Quality factor in electronic circuits

The concept of the quality factor number, or simply Q as it is usually referred to, is one that has often confused me, mostly due to the various number of applications, definitions and mathematical formula that are supposed to deal with the exact same number. I’ve found surprisingly little well formulated writing regarding how to derive the various mathematical expressions from the basic definitions, and so I thought I would write my own little article, mostly for my own good, but also for anyone else that might be confused in the same way.

The basics

The formal base definition can be found here:

The quality factor of a resonator/oscillator can be defined as the ratio of the energy stored in the oscillator, to the energy loss per radian of the oscillation cycle.

What does this say in simpler terms? We have some amount of energy stored in some kind of oscillatory circuit. The circuit will be made up of elements that are not perfectly reactive and there will be some measure of power loss for every cycle of oscillation. The quality factor is simply a measure of how good an oscillator is at maintaining an oscillation without energy decay.

For a very high value of Q, the energy stored is much greater than the loss in the oscillator. So if we excite the oscillator, the oscillations will only be slightly damped every cycle and the ringing will take a very long time to die out. Conversely, for a very low Q-value, the amplitude of the oscillations will be lowered severely for every cycle and the oscillation will die out in only a few cycles (or perhaps even not oscillate at all, as we will see).

For a very simple example of a relatively low-Q system, attach a weight to an iron spring, hang it from a hook, and drag and release the spring. The weight will bounce up and down, but the displacement of the weight will quickly diminish and the spring returns to the steady-state position. If this system was to somehow be granted an infinite quality factor (meaning no friction/mechanical/elasticity heat loss at all in the spring), the weight would continue the oscillatory behavior forever, with the movement never diminishing at all.

Electronic components and Q

In electronic circuits, we use components to create resonance circuits. Although these are often modeled as ideal complex impedances, real components will never have ideal characteristics. Therefore it is of interest to define a quality factor for a single component, often called component-Q. It is not entirely intuitive how to approach this for a single inductor or capacitor, since a single on of these components do not represent a resonance circuit on their own. The trick to doing this is to calculate what the Q of an otherwise infinite-Q resonance circuit would become, if the nonideal component in question is used.

As an example we can take an inductor. A real life inductor can be modeled fairly well by an ideal resistor in series with an ideal inductor. The resistor represents the resistive loss that occurs in the wire windings of the inductor coil. Let us assume that there is an AC current running through the circuit such that I(t)=Asin(wt). The energy stored in the inductor can be written as

E_I=\frac{1}{2}LI^2

In our circuit, the maximum current is at the peak of the current sine wave, so we have a maximal energy of E_I=\frac{1}{2}LA^2. Secondly we want to measure the power loss per cycle of the resistor. Note that the unit Watt stands for joules per second. If we calculate the dissipation in watts of the resistor and multiply it by the cycle length, we get the energy lost per cycle. Note also that we use the RMS voltage, A\sqrt{2} for calculating resistive loss.

E_R=\frac{A^2}{2}R\frac{1}{f}

The Q-factor is then given by:

Q=2\pi \frac{\frac{1}{2}LA^2}{\frac{A^2}{2}R\frac{1}{f}}=2\pi f \frac{L}{R}=\frac{\omega L}{R}

We can recognize this as the ratio of the inductor reactance to the resistor value. It should be noted that for component Q there is no specific resonant frequency, and instead Q should be expressed not as a single number, but a function of the frequency, Q(\omega).

This can be generalized to any complex series impedance Z_s=R_s+jX_s. The quality factor is defined as Q=\frac{X_s}{R_s}.

We can use the same intuitive reasoning for a component with a parallel parasitic resistor, say a capacitor. With the energy stored being E=\frac{1}{2}CV^2 and the dissipation in the resistor being P=\frac{1}{2}\frac{V^2}{R} we get:

Q=\frac{\frac{1}{2}CV^2}{\frac{1}{2}\frac{V^2}{R_p}\frac{1}{\omega}}=\frac{R_p}{1/(\omega C)}=\frac{R_p}{X_p}

This is quite intuitive, as for a parallel connection, the smaller impedance will dominate, so a high parallel resistance means that the total impedance looks more and more like an ideal reactance. So to summarize, we can use the following rule for component Q-values:

Q=\frac{X_s}{R_s}=\frac{X_p}{R_p}

For general circuitry, it can be a good approximation to assume that inductors contribute the most nonideal characteristics (due to their substantial series resistance), and ignore that of the capacitors and resistor circuits. But for precision circuitry, the approximation is best made when backed up by measurement data and experience.

