I get many
questions about power supplies. By the words 'power supply' it
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
rectification and such simple supplies are covered under the section for building a
supplies, I should begin to say that the basic wave forms should
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.
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.
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.
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
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
If we assume
that R1 = R2, and that C1 = C2, and that the ac voltage in the
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.
voltage and output dc voltage vs C
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.
dc voltage that can exist in a cap being charged by an ac wave
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
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
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
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
equations. I will base my mathematical processes upon an example of a power supply
for the 8585 amplifier :-
The above looks but it is merely repetition of basic simple ideas.
principles in the explanation of the
above can be
applied to any other tube amp 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.
doubler supply was chosen because it allowed me to wind the
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.
ratio of dcV/acV = about 2.65
for working circuits with doublers, so here
Vrms = Vdc / 2.65 = 480V / 2.65 = 181Vrms.
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
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.
of a capacitor in ohms = ZC
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.
= 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
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,
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.
voltage, Vr = Idc x
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.
current must be checked to make sure the ripple current is less
the maximum rated ripple current for the capacitors at the wanted frequency.
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 =
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.
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.
is some mechanically caused mains transformer noise due to
currents in dc supplies vibrating the winding, the added series resistances act to
limit currents and reduce the transformer noise.
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
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 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.
reg is very reliable but I don't use it much so it has lasted
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.
my early attempt to build a regulator with BU208A series pass
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
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
idle power is dissipated in the R and pass device. As current
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.
example of a solid state regulator is at my page '300watt
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
shown on this schematic are all included on the 300watt
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.
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.