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LW1DSE > TECH 10.07.12 01:21l 263 Lines 10855 Bytes #999 (0) @ WW
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Subj: LC Filters (CP437)
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From: LW1DSE@LW8DJW.#1824.BA.ARG.SA
To : TECH@WW
[¯¯¯ TST HOST 1.43c, UTC diff:5, Local time: Mon Jul 09 11:19:15 2012 ®®®]
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 satura-
tion, the choke is well designed.
RL at 30mADC = 425V / 0.03A = 14,166ê.
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 æFe 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 æFe 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 æFe was for this size of core at the
DC current level.
1.26 x Ný * Afe * æFe
L = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
1,000,000,000 * ML
1.26 * 1,600ý * 25ý * æFe
1.3H = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
1,000,000,000 * 140
æFe = 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
æFe = 100 at maximum IDC, knowing that when IDC was minimum, æFe could double.
With such a condition, we still need to get BDC less than about 0.8Tesla at
maximum IDC.
The choice of æFe = 100 minimum with maximum IDC allows some fine
adjustment of the core gap for best results, and we can allow æFe = 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 æFe = 200, N = 1,300t, then what is the
stack height for 25mm tongue?
1.26 * Ný * Afe * æFe
L = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
1,000,000,000 * ML
1.26 * 1,300ý * S * 25 * 200
16H = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 0.076 * S, therefore S = 210mm.
1,000,000,000 * 140
210mm is an entirely impractical stack height, so we can't use 25
tongue material. But if were able to double the turns, we would increase L
by 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
Rw = RL minimum / 30 = 1,417ê / 30 = 47ê
(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 = 16mm * 46mm, winding area = 12mm * 42mm = 504mmý.
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 = 504mmý / (0.47mm)ý =
= 2,281 turns,
N * TL * 0.0226êm 2,281 * 178mm * 0.0226êm
DC resistance = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 57 ê
1,000 * Dý 1,000 * (0.4mmý)
(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 * 38mm tongue E&I lams, window = 57mm * 19mm window
giving winding area = 53mm * 15mm = 795mmý
Oa dia. for wire, 2,281 turns = û (795 / 2,281) = 0.59mm, select wire
0.50 = 0.57mm oa, giving 2,440 turns.
Turn L = 212mm, DCR = 47 ê, just OK.
(7) Allowing æFe = 200, calculate inductance at IDC minimum.
1.26 * Ný * Afe * æFe 1.26 * 2,440ý * 38ý * 200
L = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 10.2H
1,000,000,000 * ML 1,000,000,000 * 212
(WARNING: dimensions in mm)
(8) Is L high enough? NO, so can stack be increased for required L?
Revised stack height for 16H = (16 / 10,2) * 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 * 38mm core, turn L = 212, DCR = 47ê,
With stack = 38mm * 62mm, turn L = 260mm, Rw DCR = 57ê, 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 æFe at min
IDC, so æFe = 100 at max IDC.
12.6 * æFe * N * IDC 12.6 * 100 * 2,281 * 0.30
BDC = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = 0.4Tesla
10,000 * ML 10,000 * 212
(10) Is BDC less than 0.8T at max IDC? YES, proceed to next check.
(11) Calculate VAC across choke = (0.67 * VDC) Vrms, 100Hz, = 0.67 * 425 =
= 284vrms.
22.6 * Vrms * 10,000 22.6 * 284 * 10,000
(12) Calculate BAC = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ = ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
Afe * N * F 100 * 32 * 1,892 * 100
(62.5?)
= 0.106Tesla
(13) Is BAC below 0.5Tesla ? YES, design is OK
(14) Check heat dissipation. Pd = Rw * IDCý = 57 * (0.3A)ý = 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 isn't 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.
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 ê.
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 * Vrms of the HT
winding at low IDC, and then, at the "critical maximum RL" where IDC is only
the bleed current, the VDC should be no more than the 1.0 * Vrms HT winding
voltage, and at twice the bleed current VDC should be 0.88 * HT Vrms. VDC sag
as iDC increases is due to the power transformer winding resistances, diode
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 ê, and series R = 100ê, it means that if IDC were to change from say
250 to 450mA in a class AB amp, the B+ would only fall from +425 to 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.
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