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To fully understand the function of this cap, one must make a Scattering
Parameter Analysis of the stages, the generator and it's load. A simple
Smith Chart is just a beginning, a tool that can induce substantial =
errors
if the assumptions made about inputs and outputs are wrong. I will =
suggest
you to use MWOffice for a comprehensive analysis of the amp. You might =
have
to make the S parameter list for the tube or perhaps the manufacturer =
had
done so already.
Good luck
Alex
http://www.sandlabs.com
ssage -----=20
From: Radio WC6W=20
To: W8JI@contesting.com ; amps@contesting.com=20
Sent: Sunday, December 02, 2001 1:47 PM
Subject: Re: [AMPS] Another Stupid Question
On Thu, 29 Nov 2001 19:12:35 -0500 "Tom Rauch" <w8ji@akorn.net> =
writes:
>=20
> > I think you'll find that if you do all the math that, for a =
typical
> > loaded Q of 10-12 in the Pi-net, the coupling cap value should be =
4X
> > the value of the plate capacitor at the lowest frequency of =
operation.
>=20
> Well let's see...
>=20
> If the anode operating impedance is 3000 ohms and we use the=20
> somewhat incorrect but common method of Rp/Xc =3D Q, then we=20
> would have 300 ohms Xc with a Q of ten. That means the blocking=20
> cap would be 75 ohms, which is without any doubt needlessly=20
> large since it is in series with a 3000 ohm source impedance!
>=20
> A blocking cap of even 500 ohms reactance would have a minimal=20
> affect on anode impedance seen by the tank, so why make it so=20
> large?
> 73, Tom W8JI
> W8JI@contesting.com=20
>=20
Hi Tom,
This was an interesting enough query to prompt the installation of =
the old Smith chart program on the new computer.
If the blocking cap is made as small as you propose, it becomes =
part of the matching network rather than just a bypass and some =
interesting things happen depending on the frequency of operation.
At higher frequencies, where given a typical tube the plate C =
represents a significant part of the plate tune cap, the effect is to =
significantly raise the plate impedance (as seen by the Pi-Net input) =
and the balance of the pi-net would need to be adjusted to maintain the =
desired Q & impedance ratio.
At lower frequencies, where the typical plate choke begins to =
become a influential part of the overall network, a similar effect =
occurs.
And at some median freq where the tube's plate C effectively =
cancels out the plate choke shunt L (i.e. the plate circuit is purely =
resistive at that point), the lowered value of the coupling cap looks =
less disruptive.
I think I'll remain with the traditional size coupling cap and =
avoid the extra math. :-)
73 & Good morning,
Marv WC6W
* =20
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<BODY bottomMargin=3D0 bgColor=3D#ffffff leftMargin=3D3 topMargin=3D0 =
rightMargin=3D3>
<DIV>To fully understand the function of this cap, one must make a=20
Scattering<BR>Parameter Analysis of the stages, the generator and it's =
load. A=20
simple<BR>Smith Chart is just a beginning, a tool that can induce =
substantial=20
errors<BR>if the assumptions made about inputs and outputs are wrong. I =
will=20
suggest<BR>you to use MWOffice for a comprehensive analysis of the amp. =
You=20
might have<BR>to make the S parameter list for the tube or perhaps the=20
manufacturer had<BR>done so already.<BR><BR>Good luck<BR>Alex<BR><A=20
href=3D"http://www.sandlabs.com">http://www.sandlabs.com</A><BR>ssage =
----- </DIV>
<BLOCKQUOTE dir=3Dltr=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV=20
style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
<A title=3Dwc6w@juno.com href=3D"mailto:wc6w@juno.com">Radio WC6W</A> =
</DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3DW8JI@contesting.com=20
href=3D"mailto:W8JI@contesting.com">W8JI@contesting.com</A> ; <A=20
title=3Damps@contesting.com=20
href=3D"mailto:amps@contesting.com">amps@contesting.com</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Sunday, December 02, 2001 =
1:47=20
PM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: [AMPS] Another =
Stupid=20
Question</DIV>
<DIV><BR></DIV>
<DIV></DIV>
<DIV> </DIV>
<DIV>On Thu, 29 Nov 2001 19:12:35 -0500 "Tom Rauch" <<A=20
href=3D"mailto:w8ji@akorn.net">w8ji@akorn.net</A>> writes:<BR>> =
<BR>>=20
> I think you'll find that if you do all the math =
that,=20
for a typical<BR>> > loaded Q of 10-12 in the Pi-net, the =
coupling=20
cap value should be 4X<BR>> > the value of the plate =
capacitor at=20
the lowest frequency of operation.<BR>> <BR>> Well let's=20
see...<BR>> <BR>> If the anode operating impedance is 3000 ohms =
and we=20
use the <BR>> somewhat incorrect but common method of Rp/Xc =3D Q, =
then we=20
<BR>> would have 300 ohms Xc with a Q of ten. That means the =
blocking=20
<BR>> cap would be 75 ohms, which is without any doubt needlessly =
<BR>>=20
large since it is in series with a 3000 ohm source impedance!<BR>> =
<BR>>=20
A blocking cap of even 500 ohms reactance would have a minimal =
<BR>> affect=20
on anode impedance seen by the tank, so why make it so <BR>> =
large?<BR>>=20
73, Tom W8JI<BR>> <A=20
href=3D"mailto:W8JI@contesting.com">W8JI@contesting.com</A> <BR>> =
<BR></DIV>
<DIV>Hi Tom,</DIV>
<DIV> This was an interesting enough query to prompt the=20
installation of the old Smith chart program on the new computer.</DIV>
<DIV> </DIV>
<DIV> If the blocking cap is made as small as you propose, =
it=20
becomes part of the matching network rather than just a bypass and =
some=20
interesting things happen depending on the frequency of =
operation.</DIV>
<DIV> </DIV>
<DIV> At higher frequencies, where given a typical tube =
the plate=20
C represents a significant part of the plate tune cap, the effect =
is to=20
significantly raise the plate impedance (as seen by the Pi-Net input) =
and the=20
balance of the pi-net would need to be adjusted to maintain the =
desired Q=20
& impedance ratio.</DIV>
<DIV> </DIV>
<DIV> At lower frequencies, where the typical plate choke =
begins=20
to become a influential part of the overall network, a similar effect=20
occurs.</DIV>
<DIV> </DIV>
<DIV> And at some median freq where the tube's plate=20
C effectively cancels out the plate choke shunt L (i.e. the plate =
circuit=20
is purely resistive at that point), the lowered value of the coupling =
cap=20
looks less disruptive.</DIV>
<DIV> </DIV>
<DIV> I think I'll remain with the traditional size =
coupling=20
cap and avoid the extra math. :-)</DIV>
<DIV> </DIV>
<DIV>73 & Good morning,</DIV>
<DIV> Marv WC6W</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>* </DIV>
<DIV></DIV></BLOCKQUOTE></BODY></HTML>
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