Cuộn cảm - Choke dễ ẹc

Mặc dù là OPT đã dễ ẹc, nhưng anh em DIY vẫn tỏ ra ngán vụ choke, khiến các loại choke quân sự cảm kháng cao cỡ 5 - 10 H gần như tuyệt chủng, đặc biệt là loại chịu dòng 100 - 200mA.
Cho nên, đành sắn tay áo luộc luôn vụ này cho rồi.
Lâu nay, nhiều anh em cho rằng quấn choke khó đạt yêu cầu, nào là phải đúng fe choke, nào là air gap ... nhưng thật ra là... hồi sau sẽ rõ

 

Re: Choke dễ ẹc - quấn cuộn cảm dễ ẹc

Quấn thì dễ. Chống rung mới khó. Chống nhiễu khi dùng choke input filter lại càng khó hơn.

 

Em đã dỡ rất nhiều choke QS thì thấy tiết diện lõi toàn là hình vuông. Ví dụ EI lưỡi 32mm thì bề dày các lá FE cũng là 32mm.
GVteam có thể giải thích vụ này giúp em được ko ah?
Xin cảm ơn

 

Chắc là bởi vì ... cho đẹp :mrgreen:

Hơn nữa các lõi nhựa đã đúc chuẩn thường hình vuông là dễ kiếm sẵn nhất chứ các hãng không rảnh như anh em DIY chúng ta ngồi dán khung giấy bobin.

 

Em thấy Mẽo toàn quấn..............khung giấy đóa.
Mấy cục choke hiệu như Tango, Tam em chưa có cơ hội phá coi tiết diện lõi hình gì.
Vụ này phiền lão Tam nghiên cứu xem sao chứ kỳ quá?
Hay là vì hình vuông gần với hình tròn nhất, mà các cuộn cảm lõi khí thường là hình tròn.........làm vậy cho nó có vẻ......khí :mrgreen:

 

Xin chào họa sỹ, không biết bác đã có triệu bông hồng chưa để tặng cô ca sỹ
Em không phải là GV team nhưng cũng mạo muội góp ý: đấy là toán học bác ạ hình vuông có diện tích lớn nhất so với hình chữ nhật nếu có cùng chu vi. Vậy lõi có thiết diện hình vuông có các thông số tối ưu nhất, bác có diện tích lõi lớn nhất và có chu vi nhỏ nhất và hệ quả của nó thì bác biết cả rồi.

 

Đúng là hình vuông tốt nhất, nhưng không nhất thiết cứ phải hình vuông, tùy theo dòng, cảm kháng là bao nhiêu mà cần sử dụng loại fe, khối lượng Fe bao nhiêu cho phù hợp

 

Re: Choke dễ ẹc - quấn cuộn cảm dễ ẹc

Lão Ham nói rỏ chổ này dùm em được không ? Em có ráp cái active Cross bằng đèn "chơi " choke input nó cứ hum trị hoài không hết

 

Hẳn là như vậy rồi, em có khẳng định là lúc nào cũng phải dùng lõi có thiết diện hình vuông đâu, em chỉ tập tẹ trả lời câu hỏi của bác sì tin thôi mà. Nếu chiều rộng lưỡi là a (có sẵn các giá trị công nghiệp đã sản xuất), chiều dày lớp xếp là b, thì cô giáo Tanhia ngày xưa dạy em là b không nên vượt quá 2a. (Nếu khi chọn vượt quá thì lấy a lớn hơn lên để giảm b đi.).
Thông thường thị trường ta chỉ có các loại lõi cũ có sẵn, kiếm được loại tốt là quí lắm rồi.
Còn nếu em có nguồn thép kỹ thuật, có máy dập, máy cắt các kích thước, máy cuốn, máy ép, ... em sẽ chỉ làm cho mình lõi thép có thiết diện hình vuông, tiết diện cửa sổ, đường dẫn từ theo yêu cầu tối ưu thôi.

 

Vô trang này:
http://www.turneraudio.com.au/audiofilterchokes.html
Tải được món này:
CHOKE DESIGN FOR AUDIO AMPLIFIERS.
... ...
The first question is 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.
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

This table allows a simple choice to find wire size for a winding around a stack of 25mm x tongue of 25mm.
If current = 270mA dc, simply choose the wire size for the current rating just above 270mA, ie, 339mA
and the wire = 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 wire current density is all slightly under 3A per sq.mm for all wire sizes.
Here are some useful formulas :-
However, I did come up with this equation for cores with equal Tongue and Stack dimensions.
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.

FOR 25MM X 25MM CORE ONLY,
Simplifying above eqtn, Rw = N x 0.00316 / d squared.

( And d = 0.056 x sq.root ( N / Rw ).

So for the 25 x 25 core, Rw with 0.4mm wire = 1,300 x 0.00316 / ( 0.4 x 0.4 ) = 25.7 ohms.

The bobbin window L x H = 12.5mm x 37.5mm, and with bobbin wall thickness and clearance
the bobbin winding area = 8.8mm x 34mm = 300sq.mm.

Turns possible = wind area / oa wire dia in mm squared, = 300 / od squared.

Here is a formula I came up with does allow the tongue size to be found
for a given wire size and wanted winding resistance for core with stack height equal to tongue width :-

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 has been chosen.

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.
A 50mm stack instead of 25mm stack gives a core area total of double the 25 x 25 square size,
and available inductance is doubled. 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.

...
so wire crossings are not at a sharp angle, and keeping the level of the wire as free as possible from forming large humps and troughs
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. Terminations on the bobbin should be provided where wire is less than 0.75mm dia.

Some of you may have the good 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 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 until just 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 won't affect the iron's magnetic qualities.
Don't cool the hot cores in water.
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.
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 (1) below, each method has the effect of lengthening the magnetic circuit length, and thus reducing
the maximum possible permeability, µ.
(1) 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, 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.
(2) 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 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 (1). The resulting µ 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 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.
(3) Es for the whole choke are grouped into say 5 piles of equal height, and stacked into the bobbin with each pile facing
alternate directions. Is are stacked into piles of the same height and each pile is butted to the open E ends.
This method allows the µe to be intermediate between being maximal with maximal interleaving,
and the µ/10 value or less achieved above in (2).
This method is rarely ever used, but applied sometimes to PP OPT or in balanced chokes 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.
For a power supply filter choke, there will ALWAYS be a full gap between all Es and Is and ONLY method (2) 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.
For all choke inductances, L = 1.26 x Nsquared x Afe x µe
1,000,000,000 x ML
where L is in Henrys,
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.
So adding a 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
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 Teslas, 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 about 1/2 of its capability.
If the iron is GOSS, it may saturate at 1.5Tesla. So we really don't need to have any added air gap perhaps.
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 1.5Teslas.
In the case of the CLC filter we have, the Vac across the choke is only 1.2Vrms.

