[R-390] 6080 in place of 6802 - RMS ???

[Barry L. Ornitz]

Unfortunately I must agree with Bill Hawkins and the 
several people who sent me private email.  The dumbing down 
of Amateur Radio is amply demonstrated by this group.  Few 
seem to understand even the most simple relationships 
between voltage, current, and power.

One more time.  We have two 6080 tubes with their filaments 
in series hooked to a 25.2 volt RMS AC source with an ideal 
diode in series with both.  But since the RMS seems to be 
confusing people, let us say we have 25.2 volts DC and a 
switch that opens and closes very rapidly with a 50 percent duty 
cycle.  The result is the same.

The tube can be viewed as a resistance of 2.52 ohms (6.3 
volts divided by 2.5 amps).  With 6.3 volts applied to the 
tube filament, the heater develops 15.75 watts (6.3 volts 
times 2.5 amps).

Now with the first half of the AC cycle (the switch is 
closed), the voltage is divided between the two tubes and 
each tube sees 12.6 volts.  From Ohm's law, it draws 5 
amps (12.6 volts divided by 2.52 ohms).  Now this power is 
63 watts (12.6 volts times 5 amps).  Note how this is 
exactly 4 times the power the filament is rated for.

You could also calculate this by using the total voltage 
and the total resistance.  25.2 volts divided by 5.04 ohms 
equals 5 amps.  5 amps times 25.2 volts equals 126 watts 
for both tubes.

With the next half cycle (switch open), the tubes receive 
no voltage.  So the heater power is zero.

Average the two and you see each tube filament is heated 
with 31.5 watts (63 watts divided by two).  This is twice 
what the filament is designed for.  For both tubes, the 
combined heater power is 63 watts (126 watts divided by 
two).

Now ask what would be the steady voltage that would produce 
the same heater power if the current flowed continuously.  
The answer is 8.909 volts.  Calculate this by the 
following:

  V^2 = 31.5 watts * 2.52 ohms = 31.5 volt-amps * 2.52 ohms 
      = 31.5 volt-amps * 2.52 volts/amp = 79.38 volt-volt

Take the square root and you get 8.909 volts.

Knowing this, calculate the tube current as 3.535 amps 
(8.909 volts divided by 2.52 ohms).

Multiply the two together to get the filament power and you 
get 31.5 watts as we calculated earlier.

This is twice the rated filament POWER for the 6080, and 
you _will_ damage the tube.  Will it burn out instantly?  
Probably not, but the tube life will be severely degraded.

Between the explanation Bill Hawkins wrote and this, if you still 
do not understand, you really need to go back and study.

        73,  Barry L. Ornitz   WA4VZQ     ornitz@tricon.net