The Q of a tuned parallel L-C circuit stands for quality factor and is a predictor to how much current flows in the tuned circuit at resonance. Here's the formula: -

Q = \$ \frac{1}{R_S}\sqrt{\frac{L}{C}}\$

\$R_S\$ is in series with the inductor, L

This basically means that if \$R_S\$ is zero then Q is infinite and the circulating currents in L and C are infinite for a non-zero voltage applied. Of course this doesn't happen because resistance is never 0 other than in a superconductor but that's another story!

Significantly for the question, if R stays the same and L is decreased by ten and C increased by ten (to retain the same resonant frequency), Q reduces by 10. In other words the high peaky tuned circuit you once had has become rather shallow and not so peaky in comparison.

Reducing inductance by ten does not usually mean \$R_S\$ reduces by ten. For a typical core wound with say 1000 turns, inductance might be (say) 1H. To reduce this to 0.1H means the turns reduce to 316 and the resistance therefore only reduces by about one-third.

Remember inductance is proportional to turns squared.

So after reducing inductance by 10 (resistance decreases by about 3) and increasing capacitance by 10, the net effect is a tuned circuit that has a Q that is 3x smaller than before and not as peaky or resonant.

Regarding the link - this is for an RLC parallel circuit where the resistance is represented as a parallel component and this is less-often used because the dominant loss in an RLC circuit is in the inductor (as series resistance). A parallel resistance *could* depict dielectric loss in the capacitor but this will be a tiny fraction (in most cases) of the inductor losses.

Holy Cow! You're running a kilowatt into a 4-inch coil and you're wondering why it get's hot? Please, rethink this one.

The reason you can run at higher frequencies is that the impedance of your coil is equal to the resistance plus a term that increases with frequency. Look up impedance and inductance. As your frequency increases, so does the coil impedance, and the total power drops. At low frequencies this doesn't apply much, and that's why you're getting hot.

Increasing the wire size is not going to help. What you need to do is something like this. Wind your coil on a piece of 4" diameter copper pipe, and solder copper tubing all the way around both ends of the coil form, as close to the windings as you can. Then run water through the tubing to get rid of the heat.

Remember, if your amplifier is putting out 1 kilowatt of power, you have to get rid of that power. If you don't heat something like cooling water or air, it all goes into the wire. Just think about how hot a 1 kilowatt light bulb would get.

## Best Answer

In an air cored inductor this problems do not arise. It is the core material that introduces these seemingly bizarre dependencies. Take a look at 3C90 material: -

Fig1 shows how the real part of permeability peaks at a frequency of about 700kHz. In fig4 as flux density increases (caused by more current for example) the permeability also peaks up. Temperature has also an effect for many ferrites and fig2 shows how permeability is affected.

It's the core that causes these problems and the only resort is to use the manufacturer's data sheets if you can get them.