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.
The inductance may be calculated for the air gapped value of µe, but
as dc flow increases from nothing to maximum allowable, µe will be reduced perhaps 50% because of the
gradual effects of increasing dc.
To really set the choke to get the best benefit, the choke MUST be tried in the power supply circuit.
The gap should be adjusted for size to get the least ripple voltage at C2 of the CLC filter with the wanted idle current flowing.
The choke is not bolted and not clamped hard, but is connected to the CLC supply, possibly
with a resistance dummy load to make a dc flow equal to when tubes will be used.
I place the choke on a block of wood to gain easy access to the gap. The Is tend to be drawn tightly
against Es when dc flows, so there isn't any need for bolts to be tight during adjustments.
But each time the gap is adjusted, turn off the power and let the B+ discharge.
We do not want to see another premature death notice in a newspaper!!!
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
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 isn't important to measure the gap, but only to find the right size for the gap.
This is all easily done in the workshop notebook accurately enough.
What one may find is that 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.
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.
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 and very difficult calculations required to be made to do it any other way.
There is most certainly NO SIMULATION software available allowing typing and a PC to do it.
One must get right away from the damn PC, and be prepared to do real and dangerous work.
Gap size can ONLY be really well determined by hands on adjustments.
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.
NOTE, When measuring the Vripple at C2 in a CLC filter, tou WILL find an ordinary
DVM quite adequate, but when Vripple is a small voltage below 10mV, you will not
often be able to read it accurately because the mains voltage level is constantly moving up and down
according to who switches appliances on and off in your house, and in all the houses down your street.
Its not unusual to find that the signal at C2 jitters up and down with low frequencies with a peak to peak
voltage = +/- 0.05V.
To remove the effects of such LF noise, a high pass two stage CR filter must be made.
This can be two CR sections with C = 0.47uF, R = 10k, and a second CR with C = 0.1uF and R = 47k, giving
a cut off of about 28Hz with about 1dB attenuation at 100Hz, but 20dB at 10Hz and 40dB at 2Hz.
Use of an oscilloscope is the best way to look at the wave form and determine its amplitude from amoung the noise residual.
How is the inductance measured when the gap has been set?
Well, you don't measure it in the test circuit at the top of the page,
because the L without DC low could be twice the L with DC flow.
You look at your work book where you recorded the final Vripple measurement
when the gap was set.
The C2 and choke L form what is known as a second order filter low pass filter with 12dB attenuation
per octave above the pole frequency of below 10Hz, and where the L and C become series resonant, Fo.
At F well above the Fo, all such simple filters will have Vac across L and across C in proportion
to ZL : ZC at the higher F, in this case, 100Hz.
In the example with the choke we have wound with 1,300 turns for 270mA,
let us suppose we find 1.2Vrms 100Hz at C1 = 470uF and 0.004Vrms at C2, 470uF.
The ZC = 3.4 ohms.
L = ZC x ( Vripple C1 - Vripple C2 )
VripleC2 x 628
= 3.4 x 1.196
0.004 x 628
= 1.618 Henrys
To double the amount of inductance for the same N and Idc, and same tongue = 25mm,
we would have to have to wind another choke with an increased stack of 50mm to double Afe.
The Bdc will stay the same, and Bac would be halved, but this Bac reduction does not matter.
What is the resonant frequency of the completed choke with C2?
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.
In this case, Fo = 5,035
sq.root ( 1,618 x 470 )
= 5.77 Hz.
This F is OK, because its below 7Hz, but to halve Fo, the C must be quadrupled, or L quadrupled.
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 250mA, its obviously easier to
add three more 470uF caps in parallel with C2, so the C = 1,880uF.
The law of diminishing returns is in operation, because reducing ripple voltage from
say 4mV to 1mV is usually utterly inaudible, and even if Vripple was 10mV,
there should be little hum from a speaker.
CAUTION There is another big issue to address with a CLC filter such as I have suggested here.
At turn on, C1 and C2 all require fast charging up to full working voltage, say +420Vdc, and the total C could be 4 x 470uF = 1,880uF.
The flow of current in the first few mains cycles at turn on during charge up is quite high.
There is magnetization of the mains transformer, and the current is only limited by the mains wiring resistance, series resistances of windings, and diodes
and dc 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.

During the first 4 seconds it takes to charge the B+ caps, a limiting resistance of say 100 ohms of 20 watt rating should be allowed to
remain in series with the HT winding and then shunted out with a relay acting after the 4 second delay circuit allows.
There should also be fixed resistances used between the HT winding and diodes.
C1 to the
For a full wave bridge, the normal peak charges to C1 can be nicely limited by using a resistance of 7 x ZC1 at 100Hz.
Because here we have ZC1 = 3.4ohms at 100Hz, R = 22 ohms, and 10 watt rated.
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 charge current limiters,
you will see these resistors cause slight de-regulation of the B+, but as i have said, its not important because
all your music is made while the tubes are in class A and there is no change of anode supply current from the power supply.
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 220uF for C1, retaining such resistances,
and place more C after the choke, say 3 x 470uF.

The use of limiting resistors does not reduce the Vripple at C1, but makes the up-going charge part of the
Vripple wave take more time to achieve, and the down-going drain part of the wave becomes a shorter time.
In other words, because the added resistance causes charge time to be longer, peak charge currents become lower
to still achieve the energy in = energy out equilibrium at C1.

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 with a full wave CT winding with two GZ34,
one on each phase of the winding for 270mA, 100uF is the safe maximum C1 value. The 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 tube, and is high,
and which limits peak charge, so added resistances have to be high to achieve any 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.

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 of such occasions hasten the death of the tube rectifier.
Such GZ34 might be OK in a Quad 22 with pathetically low value C and horrible B+ rail filtering,
but in modern amps with large amounts of C they simply do not belong.

And finally, for those who like even better filtering, a real freeby.......

The L may be set up to resonate at 100Hz with a suitable capacitor of polypropylene or polyester type and at least 400V rated.
The value is found using above resonance formula.
The cap is connected across the choke, but with a resistance also in series with the cap,
and usually between 15 and 47 ohms is approximately right.
Keen experimenters will measure the 200Hz ripple content and adjust R value for the least 2H.
This will probably make the C2 ripple voltage maybe at least 12dB or 1/4 of what it is without the RC.
With 1.6H one would need C = 1.6uF, and R about 22 ohms.
Its usual to get a 12dB reduction in Vripple at C2, so I guess I just saved you the cost of 3 more 470uF caps.
You owe me a beer at least.



