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schematic

simulate this circuit – Schematic created using CircuitLab

Is it safe to consider the resonance frequency of a second order RLC circuit to be alway equal to 1/sqrt(LC) or the transfer function has to be calculated?

When there are three active components such for example for the case of bridged-T RLC network as is shown in the figure, is the resonance frequency still 1/sqrt(LC)? If capacitances are different, would that mean that there are two resonance frequency?

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  • \$\begingroup\$ If the resonant frequency is below 1GHz, I would say an RLC circuit will perform close to the equation you listed. Above 1GHz component and board parasitics will start to play a role. Good question on the Bridged-T, I believe it will also follow the equation but I'd be interested to hear what people say. Have you tried simulating it? \$\endgroup\$
    – MadHatter
    Feb 17, 2016 at 15:36
  • \$\begingroup\$ Yes I get the same 1/sqrt(RC) resonance frequency but I was wondering if this a rule or depends on the circuit? \$\endgroup\$
    – Jack
    Feb 17, 2016 at 15:49
  • \$\begingroup\$ Ahem, it's sqrt(LC) not sqrt(RC)! for more complex circuits the math gets complex and multiple resonant frequencies are apparent. \$\endgroup\$
    – Andy aka
    Feb 17, 2016 at 15:57
  • \$\begingroup\$ Thanks i fixed it. The multiple ones are still sqrt(LC) depending on the values of L and C? \$\endgroup\$
    – Jack
    Feb 17, 2016 at 16:12
  • \$\begingroup\$ If you're asking whether there is a rule of thumb that will always apply to figure out the resonance frequency for a handful of randomly assembled passives that can replace a real analysis and understanding, the answer, unsurprisingly is "no". I can't believe that is what you are asking. If you mean to ask "when will this approximation hold?" I suggest asking it. \$\endgroup\$ Feb 23, 2016 at 13:41

5 Answers 5

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KCL: (\$v_o\$ over your output, \$v_i\$ over voltage source and \$v\$ over the internal node:

$$ \frac{v_o}{R_3} + \frac{v_o-v}{Z_{C1}}+\frac{v_o-v_i}{R_2} = 0 \\ \frac{v}{Z_L} + \frac{v-v_i}{Z_{C2}} + \frac{v-v_o}{Z_{C1}} = 0 $$

can be solved to get rid of the internal node \$v\$, setting \$Z_{C i} = \frac{1}{sC_i}\$ and \$Z_L = sL\$:

$$ \frac{v_o}{v_i} = \frac{L C_1 C_2 R_2 R_3 s^3 + L (C_1 + C_2) R_3 s^2+R_3}{ L C_1 C_2 R_2 R_3 s^3 + L(R_2+R_3)(C_1+C_2) s^2 + (R_2 R_3 C_1) s + (R_2 + R_3)} $$ which is in the form $$ \frac{s^3+n_2s^2+n_0}{s^3+d_2 s^2 + d_1 s + d_0}$$ with $$ n_2 = \frac{(C_1+C_2)}{C_1 C_2 R_2} = \frac{1}{R_2C_*}$$ $$ n_0 = \frac{1}{LC_1 C_2 R_2} $$ $$ d_2 = \frac{(R_2+R_3)(C_1+C_2)}{C_1 C_2 R_2 R_3} = \frac{1}{R_\| C_*}$$ $$ d_1 = \frac{1}{LC_2} $$ $$ d_0 = \frac{R_2+R_3}{LC_1 C_2 R_2 R_3} =\frac{1}{LC_1 C_2 R_\|}$$ where I introduced the parallel resistance and serial capacitances $$ \frac{1}{R_\|} = \frac{1}{R_2} + \frac{1}{R_3}\\ \frac{1}{C_*} = \frac{1}{C_1} + \frac{1}{C_2} $$

This is a third-order high-pass filter, so general second order LRC intuition does not apply. You have to plug in your values and show them on a Bode-plot in order to find oscillation problems.

Appendix

Maxima script:

sol  : solve([v_o/R3+(v_o-v)*s*C1+(v_o-v_i)/R2=0, 
              v/(s*L)+(v-v_i)*s*C2 + (v-v_o)*s*C1 = 0], [v_i, v_o]) $
trf  : ev(v_o / v_i, sol) $
res  : rat(trf, s);
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  • \$\begingroup\$ Using a math package helps. Maxima script: sol : solve([v_o/R3+(v_o-v)*sC1+(v_o-v_i)/R2=0, v/(sL)+(v-v_i)*sC2 + (v-v_o)*sC1 = 0], [v_i, v_o]) $ trf : ev(v_o / v_i, sol) $ res : rat(trf, s); \$\endgroup\$ Feb 23, 2016 at 19:17
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It is important to realize that the resonance frequency (i.e., the frequency at which the circuit resonates if the damping is sufficiently low) of a second-order RLC circuit never equals \$\frac{1}{\sqrt(LC)}\$ if \$R>0\$, i.e., if there is damping. The resonance frequency (in radians per second) equals \$\frac{1}{\sqrt(LC)}\$ only if you have an ideal LC-circuit with zero damping. As soon as you have damping, the resonance frequency is lowered compared to an ideal LC-circuit.

