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Ever wonder how a
T/R switch works?
Don’t use it with a
solid state
receiver!
Lets assume a 50
ohm load and 700
watts, which would
have typically been
maximum legal power
back when this
switch was made. The
peak voltage at pin
1 of the 6BL7 on the
first cycle would
have been
approximately 1.414
times the sqrt of
50*700, or 265
volts. Worse case
when the
transmission line
has a zero impedance
dc path, this
voltage would charge
C2 negative at pin
1, and positive
towards J2 and J3.
This would take
several RF cycles.
Grid current
would, after C2
stabilizes at
maximum charge level, be
determined by grid
leak R2. R2 limits
charging current
into C7 as
the resistor-capacitor network
R4 and C6 is charged negative.
Maximum steady state
grid current on pin
1 would be the
current through R2
and R3, or about
177/1,056,000 = 160 uA. Peak current
would be fairly
high, dozens of mA, but
also very
brief. Peak current
(assuming a square
rise on the input
envelope as worse
case) would be the
saturated grid-to-cathode
current of the tube. It would be so
brief that
tube grid damage would be
very unlikely. The
grid would not have
time to heat before
C2 and C7 started to
charge.
After 1 or 2
milliseconds, the delay
time set
primarily by R4 and
C8, pin 4 of the
second section would
go negative. Negative voltage
would be roughly 150
volts, the same as
the negative voltage
at pin 1 caused by
C2 charging negative
on the pin 1 side.
This further
isolates the
receiver, although
it takes a brief
time to do so.
My spice model
uses a 6SN7, the
closest low-mu
triode to the 6BL7 I
could find spice
data for.
Voltage at C8 (point
A) below:
RF voltage at R6
shown below:
This voltage would
be reduced by the
step-down ratio of
T2 in the actual
switch.
The actual
voltage at the start
would be less,
with the T/R tube
limiting most of the
power. Peak power to
the receiver would
be several watts
maximum for less
than one
millisecond.
This most
accurately shows
the time before the
tube cuts off.
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