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Let us see what Eimac says about HV glitch protection:
In cases where operation involves oxide cathode tube types
up to about 1500 watts dissipation, the addition of a series
resistor in the anode and/or screen grid high voltage lead
will limit peak currents and provide a means of dissipating energy
in the event of a tube or circuit arc.
This resistance should be as HIGH as is practical without=20
dissipating an excessive amount of power. Typically, a resistance
of 20-50 ohms is sufficient to protect the internal tube structure
in the event of an arc. It also serves as protection to the other
circuit components such as bypass capacitors, RF chokes, meters,
and diodes. For oxide cathode tubes a maximum of four Joules
total energy is permitted to be dissipated in an internal arc. An
amount in excess of four Joules will permanently damage either
the cathode or the fine wire grid structure during repetitive
arcing. In a case involving a 2000 volt power supply with a 2mfd
filter capacitor a total of four Joules is stored in the capacitor
alone, not including the energy resulting from FOLLOW-ON current
from the power supply. (Follow-on energy is the energy being delivered
from the power supply from the time the fault starts until the primary
C.B's/fuses trip/blow. This energy must be added to the energy stored
in the filter capacitor(s) in calculating the proper HV glitch =
resistor.)
Most power supplies employ much larger filter capacitors than 2 MFD,
resulting in even more stored energy. With no series resistor all this
energy is dissipated in the tube structure in the event of a tube arc.
Typically, there is about 50 volts across a tube arc and the addition of
a series resistor of 50 ohms will limit the peak fault current to 40 =
amperes.
With 1950 volts across the resistor and 50 volts across the tube arc,
less than 5% of the total energy is dissipated in the tube with 95% =
being
absorbed in the series resistor. Note that a MAJOR portion of the fault
energy may come from the FOLLOW-ON current if the primary C.B.'s/fuses
take excessive time to trip or the overload relay(s) take too much time =
to
open the primary contactor. Without protection, several low-energy arcs=20
may occur before serious degradation in tube performance is noted.
However, in cases where sufficient energy is involved, a single arc may
completely destroy the tube.
As a test for the amount of energy delivered in an arc, EIMAC suggests
that the power supply be short-circuit tested by causing an arc (in air)
from a short lead connected to the tube terminal to the surface of a
grounded sheet of 0.025mm (0.001in) thick aluminum foil. If total energy
delivered is less than four Joules, the hole burned in the foil will be
no greater than 3mm (0.120in) in diameter.
An alternative test which will verify tube protection is to =
short-circuit
test the power supply through a 6 inch length of 0.079mm diameter (#40
AWG) soft copper wire. If the total energy delivered is less than four
Joules the wire will remain intact. This test must be run at full =
operating
voltage and may be performed by using a vacuum switch or other suitable
high voltage relay to apply a short to the supply through the aluminum =
foil
or the copper wire.
Most tubes employing thoriated tungsten filaments are capable of =
withstanding
higher energy arcs than are the oxide cathode types. However, some of
the smaller types such as the 4CX1500A, 3-500Z, 5CX1500A, as well as
some larger high-mu triodes such as the 3CX3000A7, 3CX10000A7, or
3CX20000A7 while employing thoriated tungsten emitters, have very fine
wire grid and cathode structures and should be protected in much the
same manner as the smaller oxide cathode tubes.
The above design criteria and testing should take less time to perform
on a power supply than is spent measuring for proper anode cooling when
building up an amp. If the proper C.B's/fuses and glitch resistor are
selected, you will not even destroy your 6 inch #40 wire!