E.g., for a simple series RLC circuit in the underdamped case, the resonance frequency is given by

$$\omega_r=\sqrt{\frac{1}{LC}-\frac{R^2}{4L^2}}\tag{1}$$

The frequency \$\omega_0=\frac{1}{\sqrt{LC}}\$ is called the natural frequency, but in the case of non-zero damping it is just an abstract quantity, which can be used to express the resonance frequency together with the damping constant \$\zeta\$:

$$\omega_r=\omega_0\sqrt{1-\zeta^2}\tag{2}$$

Note that even for very simple configurations of RLC-circuits different from series or parallel RLC-circuits, the formulas for the resonance frequency are different from (1) (see here).

So concerning your example, you have a third order RLC-circuit, which is very different from a standard second-order RLC-circuit. As discussed above, even for second-order RLC circuits, the resonance frequency depends on the specific configuration, and it never equals the natural frequency \$\frac{1}{\sqrt{LC}}\$ as long as \$R\neq 0\$. So you can't expect that the resonance frequency of your circuit is given by any simple expression that in any way resembles the simple formula of the natural (NOT resonance!) frequency of a second-order RLC-circuit.

An exact expression for the input impedance can be derived in a fashion similar to the derivation of the transfer function in P.-K. Engstad's answer. It is given by

$$Z_i(s)=\frac{as^3+bs^2+cs+d}{es^3+fs^2+gs+1}\tag{3}$$

with

$$\begin{align}a&=L_1C_1C_2R_2R_3\\\ b&=L_1(C_1+C_2)(R_2+R_3)\\\ c&=C_1R_2R_3 \\\ d&=R_2+R_3 \\\ e&=L_1 C_1 C_2 R_2 \\\ f&=C_1C_2R_2R_3+L_1(C_1+C_2) \\\ g&=C_2R_2+R_3(C_1+C_2) \end{align}$$

Evaluating (3) for \$s=j\omega\$ and setting its imaginary part to zero gives an expression for the resonance frequency. Using above constants to define auxiliary constants

$$\begin{align}A&=af-be\\\ B&=de+bg-cf-a\\\ C&=c-dg\end{align}$$

the expression for the exact resonance frequency can be written as

$$\omega_r=\sqrt{-\frac{B}{2A}+\sqrt{\frac{B^2}{4A^2}-\frac{C}{A}}}\tag{4}$$

For the given values of \$L_1,C_1,C_2,R_2\$ and \$R_3\$, we get from (4) \$\omega_r=9.9980\cdot 10^5\;\text{rad/s}\$. Note that this value is indeed very close to the value \$\frac{1}{\sqrt{L_1C_2}}=10^6\;\text{rad/s}\$, as mentioned in LvW's answer. However, this is only true in a certain range of the parameters around the given values. Even if only the value of \$L_1\$ is changed to \$L_1=1\;\text{mH}\$ (leaving all other values unchanged), the exact resonance frequency according to (4) is \$\omega_r=2.70\cdot 10^4\;\text{rad/s}\$, whereas we have \$\frac{1}{\sqrt{L_1C_2}}=3.16\cdot 10^4\;\text{rad/s}\$. If other parameter values are also changed, the difference between the exact value and the "approximation" becomes arbitrarily large. So in general we have to use the exact value of the resonance frequency given by (4), and no "simple" formula is generally applicable.

The plot below shows the magnitude of the input impedance \$|Z(j\omega)|\$ and the magnitude of the transfer function \$|H(j\omega)|\$ for the values given in your question. At the resonance frequency the transfer function has a huge peak (higher than the plot range), and the input impedance becomes almost zero. For \$\omega\rightarrow\infty\$, the input impedance converges to the value of \$R_3=100\,\Omega\$, and the transfer function converges to \$1\$. At DC the transfer function is \$H(0)=R_3/(R_2+R_3)=\frac12\$, and the input impedance is \$Z(0)=R_2+R_3\$ (outside plot range).

