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Definition of linear power supplies.
Fig 1. Basic wave forms in rectifier circuits.
Single phase house wiring, ac waveform basics, filling the bath with water,
diode resistance, ripple voltages C vs Idc.
Cap ripple I and V ratings, ac to dc conversion ratios, doubler rectifiers.
Fig 2. Schematic of 8585 amp power supply used as example for PS calculations.
Minimum C value for reservoir C1 input cap, C reactance, Ripple voltage calcs,
peak charge currents, charge current limiting,
CRC and CLC filters, R and choke values, LC resonance, choke reactance,
LC damping resistance, CT cap values, need for chokes. DC heater supplies,
B+ regulators.
CRCRC and CLCLC filters.
Fig 3. Schematic for solid state regulator for screen supplies in 300 watt amp.
Send me your SMPS schematics for B+ supplies.

I get many questions about power supplies. By the words 'power supply' it is assumed
the supply is a "linear" type; this includes a mains power transformer, diode rectifiers,
capacitors, chokes, possibly protection circuitry and regulators. I don't build SMPS,
ie, switch mode power supplies.

In tube amps there is ac power required for ac power to heater circuits without any
rectification and such simple supplies are covered under the section for building a
power transformer.

For DC supplies, I should begin to say that the basic wave forms should be well
Fig 1.
Basic rectifier
          wave forms.

Basic mains wave forms. The above sketch shows the basic ac wave forms you
should understand where ac voltage levels vary positively and negatively when
referenced to 0V, the earth or ground reference voltage potential.
There are two wires coming into your house from the mains.
One is black, and called the "neutral" wire and is connected to ground at the
house circuit distribution board via an earthing to copper water pipes or a copper
clad stake buried in the ground and the voltage on the black is almost zero volts
in reference to earth.
The green yellow insulated wire in the 3 wire cables around a house are all joined
to the water pipe or stake connection.
The other wire is called the "active" because its voltage is moves to + 340V peak to
-340V peak at a rate of 50 Hz  and the graph of such waves is shown above as
approximate sine waves.
The active and neutral wires are connected to a circuit breakers or fuses and then
to the 3 wire cables for power and lighting.
Each wire in the cable has red insulation for active, black insulation for neutral and
green+yellow for the earth wire in Australia. 
Appliances which require  their cases to be connected to earth directly  can be
accommodated such as washing machines, but the energy carrying circuit is via
the red and black wires.
In the US the mains active is about 110Vrms, and has F = 60Hz.

In Fig 1 the top waveform X is the incoming mains single phase of ac wave
applied to the transformer primary.
The single secondary shown at left shows the wave form X also occurring.
The top left transformer is supplying AC power only to the load resistance
which could be a heater filament in a tube. No rectified dc currents flow, only ac.
It is impossible to power signal circuits with AC since the ac signal would swamp
any signal we tried to have.

Rectified wave forms,
The middle waveform shows the ac wave X shown again, but with the ripple
wave form that appears at the top of RL1, and C1. When the positive going voltages
of the ac wave go higher than the voltage in the cap C1, the diode can conduct current
in the direction of the "arrow" and the the cap is charged up to the peaks shown in the
ripple voltage wave. But no sooner does the cap get charged up and the ac wave
potential reduces and travels negatively, and the cap tends to discharge its store of
energy through RL1 much more slowly than the ac wave goes negative, so then the
ac wave voltage is less than the cap voltage and current cannot flow in the diode in
the opposite direction of the arrow, so while the ac voltage is negative the cap voltage
stays relatively positive with respect to 0V.

Bath water?
Rectifying is the converting of alternating voltage to a single polarity voltage and is
like a like a guy filling a bath with water by tipping a bucket full in at each positive
wave crest, but the bath is losing a steady flow of water out the plug hole as he fills
the bath. The water running out is like the Resistive Load connected to every power
The average bath water level is like the dc voltage level at half way between the peaks
and troughs of the ripple wave form. And so we have a dc voltage level in C1, but there
is small ac wave as shown also superimposed upon the dc level. The flow in RL1 is
mainly a DC flow, but because some ripple voltage is present at the top of the capacitor,
there is some small ac ripple current in RL1, and its frequency is the same as the ac wave
form X, or mains frequency of 50 or 60 Hz. The ripple voltage contains many harmonics
of the mains basic frequency. The current flow in the diode is shown in a hatched wave
below positive peaks in the ac voltage wave. The current only flows in the diode for a
small fraction of the ac wave form; the current flow is like bucket fulls of water being
tipped into the bath, and the peak charging current into the cap through the diode must
be higher than the ripple current measured with an rms meter, since the input charge
current x time must transfer the same energy into the cap as flows out of the cap into
the load RL in the form of Vdc x current x time. The set up as shown in the middle
transformer with R1, C1 is a half wave rectifier, since only the positive going 1/2 of
each the ac wave is converted to a DC flow.

