Transmitting AØ for long periods on 10m with the load
capacitor set for maximum C would result in very high grid-current
and almost no RF output. Under such a condition it might be possible
to overheat a grid. However, since most people tune an amplifier for
maximum output--and maximum output virtually coincides with normal
grid-current--very few people are likely to overheat a grid. Thus,
complex electronic grid-protection circuits are seemingly
unnecessary.
A disadvantage of electronic grid-protection circuits is that they
are not effective against the most common source of grid
damage--intermittent VHF parasitic oscillation.
During a major problem, the anode (plate) current meter and other
amplifier components can be subjected to a large current surge as the
HV filter capacitors discharge. The peak discharge current can exceed
1000a if a series resistor is not used to limit the short circuit
current that can be delivered by the HV filter capacitors. The
current limiting resistor is placed in series with the positive
output lead from the filter capacitors. A wire wound resistor with a
high length to diameter ratio works best. A 10 ohm, 10W wire wound
resistor is adequate for up to about 3kV & 1A. For higher
voltages, additional 10 ohm, 10W resistors can be added in series to
share the voltage drop during a glitch. Wire wound resistors with a
high length-to-diameter ratio are best for this type of service.
Since about 1985, Eimac® has recommended the use of a glitch
protection resistor in the anode supply circuit. Svetlana®
typically recommends using a 10 to 25 ohm glitch resistor.
A HV current limiting/glitch resistor may disintegrate during a major
glitch--so it should be given a wide berth with plenty of chassis
clearance. If the chassis clearance is minimal, its a good idea to
cover the chassis with electrical insulating tape. Glass-coated
(a.k.a. vitreous) wire wound resistors are the most suitable type of
resistor for this application. If a glass-coated resistor comes apart
during a major glitch, it won't be throwing chunks of shrapnel
around--like a less-expensive rectangular ceramic-cased resistor
often does. Metal-case power resistors should not be used in this
application. If a glass-coated glitch resistor is damaged during a
glitch, it should be replaced with two such resistors in series to
reduce the peak V-gradient per unit of length during a problem.
If the positive HV arcs to chassis ground--due to lint, a hapless
insect, a VHF parasitic oscillation, or moisture--the negative HV
circuit will try to spike to several kilovolts negative in the
typical 1500W amplifier. In the real world, this type of glitch is
not an uncommon occurrence. Anything that gets in the way of the
negative spike may be damaged. Since the grid-current meter is
normally connected between chassis ground and the negative HV
circuit, the meter can be exposed to kilovolts at hundreds of
amperes.
The easiest way to protect a current meter is to connect a silicon
rectifier diode across it, or across its shunt resistor. Usually,
only one diode {cathode band to meter negative} is needed in parallel
with a DC meter. In some circuits, it is best to use two diodes in
parallel [anode to cathode] with the meter movement to protect
against positive and negative surges.
It may take more than one diode to protect a meter shunt resistor. A
silicon diode begins to conduct at a forward voltage of about 0.5V.
To avoid affecting meter accuracy, the operating voltage per glitch
protection diode should not exceed 0.5V. For example, a 1 ohm shunt,
at a reading of 1A full-scale, has 1V across it. Thus, two protection
diodes in series would be needed to preserve meter accuracy.
Similarly, if the shunt resistor for a 1A full-scale meter is 1.5
ohm, the maximum shunt voltage is 1.5V--so three diodes are
needed.
Glitch protection diodes should not be petite. Big, ugly diodes with
a peak current rating of 200a or more are best. Smaller diodes--and
the meter they were supposed to be protecting--can be destroyed
during a glitch. Suitable glitch protection diodes are 1N5400 (50PIV)
to 1N5408 (1000PIV). In this application, PIV is not important. The
1N5400 family of diodes is rated at 200a for 8.3mS.
During an extremely high current surge, a glitch protection diode may
short out--and by so doing protect the precious parts. Replacing a
shorted protection diode instead of a kaput meter is almost fun.
