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

  • #1
bizuputyi
42
1

Homework Statement



Given data of the given tuned R-LC circuit:

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

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 [itex] \pm[/itex]2% from resonance.

Homework Equations



[itex] BW = \frac{f_{r}}{Q} [/itex]

[itex] f_{lower}=f_{r} \Big( \sqrt{\frac{1}{4Q^2}+1}-\frac{1}{2Q} \Big)[/itex]

[itex] f_{upper}=f_{r} \Big( \sqrt{\frac{1}{4Q^2}+1}+\frac{1}{2Q} \Big)[/itex]

[itex] f_{r}=\frac{1}{2\Pi} \sqrt{\frac{1}{LC}}[/itex]

[itex] Q=\frac{2\Pi f_{r}L}{R}[/itex]

[itex] R_{dynamic}=RQ^2[/itex]

[itex] \frac{Z}{R_{D}}=\frac{1}{1+j2Q\frac{δf}{f_{r}}}[/itex]

The Attempt at a Solution



[itex] BW=1000Hz[/itex]

[itex] f_{lower}=999.5kHz[/itex]

[itex] f_{upper}=1000.5kHz[/itex]

[itex] L=25.33 \mu A [/itex]

[itex] R=0.159Ω [/itex]

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 [itex] Z=\frac{V_{s}}{I}=166.67kΩ[/itex]?

or [itex] Y=j2\Pi f_{r}C+\frac{1}{R+j2\Pi f_{r}L}[/itex] then [itex] Z=\frac{1}{Y}[/itex]

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

or from dynamic impedance equation [itex] Z=RQ^2=159kΩ [/itex] plus [itex] R_L [/itex] again [itex] Z_t=169kΩ [/itex]

or this is a bit complicated but I found it in the textbook:
[itex] 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) } [/itex] from which I get [itex] Z=159138+j1700 Ω [/itex] plus [itex] R_L [/itex] again [itex] Z_t=169138+j1700Ω[/itex]

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 [itex] \pm [/itex] 2%:

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

Your comments are greatly appreciated.
 

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  • #2
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 RL) at fractional freequency deviation δ, and Zo is the RLC network (real) impedance at resonance.

δ = (ω - ωo)/ωo
f = frequency, Hz
ωo = 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 ωo → L → R → half-power δ → Zo → |Z|2%.
 
  • #3
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!
 
  • #4
bizuputyi said:
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!
 

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