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[AMPS] basic question about dipping the plate

To: <amps@contesting.com>
Subject: [AMPS] basic question about dipping the plate
From: 2@vc.net (measures)
Date: Sun, 22 Apr 2001 17:37:12 -0700
>
>
>-----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.
>
good reply, Steve
>
>
>--
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>
>


-  Rich..., 805.386.3734, www.vcnet.com/measures.  
end


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