The lowest set of waves show wave X and wave Y with the ripple wave at C2, R2.
In the transformer at bottom left the secondary has two windings arranged so equal turns
exist each side of where the two windings join, which is called a centre tap. Each winding
has the same turns as the half wave rectifier transformer winding. Wherever you have a
winding with a CT taken to 0V, the ac signals at each free end are of opposite phase
and 180 degrees out of phase with each other. The result is that "balanced output
voltages" exist at each free end of the two windings. There are also two diodes, and as
each wave goes positive there are alternating XYXYXY charging pulses at twice the
mains fundamental frequency of 50 or 60Hz. This arrangement is called a full wave
rectifier, and is very common in tube amps which use a tube rectifier which contains
two diodes with a commoned cathode. When silicon diodes were invented, bridge
rectifiers and voltage doubler arrangements which are shown in textbooks were rapidly
adopted because they offered much greater efficiency, lower cost and far better voltage

Diode resistance,
Tube rectifiers have considerable series resistance often above 50 ohms when conducting
current and thus dissipate heat during their function, so the tube gets quite hot as a result.
There are strict limitations on the capacitance value in uF so that peak currents do not
exceed the cathode current ability. Silicon diodes have very low "on" resistance of only
about 1 ohm and a 1N5408 can easily take 3 amps or more than 10 times the current
rating of a tube rectifier. So Si diodes tend to run cool because the heat loss is low.
Heat =
I squared x R. Because Si diode "on" resistance is so low the power supply
output voltage regulation with load current change is far better than with any tubed
rectifier power supply, and the larger current ability allows for large value electrolytics
to be used and the peak charging currents are then limited by the winding resistances
and not the diode "on" resistance.
Because peak charge currents in silicon diodes are much larger than tube rectifiers some
say there is a noise problem since power supply noise can all too easily find its way into
earth paths and feedback wires by magnetic induction or other leakage or voltage
generation in low impedance earth buss wires.

I have found that with careful design and wire layouts such noise problems never arise.

If we assume that R1 = R2, and that C1 = C2, and that the ac voltage in the half wave
rectifier winding is the same as the voltages in each half of the balanced winding, then
the dc output voltage will be slightly higher with the balanced set up because the rate
of discharge of C2 is about the same as C1, but C2 is charged twice as often as C1,
so in fact the dc voltage in C2 is slightly higher, and the ripple voltage is about 1/2
the value of that across C1. In other words, the full wave rectifier is more efficient
than the half wave rectifier.

Ripple voltage and output dc voltage vs C and Idc.
In all the above rectifiers, the higher the value of C, the less ripple voltage you get for
a given dc current output, and the closer the dc voltage output becomes to the peak
ac voltage at the winding. Or you an say the lower the dc current output for a given
value of C, the less ripple voltage is present, and the higher the dc voltage approaches
to the peak ac voltage from the winding.

The maximum dc voltage that can exist in a cap being charged by an ac wave is the
the peak voltage of the ac wave form when there is no load to drain out the voltage
in the cap. So all caps in the power supply should be able to easily withstand
1.41 x the Vrms of the HT winding. Thus where a 280Vrms secondary is used,
the actual Vrms could be +/- 10%, or between 252V and 308V due to mains voltages
variations; I have seen the mains here at 255Vrms on some days. Always design
the power supply to be able to cope with the HT winding being 10% higher than
the actual design centre value. Hence the rectified peak voltage with no load could
be 1.41 x 308 = +434V dc, so therefore caps should have a V rating well above 434V.
450V rated caps are easily available, but seriesed 250V or 350V rated caps would
be better.
Often the price of 470uF x 350V rated caps are much lower than 470uF x 450V rated
caps so using the 350V rated types in seriesed pairs with dividing resistors to equalize
the Vdc across each cap is not too expensive.

The mains ac wave is usually a sine wave, but harmonics do exist to make the mains
look like a triangular wave with flattened peaks. The harmonic voltages in the mains
supply are seldom more than 5% of the 50Hz or 60Hz wave and are usually all odd
numbered, mainly 3H and 5H. But for practical design purposes, the mains wave form
is considered to be a sine wave and the rms measurement of it is 240Vrms in Australia
and the peak voltage value of the wave crests is 1.414 x Vrms = 339.4V. 

With a given secondary winding the Vdc will reach a peak value of nearly 1.41 x Vrms.
As we drain more and more current to a load from the input cap the dc voltage tends to
drop so that in an average power supply, the conversion factor from Vrms to Vdc reduces
from a maximum of 1.414 with no load to about 1.35 with Si diodes. With tube diodes the
Vdc is often only just above the Vrms value of the transformer winding voltage.
To have 600mA at +480V with tube diodes I would have to use a full wave CT winding
for the B+ with the ac voltage at about 420V-0-420V and with at least three paralleled GZ34.
The tube rectifiers could all be replaced with just two seriesed pairs of IN5408, and
capacitors could have higher values and thus the ripple voltage would be much lower.
Half wave rectifiers have their place in all sorts of circuits where load current is low
and efficiency isn't a big problem such as deriving a grid bias voltage for output tubes.
Half wave rectifiers can use 2 diodes and two caps to make a voltage doubler so
alternatively charge one cap positively, the other negatively, and with one transformer
winding end taken to the connection of the two series caps, thus giving a dc voltage
output near twice the peak ac voltage of the winding. This voltage doubler rectifier
produces a ripple frequency same as a full wave rectifier but is not quite as efficient
as a full wave type, but with silicon diodes and small sized large value modern capacitors,
this doubler is much more efficient than any tubed rectifier arrangement. The voltage
doubler arrangement allows a more efficient transformer winding with only 1/2 the turns
and 1/2 the voltage of the bridge rectifier. Or 1/4 of the turns and 1/4 of the end to
end voltage of a CT winding used for a tube rectifier. So with a doubler used in
B+ supplies, and much lower winding voltages, control relays meant for mains
voltages may be used.

