[AMPS] Finding the Q of a Coil

[Maurizio Panicara]

You are nearly right, Q is  the reactance divided by the equivalent
resistance, that's the sum of ohmic losses plus the RF losses.
Generally speaking anyway, what counts is often the circuit Q rather than
the single component Q.
For example, after all this is the AMP reflector, one can build a coil with
Q=300 but the PI network where that coil will work could have been designed
for a loaded value Q=10 (or even 2) , thus accepting lower quality inductors
without any practical problem.
Incidentally, sometimes the current and dissipation capabilities, rather
than the Q, are much stringent parameters with inductors.

73,
Mauri I4JMY


----- Original Message -----
From: "Bill Smith" <ko4nrbs@yahoo.com>
To: <amps@contesting.com>
Sent: Sunday, October 07, 2001 5:43 AM
Subject: [AMPS] Finding the Q of a Coil


>
> Hung up trying to calculate the Q of a coil I want to
> build.  It will be 2 inches in diameter with 26 turns
> of #10 wire.  The length is 5.3 inches.  4.9 Turns per
> inch.  HAMCALC computes this at 11uh of inductance.
>
>
> If I got it right
> Inductive Reactance of this coil on 10 meters=
> 2*3.14*29000000*.000011 or 2003 ohms
>
> Q of this coil should be the Inductive Reactance
> divided by the internal resistance of the coil's wire
> (factoring in the RF).  This is where I got bogged
> down.  Couldn't come up with an appropiate number.
> Where am I going wrong?
> 73,
> Bill
>
> =====
> Bill Smith KO4NR
>
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