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18 Oct 2011

Speaker Wiring Featured

fender_wiring

Speaker wiring: There seems to be a genuine lack of knowledge and understanding of speaker wiring among the church volunteer. A lot of people think you can keep adding speakers to an amplifier as long as you have more wire. Amplifiers have a minimum rated load for the outputs. A typical pro amplifier has a minimum load of around 4 ohms. I hope this article will help people better understand how to properly calculate the load of an amplifier based on the wiring scheme that you choose.

Speakers in Series

The essence of series wiring is really quite simple: When speakers are connected in this fashion, load impedance increases – the more speakers, the higher the impedance. The most common reason for wanting to raise impedance is to lower acoustical output, as in the case of rear-fill or center-channel speakers. Speaker output declines because the amplifiers power output decreases as the load impedance increases. While you can connect any number of speakers in series, try to keep the total equivalent-load impedance for each channel below 16 ohms, since most amps are not designed to handle higher loads.

image33

Figure 1A demonstrates how to wire a pair of speakers in series. The positive output terminal from one channel of the amplifier is wired to the positive terminal of Speaker A, and the negative terminal of Speaker A is connected to the positive terminal of Speaker B. Finally, a loop is created by wiring the negative terminal of Speaker B to the negative-output terminal of the same amplifier channel. The second channel is wired the same way. 

If you’re wiring more than two speakers in series, you simply continue alternating the negative and positive wires between speakers. To wire four speakers in series, for example, you connect the negative terminal of Speaker B to the positive terminal of Speaker C (instead of back to the amp); the negative terminal of that speaker is then wired to the positive terminal of Speaker D, and the loop is completed by connecting the negative terminal of Speaker D to the amps negative-output terminal.

To calculate the load impedance for the series-wired channel in Figure 1A, add up the impedances of each speaker in the chain. You can visualize the result as a single imaginary speaker (Figure 1B), whose impedance is represented by Zt. The math involves a simple equation in which Zt stands for the equivalent-load impedance and Za and Zb represent the impedances of Speakers A and B, respectively:

Equation 1: Speakers in Series:
Zt = Za + Zb 

Consider this real-world example of series wiring. Say you have a yearning for ultra-low bass – the kind that loosens the weather stripping around your windows – and you’re determined to install four 15-inch subwoofers in your car. The amplifier you’ve reserved for this task delivers 100 watts x 2 into 4 ohms and is capable of driving a minimum load impedance of 4 ohms; the subs are rated at 4 ohms each.

Assuming there’s enough room in your car for these monsters, the only viable option – given the above scenario – is to wire two subs in series to each amplifier channel. Doing so raises the net, or equivalent-load, impedance of each channel to 8 ohms – well within our standard 16-ohm ceiling. Mathematically, you substitute 4 ohms (the impedance rating of each sub) for Za and Zb in Equation 1 and work it through as follows:

Zt = Za + Zb
Zt = 4 + 4
Zt = 8 ohms

Parallel wiring, which well discuss later, isn’t advisable here because the net impedance for each channel drops below the minimum-load rating of the amplifier.

Power Calculations

Whenever you connect more than one speaker to an amp channel, its important to gauge what effect the speakers will have on the amp and each driver in the chain. In other words, how much power will the amp deliver into each channel given the equivalent-load impedance you’ve created? And how much power will each speaker in the chain receive? Answering these questions will help you to avoid costly damage to your amp and speakers.

Referring back to the hypothetical subwoofer installation outlined above, we know that the amplifier in question is rated to deliver 100 watts x 2 into 4 ohms. To find out how much power each channel of this amplifier will deliver into the resulting 8-ohm load, we must solve Equation 2, in which Po is power output, Pr is the amps rated power, Zr is the impedance the amps output power is rated at, and Zt is the equivalent-load impedance for each channel:

Equation 2: Calculating Output Power:
Po = Pr x (Zr / Zt) 

Plugging in the appropriate numbers, the calculation is worked through as follows:

Po = 100 x (4 / 8)
Po = 100 x 0.5
Po = 50 watts

Now that we know each amplifier channel will deliver 50 watts into an 8-ohm load, we can figure out how much power will be applied to one of the subwoofers – Pa – by solving Equation 3, in which Zn stands for the rated impedance of the speaker:

Equation 3: Power Applied to Each Driver:
Pa = Po x (Zn / Zt) 

Substituting 50 for Po, 4 for Zn, and 8 for Zt, the equation works through as follows:

Pa = 50 x (4 / 8)
Pa = 50 x 0.5
Pa = 25 watts

Since both subwoofers are rated at 4 ohms, we know that the second subwoofer (Pb) would also receive 25 watts.