To reduce the chance of the negative HV circuit spiking to several
kilovolts, connect a string of glitch protection diodes from the
negative terminal on the HV filter capacitor to chassis. At 200a,
each diode will limit the surge voltage across it to about 1.5v.
Typically, three diodes are needed--thusly limiting the negative
spike to about 4.5 volts. Diode polarity is: cathode band toward the
negative HV. With one simple wiring change, the same string of diodes
can also protect the grid I meter and the anode I meter. This dual
protection technique is incorporated into the Adjustable Electronic
Cathode Bias Switch on Figure 7.
A HV arc can destroy an indirectly-heated cathode tube. Here's how
it happens: In some amplifiers, one side of the filament/heater is
grounded. The cathode is connected to the negative HV circuit. If the
negative HV spikes to several kilovolts, the cathode will often arc
to the grounded heater. At a minimum, this breaks down the insulation
between the heater and the cathode. Sometimes the heater wire burns
out--and sometimes the cathode arcs to the grounded grid. Either way
the tube is kaput.
Grounding one side of the heater is an invitation for
cathode-to-filament breakdown. Instead, let both heater wires float.
If the heater is fed through an c.40micro H bifilar RFC, one side of
the heater can be wired to the cathode. Even though this arrangement
can not protect against cathode-to-grid breakdown, it assures that
the voltage between the filament and the cathode is unlikely to rise
to dangerous levels.
Manufactured amplifiers typically use a safety device to
automatically short the +HV supply to chassis ground when the output
section cover is removed. If the cover is removed before the HV
filter capacitors have discharged, the resulting positive HV to
ground short can damage the amplifier. In most g-g amplifiers, the
only DC current path between the negative HV circuit and chassis
ground is the grid-current meter and its shunt resistor. Even if the
remaining charge in the HV filter capacitors is only 200V when the
short from positive to ground occurs, without glitch protection
diodes, the entire 200V appears across the grid-current meter shunt
and the grid-current meter. Many potentially-fatal amperes can flow
into the grid meter as the HV filter capacitors finish discharging.
If the amplifier is accidentally switched on with the cover removed,
rectifier diodes are a common casualty.
Automatic-shorting safety devices are not only dangerous to amplifier
components, they can be dangerous to operators. It is dangerous to
assume that an amplifier is safe to work on because it contains a
safety device. Even though an amplifier's HV supply is shorted, if
the amplifier is plugged in, its electric-mains circuitry is still
alive and potentially fatal. Amplifiers are inherently dangerous.
They should not be worked on casually--even if they have so-called
safety devices.
The safest quick method of discharging HV filter capacitors is
through a paralleled pair of wire wound resistors. The resistors
limit the discharge current to a safe amount. In the unlikely event
that one resistor opens, the remaining resistor will do the job. For
the average 1500W amplifier, a paralleled pair of non-adjustable 1k
ohm to 5k ohm, 50W resistors will do the job. Its always a good
idea to check the anode supply voltmeter before putting your hands
inside an amplifier.
Fuses have current and voltage ratings. For a fuse, the real test
is opening safely--not operating without opening during normal
operation. A fuse's maximum voltage rating is important. In some
amplifiers, ordinary 250V 3AG fuses are casually used in circuits
where they may be required to interrupt several kilovolts. Examples
are the anode or cathode circuit. All's well until a problem occurs.
When a 250V fuse attempts to interrupt a potentially-damaging flow of
current in such circuits, the frangible link inside the fuse parts as
it should. However, due to the available voltage, when the link
melts, a metal vapour arc forms in its place. Metal vapour arcs
typically have a voltage drop of around 20V--so the unsafe current
will continue to flow in the circuit. At some point, the fuse will
eventually explode--usually after serious damage has been done to
other components.