(2) Chokes for LC input filter power supplies.
The rectifier always must be a full wave type with CT winding and two diodes or a single winding with a four diode bridge.

The diodes do not charge up a capacitor directly as in the case of the CLC, but
charge a C through an L which is simply a second order filter with low F pole as low as we can make it,
usually below 7Hz.

The diodes establish an ac voltage of mainly 100Hz with even order harmonics all applied the choke input,
and all at an average level above 0V, allowing an average vdc level to flow, largely independent of Idc.
The action of the LC filter allows dc flow, but not much ac.
The choke input filter can give good regulation of the B+ output voltage where the current may change
tenfold from idle conditions to full power, but regulation depends on winding resistances and diode resistances.
The high peak charge of the CLC filter do not occur.
The diode charge current flows almost continuously, so power dissipated in the HT winding is 30%
less than in a typical CLC supply providing the same VA.
The choke input supply ripple voltage will be higher 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 current flow in the choke
is constant and there are no high peak charge currents to a capacitor, and for one reason or another, including whimsical preference,
the available HT supply voltage at the power transformer suits a wanted lower Vdc to be produced.
The choke used is designed to have high inductance under low current idle conditions, but when current
is increased 10 fold, there will be no core saturation despite the fall of inductance due to increase in Idc.
The Vac wanted for a given Vdc = Vdc / 0.88
The choke must be mechanically quiet, and this is the difficult aspect of chokes used in LC inputs.
The Vac in Vrms ahead of the choke from rectifier diodes and at 100Hz = 0.6 x Vdc at the capacitor
A typical amp with B+ = +425Vdc will have a HT winding of 483Vrms with a bridge,
or 966V with a CT, for use with two diodes.

The peak Vac = 680V, and this is the peak to peak voltage for Vac ahead of the choke, and allowing for the
large 2H component the 100Hz wave will be approx 230Vrms, nearly as high as the mains input 240vrms.
Choke inputs tend to hum, and vibrate, due to the high amplitude voltage of twice the mains frequency
plus harmonics, which is applied across the choke effectively.
So the choke must not saturate with combination of dc magnetization and the applied ac voltage caused magnetization.
This type of choke should be potted after being very well varnish impregnated; potting will stop much stray radiated magnetic field such chokes may generate,
as well as quieten them.
The varnishing is very important because of the high Bac at which the iron operates and the forces
produced in the windings try to make the windings move around.
In late 2004, I used a choke input filter power supply for a customer's much revised Phase Linear 700W transistor amplifier.
The Phase Linear amp had rail voltages way too high for reliability, a noisy mains tranny, and heat sinks far too small.
With a choke input filter you get a maximum Vdc rail = 0.89 x Vrms of the transformer winding instead of 1.41 x Vrms of the winding. So for a given wanted output dc voltage of +60V, you need 67Vrms transformer windings.
For the same +60Vdc with a bridge rectifier you would need only 43Vrms. But since the rectifier current flow is
constant from a choke input supply the transformer windings can use thinner wire than for the bridge.
The choke input filters much reduced the noise of the existing power transformer in this amp.
With reduced rail voltages, the amp still had a huge power capability.
I wound two chokes, for the +ve and -ve rails, and potted them in steel boxes with roof pitch.

I probably may have done better by placing the existing large Phase Linear power transformer in my open wood fire
to heat it up to red for a few minutes and let it cool for a day.
Then the lams can easily be retrieved since all plastics are vaporized by the firing, but core steel is unaffected by the heating,
perhaps even improved by the slight annealing, so the iron may be re-used, and the awful transformer re-wound.
However, after a heroic struggle, the Phase Turner has remained as a very fine sub-woofer amp for 3 years so far.

If you insist you want a choke input filter, there are rules.
Minimum choke value = RL / 900.

Where mains = 50Hz, and where RL is the Vdc output / dc current at the minimum current.
This choke value is called the critical choke value.
Choke input PSUs were often used for class B tube amps where the anode input current flowing
into the output stage varied enormously. But because the current into the caps after the choke is continuous in each 1/2 cycle,
and without a heavy peak charge current over a short % of the wave cycles, vacuum tube rectifiers
could be used, and in high power and high voltage applications, use of mercury vapour vacuum tube diodes such as 866
were good with a following class B amp of kilowatt capability.

The minimum and maximum dc current flows must be known, and usually the minimum current is taken to be 10% of the maximum.
The minimum which must be allowed to flow even when output tubes are biased right off, as in the case of a 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 voltage,
possibly stressing other components or insulation layers.
I recently revised a badly designed CR Audio Developments "Woodham" amp with a pair of PP channels with KT88.
The HT transformer voltage was 487Vac, and the cap input with Si diode bridge gave B+ = +650Vdc, for a dumbed down McIntosh
style of output stage. One could only get a brief 40 watts before smoke arose from this very poorly designed amplifier.
I found what I thought would be a choke which could be suitable. So after a few tests i potted it and got a better +425Vdc
at the at the reservoir cap after the choke at full current draw, and converted the amp to 50% UL.
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 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 = 15,000 / 900 = 16.7Henrys.
The choke only had 9H at 30mA. At 250mA, when the 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 the 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 designs, Bdc = Bac = 0.6Tesla max
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 to resonate with the L value 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 twice the maximum inductance,
and B+ soaring was prevented, and Vdc better regulated, and a low bleed current was required.
So if one wants 250mA amps max, bleed current should be at least 25mA. For best natural B+ regulation,
the power transformer and choke should have low winding resistance.
If the choke was 50ohms, the change in current of 200mA would cause an unavoidable drop of 10Vdc.
This isn't bad, but never use a choke of 500 ohms!
Let us consider the design ideas around a +425Vdc PSU for 300mA 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 should be added to the design 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.

The more maximum inductance the better, and RL@Idc min / 900 is a fair guide. 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 saturation, the choke is well
designed.
RL at 30mAdc = 425V / 0.03A = 14,166ohms.
Lcritical = 14,166 / 900 = 15.76H, so let's aim for 16H maximum.
Let us try a core of 25mm tongue, and for 300mA, use 0.4mm Cu dia wire, allowing 1,300 turns.
With a square core section, the Rw = 26 ohms.
Experience tells me the stack would need to be a lot more than 25mm.
What would be the stack height? Stack height can be worked out when we need to adjust the design for inductance.

First the dc magnetization effects must be established and the iron must be gapped to
suit a Bdc max not exceeding about 0.8 Tesla, ( 8,000 gauss ).

