tubes age 87
power, but you have to take the square root to get the equivalent voltage. Thus
the figure 0.707 or the square root of one half.
> Rule of thumb is, if you apply the voltage for half the
> cycle, then you get half the current over a full cycle (AVERAGE).
This is correct. You get half the power. But remember that RMS is the voltage
that produces the same power as if it were DC. But power is V*V/R. Note the
squaring. Half-wave rectified 26.5 volts AC produces the same filament
temperature as 18.74 volts DC or 18.74 volts RMS.
Feel free to look this up in any electrical engineering text.
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Subject: Re: [R-390] 6080 in place of 6802 - RMS ???
Date: Tue, 8 Jan 2002 21:55:47 -0500
Hey Rodney, I used a diode to drop halve the filament voltage on the 6BA6's in
my R-390A......... it worked so good I forgot it was in there......and just recently
changed it.....after over 20 years of trouble-free service.Absolutely no problem
with short tube filament life. Of course, if you are after really really really
loooonnnnnggg life, I guess it is a bad idea, like the fellas say. ;-) .
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Subject: Re: [R-390] 6080 in place of 6802 - RMS ???
Date: Wed, 9 Jan 2002 10:22:31 -0800
I follow all the logic in these arguments and see Zero Point Energy here. If we
push an AC voltage across a diode to develop a positive pulse we get more
than 1/2 the available power on the far side of the diode. If we push the same
AC voltage across a second diode to develop a negative pulse we get more
than 1/2 the available power on the far side of the diode. If we filter the positive
pulse to a DC average and If we filter then negative pulse to a DC average the
working power potential between the two DC voltage is more than the input
power. Put all the good mis-applied math away, and understand you do not get
more out than you put in. If you argue that the DC pulsed power out of diode is
greater than 1/2 in a positive polarity then you must accept that the DC pulsed
power out of diode is less than 1/2 in a negative polarity
If you apply 12.6 volts to a diode and 6.3 volt filament in series the filament
sees the correct power whether the diode provides a positive or a negative
pulse to the filament diode junction.
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