W4TV posted a satisfactory explanation of why the circulating current
goes up with Q in a high powered output circuit. This is fundamental to
understanding the trade-offs in high power amplifier design. In VHF
cavity circuits, we are always trading bandwidth and Q, for circulating
current and losses, without increasing the output voltage across the 50
ohm output point. What this says is that matching networks, such as pi,
pi-L, loops in cavities, capacitive paddles, etc. are just transformers,
that effect the efficient transfer of RF power from an active device
(transistor or tube) to a fixed load of 50 ohms. Once a matching network
is built in metal or discrete components like capacitors & inductors
then it has a fixed transformation ratio, PERIOD. If the operating point
of the active device (called 'plate impedance' for tubes) varies, this
network is no longer optimal. Reduction of drive power into a linear
amplifier changes Rp. If the plate voltage is reduced the proper amount
(this is easily calculated) then the same network becomes optimal again.
If no change is made, however, the network may no longer be terminated
in the proper R. As Joe explained, the circulating current can rise
significantly. This can cause higher reactive (out of phase) voltages
across the components in the circuit loop that includes the series
inductance and the tuning capacitance, or in the walls of a cavity
amplifier and the seals and rings of a power tube. In other words,
stored energy goes up (Q). The amplifier may still operate, but
overheating or arcing may occur if it is pushed hard in this unloaded
condition. Efficiency also suffers, but we don't watch this when backing
off drive usually.
73
John
K5PRO
W4TV posted this:
"If you change the network by changing the plate load impedance
as you do when you run 1000 watts in the high voltage position
the VALUES of the capacitors must change since the value of the
inductor is fixed.
For example at 14 MHz:
3000 Ohms (Q=12) 6000 Ohms (Q=24)
C (tune) 42 pF 41 pF
L 3.36 uH 3.36 uH
C (load) 231 pF 393 pF
3000 Ohms is approximately 2800 V @ 800 mA (1500 W PEP out)
6000 Ohms is approximately 2800 V @ 360 mA (1000 W DC in)
In order to use the SAME output at the lower power, the tube
operating conditions must be changed to approximately 2000 V
@ 500 mA. Otherwise the circulating currents, losses, and
stress on the components increase significantly.
------
Circulating current increases with network Q - that's the
primary reason for the higher losses (I2*R losses in the
coil). If you use the same inductor in a 6000 Ohm to 50 Ohm
pi network than you used in a 3000 Ohm to 50 Ohm pi network,
the network Q and the circulating current will go up."
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