If we refer to the choke design for CLC, we ended up with a core with Afe = 25mm x 25mm.
If we have 1,600 turns, and 250mA, we found the Bdc was 1Tesla, and high, leaving little room
for much ac magnetization.
With dc magnetization, it doesn't matter how high the stack is,
it will not reduce the steel magnetization if the magnetic path length and current and turns remain constant.
The stack height could be 250mm, and with the same µe and gap and dc flow,
the Bdc would still be 1Tesla, ( 10,000 gauss ).
the CLC choke has quite a negligible Bac from ac flow, but the choke for LC input
has a much larger voltage at its input, and 240Vrms at 100Hz in this example.
So the gap may need increasing to reduce µe so there is room for the Bac produced by
the large ac voltage across the choke.
Working from the choke example for CLC, I concluded that the final amount of inductance with 250mAdc present = 1.3H.

From this, we can work out what the µe was for this size of core at the dc current level.

1.3H = 1.26 x 1,600 x 1,600 x 25 x 25 x µe
1,000,000,000 x 140
µe = 90.

This is a very low figure, and it will about double for where there is no Idc.

So where we want a choke for LC input, we should start off with
allowing µe = 100 at maximum Idc, knowing that when Idc was minimum, µe could double.
With such a condition, we still need to get Bdc less than about 0.8Tesla at maximum Idc.

The choice of µe = 100 minimum with maximum Idc allows some fine adjustment of the core gap for best results,
and we can allow µe = 200 for low Idc < 1/10 of max Idc.
This condition also allows the use of low grade iron.

If we want 16H at low Idc when µe = 200, N = 1,300t, then what is the stack height for 25mm tongue?

L = 1.26 x Nsquared x Afe x µe
1,000,000,000 x ML
16H = 1.26 x 1,300 x 1,300 x S x 25 x 200 = 0.076 x S, therefore S = 210mm.
1,000,000,000 x 140

210mm is an entirely impractical stack height, so we cannot use 25 tongue material.
But if were able to double the turns, we would increase L x 4 times and use stack = 52mm for 16H.
Trouble is that dc resistance would rise and Pd exceed safe levels.
Hanna was confounded by the difficulty of working out chokes, and came up with his
solutions in his Hanna's Method.
I have found Hanna's Method as described in RHD4 to work OK for CLC chokes where Bac is always low.
RDH4 refers ppl to a lot of references with regard to LC inputs. But the references are no longer to be found.
I found Hanna works best for LC inputs if twice the wanted inductance is sought which forces more turns into the process.

Then I came up with the Turner Method for Choke Inputs.

(1) What is the range of load values supplied by power from the power supply?

Nominal wanted B+ at C = +425V, RL = Vdc / Idc.
Idc min = 30mAdc, RL = 14,166 ohms.
Idc max = 300mAdc, RL = 1,417 ohms.

(2) Wanted winding resistance, Rw, of choke = RL minimum / 30 = 1,417 / 30 = 47 ohms

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

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

(5) Design choke for maximum L at minimum Idc and wanted Rw

Try wasteless pattern E&I laminations, 32mm tongue core stack = 32mm.
Window size = 16 x 46, winding area = 12 x 42 = 504sq.mm.
Turn L = 178mm.
Magnetic Length, ML = 178mm.
Wanted dia of wire for 300mAdc = 0.40mm Cu dia wire, 0.47mm oa dia..
Turns possible = winding area / oa wire dia squared = 504 / ( 0.47 x 0.47 ) = 2,281 turns,
DC resistance = N x TL x 0.0226 / ( 1,000 x D x D ) = 2,281 x 178 x 0.0226 / ( 1,000 x 0.4 x 0.4 ) = 57 ohms.
(6) Does DC resistance exceed allowed resistance for ideal B+ regulation?

if yes, try larger core tongue size and find wire size for the same turns. Then re-check turns and DC resistance.

Try 38mm stack x 38mm tongue E&I lams, window = 57mmx 19mm window giving winding area = 53mm x 15mm = 795sq.mm
Oa dia for wire, 2,281 turns = sq.root ( 795 / 2,281 ) = 0.59mm, select wire 0.50 = 0.57mm oa, giving 2,440 turns.
Turn L = 212mm, DC R = 47 ohms, just OK.

(7) Allowing µe = 200, calculate inductance at Idc minimum.
L = 1.26 x Nsquared x Afe x µe = 1.26 x 2,440 2,440 x 38 x 38 x 200 = 10.2H
1,000,000,000 x ML 1,000,000,000 x 212

(8) Is L high enough? NO, so can stack be increased for required L?

Revised stack height for 16H = ( 16 / 10,2 ) x 38 = 59.6mm. Use 62.5mm standard size stack.

(9) if stack has been increased for additional wanted L, what is new DC resistance?
From (6), for 38mm x 38mm core, turn L = 212, dcR = 47ohms,
With stack = 38mm x 62mm, turn L = 260mm, Rw dcR = 57ohms, which is over the design aim.

(10) If the DC resistance is too high, decide if the additional excess resistance is worth
increasing the core and wire size further for the very small benefit of extra Vdc regulation.

(11) Calculate Minimum RL / Rw = 1,417 / 57 = 24.8.
If answer is above 20, proceed further..

(12) Calculate maximum Bdc at Idc maximum, allow for halving of µe at min Idc, so µe = 100 at max Idc.
Bdc = 12.6 x µe x N x Idc = 12.6 x 100 x 2,281 x 0.30 = 0.4Tesla
10,000 x ML 10,000 x 212

(10) Is Bdc less than 0.8T at max Idc? YES, proceed to next check.

(11) Calculate Vac across choke = ( 0.67 x Vdc ) Vrms, 100Hz, = 0.67 x 425 = 284vrms.

(12) Calculate Bac = 22.6 x Vrms x 10,000
Afe x N x F
= 22.6 x 284 x 10,000
100 x 32 x 1,892 x 100

= 0.106Tesla

(13) Is Bac below 0.5Tesla ? YES, design is OK

(14) Check heat dissipation. Pd = Rw x Idc squared
= 57 x 0.3 x 0.3 = 5.1 watts Size of core will easily dissipate heat.

(15) Final maximum Idc capability = 580mA.

(16) Other rational considerations.

Try starting off requiring the choke's Rw = minimum RL / 20.

This will give a choke which is much lighter, and quite acceptable for where B+ regulation is not
a main concern.

(17) Other uses for this choke.

The choke has been designed for any continuous Idc up to 300mA, and for 16H at 30mA.
Inductance will fall to approx 8H at 300mA, depending on core material to some extent.

