Maximizing Q in a series resonant L/C circuit?

I think Tony is closer to the mark than you. That's because the concept of increasing L requires more coil turns, proportional to sqrt L, which means increased R proportional to L, if the wire size is reduced to keep the same size coil, so one gets no net gain in Q.

In the end the available Q is determined by the overall coil size, independent of L. This makes sense, in a Nature's revenge sort of way. Murphy shows up here too: Rac/Rdc is much less than one, due to proximity effect, and litz wire is required, even at modest frequencies, to avoid a total meltdown of Q.

Reply to
Winfield
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** LOL

You are even further away from the point than he.

Shame how the context of the OPs query had the value of R fixed at 100 ohms.

Try reading more than one post in a thread sometime.

....... Phil

Reply to
Phil Allison

I was spot on the mark Win, except PA ignored or cut the OP's context that I was replying to. I don't know which because he is killfiled here.

It's a simulator, not real life, and below is the specific OP's question that I replied to.

What's strange is that when I run a simulation, and I use two perfect (infinite Q) L and C between 50 ohm terminations, this series resonant circuit *always* has a much sharper gain response (higher Q) when the L is high in value and the C is small in value

-- when I use a small L and a large C, the series circuit is always flat and has almost no selectivity. I wish I knew the reason for this, but it has me stumped!

I assumed that he was trying various L/C values, resonating always at the same frequency.

In which case Q = sqrt(L/C)/R was exactly the reason the OP saw what he saw in his simulations.

--
Tony Williams.
Reply to
Tony Williams

"Tony Williams"

** What a lying pile of ASD f***ed shit.
** But it does not.

Q follows the L value for fixed frequency and fixed R.

Q = wL/R is where you should have stopped.

If w/R = a constant, then Q = kL

IDIOT.

...... Phil

Reply to
Phil Allison

Actually the peak power transfer frequency gets pulled off the

1/sqrt(LC) value by a factor of 1-1/4Q^2. But to answer the OP's question directly, it doesn't take all that much mathematics to show that the bandwidth (or "sharpness") is R/L rad/sec. Note that this is the reciprocal of the dissipatory time constant, a fundamental result of simple harmonic motion, make the decay time long and you have narrow band, make it short and you have broadband. Narrowband is "sharp" and broadband is "smeared."
Reply to
Fred Bloggs

Only possible if you can add electronics here: The Q-muliplier. Basically you hang on a circuit that takes this LC as the frequency determining part. Its feedback is kept just shy of the point of oscillation. The closer you get to that point the narrower the bandwidth.

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

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