During a glitch, circuits which normally carry low voltages can spike
to several kilovolts. For example, cathode circuits normally see a
maximum voltage of 30V to 100V. Thus, it might seem appropriate to
use a 250V fuse to protect the cathode from excessive current.
However, when a glitch occurs, several kilovolts can appear across
anything that attempts to interrupt the flow of cathode current. The
safest place to use ordinary 250V fuses is in the primaries of
transformers.
There are basically two types of DC filters: inductor-input /
capacitor-output, and capacitor. Each type of filter has advantages
and trade-offs.
Capacitor filters have good transient response. Since no
inductor and resonating capacitor are used, the capacitor filter is
simple to build, compact, cost-effective, requires no tuning and it
is lightweight. The main disadvantage of a capacitor filter is that
the capacitor is charged only during a small fraction of the waveform
supplied by the transformer. No charging current flows until the
instantaneous output voltage from the rectifiers exceeds the
instantaneous voltage on the filter capacitor. This means that the
transformer is either not loaded or severely loaded at different
times during each cycle.
For example, with a electric-mains frequency of 60Hz, the duration of
a half cycle is 8.333mS. Under load, using a capacitor filter, the
capacitor charging time per half-cycle is typically only about 1mS
out of the 8.333mS. This means that the ratio between output current
and peak charging current can be 8 to 1. To combat I^2 R loss in a
capacitor filter power supply, the transformer, all circuitry in the
primary (including the electric-mains) and the filter capacitor
should have low resistance. Capacitor filters are not appropriate for
use with older-design transformers that were intended for use with
inductor filters. Typically, such transformers have high winding
resistance.
Inductor-input / capacitor filters can be of the resonated
type or the non-resonated type. A non-resonated inductor tries to
maintain a constant DC-current despite changes in the load current.
This is the nature of any inductor. It always tries to maintain
constant current by temporarily increasing the output voltage when
the load current decreases suddenly, or by decreasing the output
voltage when the load current increases suddenly.
When a conventional voltmeter is used to monitor the output voltage
from a non-resonated inductor / capacitor filter power supply, the
transient unregulation characteristic will usually not be detected
because of the damped response in the meter movement. If a DC
oscilloscope is used to monitor the output voltage while a string of
caround 5 WPM CW dashes are sent, the instantaneous output voltage
swings can be easily observed. On make, the output voltage spikes
downward. On break, the output voltage spikes upward. The amplitude
and width of the spike depends on how much filter capacitance is used
after the inductor and on the change in current. Upward and downward
voltage spikes of more than ±50% are possible during a sudden
load change on a non-resonated inductor / capacitor filter power
supply.
The resonant DC filter maintains a fairly constant output voltage
during rapid or slow changes in current demand--provided that a
minimum current passes through the inductor. This minimum current can
be the zero-signal anode-current [a.k.a. 'idling current'] of the
tube itself. The inductor is resonated with a parallel capacitor. In
actual practice, the value of capacitance used--as well as the
bleeder resistance--is that which produces satisfactory voltage
regulation. The resulting resonant frequency is usually slightly
higher than double the frequency of the electric-mains. Resonant L/C
pairs are available from Peter W. Dahl, Inc.
DC filter inductors come in two types, fixed-inductance and
swinging-inductance. A swinging-inductor changes its inductance
according to the current that is passing through it. Obviously, a
swinging inductor can not stay tuned correctly with changes in
current. Therefore, resonated-inductor filters can only use a
fixed inductor.
The disadvantages of a resonant inductor filter are:
The advantages of a resonant-inductor DC filter are:
The resonant filter is used extensively by commercial and military amplifier manufacturers. Since a resonant filter demands much less peak power from the electric-mains than a capacitor filter demands, for 120V operation, where available power is typically much more limited than with 240V operation, a resonant filter is clearly the best choice. The resonant filter is also the best choice for high duty-cycle modes such as RTTY, FM or AM.