The design procedure shows just how difficult it is to get the high inductance one requires for
choke input filters when Idc is low, yet have low winding resistance for good natural B+ regulation when Idc rises.
The size cost and weight of such a large choke is near that of a power transformer or output transformer
and it is the reason why I don't bother much with choke input filters.
At the end of the process you will find you have a B+ power supply which ends up not being any better regulated than a
plain CLC filtered supply, but costing a lot more, and requiring the Queen's Navy to move anywhere.

So if B+ voltage is higher, then both minimum and maximum RL will be greater, and critical wanted inductance
will become higher greater, and to prevent soaring at low Idc, the bleeder current will need to be increased.
If the B+ is lower, the RLs will be smaller, and there is excessive available inductance, and less bleed current
is needed. The design inductance may be easily reduced by using a smaller stack of laminations.

The Zobel network for resonance at 100Hz required for 16H is 0.158 uF + 100 ohms.

The choke is expected to halve its inductance between Idc = 30mA and Idc = 300mA,
ie, from 16H to 8H, perhaps even less depending on the grade of laminations used.
The B+ Vdc output and ripple voltage should be plotted against current on a careful graph over a wide range of Idc from 5mA to 350mA (briefly).
The B+ voltage should be quite high near 1.4 x Vrms of the HT winding at low Idc,
and the at the "critical maximum RL" where Idc is only the bleed current, the Vdc should be no more than
the 1.0 x Vrms HT winding voltage, and at twice the bleed current Vdc should be 0.88 x HT Vrms.
Vdc sag as idc increases is due to the power transformer winding resistances, diode on resistances,
and choke winding resistances, and if that all totals 1/15 of the minimum RL with loaded amplifier, you are doing well.
If RL minimum = 1417 ohms, and series R = 100 ohms, it means that if Idc were to change from
say 250mA to 450mA in a class AB amp, the B+ would only fall from +425V 415Vdc, which is only a 5% B+ drop,
and far less that actually used to happen in most good quality amplifiers made in the 1960s
by the specialist boutique makers of the day.

If the reservoir capacitor = 470uF, and inductance at 250mA = 8H, then resonance
of LC filter = 2.6Hz, and will not affect the amplifier performance.
Fig 2.


Here I have a combination of choke input plus following LC filter in the classic LCLC arrangement
used in many older designs for many large mainly class B amps.
However, perfectionists who like quiet rail voltages can use it for their class AB hi fi amp if they want
and where Ia does not vary hugely.

In Fig2, I have the choke input and following choke 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.
The actual 0V connection can ONLY be made where I have indicated, or else there 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 to
when the L1 is placed in the B+ rail.

The ripple voltage at C2 above 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 is too high, and the two LC sections are needed.

If a 470uF capacitor input is employed with bridge from a HT winding of 315Vac, the ripple at the first C = 1.2Vrms,
and with 2H and another 470uF, ripple is reduced to 3.2mV, and quite low enough for any SE amp.

As I have pointed out in my pages on power supplies, to avoid peak charge currents and higher than wanted heat dissipation
in HT windings, and noise in transformers, and noise in ground rail paths, use 220 uF as the reservoir cap,
and 940 uF after the choke. In addition to the lower input C value, use a series R of 7 times the ZC between C and diodes,
say 47 ohms, 10 watts, and the peak charge currents can usually be reduced at least to a third with 220uF.



(3) Chokes for DC sources or sinks to tube gain stages

Chokes can be effectively used to feed dc current to a tube in its anode or cathode circuit to obtain a source of dc
which does not waste energy heating a resistance and which has far impedance than the resistance ever could have.
Its called choke loading, parafeed, etc.
The supply voltage for the gain stage does not have to accommodate the voltage drop across the resistance.
Cost and weight usually prohibits choke loading but if you know what you are doing, in some situations the choke
works magic, and sounds well.

There are some fanatics I know who do all they can to get rid if capacitors and resistances throughout their audio cicuits,
and I wish them well, for they have made their lives difficult. For them magnetic coupling through transformers
is the only sure way to hi-fi nivirna. I find blocking caps and resistors do not block fine music flowing through an amplifier.
The fundamental problems associated with choke loading or transformer loading of intermediate amplifier stages are with obtaining wide
bandwidth and a flat response. Using any tiny amount of global negative feedback around a
3 stage amp with two interstage transformers is entirely prohibited because of instability from phase shift.

I will therefore not dwell long upon design of interstage transformers which I have not yet found a need to ever use.

Usually the choke is used to supply dc to an anode or from a cathode of a triode with low Ra.
High µ triodes with high Ra get into trouble with chokes because the impedance of the choke needed means inductance
must be enormous, and then because of the high source resistance, ie, Ra driving the choke, the iron distortion
is anything but tamed, and so I will NEVER use a choke to supply 1.4mA to a 12AX7. The use of any pentode tube with
choke supply is also quite pointless due to distortions produced.

I have never used a choke to supply dc to any 6CG7/6SN7 or 12AU7 power amp input triode or preamp triode stage.
I believe the use of a transistor CCS is far easier, cheaper, and sounds better, and the headroom needed
for the voltage across the CCS is always available easily, and power wasted is negligible.

But I have found plenty of need for choke loadings.

All my standard PP amps such as the 8585 have local cathode feedback windings in the output stage, and up to around 80Vrms
needs to be applied to each phase of signal to each grid of on each side of the PP output circuit.
When I began to realize that the only really good way to drive any large octal output tube was with a suitable smaller
output tube strapped in triode, typically a 6BQ5/EL84, I began having problems with resistors supplying dc to anodes,
or from cathodes to 0V or to a negative supply.
The driver stages needed to produce less than 0.1% thd at 80Vrms from each anode, because the amount of global NFB as
only 10dB or so, and the distortion would not be reduced enough.
The biasing resistances in output stages need to low at not more than 100k for all octal output tubes because grids tend to
become slightly positive with respect to the applied grid bias voltage over time, and sometimes dangerously so,
and biasing the tube with excessive bias current that raises temperature and increases this small grid current.
So a driver stage driving output tubes, especially when many are in parallel, needs to be set up so it can easily
drive the grid bias R as well as the dc supply R to the driver tube, and this R should be 1/3 of the value of the following output stage
bias R value. To satisfy the condition, never try to use a 12AX7, such as in the Mullard 520.
Even the 6SN7 used in a Williamson will struggle if there are 4 output tubes each with Rg = 100k.

