This can be made with simple formula, and by considering a few approximations
as valid.
First, the Q must be "reasonable", this means that Q will have to be at least 5
- 6,
second the frequency excursions must be "reasonably" small, otherwise the
asymmetric frequency response of the pi-network will show up.
Also, the component losses in the network are assumed to be negligible.
The following analysis regards the pi-network as one-pole resonant network,
which is
valid near the resonant frequency and for small frequency offsets.
The general expression for the transfer properties of a single-pole circuit is
E/E0 = 1/(SQR(1+ (2Q* (|f-f0|/f0))^2))
where E = output voltage after the network, E0 is the driving voltage, Q is the
network Q, f0 the resonance
(center) frequency and |f - f0| is the frequency excursion. Twice |f-f0| is the
bandwidth BW.
Through algebraic manipulations, approximate Q can be solved for as:
Q = (f0/BW) * SQR((1/((E/E0)^2)-1))
The term E/E0 contains implicitly the SWR at the network input, as it can be
transformed to a power transmission coefficent
= (1-rho^2) = (1-(SWR-1/(SWR+1))^2) = (E/E0) ^2.
SWR = 2, rho = 0,33 or return loss = 10,5 dB equals a power transmission
coefficient of 0,89 (11% of the power is reflected)
and the corresponding E/E0 would be 0,94.
If the -3 dB power transmission coefficient (0,707 voltage) is inserted; the
term under the radical sign becomes 1,
and the formula reduces to the familar Q = f0/BW
73/
Karl-Arne
SM0AOM
----- Original Message -----
From: "Bill Turner" <dezrat1242@ispwest.com>
To: <amps@contesting.com>
Sent: Thursday, January 05, 2006 3:08 AM
Subject: [Amps] Measuring Q with an SWR analyzer
> A question for the math experts:
>
> I would like to measure the Q of a pi network plate circuit directly
> using an SWR analyzer.
>
> Let's say the tube is in circuit with power off. I have connected a
> 2200 ohm resistor from anode to ground to simulate the tube's plate
> load impedance. I connect an SWR analyzer (50 ohm) to the output, set
> to 14 Mhz, and tune the tank circuit for a perfect 1:1 SWR at that
> frequency. Now I vary the frequency of the analyzer by a certain
> amount - say 1 Mhz for discussion purposes - and now the SWR reads
> 2:1, so the 2:1 SWR bandwidth is 2 Mhz at 14 Mhz.
>
> From this is it possible to calculate the Q of the tank circuit?
> Obviously, the higher the Q, the smaller the 2:1 SWR bandwidth, but
> is it possible to calculate it exactly?
>
> This would be a very useful thing to do in order to check one's
> calculations against reality.
>
> Thanks in advance,
>
> 73, Bill W6WRT
>
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