The design of the power supply isn't difficult if we follow a path through a series of
equations. I will base my mathematical processes upon an example of a power supply
for the 8585 amplifier :-
Fig 2.
8585 power supply
The above looks but it is merely repetition of basic simple ideas.

The principles in the explanation of the above can be
applied to any other tube amp supply.

B+ plate supply.
There are 8 x output tubes and each draws about 40mA of plate and screen
current with B+ at +480V. So we want 320mA dc at idle and and then we also
want enough dc for the driver / input stages, so the total idle working dc output
from the top of C16&C17 will be about 400mA. But during class AB operation we
must allow for the anode and screen currents to the output stages to nearly double,
but the increases in current demand with music is only ever temporary so
design B+ current supply = 1.5 x idle current, which is 600mA in this example.

A voltage doubler supply was chosen because it allowed me to wind the power
transformer with just one winding of thick wire of N turns, rather than have a
winding for a bridge rectifier with 2N turns of thinner wire, or a full wave winding
for two diodes with 4N turns of even thinner wire. The one winding of N turns can
more easily be fitted into the winding space than the higher turn windings because
with more turns there is more room needed for insulation.

The ratio of dcV/acV = about 2.65 for working circuits with doublers, so here

Secondary Vrms = Vdc / 2.65 = 480V / 2.65 = 181Vrms.

The RL that is effectively connected to the dc supply = Edc / ( 1.5 x idle Idc )
= 480V / 0.6A = 800 ohms.

I chose to use CLC filtered supply supply, ie, a capacitor input plus following single
LC filter to reduce ripple voltage to a low level. There is negligible winding resistance
losses in the choke and OPT primaries and voltage drop is less than 10V between the
C1 input reservoir cap and the output anodes. When higher resistance in chokes and
OPT windings exist, an adjustment of the HT secondary ac voltage must be made so
the wanted B+ appears at the anodes.
The natural regulation of a rectifier power supply depends on the reactive value of the
C being much lower than the RL value, both measured in ohms. Or if you like the higher
the C value, the better the regulation. Old time wisdom suggested that the C1 cap
charged by the diodes, the reservoir cap, have a reactance at the ripple frequency
of no more than 1/10 x RL.
Maximum reactance of capacitor, ZC, = RL / 10,
in this case ZC max = 800 / 10 = 80 ohms.

Reactance of a capacitor in ohms = ZC  =    1,000,000  
                                                                     6.28 x C x F
Where ZC = reactance, or impedance of C, measured in ohms,
1,000,000 is a constant for all equations,
6.28 = 2 x pye, a constant for all equations,
C = capacitance in Uf,
F = frequency involved.

Therfore  C =     1,000,000       =  20uF in this case.
                         6.28 x ZC x F
20 uF is a low amount of capacitance which has a high reactance and the ripple voltage
will be fairly high which may have been acceptable in 1955 because the electrolytic
caps were then not so reliable and amp makers often used quite high inductance choke
values for a CLC filter. The ratings for peak charging current of vacuum tube diodes
is quite low so in 1955 a cap value of 20uF for a 600mA supply may have been appropriate.
The cap values which are allowable after a tube rectifier for a given Idc are given in the
tube data sheets. Using a higher C1 value than allowed after a tube rectifier will soon
result result in wrecked tube.

But today we can do a lot "better" because modern electrolytic caps are available cheaply
and which have much higher ripple current ratings and low size and Si diodes allow
brutish peak charge currents up to 10 amps. So let us choose C16 and C17 as 470uF
caps which in series will make C1 = 235uF which will give less than 1/10 of the ripple
voltage of a 20 uF cap.

So for 235uF, and 100Hz ripple frequency for each cap in a doubler supply,
Capacitor reactance 
 ZC = 1,000,000 / ( 6.28 x 235 x 100 ) = 6.8 ohms.

C1 of the CLC supply consists of two 470uF caps in series. The doubler charges
each of the two caps so that a 50Hz ripple voltage frequency exists across each cap.
Because the two caps are in series the ripple voltage at the top of two caps must be
twice the mains F of 50Hz, or 100Hz because while one cap discharges energy into
RL, the other one is charged up. So we can consider that we have 235uF shunting
the 800 ohm RL and ripple F = 100 Hz, so therefore ZC that is shunting RL
= 1,000,000 / ( 6.28 x 235 x 100 ) = 6.8 ohms.

The ratio of ZC/RL is 6.8/800 = 0.0085, which is much smaller than 1/10.

Ripple voltage calculations.
If ZC/RL < 1/10, the next formula for Vripple, Vr, will work OK.
if ZC/RL > 1/10, the calculation of Vr will not be valid.

Ripple voltage, Vr  =   Idc x 2,200
                                        C uF

where Vr is in Vrms,  Idc = dc load current in Amps,
2,200 is a constant for 100Hz ripple, but must be 1,833 for 120Hz,
4,400 for 50Hz and 3,666 for 60Hz.
C is in uF shunting RL.

So in this case, Vr = 0.6 x 2200 / 235 = 5.6Vrms.

Now ripple current must be checked to make sure the ripple current is less than
the maximum rated ripple current for the capacitors at the wanted frequency.

Ripple current, Ir = Vr / ZC

So in this case Ir = 5.6V / 6.8ohms = 0.82Amps rms.