Parallel and series circuits

Sometimes we have a situation where we have a component with some series reactance and a Q-factor, but we want to use it in a parallel circuit and therefore want to transform the model into one where the parasitic resistor is in parallel instead.

Suppose we have an inductor with finite Q represented by an ideal inductor in series with a resistor, and we want to replace it with an inductor in parallel with a resistor. How do we select the new inductor and resistor values to have an exactly equivalent impedance? Let us look at the impedances for both cases. Series circuit:

Z_s = R_s + j\omega L_s

And, parallel:

Z_p = R_p||j\omega L_p=\frac{j\omega R_pL_p}{R_p+j\omega L_p}

Let’s use the complex conjugate to get separate imaginary and real parts for the parallel circuit:

\frac{j\omega R_p^2L_p-R_pj^2\omega ^2L_p^2}{R_p^2-j^2\omega ^2L_p^2}=\frac{R_p\omega ^2L_p^2}{R_p^2+\omega ^2L_p^2}+j\omega\frac{R_p^2L_p}{R_p^2+\omega ^2L_p^2}

Now, if we want the parallel model to be equivalent to the series model, how do the series impedances relate to the parallel impedances?

Remember that Q=\frac{X_s}{R_s}=\frac{R_p}{X_p}. Let’s equate the real parts first:

R_s=\frac{R_p\omega ^2L^2}{R_p^2+\omega ^2L^2}=\frac{R_p}{\frac{R_p^2}{\omega ^2L^2}+1}=R_p\frac{1}{Q^2+1}

And the imaginary parts:

L_s=\frac{R_p^2L_p}{R_p^2+\omega ^2L_p^2}=\frac{L_p\frac{R_p^2}{\omega ^2L_p^2}}{\frac{R_p^2}{\omega ^2L_p^2}+1}=L_p\frac{Q^2}{Q^2+1}

What can we deduce from this? For general components at moderate frequencies we can probably assume Q to be much higher than 1, which means that the ratio that is used for relating the inductor values is quite close to one. Therefore a good approximation is to simply replace a the series connection with an inductor of the exact same value, in parallel with a resistor chosen according to the ratio \frac{1}{Q^2+1}.

Resonance circuits

Having covered the not entirely intuitive concept of component Q, we now come to the more easily understood quality factor of a discrete resonance circuit. Suppose we have a circuit consisting of a resistor, an inductor and a capacitor in parallel. Assume that the starting condition is a voltage A over the entire circuit, with no current being drawn anywhere. For this circuit, this means that the capacitor is fully charged with the energy of the resonator, and that the inductor stores no energy at all.

What happens as time begins to pass in the circuit, is that the inductor and resistor will begin to draw current from the capacitor. At some point the inductor will be drawing a maximal amount of current, at which the voltage over the circuit will be zero. As the voltage then goes negative, the current diminishes until the circuit reaches a maximum negative voltage (with the capacitor again fully charged, and the inductor free of energy). All the while, the resistor will be diminishing some energy as heat loss. What this means is that we have an oscillator at the resonant frequency of the capacitor/inductor combination, and also that we have a finite Q-factor since the resistor slowly removes some energy from the system for each cycle.

Let us analyze the admittance of the circuit:

Y =\frac{1}{R}+\frac{1}{j\omega L}+j\omega C=\frac{1}{R}+j(\omega C - \frac{1}{\omega L})=\frac{1}{R}+j(\frac{\omega^2LC-1}{\omega L})

We can see that for a specific frequency \omega_r=\frac{1}{\sqrt{LC}} a resonance phenomenon occurs and the entire reactive component becomes zero, leaving the resistor as the only impedive element in the circuit. This is possible due to the fact that an inductor will provide a negative imaginary admittance and the capacitor will have a positive one. At resonance they will have the same absolute values, and thus sum up to zero. It is at this specific frequency that the capacitor and inductor will be trading energy, as discussed earlier, producing an oscillation.

If we imagine an ideal circuit where the shunt resistor is removed (i.e. let R be infinite), the admittance will actually be zero at resonance, meaning that the impedance is infinite. If we connect an AC voltage source producing a sine wave of this exact frequency, the circuit will draw no current when it has reached steady state.

