Encyclopedia of Electrical Engineering

Electrical Circuit Analysis >
Series and Parallel ac Circuits >
Application of Series and Parallel ac Circuits

The best reproduction of sound is obtained using a different speaker for
the low-, mid-, and high-frequency regions. Although the typical audio
range for the human ear is from about 100 Hz to 20 kHz, speakers are
available from 20 Hz to 40 kHz. For the low-frequency range usually
extending from about 20 Hz to 300 Hz, a speaker referred to as a
woofer is used. Of the three speakers, it is normally the largest. The
mid-range speaker is typically smaller in size and covers the range from
about 100 Hz to 5 kHz. The tweeter, as it is normally called, is usually
the smallest of the three speakers and typically covers the range from
about 2 kHz to 25 kHz. There is an overlap of frequencies to ensure that
frequencies arenÃƒÆ’Ã‚Â¢ÃƒÂ¢Ã¢â‚¬Å¡Ã‚Â¬ÃƒÂ¢Ã¢â‚¬Å¾Ã‚Â¢t lost in those regions where the response of one drops
off and the other takes over.
**Fig. 1: **Crossover speaker system
One popular method for hooking up the three speakers is the crossover configuration of Fig. 1. Note that it is nothing more than a parallel network with a speaker in each branch and full applied voltage
across each branch. The added elements (inductors and capacitors)
were carefully chosen to set the range of response for each speaker.

Note that each speaker is labeled with an impedance level and associated frequency. This type of information is typical when purchasing a quality speaker. It immediately identifies the type of speaker and reveals at which frequency it will have its maximum response. For now, however, it should prove interesting to determine the total impedance of each branch at specific frequencies to see if indeed the response of one will far outweigh the response of the other two. Since an amplifier with an output impedance of 8 Ω is to be employed, maximum transfer of power to the speaker will result when the impedance of the branch is equal to or very close to 8 Ω. Let us begin by examining the response of the frequencies to be carried primarily by the mid-range speaker since it represents the greatest portion of the human hearing range. Since the mid-range speaker branch is rated at 8 Ω at 1.4 kHz, let us test the effect of applying 1.4 kHz to all branches of the crossover network.

For the mid-range speaker: $$X_C = { 1\over 2 \pi f C}\\ = { 1\over 2 \pi (1.4kHz)(47 \mu F)} = 2.42 Ω$$ $$X_L = 2 \pi f L = 2 \pi (1.4kHz)(270 \mu H) = 2.78Ω$$ $$ R= 8 Ω$$ and $$ Z_{midrange} = R + j(X_L - X_C) \\ = 8Ω + j( 2.78 Ω - 2.42 Ω) \\ =8 Ω + j 0.36Ω \\ = 8.008 Ω \angle -2.58^\circ = 8 Ω \angle 0^\circ$$ In Fig. 2(a), the amplifier with the output impedance of 8 Ω has been applied across the mid-range speaker at a frequency of 1.4 kHz. Since the total reactance offered by the two series reactive elements is so small compared to the 8-Ω resistance of the speaker, we can essentially replace the series combination of the coil and capacitor by a short circuit of 0 Ω. We are then left with a situation where the load impedance is an exact match with the output impedance of the amplifier, and maximum power will be delivered to the speaker. Because of the equal series impedances, each will capture half the applied voltage or 6 V. The power to the speaker is then $V^2/R = (6 V)^2/8 Ω = 4.5 W$
**Fig. 2: **Crossover network: (a) mid-range speaker at 1.4 kHz; (b) woofer at 1.4 kHz;
(c) tweeter.
At a frequency of 1.4 kHz we would expect the woofer and tweeter
to have minimum impact on the generated sound. We will now test the validity of this statement by determining the impedance of each branch at 1.4 kHz.

For the woofer: $$X_L = 2 \pi f L = 2\pi (1.4kHz)(3.3mH) = 29.03Ω$$ and $$Z_{woofer} = R + j X_L = 8 Ω + j 29.03Ω\\ = 30.11 Ω \angle 74.59^\circ$$ which is a poor match with the output impedance of the amplifier. The resulting network is shown in Fig. 2(b).