Speakers in Parallel

Parallel wiring has the opposite effect of series wiring – load impedance drops when speakers are wired in this fashion. And the more speakers you wire in, the lower the impedance. The most common reason for wanting to lower impedance is to raise acoustical output. Speaker output increases because the amplifiers power output rises as the load impedance decreases.

The number of speakers that can be connected in parallel is limited by the minimum load impedance that the amplifier is capable of driving and the power-handling capacity of the speakers. In most cases, load impedance should be held to a minimum of 2 ohms – provided the amplifier can handle impedances that low.

Figure 2A shows how to wire a pair of speakers in parallel. A wire from the positive terminal of one channel of the amp is wired to the positive terminals on speakers A and B. (The simplest way to do this is to run a wire from the amp terminal to Speaker A and then run a second wire from that terminal to Speaker B.) Then the negative terminal of the same amp channel is wired in like fashion to the negative terminals on both speakers. The second channel is wired the same way.

image34

Calculating the load impedance for the parallel-wired channel in Figure 2A is a bit more complicated than doing so for speakers wired in series. Using Equation 4, multiply the impedances of each speaker and then divide the result by the sum of the speakers impedances. You can visualize the result as a single imaginary speaker (Figure 2B), whose impedance is represented by Zt. Zt stands for the equivalent-load impedance, while Za and Zb represent the impedances of speakers A and B, respectively.

Equation 4: Speakers in Parallel:
Zt = (Za x Zb) / (Za + Zb) 

Turning again to our subwoofer install, say you want even more oomph from your system. So you trade in the original amp for one that has the same 4-ohm power rating (100 watts x 2) but is also 2-ohm stable. Since the power output of most amps increases as impedance decreases, you could boost the amps power output and the systems bass response simply by switching to a parallel wiring scheme. Doing so would drop the net, or equivalent-load, impedance for each channel to 2 ohms. Mathematically, you substitute 4 for Za and Zb in Equation 4 and work it through:

Zt = (Za x Zb) / (Za + Zb)
Zt = (4 x 4) / (4 + 4)
Zt = 16 / 8
Zt = 2 ohms

To calculate the new amplifiers power output into 2 ohms, refer to Equation 2. Plugging in the appropriate numbers, the calculation goes as follows:

Po = 100 x (4 / 2)
Po = 100 x 2
Po = 200 watts

As you can see, by upgrading to a 2-ohm-stable amplifier and wiring the same four 15-inch woofers in parallel – two per channel – power output jumps fourfold – from 50 watts x 2 to 200 watts x 2.

Now, to find out how much power each subwoofer will receive when wired in parallel, we must use Equation 5, which is actually a scrambled version of Equation 3 (remember, well be working the equation for just one speaker (Pa)):

Equation 5: Power Applied to Each Speaker:
Pa = Po x (Zt / Zn) 

Substituting 200 for Po, 2 for Zt, and 4 for Zn, the equation works through as follows:

Pa = 200 x (2 / 4)
Pa = 200 x 0.5
Pa = 100 watts

Since both subwoofers are rated at 4 ohms, the second one (Pb) would also receive 100 watts.

Series/Parallel Wiring

Now its time to combine the two wiring methods. The most common reason for wanting to do this is to increase the number of speakers you can use in your system – perhaps to achieve greater volume and/or visual effect – and still maintain an impedance load that’s compatible with the systems amplifier. Any number of speakers can be linked using a series/ parallel wiring scheme, as long as you keep the total equivalent-load impedance between 2 and 16 ohms.

Figure 3A shows how to wire four speakers to a single channel using a typical series/parallel combination. A single wire running from the amps positive terminal runs to the positive terminals of speakers A and C. Next, the negative terminals of Speakers A and C are wired to the positive terminals of Speakers B and D, respectively. Finally, a loop is created by running a single wire from the negative terminal of the amp channel and splitting it between the negative terminals of Speakers B and D.

image35

The best way to understand the electrical implications of this wiring scheme is to conceptualize it in three stages, as represented by Figures 3A, 3B, and 3C. First, draw the entire wiring scheme for one channel on paper, following Figure 3A. Next, simplify the diagram by replacing each pair of series-wired speakers – A/B and C/D – with an imaginary equivalent speaker, as shown in Figure 3B. Well call these "combined" drivers Zab and Zcd. Now, reduce these speakers to a single, equivalent driver and call it Zt (Figure 3C). In a nutshell, we have reduced a relatively complex four-speaker system down to one imaginary driver, which represents the total load impedance created by wiring four speakers in a series/parallel combination.