Hence I don't muck around, I either use EL84 or EL86, or 6V6 and in PP amps they are used in an LTP,
with about 14mA Ia each.
In my 8585 to ensure the RL seen by the triodes is as high a value as possible for lowest distortions, I supply DC to anodes via
a choke with CT with fully interleaved E&I laminations with 25mm tongue size, but perhaps only a 10mm stack,
and with thousands of turns of fine wire, perhaps only 0.15mm Cu dia, and this makes an inductive impedance
which is hundreds of kohms, and which has negligible loading on the triodes compared to the capacitor coupled
following stage grid biasing resistances.
The choke is a 'balanced choke' and acts like a primary of a PP transformer.

I have also tried the idea on my 300 watt amplifiers with 12 x 6550 per channel.
The method works extremely well. The 300 watters have chokes with cores = 32mm stack x 25 tongue and thicker
wire, but the chokes still have ZL very high compared to the 16k or parallel Rg cap coupled to EL84 anodes.

Since my first attempts to make 300 watters with multiple output tubes, I have now moved to using a
balanced choke to bias the output tubes from 0V, so that all 12 coupling caps and separate bias resistors
may be abolished, and the ends of the choke are coupled each end to each driving anode via
100uF elcaps with 0.47 plastic bypasses. There is now 18k anode dc supply resistances acting as the dc and ac load.
This means the EL84 triodes work with exactly the same load as before, but the biasing of output tubes is properly
held constant at 0V for ALL output tubes through the low winding resistances of the choke.
There is a high enough B+ supply to obtain the wanted Vswing, and with IA = 14mA, and RLdc = 18k,
a swing of 80Vrms at each anode means a small peak current change of only +/-6.2mA Pk.
Having the balanced choke in the grid circuit means any unwanted imbalance of dc currents in the LTP halves
does not affect the choke.

The only problem is bandwidth. To avoid the shunt C and shunt L effects of the choke used for for LTP
I have a low value 4k7 in series with each end of the choke, and so the triodes' lowest load value possible is 4k7,
and phase shift caused by stray C and L is virtually eliminated.

Fig 3.

The above shows the choke with center tap in the anode circuit.
The inductance is higher because the laminations fully interleaved with balanced dc flow.
Ra-a of the two triodes = 4k4, and the choke ZL = 4k4 at 2.3Hz, so the bass will be OK.
The typical shunt capacitance can be about 300pF, which has equal ZC to Ra-a at 120kHz.
So the HF will be well extended, and the 4k7 will isolate L and C from having any effects.

Fig4

In this schematic, the choke is used for biasing the output grids instead of the two 18k, which
now have been moved to carry dc to anodes. The loading of the tubes is the same as in Fig3.
Because there is no dc flow in the choke at all, the choke can have thinner wire if desired, and the effect of
reducing wire dia from 0.25mm to 0.2mm is a doubling of inductance.
But notice the caps coupling the anodes to following choke circuit.
The LF pole for what is effectively a series resistance damped high pass filter with 50uF to 630H CL values
giving a pole at 0.9Hz. But at low levels of operation, choke L falls to
much lower than the hundreds of H when signal voltage is high, and at 4Vrms a-a the L could be
1/20 or even less than values shown so the Fo could be up to 10 times higher.
The 100uF shown as coupling caps could be excessive, and maybe 22uF is OK, but this has to be trialled
and with NFB stability at LF becomes slightly unpredictable.

The recipes for the chokes are given on the Fig3 and Fig4 schematics, so happy winding!

I am presently building a pair of SE amps with a pair of 845 in each amp to make 50 watts of pure class A.
Fig 5.


The amps will be set up to allow the use of any variety of 845 or 211. The latter require considerable class A2 drive, ie, grid current drive to
enable a full power output above the class A1-A2 threshold for handling transients.
Each 845 will have separate cathode biasing using a constant current sink as the cathode resistances, but are well bypassed to 0V with 470uF capacitors.
Variations in mains voltages will never much upset the biasing which could otherwise become troublesome if fixed bias is used.
Cathode biasing is the ONLY way SE amps should be supplied to owners who are not at all technically trained because mistakes with setting
bias on tubes with power supplies over 1,000V are usually spectacular, and disgustingly expensive.

V3, an EL34 triode cathode follower buffer drives the pair of output tube grids directly.
This buffer stage must be capable of working into grid current without causing horrendous distortions.
There is a negative 370Vdc supply, and dc is passed to 0V through L2 choke loading in the cathode circuit.

If the EL34 does try to conduct excessive current, and more than the planned 25mA maximum,
the 845 grid voltage will still be kept low because of the choke's low dc resistance. Should the 845 Ek ever rise more than 50V above the wanted
Ek levels, the active protection will shut down the amp.

The buffer is driven by a single ended V2 EL34 triode drive stage to make up to about 130Vrms of drive voltage.
The V2 EL34 has excellent linearity, gain of about 9, Ra = 1k5, and will be happy with Ia = 22mA.
The dc is supplied through choke L1 plus 1k8 resistance for a damped LR with high Z.
Alternatively, there would have to be a 22k 60 watt rated resistance to the +800V rail for 845 tubes, so I want to use
a choke to get the dc to the anode. With the choke, the ZL is huge at most frequencies
and the load is dominantly just the buffer biasing R = 220k, and the high value impedance loading this driver tube helps
to give very low distortion and beautiful sound. L1 and L2 have identical design.

What is the design for the chokes?

The Ra of the EL34 in triode with Ia = 22mA is approximately 1k5.
The wanted cut off point is at 5Hz, so inductance wanted, L = 1,500 / ( 5 x 6.28 ) = 48H.
If more inductance can be found, it is welcome.

Unlike the filter chokes mentioned above in parts1 and 2, the chokes we now are considering
need not have low dc resistance, and as long as the current density is well below 3amps/sq.mm, high winding resistance
can be used to help bias the cathode follower driver buffer.

Design Ia for EL34 is 22mA, but we must allow wire size for 100mA in the case of some overload problem.
For working Bdc, allow Ia = 22mA.
Suitable wire will be 0.20mm Cu dia, capable of a sustained 100mA long term. Oa dia = 0.25mm.
Let us try a core with tongue 25mm, GOSS, with winding area available = 300sq.mm.
Turns available = 4,800.

The wanted Vac swing will be perhaps up to 150Vrms and unlike the power choke we must be able to get down to 14Hz
without core saturation from too high a Bac. Bac would not want to be higher than 0.5Tesla.
To find the stack height :-

Bac = 0.5 = 22.6 x 150V x 10,000 = 20.2 / S, so S = 20.2 / 0.5 = 40.4mm.
25 x S x 14 x 4,800

Therefore, try a bobbin for 25mm tongue x 38mm stack.

