How Do You Calculate Total Impedance in a Tuned R-LC Circuit?

[bizuputyi]

Homework Statement

Given data of the given tuned R-LC circuit:

$Q=1000$
$f_{resonance} = 1MHz$
$I = 15 \mu A$
$V_{s} = 2.5V$
$R_{L} = 10kΩ$
$C=2nF$

Calculate the bandwidth, half-power frequencies, the value of R and L.
Estimate the impedance offered to the supply at resonance and at the frequencies of $\pm$2% from resonance.

Homework Equations

$BW = \frac{f_{r}}{Q}$

$f_{lower}=f_{r} \Big( \sqrt{\frac{1}{4Q^2}+1}-\frac{1}{2Q} \Big)$

$f_{upper}=f_{r} \Big( \sqrt{\frac{1}{4Q^2}+1}+\frac{1}{2Q} \Big)$

$f_{r}=\frac{1}{2\Pi} \sqrt{\frac{1}{LC}}$

$Q=\frac{2\Pi f_{r}L}{R}$

$R_{dynamic}=RQ^2$

$\frac{Z}{R_{D}}=\frac{1}{1+j2Q\frac{δf}{f_{r}}}$

The Attempt at a Solution

$BW=1000Hz$

$f_{lower}=999.5kHz$

$f_{upper}=1000.5kHz$

$L=25.33 \mu A$

$R=0.159Ω$

Do these calculations appear to be correct?

I'm struggling to find total impedance of the circuit, although I have some idea:

Is it simply $Z=\frac{V_{s}}{I}=166.67kΩ$?

or $Y=j2\Pi f_{r}C+\frac{1}{R+j2\Pi f_{r}L}$ then $Z=\frac{1}{Y}$

from which I get $Z=159006+j4780 Ω$ plus $R_{L}$ total impedance comes to $Z_{t}=169006+j4780 Ω$

or from dynamic impedance equation $Z=RQ^2=159kΩ$ plus $R_L$ again $Z_t=169kΩ$

or this is a bit complicated but I found it in the textbook:
$Z= \frac{R \Big( 1+\frac{(2\Pi f_r)^2L^2}{R^2} \Big) }{1+j(2\Pi f_r)\frac{L}{R} \Big( \frac{C}{L}R^2+(2\Pi f_r)LC-1 \Big) }$ from which I get $Z=159138+j1700 Ω$ plus $R_L$ again $Z_t=169138+j1700Ω$

Which of the total impedance is right? Maybe all of them are acceptable? Or a fifth one?

And as of to find total impedance at the frequencies $\pm$ 2%:

From the relevant equation I've got $Z=99-j3972Ω$ plus $R_L → Z_t=10099-3872Ω$

Your comments are greatly appreciated.

 

[rude man]

For high-Q circuits like this one you need to use approximations.

You should be able to derive the following:
|Z|/Zo ~ {1 + (2Qδ)²}^-1/2

where |Z| is magnitude of the RLC network impedance (as seen by R_L) at fractional freequency deviation δ, and Zo is the RLC network (real) impedance at resonance.

δ = (ω - ω_o)/ω_o
f = frequency, Hz
ω_o = resonant frequency
Q = quality factor at resonance.

You can then compute R and L for the RLC network and go on from there.

The sequence of computations might be ω_o → L → R → half-power δ → Z_o → |Z|_2%.

 

[bizuputyi]

Right, I think I get it. This is similar approach than mine and I get the same results. I was overthinking a bit on Z as such a high-Q circuit has a very sharp-edged graph.
That's one more tiny step towards my degree. Thank you!

 

[rude man]

Right, I think I get it. This is similar approach than mine and I get the same results. I was overthinking a bit on Z as such a high-Q circuit has a very sharp-edged graph.
That's one more tiny step towards my degree. Thank you!

Good luck to you!