BTW, on the discussion of k=1.8 vs. 1.87 in the calculations: There is less
than 4% difference between these two values. The difference can be covered by
only a small "tweak" of the loading and tuning controls.
This is not a real design issue. When we tune an amplifier for maximum power
output plus a tad heavier loading, we are tuning out all of these little
design variations. What with the inductors being tapped at one turn intervals
we
are not going to hit the design impedance exactly on target anyway.
73,
Gerald K5GW
In a message dated 3/31/2006 10:22:21 A.M. Central Standard Time,
gpatterson53@hotmail.com writes:
ok.... one guy says use the "open circuit voltage" the other guy say use the
loaded "operating voltage" ..................who is correct
>From: "Phil Clements" <philc@texascellnet.com>
>To: "'Partain, Chuck'" <Chuck_Partain@maxtor.com>,<amps@contesting.com>
>Subject: Re: [Amps] tube impedance, figuring tank circuit values
>Date: Thu, 30 Mar 2006 16:20:27 -0600
>
>
> >
> > So, when I power up my amp, I see 3910vdc on my plate. The way I have it
> > configured, I put 330vdc on the screen grid
> > and -130volts on the control grid., when I bring the control grid to
>-69v,
> > the tube pulls current. The voltage on the plate is
> > dropping to ~3600 to 3700. Is this the value I would use to figure the
> > impedance of the tube for tank calculations?
>
>
>No, the voltage and current of the amp when it is running at what you plan
>to operate it at are the values that you plug into the formula. Say that
>the
>voltage drops to 3500 volts at an operating anode current of 1 amp; your
>tank should be configured for that, not the resting anode I.
>
>(((73)))
>Phil, K5PC
>
>
>
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