[AMPS] basic question about dipping the plate

[measures]

>
>
>-----Original Message-----
>From: measures <2@vc.net>
>To: AMPS <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
>
>
>--
>FAQ on WWW:               http://www.contesting.com/FAQ/amps
>Submissions:              amps@contesting.com
>Administrative requests:  amps-REQUEST@contesting.com
>Problems:                 owner-amps@contesting.com
>
>


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


--
FAQ on WWW:               http://www.contesting.com/FAQ/amps
Submissions:              amps@contesting.com
Administrative requests:  amps-REQUEST@contesting.com
Problems:                 owner-amps@contesting.com