Turn length will be 166mm, and winding resistance = 4,800 x 166 x 0.0000226 / ( 0.20 x 0.20 ) = 450 ohms.
about 1k2 ohms may have have to be added to the choke to make it act as a large enough biasing resistance for the cathode follower
whose grid is biased by resistance from the Vdc at the bottom end of the choke.
The 1k2 ohms will help isolate the choke's shunt C and shunt L to a small but not unwelcome amount.
The EL34 Ek will thus be around +35V above the -370V rail. Some trimming of the added cathode resistance
will be required.

What µe is required to keep Bdc below a safe 0.5Tesla maximum?

Bdc = 12.6 x N x µe x Idc = 0.5 max = 12.6 x 4,800 x µe x 0.022 = 0.00094 x µe, so µe can be 0.5 / 0.00094 = 531.
10,000 x mL 10,000 x 140

The value of µe = 531 is quite high and we may not obtain it by merely butting all Is hard against all Es.
If we tried, µe might only be 300 max, depending on the iron's maximum µ when all laminations are fully interleaved.
But for any sample of iron that you may have, and if you are in doubt, use a larger stack than the minimum I have calculated,
so wind the choke, install the E&I just close butted, and measure the inductance with a 50Hz signal of 10V across the choke.
The performance of the choke in this application will always be better if the stack height is increased.
Where the core is just butted close but with no actual gap, the µe is often about 1/10 of the max µ attainable with
fully interleaved lams, so if the max µ was 3,000, when butted the µe = 300, and when 10,000, perhaps you get µe = 1,000.

The inductance with µe = 300 will be

L = 1.26 x 4,800 x 4,800 x 25 x 38 x 300 = 59H and this seems to be plenty.
1,000,000,000 x 140

But the inductance could be more if the stack was increased to 50mm.

Alternatively, we could use C-cores which will allow easier fine adjustment of the µe between say 300 and the max µ
of perhaps 10,000. Trouble is, finding a supplier of C cores is like searching for a needle in haystack.

A larger core size could be used, such as 32 tongue x 32 stack, so the turns with 0.2mm wire = 8,000.

The Bdc for 32S x 32T = 0.7Tesla max, = 12.6 x 8,000 x µe x 0.022 = 0.00125 x µe, so µe wanted = 561.
10,000 x 178
If µe = 300 only,

L = 1.26 x 8,000 x 8,000 x 32 x 32 x 300 = 139H and this is more than we need.
1,000,000,000 x 178
We could then afford to have a smaller stack of only 20mm, and still get 70H, depending on µe with dc..

If the µe was found to be only 300 after tests with such a slightly larger core, the choke would be able to carry more dc flow without saturation.
I have figured on 22mA with EL34 in triode, and I see little reason why there should be more because with a load of above 100k,
the current swing is only 1.66mArms at 166Vrms with Ea = 275V.

The sizes for cores for such good quality chokes will always be larger than what you see in mass produced amplifiers, because mass market
makers invariably try to sell you less inductance and bandwidth to reduce their costs and boost profits.
So while core sizes of 25T, 28T, and 32T seem large, they are not, and its just what good design calls for.

With fully interleaved GOSS lams, the µ is at its maximum at perhaps above 5,000, and the inductance easily over 500H,
but with dc flow the core would be solidly saturated, leading to severe distortions with ac.

The other way of gaining a higher µe over 300 which one might get with simple E&I close butting is to arrange the laminations
in close butted piles of Es and Is about 5mm high. Alternate the insertion direction when filling the bobbin so each pile of Es face
alternative direction. This should give a gradual transition between maximum possible µ and the wanted value of µe for the core.
The experimentation can allow you to obtain some surprisingly fine results, without having used more iron than you need to.

The alternative is the use of C-cores for this choke, and these have a µ max well over 5,000 when butted together without any gap material,
thus allowing some very thin gap to be placed to lower the max µ to a µe which is acceptable.

The gap in a choke core will make the choke have a more even amount of inductance regardless of voltage and frequency because
the gap dominates the µe equations. A gap filled with non magnetic material is linear, and iron is not.
With a gap, the effective magnetic path length is increased from being just that of the iron.

Magnetic length with a gap = Iron mL x [ 1 + ( µ x gap / mL ) ]
The equation for inductance has µ on the top line of the equation, and mL on the bottom as a factor µ / mL.

With a gap this becomes µ
mL x [ 1 + ( µ x gap / mL ) ]

Therefore the equations with a gap are simplified by leaving the iron mL on the bottom line,
and simply changing the top line figure for µ to µe found as :-

µe = µ
1 + ( µ x gap / mL )

For 25 tongue GOSS material, µ max without any gap might be 10,000, and mL = 140mm.
Therefore with a gap = 0.5mm,

µe = 10,000 / 1 + ( 10,000 x 0.5/140 ) = 272.

If the iron had µ = 3,000 max, µe = 3,000 / 1+ ( 3,000 x 0.5/140 ) = 256, which is very little different to the
higher µ material.

But if the gap was only 0.05mm, the higher µ material gives µe = 2,188.
and the lower µ material gives µe = 1,615, so its a considerable difference.



=> Nếu mà em biết dịch sang tiếng Việt thì đúng là đã... dễ ẹc rồi

 

Re: Choke dễ ẹc - quấn cuộn cảm dễ ẹc

Bí kíp của lão Ham thì một lão khác của vnav mượn chưa trả, nhưng cái bài của Biến Nút Nhấn chép của Tuner audio nói khá đầy đủ. Choke input filter được dùng trong các mạch tiêu thụ dòng biến thiên (AB2, B) có tác dụng ổn áp khi dòng thay đổi. Tuy nhiên do tác dụng ổn áp, cuộn choke sinh ra một từ trường rất lớn dễ gây nhiễu cho các mạch từ kế cận nếu không bọc che chắn tốt. Đối với các mạch có dòng tiêu thụ ổn định thì dùng CLC filter ít nhiễu hơn.
Các DIYer muốn hạ áp một bộ nguồn có sẵn đôi khi cũng dùng choke input filter nhưng nếu không khéo thì dễ bị nhiễu, cuối cùng phải quay trở lại cách dùng RCLC để hạ áp và lọc.

 

Tiết diện lõi hình vuông là vì để quấn cho khít hơn, ít bị khe hở không khí làm giảm hiệu suất hơn đó mà.

 

Tiếp đi bác ơi, em đang cần quấn choke, lên đây hóng hớt các tiền bối để có thể tự quấn 1 choke lọc nguồn.

