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[AMPS] basic question about dipping the plate

To: <amps@contesting.com>
Subject: [AMPS] basic question about dipping the plate
From: g8gsq@qsl.net (Steve Thompson)
Date: Sun, 22 Apr 2001 18:20:45 +0100

-----Original Message-----
From: measures <2@vc.net>
To: AMPS <amps@contesting.com>
To: <amps@contesting.com>
Date: 22 April 2001 17:38
Subject: Re: [AMPS] basic question about dipping the plate


>
>>
>>The output network becomes parallel resonant at dip and parallel resonant
>>circuits have their highest impedance at resonance.  Off resonance, the
tune
>>cap and inductor look like shunts to ground and the tube has to try to
work
>>into a lower impedance...thus the high current.


Using the term resonance can be misleading. The pi network is operating as
an impedance transform from what the antenna presents to what the plate
wants to see. This doesn't rely on  resonance in the network. The dip occurs
when the anode load impedance goes purely resistive. As you move away from
that, the resisitive part of the load gets smaller and there's reactive
current as well, so the anode current increases.

Messing around with a Smith chart program such as MIMP is an excellent way
to get a feel for it.

Steve


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