The 470 uF caps I bought for this amp have a have a ripple current rating = 5 Amps
at least. They have 5mm screws to terminate the wires to them and were made
for arduous conditions. Many 470uf caps have only a 2Amp rms rating, and in a
fault situation they could fail to a short circuit after overheating because because excess
ripple current and overheating is a real killer of electrolytic capacitors.
If double the capacitance was used, Vr would halve to 2.8Vrms, and ZC = 3.4 ohms,
so Ir would be 2.8 / 3.4 = 0.82Arms, or the same as with 470 uF, but the Ir could be
shared over two lots of paralleled caps, and so each cap would have 0.41Arms instead
of the full 0.82Arms.

On the other hand if C1 was say 20uF,  Vr = 66Vrms, and ZC = 80 ohms,
and Ir = 66 / 80 = 0.82Arms and we would need to find a reliable type of 20uF cap which
could put up with 0.82Arms, and ripple current ratings tend to fall as C gets lower,
so using 20uF electrolytics isn't a good idea unless we decided to use some polypropylene
motor start capacitors or paper+foil+oil caps which will be much larger than electros.
I don't like oil filled caps. But with low C1 values then I have a much larger ripple voltage
to filter away before connection to the amp.

Peak charge currents.

It must be remembered that the peak input charge current can be perhaps 10 times
higher than the dc current draining out from the cap if the power transformer and
mains supply has low winding resistance and source impedance, and we know silicon
diodes are less than 1 ohm when conducting, and we have large value electrolytics.
In the 8585, tube diodes could never be used unless I had paralleled about 4 x GZ34
and used a fullwave CT winding for HT.

Current limiting resistances.
In amps with a large amount of class A we can cheat a little to reduce the *peak* charge
current several times by using an R in series with the transformer winding and diodes.
Usually R = 4 x ZC is plenty. In this case  ZC = 6.8 ohms, so a 27ohm R would suffice
to reduce cap charging currents. One may measure the Vrms across such a resistance
but in fact the effective Vac is higher than measured because the wave form is not at
all like a since wave. The power rating of current limiting resistors should at least be
twice the calulated value using P = Vac squared / R in ohms. So the 27ohm R should
at least be 10W, and could be 4 parallel 100 ohm x 5W and siliconed to the chassis.
Aluminium clad R screw fixed are even better.

The resistance we add almost exactly mimics the action of more than one tube rectifier.
But we don't have to use a heater supply for a single/multiple tubed rectifier and nor
have a tube socket/s. There is much less overall wasted heat to be dealt with when
using series R because tube diodes need usually up to 15W of heater power alone
not counting the anode power. A couple of well rated resistors glued to the chassis
with silicone will better dissipate the much lower amount of heat involved.
If a fault occurs and the protection elsewhere fails, the resistor will fail by fusing open,
and a resistor is much easier and cheaper to replace than a transformer/and or
rectifier tube. The current limiting R will help prevent cap failure.

Where there is some mechanically caused mains transformer noise due to switching
currents in dc supplies vibrating the winding, the added series resistances act to
limit currents and reduce the transformer noise.

But with series R there is a less well regulated B+ and a 5-10% lower B+.
But peak charge currents and switching transient currents around the earth
paths would be much lower. The measured Vripple will not change although the
shape of the ripple wave will have more even up and down charge and discharge
strokes, so that the charging current time is about equal to the discharge time.
The series resistors can get hot and related to current squared x R, and there
is a high peak AC flow, but since the Iac peak is lowered, the dissipation is
then less in the winding resistances of the primary and secondary transformer
windings which can run cooler with the series R.

In the 8585 case I saw no need to use charge current limiting resistors in series
with high current diodes and caps, despite the low winding resistances of the
transformer I wound which is an 800 VA toroidal type. The 800 VA toroid took a couple
of days to wind using a hand held shuttle but at least I was able to get a transformer
which has an operating B at less than 0.9 Tesla which usually leads to silent transformer
 operation even with a rectifier connected. The winding resistance is 33% higher
than a transformer with the same wire size running at 1.2 Tesla.

I have now justified the use of 470 uF caps and we now have to proceed to filter
the B+ rail.

I have only ever used low value polypropylene well rated motor start capacitors in a
power supply for a class A SE amp. The 1kW rated transformer I wanted to use had
a HT winding with CT, 420V-0-420V, and would give +560Vdc at the 700mA required
for two 60 watt class A channels.
But I only wanted +470Vdc for the cathode biased 6550 and did not want to have
a hot running series resistance to lower the B+ by 90Vdc which meant PdR = 63 Watts.
So I used series 6Amp x 1,000V rated Si diodes to charge a pair of 60uF x 450V rated
polypropylene motor start caps each about 52mm dia x 100mm and in series with
220k x 2W R across each. Effective C = 30uF and Vripple = 53Vrms. The 30uF is
followed by a large 4H choke with winding R of 20 ohms which dissipated 10 watts
and then I had 6 x 470uF x 350V rated caps in series/parallel to get 705uF which gave
Vripple of 50mV, quite OK for the anode supply with CFB arrangement of 12 x 6550
in the two channels. I then had additional RC filtering to the screen supply and input
Noise at the outputs was less than 0.25mV of hum. Although ripple voltage across the
C1 30uF capacitor was high, there is no dissipation in the reactive elements of C or
following L, and I got the V drop I wanted. Pure class A amps don't need regulated power
supplies at all and their B+ voltage can be allowed to drop if there is a fault and without
causing damage.

Some may say polypropylene caps throughout a PSU will sound better but I have heard
no evidence to support the idea.

