>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?
>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:
>> 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
>> What secondary impedance do we obtain?
>> 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.
>> 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
>> 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
>> So what am I missing or what have I misunderstood that folks have
>> Jon Ogden
>> 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: firstname.lastname@example.org
>> Administrative requests: amps-REQUEST@contesting.com
>> Problems: email@example.com
>> Search: http://www.contesting.com/km9p/search.htm
>FAQ on WWW: http://www.contesting.com/ampfaq.html
>Administrative requests: amps-REQUEST@contesting.com
R. L. Measures, 805-386-3734, AG6K, www.vcnet.com/measures
FAQ on WWW: http://www.contesting.com/ampfaq.html
Administrative requests: amps-REQUEST@contesting.com