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

[Roger L Ruszkowski]

>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.

This is not true. Close but no cookie.


>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).

This passes the logic test.

>Now with the first half of the AC cycle (the switch is
>closed),

Changed subject here in logical argumnet. It was DC with a 50 percent duty
cycle.

>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).

For the 1/2 duty cycle in which DC power is applied.

>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.

This logic mixes AC and DC and is not true.
The math is fine the logic is in error.


>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).

This hypothetical problem is logically correct but nothing
I would ever implement in tubes and hardware and then apply power to.


>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:

Another hypothetical problem. Getting to the Diode part
current flow is only 50% of continuously.


With the tubes and diode all in series.
Apply twice the voltage across the proper filament
resistance for 1/2 the time and the power comes out
correct.

The AC is 1/2 wave rectifed by the diode.
The voltage is pulsed.
The current is pulsed.
The filament resistance limits the power flow.
The average results works.

Diddle this math all day. But other people report
their circuits have worked for years.

Barry,

Where did you miss the diode in this thread?

Roger KC6TRU.