Rules for CRC and CLC filters
Wherever you have a CRC  or CLC input filter system,  for there to be any real benefit
to the circuit working ( and hence the sound ), the attenuation factor of the filter after
C1 should be greater than 10, so that if Vr = 5.6Vrms at C1 there should be less than
0.56Vrms at C2.

The following RC filter should have R equal to or greater than 10 times the C1 reactance,
and C2 is then the same minimum value as C1.

So if ZC1 = 80 ohms with 20 uF, R would have to be 800 ohms lest too much charge
current into C1 just takes the short cut into C2 and the load without much attenuation
of the Vr at C1.

800 ohms would cause a huge voltage drop with 0.6 amps so there would be not
enough voltage for the tubes to work with, and if we made the HT winding voltage
much higher to compensate for the voltage drop the power dissipated in the 800
ohms = 288 watts.
With a low value C1 and high current CRC filtering cannot be used
unless RL = 10 x R.

But where we had C1 = 235uF as in the above PS, we could have 10 x ZC = 68 ohms,
and the dcV drop
would only be 68 x 0.6 = 41V. It is still too much though, and this R consumes 24 watts.

However, let's explore the idea of CRC further. Both C1 and C2 can be as big as you
like to make using silicon diodes of sufficiently high current ratings and with current
limiting series R between the diodes and caps.

For the CRC to be efficient, C2 should have ZC = 1/10 x R at least, so that we get
a reduction of ripple at C2 of ten times. So if R = 68 ohms, C could be 235 uF or
anything above. With ZC1 = 6.8 ohms, R = 68 ohms, and ZC2  = 6.8 ohms ( 235uF ),
we can get a reduction of 5.6Vrms at C1 to 0.56vrms at C2. ( If C2 was 1,000 uF,
Vr =  0.14Vrms. ) The silicon diodes will be able to charge up C1 and C2 through
its series R without too much inrush current, if I had C1 = C2 = 1,000uF.

Let us say the the design aim must limit dissipation in any series filter R to be less
than 5% of total RL power. In the 8585 case, total power from PS to tube circuits
= 600mA x 480V = 288 watts, and 5% = 14.4 watts, so R = 14.4W / 0.6A = 24 ohms.
ZC = 1.6 ohms at 100Hz, so the attenuation factor for the RC2 after C1 = 1.6 / 24
= 0.066 Ripple at C1 would be 1.3Vrms and at C2 is 0.087Vrms and V drop across
R = 14.4Vdc and power lost = 8.64 watts, perhaps acceptable enough.
could be made to work if the C values are high enough, and the use of a
choke is entirely unnecessary.

An alternative to CRC is the CRCRC if we wanted much lower Vr at the filter output.
So for two R, R = 12 ohms each. to comply with dissipation limits in R of less than
5% of 288 watts to the tube circuits. Where there are three C in the supply, each
C should have a reactance = 1/10 x R = 1.2 ohms each. Therefore the minimum
C value = 1,333 uF.
There are two RC filters, each with attenuation factors of 0.1.
So two cascaded filters give an attenuation factor = 0.1 x 0.1 = 0.01.
The Vripple at C1 at 600mA dc = 0.99Vrms, and at C2 = 0.1V and at C3 = 0.01V.

The CRCRC filter has the advantage that there is no resonance at LF.
Regulation of the power supply for SE amps or mainly class A PP amps will be entirely
sufficient with the R in series from the rectifier because the supply current to the
tubes varies only very slightly with music programme.

For an amp where Idc = 150mA, and V dc = 480V, to get the above 0.1Vrms
ripple at C3, the C values could be 1/4 of those used for 600mA and C1 = C2 = C3
= 330 uF. The R values would be four times higher at 48 ohms for each R.

But with huge values for C1, inrush current will still be considerable just to charge up
the supply from OV. Even with 235uF for C1 without charge current limiting R a current
limiting R with relay shunting it is advisable as shown in the 8585 supply schematic.
The delay circuit which controls the relay shunting R1 is not shown here but is in the
pages on the 8585 amp. It allows the B+ to rise to 2/3 of its final value when R1 is in
series with the HT winding circuit within 4 seconds, and then the R1 is shunted to
allow the B+ to rise the last 33% but without causing a second huge inrush surge
current any higher than the initial surge. With such a "current managing" circuit in the
above power supply the mains fuse can be a low enough value to give real protection
and to prevent nuisance fuse blows. To minimize charge currents without losing B+
voltage headroom I often have only 235uF for C1, but sometimes have 470uF or much
larger for C2. 100 watt mono amps I built with a six pack of EL34 amp used 1,500 uF
caps from old ex-university test gear. These caps were 75mm dia x 150mm high,
450V rated. The C2 in all such power supplies where connected to an OPT should
have the same high ripple current rating as for C1.

The choke is a far better SERIES filter element than pure resistance.
( The caps are necessary SHUNT filter elements. )

With high value capacitances of many times what were used in the past the rules
for inductor reactance still apply. To be worth using inductances the impedance
relationships for CLC filters means that ZL > 10 x ZC1, and ZC2 < 1/10 ZL.
Or more simply, the L must have at least ten times ZC1 = ZC2.
With the 8585 with C1 = 235uF, it could be possible to have ZL = 68 ohms with
only L = 108mH, and C2 = 235uF, and we would equal the filter performance
of the CRC with the 235uF and 68 ohms R, but the voltage drop across the
L would be negligible since its dcr could be made easily to be less than 4 ohms.
We could have CLCLC with two 108mH and another 235uF and achieve
Vr = 0.056Vrms.

