wiring parallel or series sensitivity rules?

[Toast_Master]

Hello

I need to clarify these rules, I know how to wire in series and parallel like anyone would but I don't know how this affects sensitivity's


Now for my example, assuming you had an amp that put out 100 watts at 4, 8 and 16 ohms, and a woofer that had an spl or whatever it called at 100 db 8 ohms at 1 metre with 1 watt

Now say we wire one speaker to the amp, its sensitivity will remain 100db/8 ohms/1 meter

Now say we wire 2 of the speakers in parallel making a 4 ohm load, I assume the sensitivity now rises to 103 db because a doubling of cone area at the same watts

Now say we wire 2 in series causing a 16 ohm load, I assume again that the spl now rises to 103 again because of the doubling of cone area,

Based on that does parallel and series sensitivity go by what the amp can out put at certain ohm loads

Or give me a realistic exmaple

Thanks

[roddyama]

The speaker's acoustic output is proportional to the RMS current through the speaker's voice coil. Since the current through each speaker's voice coil is the same (for the same power delivered by the amp) for both the series and parallel arrangement, the acoustic output (sensitivity) is the same in either case. This is of course ignoring the acoustic coupling of 2 or more speakers in proximity to one another.

[myhrrhleine]

Originally posted by Toast_Master
[B]Hello

I need to clarify these rules, I know how to wire in series and parallel like anyone would but I don't know how this affects sensitivity's

When 2 sounds are in phase, the pressures add together.
That means +3dB if the signals are identical, 0dB for random noise to each.

2 speakers in parallel will add 3dB because the sound pressures are in phase, and 3dB because you've gone from 8ohm to 4ohm,
so 6dB overall output increase from the pair.

But in series, 2 speakers will add 3dB because the sound pressures are in phase, but lose 3dB because you've gone from 8ohm to 16ohm thus cutting the power in half.
Overall, then, +3dB - 3dB=0dB

[Cal Weldon]

Originally posted by myhrrhleine
so 6dB overall output increase from the pair.
<snip>
Overall, then, +3dB - 3dB=0dB

He was asking about the sensitivity not the systems reaction to a change in voltage or overall power so 3dB seems right for both answers as we are dealing with a coupled radiating area.

[BlueWizard]

I thought Myhrrhleine did a pretty good job of explaining it. I'll just add a few side notes.

First, if an amp puts out 100w at 8 ohms, it is not going to put that same power out at 4 ohms or 16 ohms. At 8 ohms, 1 watt means a signal voltage of 2.83 volts. To keep things fair, other impedances also use that same signal voltage.

But, for this example, let's use 8 volts because it's easier to calculate.

8 volts applied to 8 ohms is P = E^2/R = 8^2/8 = 64/8 = 8watts.

8 volts applied to 4 ohms is 64/4 = 16 watts.

8 volts applied to 16 ohms is 64/16 = 4 watts.

Think of the applied signal as the position of the Volume Control. When you turn the Volume Control up, you aren't increasing the power, you are increasing the signal voltage and more power is consumed as a result.

Now keeping this voltage, current, and power relationship in context, consider that when two equal speakers are in parallel, they both get the same signal voltage but the resulting total current is divided between them. If the total voltage is 8 volts, then the total current is 2 amps, with 1 amp each going to each speaker.

P = I^2 x R = 1^1 x 8 = 1 x 8 = 8 watts to each speaker.

Or

P = E^2/R = 8^2/8 = 64/8 = 8 watt to each speaker.

But note, 8 watts to each speaker means 16 watts total power consumed.

Now in series both speaker have the same Current but the Voltage is divided between them. Since they are in series, whatever current flows through one MUST flow through the other. However, they make a voltage divider, and if they are equal then voltage is divided equally between them.

To keep the numbers simple, let's put two 4 ohms speakers in series for a total of 8 ohms. Now we apply 8 volts which equals a total current of 1 amp. Both speakers get 1 full amp of current, but each individual speaker only gets 4 volts. Consequently, the power to each individual speaker is P = E^2/R = 4^2/4 = 16/4 = 4 watts to each speaker. 8 watts total to the combined speakers.

In the above Series example, if we use 8 ohms speakers, then the power delivered to each speaker is - keeping in mind that each speaker is getting half the total voltage.

P = E^2/R = 4^2/8 = 2 watts per speaker.

Since there are two speakers, the total power seen by the amp is 4 watts.

In the 8 ohm parallel example, the speakers consume a total of 16 watts with a 8 volt input signal.

In the 8 ohms series example, the speakers consume a total of 4 watts with an 8 volt input signal.

