Crossover Networks

Electrical Circuit Analysis > Frequency Response

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The topic of crossover networks is included primarily to present an excellent demonstration of filter operation without a high level of complexity. Crossover networks are employed in audio systems to ensure that the proper frequencies are channeled to the appropriate speaker.

Although less expensive audio systems have to rely on one speaker to cover the full audio range from about $ 20 \mathrm{~Hz} $ to $ 20 \mathrm{kHz} $, better systems will employ at least three speakers to cover the low range $ (20 \mathrm{~Hz} $ to about $ 500 \mathrm{~Hz}) $, the midrange $ (500 \mathrm{~Hz} $ to about $ 5 \mathrm{kHz}) $, and the high range $ (5 \mathrm{kHz} $ and up $ ) $.

Fig. 1: Three-way, 6-$dB$-per-octave, crossover network.

The term crossover comes from the fact that the system is designed to have a crossover of frequency spectrums for adjacent speakers at the $ -3-\mathrm{dB} $ level, as shown in Fig. 1. Depending on the design, each filter can drop off at $ 6 \mathrm{~dB}, 12 \mathrm{~dB} $, or $ 18 \mathrm{~dB} $, with complexity increasing with the desired $ \mathrm{dB} $ drop-off rate. The three-way crossover network of Fig. $ 1 $ is quite simple in design, with a lowpass R-L filter for the woofer, an R-L-C pass-band filter for the midrange, and a high-pass R-C filter for the tweeter.

The basic equations for the components are provided below. Note the similarity between the equations, with the only difference for each type of element being the cutoff frequency.

For the crossover network of Fig. $ 1 $ with three $ 8-\Omega $ speakers, the resulting values are

$$L_{\text {low }}=\frac{R}{2 \pi f_{1}} \quad L_{\text {mid }}=\frac{R}{2 \pi f_{2}}$$

$$C_{\text {mid }}=\frac{1}{2 \pi f_{1} R} \quad C_{\text {high }}=\frac{1}{2 \pi f_{2} R}$$

$$\begin{split} L_{\text {low }}&=\frac{R}{2 \pi f_{1}}=\frac{8 \Omega}{2 \pi(400 \mathrm{~Hz})}=3.183 \mathrm{mH} \longrightarrow 3.3 \mu \mathrm{H} \text { (commercial value) }\\ L_{\text {mid }}&=\frac{R}{2 \pi f_{2}}=\frac{8 \Omega}{2 \pi(5 \mathrm{kHz})}=254.65 \mu \mathrm{H} \longrightarrow 270 \mu \mathrm{H} \text { (commercial value) }\\ C_{\text {mid }}&=\frac{1}{2 \pi f_{1} R}=\frac{1}{2 \pi(400 \mathrm{~Hz})(8 \Omega)}=49.736 \mu \mathrm{F} \longrightarrow 47 \mu \mathrm{F}\\ C_{\text {high }}&=\frac{1}{2 \pi f_{2} R}=\frac{1}{2 \pi(5 \mathrm{kHz})(8 \Omega)}=3.979 \mu \mathrm{F} \longrightarrow 3.9 \mu \mathrm{F} \text { (commercial value) }\\ \end{split} $$

as appearing on Fig. $ 1 $.For each filter, a rough sketch of the frequency response is included to show the crossover at the specific frequencies of interest. Because all three speakers are in parallel, the source voltage and impedance for each are the same.

The total loading on the source is obviously a function of the frequency applied, but the total delivered is determined solely by the speakers since they are essentially resistive in nature. To test the system, let us apply a 4-V signal at a frequency of $ 1 \mathrm{kHz} $ (a predominant frequency of the typical human auditory response curve) and see which speaker will have the highest power level.
At $ f=\mathbf{1} \mathbf{k H z} $

$$X_{L_{\text {low }}}=2 \pi f L_{\text {low }}=2 \pi(1 \mathrm{kHz})(3.3 \mathrm{mH})=20.74 \Omega$$

$$\begin{aligned} \mathbf{V}_{o} &=\frac{\left(\mathbf{Z}_{R} \angle 0^{\circ}\right)\left(V_{i} \angle 0^{\circ}\right)}{\mathbf{Z}_{T}}\\ &=\frac{\left(8 \Omega \angle 0^{\circ}\right)\left(4 \mathrm{~V} \angle 0^{\circ}\right)}{8 \Omega+j 20.74 \Omega} \\ &=1.44 \mathrm{~V} \angle-68.90^{\circ}\\ X_{L_{\text {mid }}}&=2 \pi f L_{\text {mid }}=2 \pi(1 \mathrm{kHz})(270 \mu \mathrm{H})=1.696 \Omega \\ X_{C_{\text {mid }}}&=\frac{1}{2 \pi f C_{\text {mid }}}=\frac{1}{2 \pi(1 \mathrm{kHz})(47 \mu \mathrm{F})}=3.386 \Omega \\ \mathbf{V}_{o}&=\frac{\left(\mathbf{Z}_{R} \angle 0^{\circ}\right)\left(V_{i} \angle 0^{\circ}\right)}{\mathbf{Z}_{T}}=\frac{\left(8 \Omega \angle 0^{\circ}\right)\left(4 \mathrm{~V} \angle 0^{\circ}\right)}{8 \Omega+j 1.696 \Omega-j 3.386 \Omega} \\ &=3.94 \mathrm{~V} \angle 11.93^{\circ} \\ X_{C_{\text {high }}}&=\frac{1}{2 \pi f C_{\text {high }}}=\frac{1}{2 \pi(1 \mathrm{kHz})(3.9 \mu \mathrm{F})}=40.81 \Omega \\ \mathbf{V}_{o}&=\frac{\left(Z_{R} \angle 0^{\circ}\right)\left(V_{i} \angle 0^{\circ}\right)}{\mathbf{Z}_{T}}=\frac{\left(8 \Omega \angle 0^{\circ}\right)\left(4 \mathrm{~V} \angle 0^{\circ}\right)}{8 \Omega-j 40.81 \Omega} \\ &=0.77 \mathrm{~V} \angle 78.91^{\circ}\end{aligned}$$

Using the basic power equation $ P=V^{2} / R $, the power to the woofer is

$$P_{\text {low }}=\frac{V^{2}}{R}=\frac{(1.44 \mathrm{~V})^{2}}{8 \Omega}=\mathbf{0 . 2 5 9} \mathbf{W}$$

to the midrange speaker, and to the tweeter,

$$P_{\text {high }}=\frac{V^{2}}{R}=\frac{(0.77 \mathrm{~V})^{2}}{8 \Omega}=\mathbf{0 . 0 7 4} \mathbf{~ W}$$

$$P_{\text {mid }}=\frac{V^{2}}{R}=\frac{(3.94 \mathrm{~V})^{2}}{8 \Omega}=\mathbf{1 . 9 4} \mathbf{W}$$

resulting in a power ratio of $ 7.5: 1 $ between the midrange and the woofer and $ 26: 1 $ between the midrange and the tweeter. Obviously, the response of the midrange speaker will totally overshadow the other two.

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