enter image description here

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  • \$\begingroup\$ Thanks Matt, for saving me from doing this calculation! However, do check your transfer function, because your \$c\$-value doesn't quite correspond to mine (my \$n_1 = 0\$). \$\endgroup\$ Feb 24, 2016 at 1:19
  • \$\begingroup\$ @Pål-KristianEngstad: It's not the same transfer function as yours, it's the input impedance of the circuit. The numerator of the input impedance is equal to the denominator of the transfer function, and comparing the two this is indeed the case. \$\endgroup\$
    – Matt L.
    Feb 24, 2016 at 6:45
  • \$\begingroup\$ Simulation of the input impedance phase response confirms that at wo=999.8 krad/s the phase crosses the 0 deg line. \$\endgroup\$
    – LvW
    Feb 24, 2016 at 13:32
  • \$\begingroup\$ @LvW: Thanks, I also carefully checked the result and I'm pretty convinced that it is correct. \$\endgroup\$
    – Matt L.
    Feb 24, 2016 at 13:54
  • \$\begingroup\$ Yes - it is. Two days ago, I also arrived at the input impedance function - however, I stopped at the point where I had to split the whole function into a real and an imginary part. The problem was, of course, that also the denominator was a complex function. I think, you have found a nice trick in defining the auxiliary constants A, B and C. By the way, according to my simulation the output voltage maximum is somewhat larger: 999.9 krad/s. \$\endgroup\$
    – LvW
    Feb 24, 2016 at 14:45
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The resonant frequency you are describing is for an LC circuit in parallel or in series, but you have an RLC network here. Consider the resonant frequency of Colpitts oscillator for example, and notice the derivation isn't so straightforward as simply applying sqrt(LC).

Use Kirchhoff's voltage and current laws to determine the relationships between the different components.

VR2 = VC1 + VC2

Vin = VR1 + VR2

Vin = VC2 + VL1

Likewise the sum of the currents at the node where C1, C2, and L1 meet is zero. For a transfer function you need to define where the output is, where is the output?

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  • \$\begingroup\$ the voltage output is across R3 \$\endgroup\$
    – Jack
    Feb 22, 2016 at 12:08
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schematic

simulate this circuit – Schematic created using CircuitLab

This is a complete revision of my answer. It is possible to give the exact resonant frequency without any calculation. This becomes obvious after redrawing the circuit.

Let me explain: The source V1 drives a current through the series combination C2-L1 and at the same time (in parallel) through R2-R3. The capacitor C1 is positioned between both nodes where the two elements of both branches are connected. This is nothing else than a classical bridge circuit with a capacitor in the "detector branch". In case of a balanced bridge there will be no current through C1.

In this context, it is important to realize that this bridge is balanced if L1 is in resonance with C2 (equal voltage drop across these elements) because we have two equal resistors (R2, R3) in the other branch of the bridge.

With other words: In case of resonance, the capacitor C1 does not contribute at all to the current-voltage distribution within the circuit. Hence, the angular resonant frequency simply is wo=SQRT(L1*C2).

This result was confirmed with simulation. It can be shown that the angular frequency wo where resonance occurs (input resistance minimum, current maximum, zero phase between voltage and current, voltage peak between R2 and R3) does not depend on C1 as long as the bridge is symmetrical with R2=R3. It is clear, that the actual values R2=R3 do not influence the resonance point.

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  • \$\begingroup\$ Something wrong with the definition (because of down-voting)? According to my knowledge - resonance is not ALWAYs defined as a voltage peak. It involves the phase between voltage and current!. \$\endgroup\$
    – LvW
    Feb 22, 2016 at 17:32
  • \$\begingroup\$ I have spent about 60 minutes for deriving the expression for the input resistance and some more time for identifying the imaginary part (and setting it to zero) of this expression. In addition, I have found out that only L1 and C2 dominate the resonant frequency and all other parts have only minor influence. And - what is the disappointing result? Downvote! I am really interested to learn WHY! \$\endgroup\$
    – LvW
    Feb 22, 2016 at 20:56
  • \$\begingroup\$ I didn't downvote, but the question does not seem specific to THIS circuit to me, just some combo of passives. \$\endgroup\$ Feb 23, 2016 at 17:05
  • \$\begingroup\$ I like. Complex made simple. When I initially looked at the first circuit, I wondered why both resistors were the same. Good job! \$\endgroup\$ Feb 23, 2016 at 19:10
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    \$\begingroup\$ @MattL, yes-I agree. I think, you have found the correct expression based on the criterion: Input resistance real at w=wo. Good job! \$\endgroup\$
    – LvW
    Feb 24, 2016 at 9:14
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Resonance draws the maximum or minimum of energy, and there is not more energy you can draw from the voltage source than by shortcircuiting it. L1/C2 will be a shortcircuit in this diagram when in resonance so any other elements have no influence when L1/C2 are in resonance. But this is not a realistic case since the resonant circuit connected to the voltage source only contains idealized elements with zero resistance, being completely undampened.

The usual case will involve real elements with finite resistance. In that case adding additional elements will affect resonance frequencies, and the resonance will not be catastrophic like it is here.

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  • \$\begingroup\$ This too simplistic. C1 (via the two resistors) will lower the Q of the series resonant circuit and it won't be a perfect short circuit. If another inductor (1uH) were placed parallel to C1 then I agree but that's a different question. \$\endgroup\$
    – Andy aka
    Feb 17, 2016 at 17:44

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