This is just a complex way of saying what Myhrrhleine said, but it explains why the amp power is lower when the speakers are in series.

But again, this is all assuming every combination gets the same input voltage. There is a Volume Control knob on the front of your amp for a reason. If it is not loud enough, you just turn it up.

I think more important than sensitivity or efficiency (100db at 1 watt at 1 meter) is getting the individual component of a given speaker system balanced. Making sure the woofer, midrange, and tweeter are all putting out close to the same perceived output level. Once you have a balanced speaker system, you can just turn the volume control to where ever you need it to be to get the sound intensity you want.

That's probably a lot more than you wanted to know, but to understand what is going on, I think you needed to know it.

If it really is way more than you wanted to know, then just listen to Myhrrhleine and RoddyYama, and ignore me.

Steve/bluewizard

[myhrrhleine]

Originally posted by Cal Weldon


He was asking about the sensitivity not the systems reaction to a change in voltage or overall power so 3dB seems right for both answers as we are dealing with a coupled radiating area.

perhaps you should reread my answer.
When 2 sounds are in phase, the pressures add together.
That means +3dB if the signals are identical, 0dB for random noise to each.


===============

2 speakers in parallel will add 3dB because the sound pressures are in phase, and 3dB because you've gone from 8ohm to 4ohm,
so 6dB overall output increase from the pair.

But in series, 2 speakers will add 3dB because the sound pressures are in phase, but lose 3dB because you've gone from 8ohm to 16ohm thus cutting the power in half.
Overall, then, +3dB - 3dB=0dB


This is important when building a system.
while either case adds 3dB, one case doubles the impedence, and one halfs it resulting in a different output
it includes sensitivity for 2.83volts, not just for 1 watt, because amps are mostly constant voltage

[forr]

Hi,

When the distance between drivers fed by the same signal is much lower than the wavelength, they are is said in mutual coupling.

When two drivers are identical and in mutual coupling, the whole behaves like a single driver having an emitting area equal to twice the area of a single driver.

For an unchanged displacement of each driver, their mutual coupling has the effect of doubling the volume velocity and, then, the acoustic power is mutiplied by 4 (+6 dB).


Drivers in parallel
¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨
If the drivers are connected in parallel to a voltage source (which means an amplifer having a low impedance, the most common case), the displacement of each one is the same as if it was alone.

So, with two drivers in parallel driven by an unchanged voltage gives an acoustic power of +6 dB. The sensitivity for a given voltage (never express sensitivity for 1 W !) has increased by +6 dB.

However, as the impedance seen by the amplifier is halved, its output current, and then the electric power, Pe, is doubled, meanwhile the acoustic power, Pa, is multiplied by 4.
Efficiency, being equal to Pa/Pe, is multplied by 2.
The efficiency of the two drivers in parallel is doubled.



Drivers in series
¨¨¨¨¨¨¨¨¨¨¨¨¨¨
Each driver now sees half of the voltage output by the same amplifier and delivers an acoustic power divided by 4.

However, the mutual coupling of the two drivers gives an acoustic power Pa equal to 4 times the acoustic power of a single driver which has been previously divided by 4.

The sensitivity of identical two drivers in series is unchanged compared to a driver alone.

However, the impedance seen by the amplifier is doubled, the output current is halved and then the electric power, Pe, is halved.
The ratio Pa/Pe is then doubled.
The efficiency of the two drivers in series in doubled.

[forr]

Hi Toast_master,

Realistic example

---woofer that had an spl or whatever it called at 100 db 8 ohms at 1 metre with 1 watt----

Let's say
100 dB/2.83 V at 1 m


---Now say we wire 2 of the speakers in parallel making a 4 ohm load, I assume the sensitivity now rises to 103 db because a doubling of cone area at the same watt----

Forget watts, think volts
Doubling the cone area, four times the acoustic power :
106 dB/2.83V at 1m


---Now say we wire 2 in series causing a 16 ohm load, I assume again that the spl now rises to 103 again because of the doubling of cone area---
100 dB/W at 1 m


In the axial response, the reasoning in dB is still valid at higher frequencies.

[Cal Weldon]

Originally posted by forr
The efficiency of the two drivers in parallel is doubled.

The efficiency of the two drivers in series in doubled.

Thank you, I couldn't agree more, but isn't the term sensitivity?

[roddyama]

Originally posted by forr
... (never express sensitivity for 1 W !) ...

From Rane's Audio Dictionary:
"loudspeaker sensitivity The standard is to apply one watt and measure the sound pressure level (SPL) at a distance of one meter. [IEC 60268-5]"
Note The reference is the IEC Standards.