 

hic, bác đào sâu ghê đó chứ. quấn choke thì bác cứ kiếm cái bobin rồi quấn đầy vào là được. quấn sao mà bác thấy sảng khoái là cái amply nó hót cũng hay à.

 

Công nhận dễ ẹc, nhưng với trình gà của em thôi thì cứ xin các bác vài ví dụ rồi làm theo. Đơn cử như loại 5H - 250mA thì lõi...., dây.... bao nhiêu vòng; 10H - 50mA thì lõi... dây....., số vòng. Giống như xưởng chế tác của Bác Trà Xanh thì Member chúng em cảm ơn lắm lắm.

 

Em cũng đang dài cổ đợi mấy bác chỉ cho cách tính toán, em nghe nói trên đây nhiều bác pro lắm

 

Hồi sau là đến năm Công Gô nào ơi hỡi bác GVTeam ???

 

Chào các A/E "mê Choke".

Quấn choke dễ, nhưng không hiểu đúng về nó thì có thể cho kết quả... dở ẹc

Tôi thì cũng đang đói choke quân sự và vintage để phục vụ khách hàng sửa chữa và nâng cấp ampli. Nên đang tiến hành DIY một ít để dùng. Nhân thấy các A/E cũng đang quan tâm vấn đề nầy tôi mạo muội tham gia chia sẽ ít tài liệu kỹ thuật tiếng Pháp (Transformateur RADIO- Charles GUILBERT)do KS Hồ Vĩnh Thuận, giảng viên ĐHBK TP.HCM, nguyên phó GĐ đài TH. TP.HCM dịch tặng shop Restore.

*Nếu các A/E không nóng vội! tôi sẽ cố gắng dành ít thời gian trích lược lại tài liệu nầy cho A/E có thêm một góc nhìn về "CHOKE".
Nhưng không lâu đến Tết Cong gô đâu!

*Sách gồm 06 chương liên quan đến cách tính toán, thiết kế, thực hành quấn các biến thế : nguồn, cuộn lọc( choke ), OPT cho các dự án ampli đèn cụ thể. Nhưng ở đây tạm thời tôi sẽ trích lược đoạn nói về Thực hiện cuộn cảm bộ lọc mà thôi.

Mr Chánh.

 

Cái này hay quá bác à. Khi nào bác dịch xong em sẽ down về đọc .

 

Chào các A/E "mê Choke".

Quấn choke dễ, nhưng không hiểu đúng về nó thì có thể cho kết quả... dở ẹc

Tôi thì cũng đang đói choke quân sự và vintage để phục vụ khách hàng sửa chữa và nâng cấp ampli. Nên đang tiến hành DIY một ít để dùng. Nhân thấy các A/E cũng đang quan tâm vấn đề nầy tôi mạo muội tham gia chia sẽ ít tài liệu kỹ thuật tiếng Pháp (Transformateur RADIO- Charles GUILBERT)do KS Hồ Vĩnh Thuận, giảng viên ĐHBK TP.HCM, nguyên phó GĐ đài TH. TP.HCM dịch tặng shop Restore.

*Nếu các A/E không nóng vội! tôi sẽ cố gắng dành ít thời gian trích lược lại tài liệu nầy cho A/E có thêm một góc nhìn về "CHOKE".
Nhưng không lâu đến Tết Cong gô đâu!

*Sách gồm 06 chương liên quan đến cách tính toán, thiết kế, thực hành quấn các biến thế : nguồn, cuộn lọc( choke ), OPT cho các dự án ampli đèn cụ thể. Nhưng ở đây tạm thời tôi sẽ trích lược đoạn nói về Thực hiện cuộn cảm bộ lọc mà thôi.

Mr Chánh.[/quote]

Xin chào Mr Chánh ! Nghe tiếng bác đã lâu,và cũng ko ngờ đc thấy bác mở lòng với ae,xin cám ơn bác trước nha!
Tôi chỉ xin đ/nghi với bác là ae nói chung là amate vì thế bác chỉ cần trích lược những vấn đề chính mà quan trọng nhất là phần thực hành là tốt lắm rồi. Tôi xin đc ví dụ như : nguyên lý hoạt động và tác dụng của choke; cách đo đạc và nhận biết giá trị của cục choke và sau cùng là phần thực hành chọn fe,lựa dây,cách quấn và lắp ghép tạo aigap để đạt đc mục tiêu. Tks bác !

 

Vâng,vậy xin mời Mr Chánh,bớt chút thời gian tiếp tục đăng đàn đi ạ.Tks!

 

Trong lúc chờ các cao thủ vào chỉ giáo thêm em xin có chút ý kiến.
Quấn Choke và OPT nó hơi khác chút nên không thể áp dụng một sơ đồ quấn chung đc bác ạ.
Với OPT thì bác quấn số vòng nhất định nhưng cảm kháng cuộn sơ cấp cho phép tổi thiểu là bao nhiêu đó, lớn hơn chút nữa thì càng tốt.
Với Choke thì múc đích lại là trị số cảm kháng rồi đến dòng chịu vì vậy với mỗi lõi có chất lượng khác nhau sẽ có số vòng quấn khác nhau. Lõi tốt thì quấn ít vòng và đồng thời sẽ tăng cỡ dây để giúp chịu tải tốt đồng thời giản trở kháng thuần của cuộn dây.
Vì vậy quấn Choke nên tính toán theo thực nghiệm.
Cách làm:
Với lõi đã có, ta quấn thử khoảng 1000 vòng vào lõi rồi vào Fe, điều chỉnh khe hở 0.2-0.4mm => Đo cảm kháng. Tử kết quả thực nghiệm ta tính ra số vòng càn quấn để đạt được cảm kháng dự đinh quấn ban đầu.
Ví dụ thực tế em đã làm:
Với lõi Fe dạng chữ M của Đức, cỡ lưỡi 20mm, lưỡi dài 44mm, xếp dày 27mm, khe hở của lưỡi có sẵn ~0.4mm.
Quấn 1000 vòng dây 0.5mm => đầy cửa sổ đo được 1.45H
Dự tính quấn cục Choke 10H.
Tính số vòng dây cần quấn để ra 10H: N = [(10/1.45) ^0.5 ] x 1000 = ~2626 vòng.
VỚi 2626 vòng ta sẽ chọn cỡ dây lớn nhất có thể quấn: d = [(1000/2626)^0.5] x 0.5 =~ 0.308 => chọn dây 0.3mm (cả emay)

=> Action

 

với VD trên choke này chịu dòng bao nhiêu (ma) bác ơi ?

 

quấn dây 0.3 được 180mA , dây 0.5 được 500mA anh ạ