But resonance of the last LC section is at 32 Hz, way too high, so I would
not use
the double LC filter with such low L values.

Vripple at the CT of push-pull output transformers should be less than
and for SE amp output transformers less than 20mV.

The choke in a CLC filter is the only way to get low power losses in a series
filter element where C values are moderate. Its dc resistance can be less than
20 ohms, and its ac impedance at 100Hz can be perhaps thousands of ohms
and still not be too large or heavy.

So suppose we have C1 = C2 = 235uF with ZC = 6.8 ohms.
Vripple at C1 with 0.6Adc = 5.6V, and we want Vr < 0.1V at C2 for the 8585
PP amp, so the ripple attenuation factor required = 0.1 / 5.6 = 0.0178, so the

CLC ZL reactance value = ZC2 / atten factor = 6.8 / 0.0178 = 380 ohms.

L = ZL / 628 for 100Hz ripple, ( or ZL / 523 for 120Hz ) .
So for the 8585 supply,  L = 0.6Henrys is just enough.

A check on resonance must be done.

Frequency of resonance for any LC circuit,
Fo  =  
          square root of ( L x C )

where Fo is in Hz,
5,035 is a constant for all equations to work,
C is in uF, and L is in milliHenrys.
For 235uF and 0.6H, Fo = 13.4 Hz.

Fo should be less than 7Hz!
Therefore resonance governs the choice of L value which must be larger to
reduce Fo < 7Hz

Where C2 value is known, L = 
5,035 x 5,035
                                                     Fo x Fo x C
L in millihenrys, 5,035 is a constant, Fo in Hz, C in uF.

In the 8585 L = 2,200mH, or 2.2H for Fo = 7Hz.

I wound the L in the 8585 with a 40mm stack of 25mm tongue GOSS material
using 0.55 dia copper wire filling the bobbin and I got 2Henrys at 0.6A.

Inductor impedance in ohms, ZC =  6.28 x L x F
where ZC is reactance in ohms at the F, 6.28 = 2 x pye, a constant for all
equations, and L is in Henrys, and F is in Hz.

So for 2Henrys at 100Hz, ZL = 6.28 x 2.0 x 100 = 1,256 ohms.

The C chosen was 235uF ( see C18, C19 ), so the ZC = 6.8, so the attenuation
factor is ZC / ZL = 6.8 / 1,256 = 0.0054, (or 46dB.)

Ripple voltage at the CT = attenuation factor x ripple voltage at input
C = 0.0054 x 5.6 = 0.03Vrms which is a lot less than the 0.1Vrms aimed for.
In fact at idle, where B+ Idc is lower than the 600mA design condition,
Vr at the CT is more like 0.024V, but AB current demands and mains voltage
undulations make the Vr vary somewhat.

Power supply resonance
Since we have an LC filter, it will have a series resonance and the RL of 800
ohms we have is not a low enough R value to damp the resonance. For the power
supply to have a flat response without a peak at Fo RL must be = 1.41 ZC or ZL
( both equal reactance values at Fo ). This is something you need to remember for
all LC or CL low pass or high pass second order filters.
The LC filter is a low pass second order filter and behaves just like any other LC filter
you may use at RF or at AF in a speaker crossover. The frequency response of an
under-damped LC filter will be flat at a couple of octaves below the Fo, and then
a peak occurs at Fo, and then the response rapidly rolls off after Fo at an ultimate
rate of 12dB/octave. 

In this case for 235uF and 2H, Fo = 7.3 Hz.

ZL at 7.3Hz = 6.28 x 2.0 x 7.3 =  92 ohms.

The required damping  R to prevent any 7.3Hz oscillations when Idc flow rapidly
changes = 1.41 x ZL or ZC = 1.41 x 92 = 130 ohms.

But we have RL = 800 ohms, so there is very little damping of the LC supply filter.
However, the Fo is well below 10Hz, and the Q of such a circuit isn't high, and in
practice the filter works very well without giving huge B+ ripples at 7Hz when
transient changes from class A to AB occur. In fact there is yet more to consider.
The figure of 800 ohms being the RL connected to the CLC output is only the
dc current load.
In the 8585 there are 8 x 6550 tubes all in parallel. If the tubes were running in triode,
the dynamic anode resistance, Ra = 1,100 ohms each so with 8 tubes the Ra total
is only 137 ohms, so in fact the damping R component of LC filter would not be much
more than the critical damping value of 92 ohms and the LF peaking at Fo during
transients will thus be negligible. But the 8585 has a fixed screen supply, and
the Ra of each tube would be a lot higher than 1,100 ohms, so the actual R across
the C part of the LC filter could be even higher than the 800 ohms we have designed for.
Hence the need to keep Fo low to ensure it is well under the lowest audio frequency.
The 8585 displays very little resonant behaviour in its PS.

Thus slight and slow undulations of anode supply currents don't cause much
B+ voltage change. This amp works mainly in class A, and in fact rarely moves
into class AB and while in class A the "common mode" working of the push
pull output stage rejects nearly all ripple voltage, and slight B+ undulations.

So as long as Fo for LC filters is less than 10Hz, the amplifier and its music will
not be compromised by the power supply.

