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[AMPS] Re: Input Impedance.

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Subject: [AMPS] Re: Input Impedance.
From: (Phil Clements)
Date: 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!


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

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