The total load on the source of 12 V is $$Z_T = 8 Ω + 8 Ω + j 29.03 Ω \\ =16 Ω + j 29.03 Ω \\ = 33.15 Ω \angle 61.14^\circ$$ and the current is $$ I = { E \over Z_T} = { 12 V \angle 0^\circ \over 33.15 Ω \angle 61.14^\circ}\\ = 362mA \angle -61.14^\circ$$ The power to the 8-Ω speaker is then $$P_{wooper} = I^2 R = (362 mA)^2 8 Ω = 1.048 W $$ or about 1 W.

Consequently, the sound generated by the mid-range speaker will far outweigh the response of the woofer (as it should).

For the tweeter: $$ X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (1.4kHz)(3.9 \mu F)} = 29.15 Ω$$ and $$Z_{tweeter} = R - j X_C = 8 Ω - j 29.15Ω\\ =30.23 \angle -74.65^\circ$$ which, as for the woofer, is a poor match with the output impedance of the amplifier. The current $$ I = {E \over Z_T} = { 12 V \angle 0^\circ \over 30.23 \angle -74.65^\circ}\\ = 397 mA \angle 74.65^\circ$$ The power to the 8 Ω speaker is then $$P_{tweeter} = I^2R = (397 mA)^2(8 Ω) = 1.261 W$$ or about 1.3 W.

Consequently, the sound generated by the mid-range speaker will far outweigh the response of the tweeter also.

All in all, the mid-range speaker predominates at a frequency of 1.4 kHz for the crossover network of Fig. 1.

Just for interest sake, let us now determine the impedance of the tweeter at 20 kHz and the impact of the woofer at this frequency.

For the tweeter: $$ X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (20 kHz)(3.9 \mu F)} = 2.04 Ω$$ with $$Z_{tweeter} = R - j X_C = 8 Ω - j 2.04 Ω\\ =8.26 \angle -14.31^\circ$$ Even though the magnitude of the impedance of the branch is not exactly 8 Ω, it is very close, and the speaker will receive a high level of power (actually 4.43 W).

For the woofer: $$X_L = 2 \pi f L = 2\pi (20 kHz)(3.3mH) = 414.69 Ω$$ with $$Z_{woofer} = 8 Ω - j 414.69 Ω\\ = 414.77 Ω \angle 88.9^\circ$$ which is a terrible match with the output impedance of the amplifier. Therefore, the speaker will receive a very low level of power ($6.69 mW = 0.007 W$). For all the calculations, note that the capacitive elements predominate at low frequencies, and the inductive elements at high frequencies. For the low frequencies, the reactance of the coil will be quite small, permitting a full transfer of power to the speaker. For the high-frequency tweeter, the reactance of the capacitor is quite small, providing a direct path for power flow to the speaker.

Note that each speaker is labeled with an impedance level and associated frequency. This type of information is typical when purchasing a quality speaker. It immediately identifies the type of speaker and reveals at which frequency it will have its maximum response. For now, however, it should prove interesting to determine the total impedance of each branch at specific frequencies to see if indeed the response of one will far outweigh the response of the other two. Since an amplifier with an output impedance of 8 Ω is to be employed, maximum transfer of power to the speaker will result when the impedance of the branch is equal to or very close to 8 Ω. Let us begin by examining the response of the frequencies to be carried primarily by the mid-range speaker since it represents the greatest portion of the human hearing range. Since the mid-range speaker branch is rated at 8 Ω at 1.4 kHz, let us test the effect of applying 1.4 kHz to all branches of the crossover network.

