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
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
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.
Toast_Master said:
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
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
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
Re: Re: Re: wiring parallel or series sensitivity rules?
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
Cal Weldon said:
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
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.
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.
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.
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.
Here is the odd thing about that sensitivity standard, if you really are using one watt for all impedances, then you are applying a different signal voltage in each of the three cases we have discussed.
Working backwards, knowing impedance and power, we can find voltage.
E = SqRt(P x R) where SqRt() = square root
8 ohms
E = SqRt( 1w x 8 ohms) = SqRt(8) = 2.828 volts
4 ohms
E = SqRt(1 x 4) = SqRt(4) = 2 volts
16 ohms
E = SqRt (1 x 16) = SqRt(16) = 4 volts
That brings up the question, which is more fair, to test all speakers with the same signal voltage, or to test all speaker with the same power consumption?
Actually, various manufacturers do it differently, some always use 2.83 volts, other use 1 watt.
Personally, I think the same signal level is the more accurate test. It is going to give me a sensitivity relative to a fix setting of the volume control.
Still for this question, maybe actually power is the better test.
Confusing.
Steve/bluewizard
Working backwards, knowing impedance and power, we can find voltage.
E = SqRt(P x R) where SqRt() = square root
8 ohms
E = SqRt( 1w x 8 ohms) = SqRt(8) = 2.828 volts
4 ohms
E = SqRt(1 x 4) = SqRt(4) = 2 volts
16 ohms
E = SqRt (1 x 16) = SqRt(16) = 4 volts
That brings up the question, which is more fair, to test all speakers with the same signal voltage, or to test all speaker with the same power consumption?
Actually, various manufacturers do it differently, some always use 2.83 volts, other use 1 watt.
Personally, I think the same signal level is the more accurate test. It is going to give me a sensitivity relative to a fix setting of the volume control.
Still for this question, maybe actually power is the better test.
Confusing.
Steve/bluewizard
Toast_Master said:
This issue is covered very well by Martin J King, in his articles. It is actually very straightforward :-
SPL is a vector quantity (like voltage), not a scalar quantity (like power). So, if you feed the same signal to two drive units in phase, you will get an SPL 6dB higher than either of them individually.
So, in a parallel connection, you get +6dB more.
But in a series connection, the available voltage is divided equally between them, giving a 6dB loss, so you are then back to 0dB overall.
Actually using either voltage or power as the fixed source is valid as long as it is stated and is common in any comparisons made. There are upsides and down sides to using either.
Power is used in the IEC standards and power is more "real world" since the typical SS amplifiers deliver power to a speaker in proportions to a given input voltage regardless of the speakers impedance. Also flat response from a speaker is in response to a fixed power input not voltage. Chances are slim that you will ever apply 2.83V to your speakers at home unless your amp is broken.
However, design engineers want a vector quantity such as voltage to use in their calculations to deal with phase. This becomes necessary in multi-speakers systems where phase is a crucial components of the end design. It is also nice for the marketing departments that get a few extra db in the sensitivity specifications when 2.83V is used instead of 1W. This makes the speaker look more efficient when they're trying to sell speakers with less than 8ohms at the spec frequency (which is most commercial speakers today).
If you're going to make these measurements at home using the voltage standard you will need an amplifier on which you can adjust the output voltage to 2.83V. In any case, make sure you know what standard is being used in any comparisons you make.
Power is used in the IEC standards and power is more "real world" since the typical SS amplifiers deliver power to a speaker in proportions to a given input voltage regardless of the speakers impedance. Also flat response from a speaker is in response to a fixed power input not voltage. Chances are slim that you will ever apply 2.83V to your speakers at home unless your amp is broken.
However, design engineers want a vector quantity such as voltage to use in their calculations to deal with phase. This becomes necessary in multi-speakers systems where phase is a crucial components of the end design. It is also nice for the marketing departments that get a few extra db in the sensitivity specifications when 2.83V is used instead of 1W. This makes the speaker look more efficient when they're trying to sell speakers with less than 8ohms at the spec frequency (which is most commercial speakers today).
If you're going to make these measurements at home using the voltage standard you will need an amplifier on which you can adjust the output voltage to 2.83V. In any case, make sure you know what standard is being used in any comparisons you make.
sonic difference?
hi, i'm sorry to go off topic somewhat, but are there any advantages other than sensitivity and ohms rating by wiring speaker in parallel vs in series, are there any sonic advantages between series and parallel wiring?
i have 4 saba greencone midrange drivers, and i dont know if i should just use one driver per side or 2 drivers per side wired in series. i can not wire it in parallel because they are 4ohms each.
thanks for all the expert help.
hi, i'm sorry to go off topic somewhat, but are there any advantages other than sensitivity and ohms rating by wiring speaker in parallel vs in series, are there any sonic advantages between series and parallel wiring?
i have 4 saba greencone midrange drivers, and i dont know if i should just use one driver per side or 2 drivers per side wired in series. i can not wire it in parallel because they are 4ohms each.
thanks for all the expert help.
And just for a bit of fun, the link is to a little Excel file that calculates impedance and gain of sets of drivers wired in series then paralleled. ie for a line array.
http://members.optusnet.com.au/~gradds55/xls_files/array calc V3.xls
right click and save as
ps.. if you want to view the formulas etc, just "unlock" the protection.
http://members.optusnet.com.au/~gradds55/xls_files/array calc V3.xls
right click and save as
ps.. if you want to view the formulas etc, just "unlock" the protection.
Re: sonic difference?
The only real issue is if the drivers in series are significantly different in spec. eg quality control. They may then start trying to influence each other. This may also apply if wired in parallel, where impedance difference between drivers may alter the balance at different frequencies etc.
Another benefit of a pair wired in series or in parallel is that each does less work.
Doing less work is good !!
rl guy said:
The only real issue is if the drivers in series are significantly different in spec. eg quality control. They may then start trying to influence each other. This may also apply if wired in parallel, where impedance difference between drivers may alter the balance at different frequencies etc.
Another benefit of a pair wired in series or in parallel is that each does less work.
Doing less work is good !!
differet size drivers (aka wilson speakers) 12 & 10) in parallel
interested in creating a woofer system to play up to 200hz... would like to combine SS 13inch (90db) with SS 10inch (88.5db) ... they have same 8ohm... what i want to do is have these in the same box connected in parallel, with a highpass on 10inch so that it works 60hz up to 200hz and have either a low pass on the 12 at 100hz or just let it also get the same 200hz upper cut-off. How difficult is that?
interested in creating a woofer system to play up to 200hz... would like to combine SS 13inch (90db) with SS 10inch (88.5db) ... they have same 8ohm... what i want to do is have these in the same box connected in parallel, with a highpass on 10inch so that it works 60hz up to 200hz and have either a low pass on the 12 at 100hz or just let it also get the same 200hz upper cut-off. How difficult is that?
zead,
There is no such school like doing it actually and
process the data afterwards. If you're into making
something like Wilson, I say go for it. If not, you can
do very well by just using single 13" woofer and with
its spl one should be more than pleased.
Mr.Wilson likes to make impressions by building
speakers with multiple woofers and it certainly has its
effect. We at diy community have the means, know how
and imagination to do the same if not better and be happy
not to have to spend a fortune to achieve similar results.
There is no such school like doing it actually and
process the data afterwards. If you're into making
something like Wilson, I say go for it. If not, you can
do very well by just using single 13" woofer and with
its spl one should be more than pleased.
Mr.Wilson likes to make impressions by building
speakers with multiple woofers and it certainly has its
effect. We at diy community have the means, know how
and imagination to do the same if not better and be happy
not to have to spend a fortune to achieve similar results.