I found an online calculator
http://chemandy.com/calculators/return-loss-and-mismatch-calculator.htm
A resistive load of 300 Ohm in a 50 Ohm System represents about a 3dB
loss over a 50 Ohm load. A resistive load of 300 Ohms is 6:1 SWR. I
also tried a load of 150+150J, which gave a similar 3dB loss and SWR 6:1.
With a known test circuit, my MFJ259 shows about 4:1 at the 3dB points.
Close enough I think.
- Paul
At 01:15 PM 5/25/2018, Fuqua, Bill L wrote:
It is not an easy comparison.
A pi network is modeled by 2 back to back L-networks. Each having a Q.
So the bandwidth of these are somewhat smaller than a single LC network.
Also, 2:1 points in SWR are not the same as the -3 dB points. -3
dB points are
where only 1/2 of the power passes thru the filter. 2:1 SWR
represents about 89% power
passed thru the system.
I went thru this exercise many years ago and but it has been to
long since I had done the
math.
73
Bill wa4lav
----------
From: Amps <amps-bounces@contesting.com> on behalf of Paul Baldock
<paul@paulbaldock.com>
Sent: Wednesday, May 23, 2018 6:09 PM
To: amps@contesting.com
Subject: [Amps] Relationship between Q and SWR in PI network.
Does anybody know the relationship between Q and SWR for a PI network?
For example let's say I design my amp PI output network for 7.0MHz. I
simulate the Tube Impedance with a 1K resistor. I put my MFJ259 on
the output of the PI and tweak the load and tune caps for a perfect
1:1 (50 Ohm) SWR. I now leave the caps where they were and adjust the
frequency to show the upper and low 2:1 SWR points and they are
7.2MHz and 6.8MHz. (2:1)=Fo/(Fu-Fl)=7.0/(7.2-6.8)=17.5. In practice
this is in fact about 1.5 times my design Q, but I would like to know
what theory says.
- Paul KW7Y
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