[SOLVED] Basic question about RLC

Hello guys,

Could you please, help me with the following basic question about RLC? (Book Razavi pag. 488).

(a) For the transfer function bellow, it seems that for no frequency, the impedance will be purely resistive. So is there a resonance frequency? If yes, what is it? And why?

(b) It is said that the conversion of a tank to three parallel components is only valid for a narrow frequency range. Why?

Thank you very much,

1. From what you attached you cannot say that the Zeq is purely resistive or not as it is the square of Zeq and does not contain any complex element. What you can say is that the square of Zeq does not diverge to infinity even at the resonance frequency. By examining the eq 14.27. you can see that for at least one frequency the Zeq is purely resistive. You can do that by just writing jw for s and do the required calculation to seperate the complex and real parts. The frequency that does not have any complex part is the frequency of resonance. The point of these equations is to show you that when elements in resonance circuit become lossy the equivalent impedance will not be infinite for parallel resonance circuit and will not be zero for series resonance circuit. The squareroot of square of Zeq is the magnitude of the impedance at one frequency.
2. The Q factor is defined only for the resonance phenomenon (Did I spell this right? ). For frequencies away from this frequency the calculation of your Rp will not hold. Actually the point of doing this is to ease the gain calculations in many cases. If you are using an LC structure you are tuning something at some frequency so only at that frequency the gain is important for your application. For other frequencies you can just solve the Zeq equation and find the gain at every frequency.

Edit: By narrow frequency range let's be clear; The Rp, the equivalent resistance at the resonance frequency, will not be equal to what you calculated by using Q^2*L for frequencies other than resonance frequency. But as three dB points are the ones that matter to that end you can say that it may still hold true. 3 dB points of a tuned system are just the frequencies away from the resonance frequency by a division of frequency of operation and the Q factor of tank circuit. I've blabbered a lot sorry .

Edit 2: Sorry I forgot to mention that the conversion of Rs to Rp changes the resonance structure's phase response. Be careful in closing the loop, the real circuit's response holds for it.

Hello Kemiyun,
Your answer was great… I complety understood. Thank you very much for sharing.
Regarding your edit 2. Why does it change the phase?
If (L1*S+RS)=( RP*LP*S)/(RP+LP*S), that is, impedance in series = impedance in parallel, so both complex numbers are equal, and therefore, both phases are equal, aren’t it?

You should write Q^2*L1 for RP which changes with frequency. But you are right they surely have the same "type" of response. In order to get exact solutions you should use the Zeq equation. There are good references for these kind of circuits, but I've read only two of them: Solid State Radio Frequency Circuits (I forgot the author sorry but it's an old book) and Fundamentals of High Frequency CMOS Analog Integrated Circuits written by Duran and Yusuf Leblebici. I am answering your questions with what I've read from those books. I am happy if these help you.

Hello kemiyun,

After some thoughts, I realized that I still do not understand why the conversion (series combination to paralllel ) is only valid to a small narrow band.

For example, why equation 14.33 is not valid for other frequency? I can not see why.
Let's forget Q in this discussion.

Thank you,

Hi there,

Surely that is the same expression as the expression before the conversion. But when you put a resistance of constant value Rp it will not be valid for all frequency range. To be valid at all the frequency range the resistor Rp's value must depend on frequency as the value of resistance is (I'm leaving Q out of it ) approximately equal to L1^2 * W^2 over Rs. If you calculate this value for every frequency (I forgot to mention Lp you must calculate it also.) it will be valid for every frequency (Or use a frequency dependent resistor [Off Topic: This definition may sounded weird at first but in switched cap circuits frequency dependent resistors can be realized but not for this purposes.]). But this is the same as not converting it in the first place . As our aim is to make life easier by happily calculating the gain at the tuning frequency, nobody cares unless some problem occurs.

Think of it like this; calculating new Lp for a new frequency changes the tuning frequency of the ideal part of the circuit. Calculating new Rp tells the impedance at that frequency.

I hope this helps. Actually I've prepared the simulations for this but my student version SPICE crashed and design files corrupted.

Hello kemiyun,

It is completely clear now. It seems so easy now. :-D
Thank you very much for this discussion!

best regards,

Hello kemiyun,

Do you have the pdf of book you have suggested? Fundamentals of High Frequency CMOS Analog Integrated Circuits written by Duran and Yusuf Leblebici.
Could you send me?

Thank you,

I do have the pdf, but Duran Leblebici happens to be our teacher and he has sent us the pdf's for class use only. I'll ask him as soon as I see him at school if he permits. I am very sorry.