I know that for parallel RLC circuits, the $Q$ factor is given by:

$$ Q = R \sqrt {\frac{C}{L}} $$

But now suppose it is connected in series to a resistor $R_2$ and capacitor $C_2$. Would the $Q$ factor be changed?

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  • $\begingroup$ Yes, it will change since there is dissipation in $R_2$ as well. The general definition according to en.wikipedia.org/wiki/Q_factor, Definition of the quality factor, allows you to calculate its value. $\endgroup$
    – Urgje
    Jul 7, 2014 at 11:12
  • $\begingroup$ How do I obtain an expression for that? Is it simply $$\frac{1}{Q} = \sqrt{\frac{L}{C}} \left ( \frac{1}{R} + \frac{1}{R_2} \right)$$ ? $\endgroup$
    – user44840
    Jul 7, 2014 at 11:25
  • $\begingroup$ I do not think so. You have to calculate the quantities mentioned in the Wikipedia article or consult a book on electronic circuit theory. $\endgroup$
    – Urgje
    Jul 8, 2014 at 9:31
  • $\begingroup$ @Urgje: See my answer. The result can be written as a "parallel combination of $Q$ values", but the $Q$ value introduced by $R_2$ isn't quite what user44840 guessed in his comment. There's a factor coming from the ratio of $C/C_2$, as explained below. $\endgroup$
    – DanielSank
    Oct 8, 2014 at 3:14

1 Answer 1


There's a really awesome trick for problems like this. This is going to be a long post but the method presented makes problems like this really easy.

The idea is to turn the series branch $C_2$, $R_2$ into an effective parallel $R$ and $C$. See the diagram. The effective parallel values are denoted $C_{2,p}$ and $R_{2,p}$. Parallel capacitances just add, so the total capacitance is now $C+C_{2,p}$. Parallel resistances add in parallel so the total resistance is now $R||R_{2,p} = \left( 1/R + 1/R_{2,p} \right)^{-1}$. Since we now have a purely parallel circuit, you can stick these values into your formula for $Q$ (which was wrong in the OP, by the way, but I edited it).

Of course, to actually do any of this we have to understand how to solve for $R_{2,p}$ and $C_{2,p}$.

Before we do that I want to simplify some notation. It is extremely useful to define $Z_{LC} = \sqrt{L/C}$. This is the "characteristic impedance" of a resonant mode, and it will show up all over the place. With this definition, the equation for the $Q$ of a parallel $RLC$ resonator is $$Q = R/Z_{LC}$$ which is really easy to remember: if $R\rightarrow \infty$ then no current flows through the resistor so there's no energy loss, and as we can see $Q\rightarrow \infty$.

Series/parallel equivalence

Suppose we have a series resistance $R_s$ and reactance $X_s$. The total series impedance is $$Z_s = R_s + i X_s .$$ We want to find the equivalent parallel circuit. The impedance of a parallel resistance $R_p$ and reactance $X_p$ is $$Z_p = \frac{iR_pX_p}{R_p + iX_p} = \frac{R_pX_p^2 + iR_p^2X_p}{R_p^2 + X_p^2}$$ Now define a new symbol $Q_p \equiv R_p/X_p$. Using this we can rewrite $Z_p$ as $$Z_p = \frac{R_p}{1+Q_p^2} + iX_p \frac{Q_p^2}{1+Q_p^2}.$$ Since this is now just a sum of a real number and an imaginary number, it's obvious what the equivalent series values are: $$R_s = R_p \frac{1}{1+Q_p^2} \qquad X_s = X_p\frac{Q_p^2}{1+Q_p^2} . \qquad (*)$$ Unfortunately we have solved the problem in the wrong direction: we found series values in terms of parallel ones instead of the other way around. To solve this problem, define $Q_s \equiv X_s/R_s$, and divide the two equations in $(*)$ by one another to find $$Q_s = Q_p .$$ Now we can easily invert $(*)$ to find $$R_p = R_s(1 + Q^2) \qquad X_p = \frac{1 + Q^2}{Q^2} X_s$$ where we now write $Q$ instead of $Q_s$ or $Q_p$ because we just showed that they are equal. We now have the parallel values in terms of the series values. The best part is that almost always when you have a circuit like the one in the original post, you have $Q \gg 1$, which simplifies the transformation equations considerably to $$R_p \approx R_s Q^2 \qquad X_p \approx X_s . $$ The take-home message is that the equivalent parallel resistance is transformed to a much larger value, and the equivalent reactance is basically the same as the series value.

