Why do VTTCs need huge grid feedback resistors?

[megabyte]

It seems even small VTTCs are designed with high-value grid feedback resistors with high power ratings. E.g. 2K Ohms and 100W.

If we're burning 100 watts to drive the grid, why not run a few less turns and use a smaller resistor? Surely this would sap less energy out of the magnetic fields too?

[Duane B]

Well, I'll try to answer your question.

The grid resistor provides the grid operating point for the tube. If a particular tube requires -200 volts on the grid, then that resistor has to have 200 volts across it. The voltage across this resistor is developed from the current in the grid when the grid is driven positive, thus it is a DC voltage. The shunt capacitor across the resistor acts like a filter capacitor and smooths out the pulsating current to make a pure DC value.

The tube also likes to run at a certain current, and this determines the value of the resistor. For example, a certain tube likes to operate at a grid voltage of -200 volts and a grid current of 66.6 ma. The resistor will need to be V/I or 200 V/66.6 ma = 3 k ohms. The resistor power dissipation will be I times V = 13.3 watts. A good wattage value to choose would be 25 watts.

If you happen to choose a smaller value resistor to save power, you would run into problems. For example, say you choose a grid resistor of 500 ohms. The tube still requires a grid bias of -200 volts, so the grid current needed to achieve this is now 200 V/500 ohms = 400 ma. This current is way too high and the tube would quickly destroy itself. By the way, if the tube did not burn itself up the resistor would dissipate 80 watts (200 v times .4 amp).

If you try to use less turns on the feedback coil the drive will be smaller and the output power will also be smaller. I hope it makes sense to you now why the grid resistor is a high value and a relatively high wattage.