Calculating the load impedance of the series/parallel-wired channel in Figure 3A is a three-step process. First use Equation 1 (Zt = Za + Zb) to find the equivalent-load impedance of Speakers A and B, which are wired in series. Then repeat the process for speakers C and D, changing the variables in Equation 1 (Zt = Zc + Zd). Finally, to find a single, total equivalent-load impedance for the "combined" speakers Zab and Zcd, substitute new variables into Equation 4 [Zt = (Za x Zb) / (Za + Zb) becomes Zt = (Zab x Zcd) / (Zab + Zcd)].

To work through this series of equations, well take our hypothetical subwoofer installation yet another step further. Say the four subwoofers wired in parallel to your new 2-ohm amplifier are no longer good enough. So you buy four more subs – that’s a total of eight. Just to show that it can be done, you decide to stick with the same 2-ohm-stable amplifier, which is rated at 100 watts x 2 into 4 ohms; the new subs are also rated at 4 ohms apiece.

Now what? You could wire four speakers in series to each channel, but this would yield a 16-ohm load. Another option is to wire four speakers in parallel to each channel, but this would yield a dangerously low 1-ohm load. The only practical option, therefore, is to combine the two wiring methods in accordance with Figure 3A. First, you connect two subwoofers in series and then wire that pair in parallel to a second pair, which is also connected in series. Follow the same procedure for the other channel.

The first step in determining the total equivalent-load impedance for each channel is to plug the appropriate impedance values into Equation 1 and work it through for each series-wired speaker pair, A/B and C/D. Plug in the values for speakers A and B and solve Equation 1 as follows:

Zt = Za + Zb
Zt = 4 + 4
Zt = 8ohms

Then repeat the calculation using speakers C and D. Since each of the speakers is rated at 4 ohms, the equivalent-load impedance for each series-wired speaker pair is 8 ohms.

Redraw the circuit diagram and replace speakers A and B with Zab to represent the new equivalent-load impedance. Do the same for speakers C and D to create Zcd. The result should be a one-channel diagram that resembles Figure 3B, with the label "8 ohms" in place of Zab and Zcd.

The next step is to find the total equivalent-load impedance of the channel by plugging the new 8-ohm values for Zab and Zcd into Equation 4, as follows:

Zt = (Zab x Zcd) / (Zab + Zcd)
Zt = (8 x 8) / (8 + 8)
Zt = 64 / 16
Zt = 4 ohms

Sketch a new one-channel diagram showing the total equivalent-load impedance. The drawing should look like Figure 3C, with the label "4 ohms" in place of Zt.

Now we know that four speakers connected to one amplifier channel with series/parallel wiring creates a 4-ohm load. Since the amplifier is rated to deliver 100 watts x 2 into a 4-ohm load, we know that each channel will receive 100 watts. To verify this, we can plug the appropriate numbers into Equation 2 and work it through as follows:

Po = Pr x (Zr / Zt)
Po = 100 x (4 / 4)
Po = 100 x 1
Po = 100 watts

As anticipated, each amplifier channel will pump out 100 watts.

To find out how much power each series-wired speaker pair (equivalent-load speakers Zab and Zcd) will receive, plug in the appropriate numbers and solve Equation 5. We use this equation because Zab and Zcd are parallel-wired to one another. Working with Zab, substitute 100 for Po, 4 for Zt, and 8 for Zab. The calculation goes as follows:

Pab = Po x (Zt / Zab)
Pab = 100 x (4 / 8)
Pab = 100 x 0.5
Pab = 50 watts

The math for Zcd is identical, since both speakers are rated at 4 ohms, so Zab

and Zcd each receive 50 watts of power.

But Zab and Zcd are imaginary drivers, each of which represents a series-wired speaker pair. To figure out how much power each real speaker will receive, work through Equation 3, substituting 50 for Po (the amplifiers output power into Zab and Zcd), 4 for Zn (the speakers rated impedance), and 8 for Zt (the equivalent impedance of Zab and Zcd). The calculation goes as follows:

Pn = Po x (Zn / Zt)
Pn = 50 x (4 / 8)
Pa = 50 x 0.5
Pa = 25 watts

Technically, you need to repeat this process for each driver – B, C, and D – but since each driver in our example is rated at 4 ohms, you’ll get the same results.

Last modified on Saturday, 11 August 2012 11:27
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