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.