How can the circuit draw no current even though there are two components sitting there, forming a connection to ground? If we look at the capacitor voltage, it will show a sinusoidal waveform of the resonant frequency. If the circuit has infinite Q, then the amplitude of the oscillation will never drop and the circuit will not need any additional energy to be added in order to maintain the oscillation. So once the oscillation is at steady-state, the capacitor will always be at the same voltage as the AC source already. If the capacitor is already at the source voltage, no charging current can flow. The same line of thinking can be used for the inductor. Together they present an infinite impedance at the resonance frequency, since they will never require any additional current to maintain oscillation.

What is the Q-factor of the circuit if R is not infinite? We can use similar calculations as for the Q of a single circuit (in fact, the derivation of the component Q is really just a way of analyzing a single component yet in the context of an RLC resonator). Remember that when the resonator is at maximum voltage, the capacitor stores all the energy since the current is zero. Likewise, at maximum current, the inductor stores all energy since the voltage is zero. Remember also that the resonant frequency is \omega_r=\frac{1}{sqrt{LC}}.

Q=2\pi \frac{\frac{1}{2}CV^2}{\frac{V^2}{2R_p}\frac{1}{f_r}}=\omega _rCR_p=R_p\sqrt{\frac{C}{L}}

For a series resonance circuit we get the similar formula of:

Q=\frac{1}{R_s}\sqrt{\frac{L}{C}}

Example calculation

Let’s calculate the Q of an LC circuit. We have a source driving a load, in addition to an LC circuit consisting of an inductor and capacitor in parallel to ground.

  • Source impedance: 500 ohm
  • Inductor: 1 µH, Q of 100
  • Capacitor: 10 nF, assume infinite Q
  • Load impedance: 1 kohm

First we calculate the resonant frequency: f_r=\frac{1}{2\pi sqrt{LC}}=\frac{1}{2\pi }\times10^7\approx 1.6 MHz

Note that in order to calculate the resonant frequency we must make the assumption that the inductor as a parallel model has an equal inductance to that of its series model. Since its component Q is 100, the transforming coefficient is \frac{100^2}{100^2+1}=0.9999 which makes it a good assumption.

What is the parallel parasitic resistance of the inductor?

R_p=\frac{1}{Q^2+1}R_s=(Q^2+1)\frac{\omega _rL_s}{Q}\approx 1 kohm

For calculating the total parallel resistance to ground, we can consider the source a short to ground.

R=R_{source}||R_{load}||R_p=250 ohms

We use the formula derived above for the parallel RLC Q-factor:

Q=Rsqrt{\frac{C}{L}}=\frac{1}{100}\sqrt{\frac{10\times 10^{-9}}{1\times 10^{-6}}}=25

Our entire circuit has a Q of 25.

Linear systems and feedback

The quality factor can be quite intuitively seen in feedback systems and LTI-transfer functions. Simple second order filters (or filter models) are often written on the form of:

H(s)=\frac{\omega_0^2}{s^2+\frac{\omega_0}{Q}s+\omega_0^2}

The denominator here is the interesting part. Whenever we have a complex pole pair in an LTI system transfer function, their complex value signifies an output oscillation frequency with some exponential decay, which the system exhibits when excited by a transient.

If the pair is completely imaginary, ie s=\pm j\omega_0, an input impulse will cause a never-ending oscillation at the resonant frequency, equivalent to an infinite Q. If the pole is shifted out slightly towards the left half-plane, this oscillation will instead exhibit an exponential decay in amplitude, and this signifies some finite Q-value, where there is a loss for each oscillation cycle. We look at the pole solutions for the denominator:

s^2+\frac{\omega_0}{Q}s+\omega_0^2 \Rightarrow s=-\frac{\omega_0}{2Q} \pm\sqrt{\omega_0^2-\frac{\omega_0^2}{4Q^2}}=-\frac{\omega_0}{2Q} \pm j\omega_0 \sqrt{1-\frac{1}{4Q^2}}

Of interest is what happens for Q=1/2. The contents of the square-root term are eliminated and we are left with a double real pole at s=-\omega_0. This situation is usually called critically damped, which means that for an input transient to such a system, the step response will be as fast as possible without any ringing/overshoot (since such behavior requires an imaginary part). Recall from feedback theory that the step response speed is determined by how far away the dominant pole is from the origin. When we decrease Q below 1/2, the double pole will split, with one pole moving towards the origin and one pole moving into the half plane. This will cause the system to react slower, which is referred to as the system being overdamped.

Conversely, any Q above 1/2 makes for an underdamped system, which signifies that some ringing/oscillation will occur. If overshoot is not critical, it is often a good idea to slightly underdamp a system, since the step response will be faster.