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We need a drawing. Here's the L network. Use a fixed point font to see =
it.
___________ =20
|_ | 50 Ohm Output
_) |
_) L \
_) / R
_________| \
_|_ | /
/__|_ \ __|__ |
( - - - } _____ C |
| _____ | | |
\|___ / _|_ _|_=20
_|_ _ Gnd _ =20
_
At "resonance" the network will form series tuned network to ground for =
the 50 Ohm output port...and creates the near-zero impedance mentioned =
previously. But from the tube's point of view, with a 50 Ohm load on =
the output, the network looks like a parallel tuned circuit, thus high =
impedance at the tube port. How high, depends on the circuit, but its =
supposed to match the tube. For this L network example, the design must =
have a specific Q, designed create the correct circulating current, =
because the load is in series with the tank coil. That's why L networks =
have only one solution for matching...the network must have a specific Q =
to make the match.
________________ =20
|_ | | 50 Ohm Output
_) | |
_) L | \
_) | / R
_________| __|__ \
_|_ | _____ C /
/__|_ \ __|__ | |
( - - - } _____ C | |
| _____ | | | |
\|___ / _|_ _|_ _|_=20
_|_ _ Gnd _ _ =20
_
Now we add the output load C and it becomes a Pi-net. Traditionally, =
this circuit is viewed at two L networks back to back, primarily because =
that view simplifies the computation of values (in the days before =
computers). But, I personally think its reasonable to think of what's =
going on differently. The Tune C and L create a tuned circuit with some =
Q (higher than what's needed for the L network)...which creates =
circulating currents. The load C shunts some of the circulating current =
back to the tank but lets some of the current flow in the output. The =
less load C, the higher its impedance and the more circulating current =
flows in the output load. Note that traditional pi-networks use a =
variable C for the load control, but it's also possible to use a =
variable L for the same effect. It's just harder to build a variable L =
than a variable C. The presence of the extra load reactance in the =
circuit requires readjustment of the tune C. That's why you need to =
re-dip the tune control for each adjustment of the load control.
jeff, wa1hco
----- Original Message -----=20
From: Michael Tope <W4EF@dellroy.com>
To: measures <2@vc.net>; AMPS <amps@contesting.com>
Sent: Sunday, April 22, 2001 9:45 PM
Subject: Re: [AMPS] basic question about dipping the plate
>=20
> Rich,
>=20
> I believe that is only true if there is no load connected across the =
shunt side of the L-network.=20
> In that case the L-network simply becomes a hi-q series resonant trap =
to ground (e.g. a near
> dead short). Connect a finite resistive load across the shunt element =
and that is no longer true.
> Under this condition, the currents through the inductor and capacitor =
are no longer equal. Thus=20
> their respective voltage drops are no longer in anti-phase (e.g. no =
more dead short).
>=20
> 73 de Mike, W4EF.....................
>=20
> > ? Murphy said that things are more complicated than they look. A=20
> > pi-network is two L-networks in series. At resonance, an L-network =
is a=20
> > dead short. Thus, L-networks are not oprated at resonance. When my =
> > SB-220 is tuned for max output on 3820KHz, a dipmeter shows that the =
tank=20
> > resonates several hundred KHz lower. Pi-networks are not like =
parallel=20
> > resonant tanks in days of yore. =20
> >=20
> > cheers, Jeff
>=20
>=20
>=20
> --
> FAQ on WWW: http://www.contesting.com/FAQ/amps
> Submissions: amps@contesting.com
> Administrative requests: amps-REQUEST@contesting.com
> Problems: owner-amps@contesting.com
>=20
>=20
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<DIV><FONT face=3DArial size=3D2>We need a drawing. Here's the L =
network. Use=20
a fixed point font to see it.</FONT></DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =20
&=
nbsp;___________ =20
</FONT></DIV>
<DIV><FONT face=3D"Courier New" =
size=3D2> =20
|_  =
; =20
| 50 Ohm Output</FONT></DIV>
<DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> =20
_)  =
; |</FONT></DIV></DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> &nbs=
p; =20
_) L =
\</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =20
=
_) / R</FONT></DIV>
<DIV><FONT face=3D"Courier New" =
size=3D2> =20
_________| =20
\</FONT></DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> _|_ =
=20
| =
/</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =20
/__|_ \ __|__ &n=
bsp; =20
|</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> ( -=20
- - } _____ C =20
|</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> |=20
_____ | | =
=20
|</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =
\|___ / =20
_|_ _|_ =
</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =
_|_ =20
_ Gnd =20
_ </FONT></DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> _</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>At "resonance" the network will form =
series tuned=20
network to ground for the 50 Ohm output port...and creates the near-zero =
impedance mentioned previously. But from the tube's point of view, =
with a=20
50 Ohm load on the output, the network looks like a parallel tuned =
circuit, thus=20