(((73)))
Phil, K5PC
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<DIV>Let us see what Eimac says about HV glitch protection:</DIV>
<DIV> </DIV>
<DIV>In cases where operation involves oxide cathode tube types</DIV>
<DIV>up to about 1500 watts dissipation, the addition of a series</DIV>
<DIV>resistor in the anode and/or screen grid high voltage lead</DIV>
<DIV>will limit peak currents and provide a means of dissipating =
energy</DIV>
<DIV>in the event of a tube or circuit arc.</DIV>
<DIV> </DIV>
<DIV>This resistance should be as HIGH as is practical without </DIV>
<DIV>dissipating an excessive amount of power. Typically, a =
resistance</DIV>
<DIV>of 20-50 ohms is sufficient to protect the internal tube =
structure</DIV>
<DIV>in the event of an arc. It also serves as protection to the =
other</DIV>
<DIV>circuit components such as bypass capacitors, RF chokes, =
meters,</DIV>
<DIV>and diodes. For oxide cathode tubes a maximum of four Joules</DIV>
<DIV>total energy is permitted to be dissipated in an internal arc. =
An</DIV>
<DIV>amount in excess of four Joules will permanently damage =
either</DIV>
<DIV>the cathode or the fine wire grid structure during repetitive</DIV>
<DIV>arcing. In a case involving a 2000 volt power supply with a=20
2mfd</DIV>
<DIV>filter capacitor a total of four Joules is stored in the =
capacitor</DIV>
<DIV>alone, not including the energy resulting from FOLLOW-ON =
current</DIV>
<DIV>from the power supply. (Follow-on energy is the energy being=20
delivered</DIV>
<DIV>from the power supply from the time the fault starts until the=20
primary</DIV>
<DIV>C.B's/fuses trip/blow. This energy must be added to the =
energy=20
stored</DIV>
<DIV>in the filter capacitor(s) in calculating the proper HV =
glitch=20
resistor.)</DIV>
<DIV> </DIV>
<DIV>Most power supplies employ much larger filter capacitors than 2 =
MFD,</DIV>
<DIV>resulting in even more stored energy. With no series resistor all=20
this</DIV>
<DIV>energy is dissipated in the tube structure in the event of a tube=20
arc.</DIV>
<DIV>Typically, there is about 50 volts across a tube arc and the =
addition=20
of</DIV>
<DIV>a series resistor of 50 ohms will limit the peak fault current to =
40=20
amperes.</DIV>
<DIV>With 1950 volts across the resistor and 50 volts across the tube =
arc,</DIV>
<DIV>less than 5% of the total energy is dissipated in the tube with 95% =
being</DIV>
<DIV>absorbed in the series resistor. Note that a MAJOR portion of the=20
fault</DIV>
<DIV>energy may come from the FOLLOW-ON current if the primary=20
C.B.'s/fuses</DIV>
<DIV>take excessive time to trip or the overload relay(s) take too much =
time=20
to</DIV>
<DIV>open the primary contactor. Without protection, several low-energy =
arcs=20
</DIV>
<DIV>may occur before serious degradation in tube performance is =
noted.</DIV>
<DIV>However, in cases where sufficient energy is involved, a single arc =
may</DIV>
<DIV>completely destroy the tube.</DIV>
<DIV> </DIV>
<DIV>As a test for the amount of energy delivered in an arc, EIMAC=20
suggests</DIV>
<DIV>that the power supply be short-circuit tested by causing an arc (in =
air)</DIV>
<DIV>from a short lead connected to the tube terminal to the surface of =
a</DIV>
<DIV>grounded sheet of 0.025mm (0.001in) thick aluminum foil. If total=20
energy</DIV>
<DIV>delivered is less than four Joules, the hole burned in the foil =
will=20
be</DIV>
<DIV>no greater than 3mm (0.120in) in diameter.</DIV>
<DIV>An alternative test which will verify tube protection is to=20
short-circuit</DIV>
<DIV>test the power supply through a 6 inch length of 0.079mm diameter=20
(#40</DIV>
<DIV>AWG) soft copper wire. If the total energy delivered is less than=20
four</DIV>
<DIV>Joules the wire will remain intact. This test must be run at full=20
operating</DIV>
<DIV>voltage and may be performed by using a vacuum switch or other=20
suitable</DIV>
<DIV>high voltage relay to apply a short to the supply through the =
aluminum=20
foil</DIV>
<DIV>or the copper wire.</DIV>
<DIV> </DIV>
<DIV>Most tubes employing thoriated tungsten filaments are capable of=20
withstanding</DIV>
<DIV>higher energy arcs than are the oxide cathode types. However, some =
of</DIV>
<DIV>the smaller types such as the 4CX1500A, 3-500Z, 5CX1500A, as well =
as</DIV>
<DIV>some larger high-mu triodes such as the 3CX3000A7, 3CX10000A7, =
or</DIV>
<DIV>3CX20000A7 while employing thoriated tungsten emitters, have very=20
fine</DIV>
<DIV>wire grid and cathode structures and should be protected in much =
the</DIV>
<DIV>same manner as the smaller oxide cathode tubes.</DIV>
<DIV> </DIV>
<DIV>The above design criteria and testing should take less time to=20
perform</DIV>
<DIV>on a power supply than is spent measuring for proper anode cooling=20
when</DIV>
<DIV>building up an amp. If the proper C.B's/fuses and glitch =
resistor=20
are</DIV>
<DIV>selected, you will not even destroy your 6 inch #40 wire!</DIV>
<DIV> </DIV>
<DIV>(((73)))</DIV>
<DIV>Phil, K5PC<BR></DIV></BODY></HTML>
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