RE: A question about LCR circuits

To: tesla-at-poodle.pupman-dot-com
Subject: RE: A question about LCR circuits
From: Tesla List <tesla-at-stic-dot-net>
Date: Wed, 14 May 1997 20:12:14 -0500
Approved: tesla-at-stic-dot-net

Subject: RE: A question about LCR circuits
  Date:  Wed, 14 May 1997 12:26:43 -0400
  From:  Heinz Wahl <hwahl-at-jtc-campus.moric-dot-org>
    To:  "'Tesla List'" <tesla-at-pupman-dot-com>

Malcolm,

I know you didn't come out and say "the circuit cannot ring at and below
DC" but you equated f=0, which I read as D.C. It
might be easier to state that lower values of Q are easier to tune
because they ring in a broader BW. However, if Q < 10 then
you don't use the simple formula fr=1/(2pi sqrt(LC)) because the angle
between Il and Vt = 84 degrees and changes more so
below Q = 10. Of course at this point the circuit is so sloppy it'll
resonate at any frequency but dampen rapidly. When Q falls
below 10 try the formula fr= 1/(2pi sqrt(LC)) sqrt(1-(CRl^2/L)). I read
ahead in the e-mail and see that you have performed
and experiment in regard to this. 

Regards,

Heinz

Heinz,
         Given the value of Q=0.5, take any LCR circuit and arrange 
the values in it so that it equates. It certainly can't ring at any 
frequency. 0.5 is the value for critical damping for any filter is 
it not? This is standard filter theory according to the texts I've 
read on the subject.
    I did not say "the circuit cannot ring at and below DC" did I? I 
said "a circuit with a Q <= 0.5 can't ring".

Malcolm

> If you equate f=0 your talking D.C.. "at and below which" your circuit
> can't ring.
> 
> Heinz
<snip>