For the mid-range speaker: $$X_C = { 1\over 2 \pi f C}\\ = { 1\over 2 \pi (1.4kHz)(47 \mu F)} = 2.42 Ω$$ $$X_L = 2 \pi f L = 2 \pi (1.4kHz)(270 \mu H) = 2.78Ω$$ $$ R= 8 Ω$$ and $$ Z_{midrange} = R + j(X_L - X_C) \\ = 8Ω + j( 2.78 Ω - 2.42 Ω) \\ =8 Ω + j 0.36Ω \\ = 8.008 Ω \angle -2.58^\circ = 8 Ω \angle 0^\circ$$ In Fig. 2(a), the amplifier with the output impedance of 8 Ω has been applied across the mid-range speaker at a frequency of 1.4 kHz. Since the total reactance offered by the two series reactive elements is so small compared to the 8-Ω resistance of the speaker, we can essentially replace the series combination of the coil and capacitor by a short circuit of 0 Ω. We are then left with a situation where the load impedance is an exact match with the output impedance of the amplifier, and maximum power will be delivered to the speaker. Because of the equal series impedances, each will capture half the applied voltage or 6 V. The power to the speaker is then $V^2/R = (6 V)^2/8 Ω = 4.5 W$

For the woofer: $$X_L = 2 \pi f L = 2\pi (1.4kHz)(3.3mH) = 29.03Ω$$ and $$Z_{woofer} = R + j X_L = 8 Ω + j 29.03Ω\\ = 30.11 Ω \angle 74.59^\circ$$ which is a poor match with the output impedance of the amplifier. The resulting network is shown in Fig. 2(b).

The total load on the source of 12 V is $$Z_T = 8 Ω + 8 Ω + j 29.03 Ω \\ =16 Ω + j 29.03 Ω \\ = 33.15 Ω \angle 61.14^\circ$$ and the current is $$ I = { E \over Z_T} = { 12 V \angle 0^\circ \over 33.15 Ω \angle 61.14^\circ}\\ = 362mA \angle -61.14^\circ$$ The power to the 8-Ω speaker is then $$P_{wooper} = I^2 R = (362 mA)^2 8 Ω = 1.048 W $$ or about 1 W.

Consequently, the sound generated by the mid-range speaker will far outweigh the response of the woofer (as it should).

For the tweeter: $$ X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (1.4kHz)(3.9 \mu F)} = 29.15 Ω$$ and $$Z_{tweeter} = R - j X_C = 8 Ω - j 29.15Ω\\ =30.23 \angle -74.65^\circ$$ which, as for the woofer, is a poor match with the output impedance of the amplifier. The current $$ I = {E \over Z_T} = { 12 V \angle 0^\circ \over 30.23 \angle -74.65^\circ}\\ = 397 mA \angle 74.65^\circ$$ The power to the 8 Ω speaker is then $$P_{tweeter} = I^2R = (397 mA)^2(8 Ω) = 1.261 W$$ or about 1.3 W.

Consequently, the sound generated by the mid-range speaker will far outweigh the response of the tweeter also.

All in all, the mid-range speaker predominates at a frequency of 1.4 kHz for the crossover network of Fig. 1.

Just for interest sake, let us now determine the impedance of the tweeter at 20 kHz and the impact of the woofer at this frequency.

For the tweeter: $$ X_C = { 1 \over 2 \pi f C} = { 1 \over 2 \pi (20 kHz)(3.9 \mu F)} = 2.04 Ω$$ with $$Z_{tweeter} = R - j X_C = 8 Ω - j 2.04 Ω\\ =8.26 \angle -14.31^\circ$$ Even though the magnitude of the impedance of the branch is not exactly 8 Ω, it is very close, and the speaker will receive a high level of power (actually 4.43 W).

For the woofer: $$X_L = 2 \pi f L = 2\pi (20 kHz)(3.3mH) = 414.69 Ω$$ with $$Z_{woofer} = 8 Ω - j 414.69 Ω\\ = 414.77 Ω \angle 88.9^\circ$$ which is a terrible match with the output impedance of the amplifier. Therefore, the speaker will receive a very low level of power ($6.69 mW = 0.007 W$). For all the calculations, note that the capacitive elements predominate at low frequencies, and the inductive elements at high frequencies. For the low frequencies, the reactance of the coil will be quite small, permitting a full transfer of power to the speaker. For the high-frequency tweeter, the reactance of the capacitor is quite small, providing a direct path for power flow to the speaker.

Home Wiring
Previous | Next
Speaker Systems