>
>Jon,
>I tried this idea with a YC156 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?
>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 Ctune has a minimum of <10pF. //
It seems to me that those who propose using an Lmatch between the anode
and Ctune, fail to consider that the added L also decreases the anode
vhfresonance. This could be problematic, depending on where the
RFgrounded screen (or groundedgrid in cath. driven config.) resonance
happens to fall.
 later, Terry
>
>Jon Ogden wrote:
>>
>> OK,
>>
>> In order to match the high impedance of the 41000A 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 Lnetwork to step
>> the impedance down to a more workable level. This Lnetwork 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 41000A's plate impedance is 4600 Ohms = R1 (from
>> Eimac Specs)
>> The Cplate of the 41000A 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 Lnet Q=R1/Xp
>>
>> At 28 MHz: Xp=1/(2*pi*28E6*7.6E12)= 747.91 Ohms
>>
>> Therefore Q=4600/747.91 = 6.15
>>
>> R2 = Xs/Q (R2 = output impedance)
>>
>> At 28 MHz: Xs=2*pi*28E6*1E6 = 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."
>>
>> 
>> FAQ on WWW: http://www.contesting.com/ampfaq.html
>> Submissions: amps@contesting.com
>> Administrative requests: ampsREQUEST@contesting.com
>> Problems: owneramps@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: ampsREQUEST@contesting.com
>Problems: owneramps@contesting.com
>Search: http://www.contesting.com/km9p/search.htm
>
Rich...
R. L. Measures, 8053863734, AG6K, www.vcnet.com/measures

FAQ on WWW: http://www.contesting.com/ampfaq.html
Submissions: amps@contesting.com
Administrative requests: ampsREQUEST@contesting.com
Problems: owneramps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