Additional filtering ideas, CLCLC for B+ tube amp supplies.
Above I showed that the resonance in a power supply LC filter would determine
the choke size. One could realize the design for 8585 with just a CLC filter with
Si diodes but what if someone wanted to use a tube rectifier?
The value for C1 would need to be kept below 150uF with four GZ34 in parallel
for +480V at 600mA. Then a CLCLC filter may be achievable with a low Fo < 7Hz
for the second LC filter. Resonance in the first LC filter could be higher since
any signals higher than 7Hz before the second filter are much attenuated.
So the 235uF and 2.2H for L2 and C3 would still need to be used to give a
good anchor for the CT, and freedom from resonance above 7Hz.
Let us try having C1 = C2 = 110uF ( two 220uF in series ).  ZL1 should be at
least 10 x ZC1 or greater = 10 x 14.5 ohms = 145 ohms, so L1 could be as
low as 0.23H for an attenuation factor of 0.1.
Vr at C1 =  12V, at C2 = 1.2V, and at C3 = 0.0059Vrms, or 5.9mV.  
The first LC filter could simply be the same as the second, but with C1
= 110uF simply to suit the tube rectifiers to give a final ripple voltage of
only 0.29mV.

For SE amplifiers, a low ripple voltage for the B+ is essential. SE pentode
amps have Ra which is high so the B+ ripple voltage does not cause much
hum in a speaker. But nobody builds SE pentode amps; nearly all SE amps are
triode and the Ra is usually low. So with a 300B with Ra of 800 ohms, and an
RL of say 5k, any Vr at the B+ supply is divided down by the effective RL and Ra so
that only 0.137 of the Vr appears at the anode and the rest is across the OPT primary.
So if there was 10mV of B+ Vr, then there is 1.37mV at the anode and 8.63 mV across
the primary of the OPT. If the OPT ZR = 5k : 8 ohms, then the TR = 25 : 1 so there
is 0.34mV at the speaker connection. If the speaker is a super sensitive LF horn or
even a large dia sensitive reflex box like an old Tannoy or Altec Lansing or JBL with
15" driver, 0.34 mV of hum may be heard. It is thus my only advice to spend enough
on inductance and capacitance in SE power supplies to keep the noise low, especially
since there is often no loop NFB used which will lower noise. SET amps often have
problems with 50Hz hum from poorly adjusted cathode heater centering potentiometers.
Using well filtered DC on the cathode circuits will cure the hum problem in 300B tubes
and other DH triodes.

CT capacitance should be high value for class AB.
The reason for having a large value of capacitance at the CT is to anchor the CT
with a low impedance to eliminate intermodulation distortions cause by low F
modulating amplitudes of higher F when the amp begins to work in class AB.
The C thus needs to be a high value to anchor the CT to ground by way of offering
a low impedance shunt path to signals between CT and 0V.
In some amps like Quad II the CT of the OPT was held down by only 16 uF without
any further filtering of the B+ there was 17Vrms of ripple at the OPT anode
winding CT! This was pretty dreadful engineering.
The ZC of 16 uf at 50 Hz = 200 ohms. The signal artifacts due to PS interactions
in Quad-II with the rest of the output stage are a quantity about equal to the tube
THD, and when such amps are forced to work with less than ideal low impedance
and low sensitivity modern speakers the PS noise gets into the signal path
and causes artifacts much higher than the thd of the tubes.
I would typically use 470uF even in a 20 watt PP amp so ZC is 6.8 ohms at 50Hz,
so the betterment factor new vs old = 29 times, but in renovations to Quad II I also
have 2 stage filtering, 100uF with Si diodes, R = 200 ohms, then 470uF at the CT,
so ripple is reduced from 17Vrms to 50mV, an improvement of 340 times.

Once you move away from using tube rectifiers and being limited by their poor
abilities, one then can enjoy better music.

The rest of the 8585 power supply will explain itself along the lines of the
general above principles if anyone wants to check it out.

There is only a choke needed in B+ supplies where the Idc is high.
It is possible to have B+ active regulation which also acts as good filtering but
it is much more prone to failure and is not necessary. In all other areas of the
B+ supply, RC filters can give enough ripple attenuation and good regulation
without resorting to solid state regulation or chokes. More complexity increases
the probability that something will go wrong. I do use some strings of zener
diodes for shunt regulating screen voltages and input stage rails, but that's
all except for my 300 watt amp screen supplies which do use a high voltage
transistor emitter follower regulator.

DC heater supply
In the case of the positive voltage dc heater supply, at the input cap C31, 4,700uF,
Vout is about +17V, I out = 1.2 amps, so RL = 17/1.2 = 14.1 ohms.
ZC33 = 0.34 ohms, so RL/ZC = 41.1 = a good ratio.
Vr = 1.2 x 2,200 / 4,700 = 0.56vrms, which is less than 4% of the Vdc.
The filter of R53, 6.4 ohms, and C33, 4,700 uF has an attenuation factor
of 0.34 / 6.4 = 0.053, so ripple at C33 = 30mV, and low enough, and only 9.2
watts is lost as heat in R53. We can afford such heat losses considering the total
heat losses in the amp, and a choke would have to be 10mH, and this would have to
be large and bulky and dissipate several watts. The alternative to CRC in dc heater
supplies is the use of 12V 3 pin fixed voltage regulators with a diode in the ground
lead to increase the output voltage to 12.6V. Most are only rated for 1A and you
may need to use more than 1 to supply input and driver tubes with DC. The input
voltage to chip regulators must be more than 3V above 12.6V ( or 6.3V if used )
for regulation to be guaranteed. With regs you only need C1 to get ripple down
to less than say 5% of the dc output voltage. A 0.1uF across the reg output is all that
is needed.

