The current circulating in a pi-net tank coil divides between
the load capacitor and the coax because they're effectively
in parallel. My old ARRL handbook shows (for 3.5MHz)
an 875 pf load C for a Q of 13 when transforming from 5000
to 50 ohms. 875 pf has an impedance of 51 ohms. So, for
this example, the tank coil current divides about equally
between the output coax and the load capacitor. At 1500W,
the coax will take 5.5A RMS and the load C will take the
same amount. So, the circulating current will be about 11 Amps.
If the Q was higher...corresponding to a larger circulating
current, then the load C would be larger (lower impedance)
and a larger fraction of the circulating current flows in the
load C to keep 5.5A flowing to the coax.
In an L network, the load C goes away and all the circulating
current flows to the coax...which means that the designer
must select the right Q to develop the right circulating current.
As a rule of thumb for a Pi-net at HF, it looks like the
circulating current runs about twice the coax current.
Assuming you design for generally recommended Q's.
When designing for higher frequencies, Q's tend to run
much higher.
This is just an interesting way to think about it...certainly
not the only way.
jeff, wa1hco
----- Original Message -----
From: G SEVEN <superiorwavelength@prodigy.net>
To: jeff millar <jeff@wa1hco.mv.com>; <amps@contesting.com>; Phil (VA3UX)
<phil@vaxxine.com>
Sent: Thursday, April 05, 2001 4:19 AM
Subject: Re: [AMPS] 4-1000A Project
>
> While we are on the subject can you explain to me how I can calculate the
> circulating current in a tank?
>
> thanks,
> KRS
>
> ----- Original Message -----
> From: jeff millar <jeff@wa1hco.mv.com>
> To: <amps@contesting.com>; Phil (VA3UX) <phil@vaxxine.com>
> Sent: Wednesday, April 04, 2001 10:13 PM
> Subject: Re: [AMPS] 4-1000A Project
>
>
> >
> > Here's another way to look a Q that I've never seen discussed
> > before. (There's probably a reason for that but maybe is helps
> > someone)
> >
> > You don't have to look at a pi-net as impedance transformation.
> > The goal is to create enough current to feed the coax. 1500W
> > requires 5.5A rms and a tube doesn't produce nearly that much.
> > Since the feed line connects to the tank, then more than 5.5A
> > must be circulating in the tank for the coax to pick off. So,
> > design the tank to provide enough Q to get more than 5.5A
> > circulating.
> >
> > The impedance of the Pi-net's output load capacitor forms a
> > voltage divider with the tank coil. The less load C, the
> > higher its impedance and the more of the tank's circulating
> > current goes into the feedline. If you replaced the output C with
> > a variable L, it would work very similarly...but variable L isn't
> > easy to make.
> >
> > The standard formula, Q = sqrt( Zin / Zout ), defines the
> > current multiplication ratio. For example, for Zin=5000 Zin
> > and Zout=50, that's ratio=100 or Q=10. Impedance ratio is
> > voltage (or current) squared. In other words, the tank
> > multiplies the (RMS RF) plate current by 10 and gets it to
> > the coax. The circulating current divides between the load
> > C and the coax so the Q must go a bit higher.
> >
> > It's perfectly possible to use a higher Q than the minimum.
> > This means more circulating current and more resistive losses.
> > The output load C will have a higher value (lower impedance) to
> > route a smaller fraction of the circulating current to the coax.
> >
> > The only other reason to pick a particular Q comes from a
> > bit of simulation I did. At just the right Q, changing the loading
> > capacitor doesn't require adjustment of the tune C. I remember
> > using a Henry 2K on 75M that acted that way. Crank
> > the load up and down and the tuning remained perfect.
> >
> > If anyone's interested I can drag out the work
> >
> > jeff, wa1hco
> >
> > ----- Original Message -----
> > From: Phil (VA3UX) <phil@vaxxine.com>
> > To: <amps@contesting.com>
> > Sent: Wednesday, April 04, 2001 9:44 PM
> > Subject: Re: [AMPS] 4-1000A Project
> >
> >
> > >
> > > At 09:14 PM 4/4/2001 -0400, you wrote:
> > > >
> > > >
> > > >A Q of 12 is NOT some kind of magic number. It is a suggestion,
> > > >and efficiency varies only a small amount if the Q is 20 or 6 if it
is
> > > >done with good tank components.
> > > >
> > > >12 is a number that makes people feel good, more than making
> > > >their amps work good. The best Q might be 6 or it might be 15...it
> > > >all depends.
> > > >
> > >
> > > "It all depends ".... on what ? Finish the thought for the benefit of
> the
> > > poor guy out there that's trying to learn.
> > >
> > > This discussion has completely overlooked the attenuation of harmonics
> as
> > a
> > > function of tank Q. You guys need to keep in mind that this thread
> > started
> > > with a request for advice from someone wanting to build their first
> 4-1000
> > > amp. I haven't seen all that much good advice so far; just a bunch of
> > > incomplete thoughts and a 4-1000 "expert" that has repeatedly
> "bespoken"
> > > himself.
> > >
> > > Phil
> > >
> > >
> > > >As Phil said, play with a roller and watch what happens. If you are
> > > >looking for a noticeable efficiency peak...it won't happen.
> > > >
> > > >What you do with the input circuit can have more to do with
> > > >efficiency than tank Q in many cases.
> > > >73, Tom W8JI
> > > >W8JI@contesting.com
> > > >
> > > >--
> > > >FAQ on WWW: http://www.contesting.com/FAQ/amps
> > > >Submissions: amps@contesting.com
> > > >Administrative requests: amps-REQUEST@contesting.com
> > > >Problems: owner-amps@contesting.com
> > > >
> > > >
> > > >
> > >
> > >
> > > --
> > > FAQ on WWW: http://www.contesting.com/FAQ/amps
> > > Submissions: amps@contesting.com
> > > Administrative requests: amps-REQUEST@contesting.com
> > > Problems: owner-amps@contesting.com
> > >
> > >
> >
> >
> > --
> > FAQ on WWW: http://www.contesting.com/FAQ/amps
> > Submissions: amps@contesting.com
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> > Problems: owner-amps@contesting.com
> >
>
>
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>
>
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