>>Is it perhaps that the Q of the tank circuit is very large?
>
>The Q in the typical ham amplifier is nowhere near very large. 10 to 12
>is common.
>
>>Larger Q will cause the voltages to be higher. And perhaps at the same time
>>the
>>switch that is used is really under-rated?
>
>The output bandswitches in the TL-922, AL-80, and SB-220 have a
>withstanding ability of roughly 6000v.
OK....I think the original question Carl posted though was not about the
AL-80 but about another Ameritron amp. Is it possible that Ameritron
uses a different, bandswitch..
So I still have my question on bandswitches that I posted the other
night, Rich:
If the Pi-network in a PA rejects VHF/UHF signals how does the VHF
voltage of a parasitic get high enough to destroy a bandswitch? Consider
this: The tube has less gain at VHF/UHF, the pi-network is not matched
at VHF-UHF. So how does it happen?
The Pi-network does not "absorb" out of band energy as someone suggested.
If that's the case RF theory of ideal components would break down.
Unless I am really, really missing something VHF/UHF energy will be
reflected at the input of the pi-net *back* into the anode circuitry.
This is how "lossless" networks "attenuate" energy.
If VHF/UHF energy was allowed to freely roam in and out of the pi-net,
then we wouldn't get harmonic attenuation.
No one responded to my earlier post on this. Anyhow, after mulling this
over some more I have another question:
* Just because the VHF/UHF energy is reflected from the pi-net back to
the anode, does this mean that there is no VHF/UHF voltage across any of
the components in the pi-network? I wish I had the capability to do a
SPICE model on this because if you build up a voltage potential, then
maybe a parasitic could break down the switch. But I can't prove this
one way or the other.
73,
Jon
KE9NA
--------------------------------------------------------------------------
Jon Ogden
jono@webspun.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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