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.

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 grounf 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 al 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 accomodated such as washing machines, but the energy carrying circuit is via the red and black wires.
In the US the mains active is 117Vrms, 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 waveX 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 +ve 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 50/60Hz.
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 usually above 50 ohms when conducting current and thus dissipate heat during their function, so the tube gets quite hot as a result, and there are limitations on what sized capacitor follows the rectifier 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
of current, about 10 times that of a tube rectifier. So they tend to run cool because the I squared x R power heat loss is low, and since their series resistance is so low the power supply output voltage regulation with load changes 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.
Beacuse peak charge currents in silicon diodes are much larger than tube rectifiers some say that is a 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, but 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 then the half wave.

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 dsign 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 rated caps would be better.

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 is seldom more than 5% of the 50Hz or 60Hz wave.
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 for the B+ of a tube amp 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 dcV reduces from a maximum of 1.4 with no load to about 1.35 when the design load is connected using well designed PS with Si diodes.
With tube diodes the Vdc is often only just above the Vrms value of the transformer ac rms 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 maybe at least a pair or quad of GZ34.

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 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, it is much more efficient than any tubed rectifier arrangement.
The voltage doubler arrangement allows a more efficient transformer winding with 1/2 the turns of the bridge rectifier and half the voltage, allowing control relays meant for mains use to be used. The doubler winding uses 1/4 of the turns needed
for the full wave CT winding so commonly used in the past.

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 of mine:-
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 betaeen the C1 input resevoir 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 resevoir 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 a fairly high figure, 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" than that because modern electrolytic caps are available cheaply which have high ripple current ratings and low size and Si diodes allow brutish peak charge currents up to 10 amps if we wish so let us choose C16 and C17 as 470uF caps which in series will make C1 = 235uF
which will give over ten times lower ripple voltage than 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, 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 excess ripple current 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 in oil caps which will be much larger than electros. But then 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 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 charge current several times
by using an R in series with the transformer winding and diodes. Usually R = 4 x ZC is plenty, so in this case  ZC = 6.8 ohms, so a 27ohm R would suffice to reduce cap charging currents.
One may say well why not just use a tube rectifier, or paralleled tube rectifiers?
The resistance we add almost exactly mimics the action of tube rectifiers. But we don't have to use a heater supply for a tubed rectifier/s, nor have a tube socket/s, and a couple of well rated resistors glued to the chassis with silicon to better dissipate their heat is still a better option, and if a fault occurs and the protection elsewhere fails, the resistors will fail by fusing open, and a couple of resistors are much easier and cheaper to replace than a transformer/and or rectifier tube. 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 scorching hot since their dissipation can be high 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 P and S windings and the series R tends to reduce heating in the transformer windings. And if a fault happens in the amp and a larger Idc flow occurs, the current limiting R will help prevent cap failure, because the R will fuse open. R is cheaper than C which is cheaper than transformer replacement.

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.

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 exercizes
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 favourite 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
cane 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 soid 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 dcV 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 HF 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.