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## [AMPS] Re: Input Impedance.

 To: [AMPS] Re: Input Impedance. philk5pc@connect.net (Phil Clements) Tue, 24 Mar 1998 08:18:48 -0600
 ```At 01:29 PM 3/24/98 -0000, you wrote: >Snip ..... > >>Since input impedance equals drive power divided by peak >>cathode current squared, you can see how the input impedance >>stays fairly close throughout the operating range. > >Snip ..... > >>The input impedance is the ratio of input drive power to peak >>cathode current. > >So which one is it? One is the formula and the other is the explanation. I cannot type math symbols on this keyboard! Sorry. >>When you come up with a value from the above formula, you >>can then substitute a non-inductive resistor of the calculated >>value for the tube and use your noise bridge or MFJ analyser >>to park your tuned input in the middle of the band. > >I wouldn't substitute, but put in parallel. As long as the tube ain't lit or is in cuttoff, it makes no difference. I usually have this chore done before I take the tube out of the box! (((73))) Phil, K5PC "To do is to be".....Sartre "To be is to do".....Aristotle "To be or not to be".Shakespere "Do be do be do".....Sinatra -- 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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