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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.

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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

understood....

Fig 1.

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

supply.

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

regulation.

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.

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.

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.

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

stages.

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.

The CRC 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

100mV, 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 = 5,035

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,

SEUL 22W,

SE35 watt,

Quad II power amp mods,

Leak amp mods,

Dynaco ST70 mods,

Preamps2,

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.

Regulators

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

regulator.

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,

sheets3,4&8'.

Fig 3.

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.

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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