Solve the original problem

In the original problem we have $$ \begin{align} R_s &= R_2 \\ X_s &= \frac{1}{\omega C_2} \\ Q_e &= \frac{X_s}{R_s} = \frac{1}{\omega C_2 R_2}. \end{align} $$ where I've written $Q_e$ to indicate that this is the $Q$ of the "external" circuit. The equivalent parallel values are $$ \begin{align} R_{2,p} &\approx R_2 Q_e^2 \\ X_p &\approx X_s \rightarrow C_{2,p} \approx C_2 \end{align} $$ We now have a new fully parallel circuit with $$ \begin{align} \text{resistance} &= R||R_{2,p} \\ \text{capacitance} &= C + C_{2,p} \approx C \qquad \text{assuming }C \gg C_2 \\ \text{inductance} &= L \end{align} $$ The $Q$ of the circuit is $$ \begin{align} Q &= \text{resistance} / Z_{LC} \\ &= \left(\frac{1}{R} + \frac{1}{R_2 Q_e^2} \right) ^{-1} / Z_{LC} \\ \frac{1}{Q} &= \frac{Z_{LC}}{R} + \frac{Z_{LC}}{Q_e^2 R_2} \\ &= \frac{1}{Q_i} + \frac{1}{Q_c} \end{align}$$ where we've defined $Q_i \equiv R / Z_{LC}$ which is the internal $Q$ of the circuit without the external series branch, and $Q_c \equiv Q_e^2 R_2 / Z_{LC}$ is the extra $Q$ induced by the coupling. In other words, when you add the series branch, the total $Q$ of the resonance winds up being a parallel combination of two components:

  1. $Q_i$: The $Q$ you would have without the coupling to the series branch.

  2. $Q_c$: The $Q$ you would have if $R$ were absent. This part comes from the coupling to the series branch.

This is, of course, just a result of the fact that $R$ and the effective parallel resistance of the series branch $R_{2,p}$ add in parallel.

We now write down a useful expression for $Q_c$. First write $$ Q_e = X_s / R_s = \frac{1}{\omega R_2 C_2} .$$ Since we're talking about properties near resonance, we take $\omega \approx 1/\sqrt{LC}$ giving $$Q_e = \frac{\sqrt{LC}}{R_2 C_2}.$$ Then for $Q_c$ we get $$ \begin{align} Q_c &= \frac{Q_e^2 R_2}{Z_{LC}} \\ &= \frac{R_2 L C \sqrt{C}}{R_2^2 C_2^2 \sqrt{L}} \\ &= \frac{Z_{LC}}{R_2}\left( \frac{C}{C_2} \right)^2 . \end{align} $$ The constitutes a full solution to the problem.


$$\frac{1}{Q} = \frac{1}{Q_i} + \frac{1}{Q_c}$$ where $$Q_i = \frac{R}{Z_{LC}}$$ and $$Q_c = \frac{Z_{LC}}{R_2}\left( \frac{C}{C_2}\right)^2.$$ The approximations made here are that $C_2 \ll C$ and $Q_s \gg 1$. The approximation $Q_s \gg 1$ is pretty good for $Q_s>3$.

  • $\begingroup$ Let's say that for $C = 10^{-10} F$ and $C_2 = 0.1 C = 10^{-11} F$ and $L = 10^{-9} H$ and $R_2 = 50 \Omega$, this would give a negative value of $R$.. $\endgroup$
    – user44840
    Dec 15, 2014 at 11:50
  • $\begingroup$ @user44840: $C=10^{-10}F$ and $L=10^{-9}H$ gives $\omega_0/2\pi=503\,\text{MHz}$. With $C_2=10^{-11}F$ and $R_2=50\Omega$ we find $Q_e = 1/\omega_0 R_2 C_2 = 0.6$. This is therefore not in the limit $Q_s \gg 1$ which was an assumption of the approximations made in my calculation. We can discuss this further in comments, but you can also either drop me a note in the chat room (see bottom of page for a link) or look up my email address by going to my user profile and following the links there. $\endgroup$
    – DanielSank
    Dec 15, 2014 at 17:53
  • $\begingroup$ @user44840: If you use the full expression $R_p = R_s(1+Q^2)$ from my post you get $R_p = 50\Omega(1 + 0.6^2)=68\Omega$. This is not negative, so I'm not sure how you got a negative resistance. Could you explain? $\endgroup$
    – DanielSank
    Dec 15, 2014 at 18:10

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