CRC filtering of the ripple of DC supplies must reduce the upper ripple harmonics
which are most likely to be capacitively injected into signals at cathodes due to the
capacitance between cathodes and heaters. Fortunately the upper harmonics are
increasingly attenuated by the large C values involved in CRC filters.

There are more power supply schematics of working amplifiers in the listed web
pages :-
300 watt mono blocs,
100 watt mono blocs,
5050 integrated,
SE35 watt,
Quad II power amp mods,
Leak amp mods,
Dynaco ST70 mods,

I hope the power supply detail shortage during previous years has been addressed.

Choke input filters are covered in my pages on power transformers and chokes.

I built a bench top regulated power supply about 10 years ago when I was learning.
It can provide +200V to +500V in 50V steps and has switched input voltages to the
series pass regulator using a 6BX6 gain pentode with 2 x 6080 pass element tubes.
The maximum output current is about 500mA and to avoid frying the tubes at low
voltages and high output currents the primary supply with CLC filtering can have
its C1 switched out to give an LC input network and a much lower B+ supply
ahead of the regulator.

This series reg is very reliable but I don't use it much so it has lasted well. It becomes
useful when I want to set up a test for some output tubes. It had a very real value  as
one of the many learning exercises I engaged myself with for 5 years before I started
spending all this time using a PC to explain to the world what I find to be valid
voltage/current management techniques.

It replaced my early attempt to build a regulator with BU208A series pass elements.
I fused several bjts in the learning process, and finally the tube regulator was far more
reliable for tests where something I did caused an ""oops"" event and some smoke.
The tubes could take the abuse with a smile, but the solid state just fried.

I have since perfected the BU208A/108A regulator for use for screen voltage regulators
as seen in my 300 watt amp pages, and I have learnt how to perfect the protection of
such circuits so that for the last 5 years I have yet to have lost a solid state high voltage
My favorite type of regulator for tube amp B+ rails is just to have the BU208A or BU108A
with an MJE340 driver transistor connected as a darlington pair to work as an emitter
follower. Since all tube amps have high voltages, as long as the base voltage applied to
the pass transistor can be kept steady, the output emitter voltage cannot sag more than
about 1.2V between no load and say 3 amps. Of course if there is say 50V across the
BU208A and a 3 amp flow, then you would have 3 x 50 = 150 watts of heat in the device
so the device will just fail to become a sullen short circuit. Its how solid state devices fail.
Tubes and resistors usually fail to an open circuit. Not bjts, unless they explode, which
isn't unusual, and then maybe they are open. When failing, they do so to protect a fuse.....
So, the rule for series regulators is to always have a well rated resistor in series so
that at the wanted current there is an equal voltage across the R and between collector
and emitter.

Thus equal idle power is dissipated in the R and pass device. As current increases,
V across the R increases, V across the device reduces, so the R gets hotter, the active
pass device gets cooler. When current doubles, the device is conducting, but has only a
volt across it, regulation stops and the output voltage is allowed to sag. But I always place
some diodes around the regulator to prevent excess output currents and backward flows
of currents which are truly sudden death for any bjt.

A good example of a solid state regulator is at my page '300watt amplifier power supplies,

Fig 3.
Schematic sheet
          4, 300W amp.
Notice the BU108A and MJE340 solid state screen voltage regulator in the centre of
the schematic. The two easily available discrete BJT devices are connected as a simple
darlington emitter follower pair. BU208 is ideal. Should excessive screen currents ever
flow, R5 will develop a voltage across it reducing the collector-emitter voltage to 1V
across the BU108, and it will stop regulation and allow the screen voltage to fall if any
screen decides to conduct way too much current in a fault condition. Excessive output
currents are also limited by R6, so that if too much emitter current should suddenly flow
in R6, then the 4 diodes from the top of C5 to MJE340 base will conduct and the base
voltage reduced quickly, and thus output voltage. The circuit is protected against reverse
flow of supply voltage applications by the diode from the output to collectors.
But under normal operation, the screen voltages held very steady between low and
clipping when dc screen currents vary considerably with signal levels.

The supply to the input stage is shunt regulated by the string of zeners near C1.

The parts shown on this schematic are all included on the 300watt amplifier chassis which
is separate to the power supply enclosure, connected to the amp chassis with umbilical
cable. If the cable is removed from the power supply with the amp turned on, the pins
of the plug of the cable carrying high Vdc levels become exposed and could make skin
contact so to minimize any risk of shock a fast B+ discharge path has been established
via R9 to 0V through the action of the relay if the 'red' cable is disconnected
from the PS.

Switch Mode Power Supply?
I do not make switch mode power supplies. I would like to use them since they would
save a lot of weight, but there are problems such as RF noise, reliability and complexity
which need to be addressed, and so far, I have yet to see any commercially or privately
made tube amps with high voltage B+ and low voltage heater dc derived from SMPS.
The reason probably is that if you have heavy output transformers in a tube amp one
may as well have a heavy power transformer and chokes.
The problems of having a switchmode supply producing +500V at 1 amp mean that the
RF energy is considerable and requires some careful engineering which probably would
end up being more costly to produce even though lighter. And all the heater voltages
also add to the problem.

I have gained C-tick accreditation from the Australian Communications Authority
for the linear power supplies I use in the amplifiers which I occasionally supply to
customers. I would have to address much more stringent and costly
regulatory requirements if I were to make my own SMPS.

But if any of you want to make a SMPS then please feel free to try.
Feel free to lend me your schematics,
because I have lent you all my schematics !!!

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