Series resistance in parallel resonance circuits

Hi,

At phase resonance the impedance of a parallel tuned circuit is equal to L/CR which is a pure resistance. 'Phase Resonance' incidentally is the point where the current in both arms are equal and out of phase. For practical purposes this occurs for Q-values of 10 or greater.

Cheers - Joe

I know you said you can find the resonant frequency and transform the inductor with winding resistance to an inductor with parallel resistance but I'd like to start with a review of that anyway. You didn't provide any values so let's say Xl = 100 and the series winding resistance is 1. To transform into parallel values: Rp = (Rs^2 + Xls^2)/Rs = 10001 Xlp = (Rs^2 + Xls^2)/Xls = 100.01

To convert back again... Rs = (Rp^2 + Xlp^2)/Rp Xls = (Rp^2 + Xlp^2)/Xlp

To be parallel resonant Xcp must be equal to -Xlp or -100.01

The other R in the circuit which I'll call R1s doesn't affect either the resonant frequency or the Q of the tuned circuit. R1p = 1/R1s.

The equivalent circuit for a current source would look like this. Note all series components become shunt components and vice versa.

____________ | | | | | Cp ^ I R1p |_____ | | | | | | Lp R2p |_____|______|_____|

Thank you for the replies. Although I'm a little confused by the comment that the other R does not affect the Q of the tuned circuit? That other R definitely effects the Bandwidth of the circuit as shown in the lab we did. By tuned circuit do you mean just the tank circuit portion?

My apologies.

You're right, it does affect the bandwidth but not the resonant frequency of the circuit.

Although I have given you one expression for the dynamic impedance at resonance, you could also use Zd = (omega x L)^2 / R(winding).

Further, remembering that an ideal voltage source has zero resistance, where does that place the series input resistance with respect to the tuned circuit? How does that then affect the Q?

And of course, bandwidth = fo/Q.

Cheers - Joe

Thanks guys. I think I'm clear on this now for the most part. My mistake was putting both resistors in series with each other as pointed out by Joe. They should be in parallel with each other with regard to the calculations for either a voltage or a current source.

I'm glad I got this figured out before our test since this exact circuit was on it. :)

Alan

Please distinguish which R you're talking about. Let's assign, say, R1 to the resistor from the signal source to the "resonant circuit", which consists of R2 in series with L1, this series string in parallel with C1, and that whole thing in series with R1.

R1 doesn't affect Q at all. There's some equation for figuring out Q from the component values, but I've long since forgotten even where to look up such a thing.

I remember in tech school, the first time we did this experiment with real inductors and capacitors, and the instructor didn't tell us that the inductor in the bench setup had about 133R series resistance. :-)

Cheers! Rich