high impedance at the tube port. How high, depends on the circuit, =
but its=20
supposed to match the tube. For this L network example, the design =
must=20
have a specific Q, designed create the correct circulating current, =
because the=20
load is in series with the </FONT><FONT face=3DArial size=3D2>tank =
coil. =20
That's why L networks have only one solution for matching...the network =
must=20
have a specific Q to make the match.</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>
<DIV><FONT face=3D"Courier New" size=3D2> =20
&=
nbsp;________________ =20
</FONT></DIV>
<DIV><FONT face=3D"Courier New" =
size=3D2> =20
|_  =
; | =20
| 50 Ohm Output</FONT></DIV>
<DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> =20
_)  =
; | |</FONT></DIV></DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> &nbs=
p; =20
_) L | =20
\</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =20
=
_) | /=20
R</FONT></DIV>
<DIV><FONT face=3D"Courier New" =
size=3D2> =20
_________| __|__ &nbs=
p;=20
\</FONT></DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> _|_ =
=20
| _____ C =
/</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =20
/__|_ \ __|__ &n=
bsp; | =20
|</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> ( -=20
- - } _____ C=20
| |</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> |=20
_____ | | =
| =20
|</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =
\|___ / =20
_|_ _|_ &=
nbsp;_|_=20
</FONT></DIV>
<DIV><FONT face=3D"Courier New" size=3D2> =
_|_ =20
=20
_ Gnd _ </FONT><FONT=20
face=3D"Courier New" =
size=3D2> _ =20
</FONT></DIV>
<DIV><FONT face=3D"Courier New"=20
size=3D2> _</FONT></FONT></DIV></DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>Now we add the output load C and =
it becomes a=20
Pi-net. Traditionally, this circuit is viewed at two L networks =
back to=20
back, primarily because that view simplifies the computation of values =
(in the=20
days before computers). But, I personally think its reasonable to =
think of=20
what's going on differently. The Tune C and L create a tuned =
circuit=20
with some Q (higher than what's needed for the L network)...which =
creates=20
circulating currents. The load C shunts some of the circulating =
current=20
back to the tank but lets some of the current flow in the output. =
The less=20
load C, the higher its impedance and the more circulating current flows =
in the=20
output load. Note that traditional pi-networks use a variable =
C for=20
the load control, but it's also possible to use a variable L for the =
same=20
effect. It's just harder to build a variable L than a =
variable=20
C. The presence of the extra load reactance in the circuit =
requires=20
readjustment of the tune C. That's why you need to re-dip the tune =
control=20
for each adjustment of the load control.</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>jeff, wa1hco</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>----- Original Message ----- </FONT>
<DIV><FONT face=3DArial size=3D2>From: Michael Tope <<A=20
href=3D"mailto:W4EF@dellroy.com";>W4EF@dellroy.com</A>></FONT></DIV>
<DIV><FONT face=3DArial size=3D2>To: measures <<A=20
href=3D"mailto:2@vc.net";>2@vc.net</A>>; AMPS <<A=20
href=3D"mailto:amps@contesting.com";>amps@contesting.com</A>></FONT></D=
IV>
<DIV><FONT face=3DArial size=3D2>Sent: Sunday, April 22, 2001 9:45 =
PM</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>Subject: Re: [AMPS] basic question =
about dipping=20
the plate</FONT></DIV></DIV>
<DIV><BR></DIV><FONT face=3DArial size=3D2>> <BR>> Rich,<BR>> =
<BR>> I=20
believe that is only true if there is no load connected across the shunt =
side of=20
the L-network. <BR>> In that case the L-network simply becomes a hi-q =
series=20
resonant trap to ground (e.g. a near<BR>> dead short). Connect a =
finite=20
resistive load across the shunt element and that is no longer =
true.<BR>>=20
Under this condition, the currents through the inductor and capacitor =
are no=20
longer equal. Thus <BR>> their respective voltage drops are no longer =
in=20
anti-phase (e.g. no more dead short).<BR>> <BR>> 73 de Mike,=20
W4EF.....................<BR>> <BR>> > ? Murphy said that =
things=20
are more complicated than they look. A <BR>> > pi-network is =
two=20
L-networks in series. At resonance, an L-network is a <BR>> =
> dead=20
short. Thus, L-networks are not oprated at resonance. When =
my=20
<BR>> > SB-220 is tuned for max output on 3820KHz, a dipmeter =
shows that=20
the tank <BR>> > resonates several hundred KHz lower. =
Pi-networks=20
are not like parallel <BR>> > resonant tanks in days of =
yore. =20
<BR>> > <BR>> > cheers, Jeff<BR>> <BR>> <BR>> =
<BR>>=20
--<BR>> FAQ on=20
WWW: &nb=
sp; =20
<A=20
href=3D"http://www.contesting.com/FAQ/amps";>http://www.contesting.com/FAQ=
/amps</A><BR>>=20
Submissions: &=
nbsp; =20
<A href=3D"mailto:amps@contesting.com";>amps@contesting.com</A><BR>>=20
Administrative requests: <A=20
href=3D"mailto:amps-REQUEST@contesting.com";>amps-REQUEST@contesting.com</=
A><BR>>=20
Problems: &nbs=
p; =20
<A =
href=3D"mailto:owner-amps@contesting.com";>owner-amps@contesting.com</A><B=
R>>=20
<BR>> </FONT></BODY></HTML>
------=_NextPart_000_000C_01C0CB3B.EBDFDE00--
--
FAQ on WWW: http://www.contesting.com/FAQ/amps
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
|