Rich Measures wrote:
>
> >
> >Jon,
> >I tried this idea with a YC-156 tube I have been experimenting with to
> >try to get it to play on 10 meters....its Cout is about 36PF. This is
> >the idea of a series inductor between the tube and C1 ??? To step down
> >the plate load X... so that you would need more C, less Xc at C1 ???
> >Talk about laying an egg!
>
> ? What happened?
Nothing ! Made no difference.
W6RU
>
> >I would sure like to talk to someone that has
> >done this successfully if they exist.
> >
> ? 36pF of Cout should pose no problem for a conventional tank -
> Provided the anode current is max. and C-tune has a minimum of <10pF. //
> It seems to me that those who propose using an L-match between the anode
> and C-tune, fail to consider that the added L also decreases the anode
> vhf-resonance. This could be problematic, depending on where the
> RF-grounded screen (or grounded-grid in cath. driven config.) resonance
> happens to fall.
>
> - later, Terry
> >
> >Jon Ogden wrote:
> >>
> >> OK,
> >>
> >> In order to match the high impedance of the 4-1000A to a lower impedance
> >> so that I have better loaded Q in my tank circuit, I decided to take the
> >> advice many people had given me and that is to make an L-network to step
> >> the impedance down to a more workable level. This L-network would
> >> consist of the output capacitance of the tube as the shunt (parallel)
> >> element and a small inductor as the series element.
> >>
> >> OK, so at 5KV, the 4-1000A's plate impedance is 4600 Ohms = R1 (from
> >> Eimac Specs)
> >> The Cplate of the 4-1000A is 7.6 pF (from the ARRL handbook)
> >>
> >> Let's choose L (the series inductor) to be 1 uH (a suggested value by
> >> others).
> >>
> >> What secondary impedance do we obtain?
> >>
> >> OK:
> >>
> >> In an L-net Q=R1/Xp
> >>
> >> At 28 MHz: Xp=1/(2*pi*28E6*7.6E-12)= 747.91 Ohms
> >>
> >> Therefore Q=4600/747.91 = 6.15
> >>
> >> R2 = Xs/Q (R2 = output impedance)
> >>
> >> At 28 MHz: Xs=2*pi*28E6*1E-6 = 175.9 Ohms
> >>
> >> R2 = 175.9/6.15 = 28.6 Ohms
> >>
> >> Now 28.6 Ohms is unrealistically small in order to properly get a good
> >> match with realistic components.
> >>
> >> OK, so what would a good R2 be? Let's try for 2500 Ohms as this will
> >> give decent Qs with the component values I have for the tank circuit.
> >>
> >> Therefore:
> >>
> >> Q is still 6.15 since neither R1 nor Xp have changed.
> >>
> >> Xs = R2*Q = 2500*6.15 = 15375 Ohms
> >>
> >> At 28 MHz: Ls = Xs/(2*pi*28E6) = 15375/(2*pi*28E6) = 87.4 uH
> >>
> >> This won't work either since it is a HUGE coil.
> >>
> >> OK, what gives? Am I missing something here or doing a calculation
> >> wrong?
> >>
> >> Everyone says they've seen this "trick" done in the ARRL handbook and
> >> other texts. Can someone be specific as to what handbook (what year)? I
> >> sure can't find it yet in the 1999 issue or in my 1989 issue.
> >>
> >> This idea won't work unless I am totally missing the concept of how to do
> >> this.
> >>
> >> So what am I missing or what have I misunderstood that folks have
> >> suggested?
> >>
> >> 73,
> >>
> >> Jon
> >> KE9NA
> >>
> >> -------------------------------------
> >> Jon Ogden
> >> KE9NA
> >>
> >> http://www.qsl.net/ke9na <--- CHECK IT OUT! It's been updated!!!!!
> >>
> >> "A life lived in fear is a life half lived."
> >>
> >> --
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> >
> >--
> >FAQ on WWW: http://www.contesting.com/ampfaq.html
> >Submissions: amps@contesting.com
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> >
>
> Rich...
>
> R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
>
> --
> FAQ on WWW: http://www.contesting.com/ampfaq.html
> Submissions: amps@contesting.com
> Administrative requests: amps-REQUEST@contesting.com
> Problems: owner-amps@contesting.com
> Search: http://www.contesting.com/km9p/search.htm
--
FAQ on WWW: http://www.contesting.com/ampfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
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