reducing 100Hz VR at C2 by a maximum of approximately

-12dB, ie, 1/4 of VR without the added resonant C. The attenuation of 100Hz harmonics

should remain the same but these will be at a lower level compared to the choke without

a 100Hz resonance. Using a 4H choke with Rw = 30 ohms can be made partially resonant

at 100Hz with C = 0.63uF and R = 68 ohms, 5W.

15. Compare CRCRC filter to CLC filter.

Let us first consider the power supply with diodes charging a C of 470uF then with TWO

following RC filter sections, with all C = 470uF and each R = 50 ohms, using 2 x 100r x 10W.

Let us suppose Idc = 270mA, and total R in a CRCRC = 100 ohms, and there would be

13.5 Vdc across each 50 ohms. Heat generated in all R = 0.27 x 0.27 x 100 = 7.3 Watts

with 3.64 W in each 50 ohms. To escape damage during faults or overloads, each 50 ohms

is rated for 20 Watts and perhaps mounted on a heatsink if chassis space allows, glued

with a bed of hi-temperature silicone sealant 'Selleys 401'. Or we might use a pair of

aluminium clad wire wound 47 ohm resistors bolted with heat paste to the chassis or a

heatsink, or we might use Welwyn vitreous enameled resistors slung between turrets

and hanging in the passing air flow above vent holes in the chassis bottom plate.

With 270mA, C1 of 470uF will have ripple voltage = 1.3Vrms at 100Hz. The following pair

of R&C filters using 50 ohms plus 470uF will each have an attenuation factor = XC / R

= 3.4 / 50 = 0.068. Two cascaded filter section have an attenuation factor = 0.068 x 0.068

= 0.0046 so ripple voltage at the output of C3 = 1.3V x 0.0046 = 0.006 Vrms, ie, 6mVrms.

This could sound a little over zealous, and for an RC filter with even lower heat losses in

resistances, the R value could have a minimum of 10 x XC, or about 33 ohms. Thus an

overall attenuation factor at 100Hz of 0.01 is available to give Vr = 13mVrms. Three

such RC sections after C1 will give attenuation factor = 0.001 and Vr = 1.2mVrms.

Total R for 3 RC filter sections with R = 33 ohms each will perform better than 2 RC

sections with total R = 100r. There are no resonance effects to worry about.

let us suppose we wished to match the performance of a CRCRCRC filter with a CLC filter.

Let us assume we have C1 = C2 = 470uF. For an attenuation factor of 0.001, we

must have XL = 1,000 x XC at 100Hz. With each C = 470uF, XL = 3,400 ohms.

The inductance required = XL / (6.28 x F) = 3,400 / 628 = 5.4Henrys.

The winding resistance may easily be made less than 100 ohms, but may be 50r, so

the choke would have Pd = 3.65 watts, The frequency resonance, Fo, between L and the

C2 must be checked. Fo = 5,035 / sq.root ( L x C ), = 5,035 / s.rt ( 5,400 x 470 )

= 3.16Hz. This is below 4Hz and acceptable for any PP or SE amplifier.

At 3.2 Hz, XC = XL = 106r and to obtain a low Q for the resonance peak the series R

between L and C2 should be under 106 ohms. If we used a 50r added resistance to

the Rw of 50r we would have good damping of the LC resonance Q.

To make the LC equal the performance of resistors in every way including a cure for

resonance the series resistance of both CLC filter and CRCRC will be about the same.

If we were happy with Vr being say 4 times higher at 4.8 mVrms, the choke value could be

quartered to 1.35H. But then with C2 = 470uF, Fo increases to 6.3Hz and it is getting too

close to the audio band. The damping of the resonance Q would require some additional

series resistance. However, if 1.35H was all we could afford, we might double C2 value

to 940uF and achieve Fo = 4.4Hz, probably acceptable, while reducing Vr to 2.4mVrms.

A 1.5H choke would be fine and much easier to make than a 5.4H choke.

In the case of where the available electrolytic caps have ratings of 450Vdc working

and B+ = +420Vdc working and peak with no load at less than +450Vdc, all is well.

But where B+ exceeds +430Vdc when loaded it is safer to use seriesed electrolytic

caps each rated for +350Vdc working. The C1 = C2 = 235uF, and we would want L

to be at least 5.4H so that Fo = 4.5Hz, barely acceptable. Vr would be 4.8mV at C2.

Added R for suppressing resonance would be about 100r.

For CRCRCRC with each R = 68r, total for 204r total to get a final Vr attenuation

factor of 0.001. Heat Pd = 15 Watts.

Both CLC and CRCRCRC will dissipate heat in resistance, although the choke plus

R will total about 150r so heat = 11 Watts.

If our resources allow us to easily make a choke, or we have bought one cheaply,

or find a suitable type in a pile of junk, we should use that choke. But to improve an

old amp, or for where little spare space is available for a choke, then RC filtering

with high cap values is easier.

If someone you know is just plain infatuated with chokes, well OK, leave them to

their choking experience. If you don't like chokes, then use bagfulls of capacitors

to minimize series resistance values. A 470uF x 350V rated electro is about $7.00,

and about 1/2 the price of a McDonald's large hamburger that may make you feel fat,

sick, and worse than had you bought a capacitor. But a 5.4H choke might cost the

same as 7 hamburgers. People argue about costs of audio gear parts. But I don't

know why they spend so much on garbage food and 101 meaningless other things.

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