Transformers are available in two basic types: E-I (conventional)
core and toroidal core. The E-I core is made from a stack of thin
E-shaped and I-shaped iron plates. When placed together they form a
rectangle with two windows for the windings. A stack of E-I
rectangles make the completed core. The toroidal core is made from a
continuous tape of grain-oriented material that contains iron and
silicon plus other elements that increase the permeability of the
core and decrease loss. This core material is known as Hipersil.
Westinghouse Corp. was the original patent and copyright holder.
Their patent expired decades ago. There are different grades of
Hipersil tape. Grade 5 has the highest performance. Grade 22 has the
lowest performance.
Higher permeability means that fewer turns are needed to achieve the
required inductance in each winding. This means that larger diameter
wire can be used. The end result is a transformer with low resistance
and high efficiency. The Hipersil core is so efficient that the
principal loss factor is the resistive loss in the copper wire.
Hipersil core transformers are capable of producing extremely high
peak currents. Thus, the Hipersil core transformer is ideally suited
for capacitor filter power supplies.
It is difficult and time-consuming to thread a continuous tape core
through the completed transformer windings--so someone came up with a
faster way of uniting the core with the windings. Here's how it's
done: The tape is wound on a form of the appropriate dimensions. The
tape is spot welded together, removed from the form, and annealed at
about 700 degrees C to relieve internal stresses. After cooling, the
core is varnished and dried. Then the core is cut in half with a
machine that makes a precise square cut. The faces of the cut are
then polished flat. Thus, the halves of the core can slip into the
completed windings, contact each other closely--restoring nearly
perfect magnetic coupling between the halves of the core. The matched
halves of the core are marked so that they can not be inadvertently
mixed up with other core halves. The reunited halves of the core are
held together tightly by steel bands like those used for binding
heavy cartons and crates. If future access to the primary and
secondary windings is needed, a Hipersil transformer can be
disassembled by cutting the steel bands and removing the core
halves.
Hipersil is no longer the most efficient type of core material. The
new amorphous core transformer is starting to come into use by
electric utilities. An amorphous core transformer is so efficient
that if the secondary is unloaded and the primary is disconnected
from the electric-mains, the collapsing magnetic field generates a
voltage spike that can destroy the transformer. To avoid this
problem, one winding is paralleled with a suitable voltage surge
absorber.
Transformers are commonly rated in maximum "volt-amperes" [VA].
Maximum VA are roughly equal to maximum RMS watts when the rated RMS
current is flowing in each transformer winding and the transformer is
operated from the rated input voltage at the design frequency. If the
electric-mains voltage is reduced, the VA capability of the
transformer decreases.
For SSB and CW operation, a lighter transformer may do the job just
as well as a much heavier and more costly transformer. Manufactured
1500W amplifiers typically use a HV transformer with a continuous
capability of roughly 600W--or VA. Such transformers are completely
satisfactory for normal SSB operation. Such transformers are also
capable of handling brief FM and RTTY transmissions--provided that
the lower voltage tap is used.
If a power supply's DC output voltage drops more than about 10% under
modulation, it's a fairly safe assumption that a more capable
transformer is needed. Of course, not using enough filter capacitance
or excessive electric-mains resistance can also cause poor
regulation.
Increasing the current in any conductor causes a square-law
increase in the amount of power dissipated in the conductor. Since
P=I^2 x R, doubling the current causes a 4 times increase in
dissipation. This is an especially important consideration with
transformers because they have considerable difficulty dissipating
the heat that is generated deep inside their windings. This problem
is compounded because copper has a positive, resistance versus
temperature, coefficient. Thus, as the copper heats up, its
resistance increases--which increases the dissipation--which
increases the resistance, et cetera. This can lead to thermal runaway
and transformer failure.
