Equation for Maximum Q in resonant circuits

M. Hamed · 2015-01-10T04:51:33+00:00

It always bothered me that in a resonant circuit, there is a bunch of inter dependent relations of circuit parameters relating to loaded Q. For example if you want to increase loaded Q, you need a much smaller reson ance reactance than load resistance, so you can pick a small L and high C, but once you reduce the reactance, you have decreased the unloaded Q which would affect the equivalent resistance parallel to the tank, which would re duce the loaded Q. So I tried to come up with some general formula to maximize loaded Q. In a previous post here (long thread about how clueless I am building oscillator s): sion one of the experts here suggested that L and C should really be determined once you have some known loaded Q and tank impedance. However suppose you w ant to maximize the Q knowing RL and Rs where RL is the load impedance and Rs is the equivalent series resistance of the inductor. My attempt at this resulted that, assuming Rs is constant, the optimum load ed Q would happen when the reactance at resonance is equal to the square ro ot of RL*Rs. Assuming Rs is constant may not be a good assumption since it will change w ith changing inductance, but once you have starting value, some tweaking is possible, especially that Rs changes slower than changes in L. My question is: is this result a well known result? Would I find this in an Electronics text? Thanks.

Jan 10, 2015 21 Replies

Rate this thread:

John S 11 years ago

I recommend the book "RF Circuit Design" by Chris Bowick. He covers filters, Q, resonant circuits along with other great stuff. One of my favorites.

Vote

M. Hamed 11 years ago

Rp = X^2/Rs Req = Rp // RL = Rp.RL/(Rp+RL) = X^2*RL/(X^2+RL*Rs) QL = Req/X = XRL/(X^2+RL*Rs) dQL/dX = RL*(RL*Rs-X^2)/(X^2+RL*Rs)^2 at QLmax => dQL/dX = 0 RL*Rs-X^2 = 0 => X = sqrt(RL*Rs) or at QLmax = X^2 = RL*Rs => X^2/Rs = RL => Rp = RL

Q.E.D

Vote

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required