[TenTec] Front End Protection

[rohre]

Ken,
It is simple (AC-RF) ohms law; consider your 50 ohm antenna system.  It is
going to be in parallel with the 100,000 ohms, but 50 ohm antenna will take
most of the 100 watt power by the parallel resistance formula. The 100,000 ohm
resistor takes a fraction of the power of the lower impedance.  The lowest
impedance in parallel (50 ohms) soaks up the most power.  Look at the ARRL
Handbook.  The ratio of power is in the ratio of resistances, which is this
case is an impedance ratio 50 over 100,000.  That is 5 over 10,000, so
5/10,000 of 100 watts is 5/100 of a watt resistor dissipation.  The only
critical issues are to use a non inductive, (carbon) resistor, and to use a
larger wattage than you need, to insure the voltage arc over rating of the
ends of the resistor is adequate for the 70.7 volts of theoretical RF at 50
ohms at the rig end of the tuner.

73, Stuart K5KVH

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