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

[James Miller]

And what about the changes in filament resistance as a function of
temperature caused by pulsating filament current?  Would the filament be
cooler on average, hence having a lower resistance on average, further
compounding the problem?

"Barry L. Ornitz" wrote:
> 
> 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
> 
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