If a transformer has a secondary rating of 1A RMS, it means 1A with a
resistive load. If connected to a rectifier and DC filter, the 1A
rating does not necessarily apply. For example, if a fullwave-bridge
rectifier, resonant filter circuit is used, the RMS current rating
can be multiplied by at least 1.2. The DC output voltage will be
about 0.85 times the RMS voltage. If a fullwave-bridge rectifier,
capacitor-filter circuit is used, the loaded DC output voltage will
be about 1.3 times the RMS voltage. A 30% increase in voltage sounds
good, but obviously you don't get something for nothing. The
trade-off for the increase in voltage is a decrease in current
capability. The high peak current demanded by the capacitor filter
translates into a substantive current capability decrease. .
Any formula for converting a transformer's RMS current rating to a DC
output current rating is bound to be problematic due to the large
number of variables. Here's a rule-of-thumb that is fairly accurate.
If, after about an hour of typical operation, the outside of the
transformer is uncomfortably hot for one's thumb, the internal parts
of the transformer are probably deteriorating. Reducing the average
load current slightly will greatly reduce transformer heating because
of the square-law relationship between current and power dissipation.
For example, reducing the current by 30% will reduce winding
dissipation by about 50%.
There is a simple, reasonably accurate, 2-step approximation for
determining the safe SSB, maximum current rating for a specific
transformer for use with a capacitor filter and a full wave bridge
rectifier. A slightly different approximation is used for a full wave
voltage-doubler. These approximations are based on the DC resistance
and the AC-voltage of the transformer's secondary winding. These
approximations are useful when shopping around surplus stores or swap
meets. All that's needed is an ohm-meter and a clip-lead. The
clip-lead is used to short the primary of the transformer. This
dampens the inductive voltage spike that occurs when the ohm-meter is
disconnected.
The fullwave-bridge, capacitor filter approximations are: Multiply
the secondary winding resistance by 70 to find the minimum
intermittent load resistance that can be placed on the power supply.
To find the DC output voltage under load, multiply the secondary
RMS-voltage by 1.3. To find the safe intermittent current rating for
SSB service, use Ohm's law and divide the output voltage by the
minimum load resistance. For a more accurate evaluation, use the
appropriate graphs in this book.
For example, a 2000V RMS secondary winding has a DC-resistance of 60
ohms. A full wave bridge rectifier, capacitor filter, circuit will be
used. The safe, minimum, intermittent load resistance is
approximately 70 x 60 ohm = 4200 ohms. The approximate voltage
delivered under load would be 1.3 x 2000V = 2600V DC. Thus, the
maximum intermittent load current is 2600V ÷ 4200 ohms =
0.62a.
Another approximation can be used to find the amount of filter
capacitance needed. The approximation is 50,000 divided by the
minimum load resistance. In the above example this is 50,000 ÷
4200 = 12micro F.
For a full wave voltage-doubler, capacitor filter, power supply, the
SSB-service approximations are: Minimum intermittent DC-load
resistance equals 300 times the winding resistance; DC-output
voltage, under load, equals 2.5 times the secondary RMS voltage.
For example: A 1000VRMS transformer has a winding resistance of 10
ohms, the minimum load resistance for full wave doubler operation
would be 300 x 10 ohms=3000 ohms and the output voltage would be 2.5
x 1000V=2500V. The maximum intermittent load current is 2500V ÷
3000 ohms = 0.83A.
The amount of filter capacitance needed for each half of the full
wave voltage-doubler circuit is approximately 200,000 divided by the
minimum load resistance. In the above example each of the two
capacitors should have a minimum of 200,000 ÷ 3000 = 67micro
F.
There is more to transformer performance than secondary resistance.
If a Hipersil® core is used, core loss is minimal and the maximum
intermittent power capability increases. Primary resistance is
another factor to consider since it is effectively in series with the
electric-mains resistance. Electric-mains resistance can cause a
voltage drop problem if the amplifier is a fair distance from the
service entrance box and you are using a capacitor filter power
supply. One solution is to use larger diameter wires than the
electric code requires. Another solution is to install the power
supply near the service box and bring the HV DC to the amplifier.
END OF PART 2