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The value of C1 is 680pF and C2 is 220pF.

I think this is a series resonant circuit with a parallel capacitor C2 to the inductor.

My question is how can I calculate the resonant frequency from this circuit? And what is the effect of the capacitor C2 to the circuit? Is C2 increases the inductance?

It would be appreciated if any one can help.

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  • \$\begingroup\$ In any circuit, all possible energy storage LC pairs must be considered, with losses in that pair controlling its effect. The parallel LC has NO obvious losses, and will inject a deep NOTCH in the input-output frequency response. \$\endgroup\$ Apr 30, 2019 at 2:05

3 Answers 3

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The frequency response for the circuit is given by the transfer function which is equal to the load resistance divided by the total impedance of the circuit.

$$F(s) = \frac{R}{Z}$$

The impedance of the circuit is given by

$$Z = X_{C_1} + (X_{L} \parallel X_{C_2}) + R$$

If I have done my math correctly (see appendix) this is equal to

$$Z = R + \frac{1}{j\omega C_1}\frac{\omega_p^2(\omega_s^2-\omega^2)}{\omega_s^2(\omega_p^2-\omega^2)}$$

where \$\omega_p\$ is the resonant frequency (in radians per second) of L and C2 and is given by

$$\omega_p = \frac{1}{\sqrt{LC_2}}$$

and \$\omega_s\$ is the resonant frequency of C1 together with the residual inductance of L and C2 together at a frequency below \$\omega_p\$.

$$\omega_s = \frac{1}{\sqrt{\frac{1}{\omega_p^2} + LC_1}}$$

From the formula for the total impedance, it becomes apparent that when \$\omega = \omega_p\$, the impedance will peak, and hence there will be a notch in the frequency response. It also becomes apparent that when \$\omega = \omega_s\$, the impedance will be at a minimum, and hence there will be a maximum in the frequency response.

Using the values:

\$R = 50 \;\Omega\$

\$L = 810 \text{nH}\$

\$C_1 = 680 \text{pF}\$

\$C_2 = 220 \text{pF}\$

gives the resonant frequencies

$$\omega_p = \frac{1}{\sqrt{810\times10^{-9}\times 220\times10^{-12}}} = 7.49\times10^7\text{rad/sec} = 11.92 \;\text{MHz} $$

$$\omega_s = \frac{1}{\sqrt{\left(\frac{1}{7.49\times10^7}\right)^2 + \left(810\times10^{-9}\times680\times10^{-12})\right)}} = 3.7\times10^7\text{rad/sec} = 5.89 \;\text{MHz}$$

Verifying this a simulation gives

schematic

simulate this circuit – Schematic created using CircuitLab

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The simulation seems to match our calculations closely.


Appendix

Deriving the impedance of an L and C in parallel

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The formula for impedances in parallel is given by

$$Z = Z_1 \parallel Z_2 = \frac{{Z_1}{Z_2}}{Z_1+Z_2}$$

Assuming ideal L and C components, this gives

$$Z = \frac{sL \cdot \frac{1}{sC}}{sL + \frac{1}{sC}}$$

refactoring gives

$$Z = \frac{sL}{s^2LC + 1}$$

Using the standard substitution \$s=j\omega\$ we have

$$Z = \frac{j\omega L}{1-\omega^2LC}$$

The impedance has a pole when \$\omega^2LC = 1\$

So, we can define the resonant frequency of a parallel LC circuit as

$$\omega_0=\frac{1}{\sqrt{LC}}$$

The impedance then becomes:

$$Z = \frac{j\omega L}{1 - \frac{\omega^2}{\omega_0^2}}$$

Deriving the impedance of a capacitor in series with LC in parallel

enter image description here

The impedance of an ideal capacitor in series with a parallel ideal LC circuit is given by

$$Z = X_{C_1} + (X_{L} \parallel X_{C_2})$$

Defining the resonant frequency of \$X_{L} \parallel X_{C_2}\$ as

$$\omega_p = \frac{1}{\sqrt{LC_2}}$$

gives the total impedance of the circuit as

$$Z = \frac{1}{j\omega C_1} + \frac{j\omega L}{1 - \frac{\omega^2}{\omega_p^2}}$$

The resonant frequency occurs when the imaginary parts vanish. Call this resonant frequency \$\omega_0\$. So,

$$\frac{1}{j\omega_0 C_1} + \frac{j\omega_0 L}{1 - \frac{\omega_0^2}{\omega_p^2}}=0$$

Cross multiplying gives

$$1 - \frac{\omega_0^2}{\omega_p^2} + j^2\omega_0^2 LC_1=0$$

Further manipulation gives

$$1 - \frac{\omega_0^2}{\omega_p^2} - \omega_0^2 LC_1=0$$ $$\frac{\omega_0^2}{\omega_p^2} + \omega_0^2 LC_1=1$$ $$\omega_0^2 \left(\frac{1}{\omega_p^2} + LC_1\right) = 1$$

$$\omega_0^2 = \frac{1}{\left(\frac{1}{\omega_p^2} + LC_1\right)}$$

The resonant frequency of the circuit is thus

$$\omega_0 = \frac{1}{\sqrt{\frac{1}{\omega_p^2} + LC_1}}$$

Plugging this into the impedance equation, and some math which I hope I did correctly, gives

$$Z = \frac{1}{j\omega C_1}\frac{\omega_p^2(\omega_0^2-\omega^2)}{\omega_0^2(\omega_p^2-\omega^2)}$$

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With a voltage applied to the left, the 50 ohm resistor shows this response

enter image description here

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  • \$\begingroup\$ Thank you for your response. As I know, parallel LC circuit can acts as a band-stop filter. But why it induces a gain at 11.92MHz frequency? \$\endgroup\$ Apr 30, 2019 at 2:33
  • \$\begingroup\$ This circuit produces an attenuation, not a gain. The dip (I zoomed in with the tool) is exactly 11.92MHz. And the resonance of 0.81uH and 220p is also ---- exactly at 11.92MHz. The lack of losses inside that circulating-energy loop seems to ignore the 680pF. \$\endgroup\$ Apr 30, 2019 at 2:48
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My question is how can I calculate the resonant frequency from this circuit?

The network has a series resonance depending upon L and C1+C2 (5.895MHz) and a parallel resonance depending on L and C2 (11.922MHz.)

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  • \$\begingroup\$ I think it would be more helpful if you show "how OP can calculate the resonant frequency" and not just give the numbers! \$\endgroup\$
    – MrGerber
    May 19 at 10:51

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