Hello Everyone. This is a follow-up to a similar post today. I do not want to hi-jack anyone's thread.
OK, ignoring the effects of the amp., if you wire two peakers in parallel, you get +3db gain, correct? What if you wire two speakers in series? 0 db gain? -3 db gain? I know I have read it somewhere but can't find it.
OK, ignoring the effects of the amp., if you wire two peakers in parallel, you get +3db gain, correct? What if you wire two speakers in series? 0 db gain? -3 db gain? I know I have read it somewhere but can't find it.
The 1W sensitivity won't change, but the 2.8V sensitivity will, simply because 2.8V will deliver more power to a lower impedance speaker.
Anyway, it's +3dB for parallel (since impedance is cut in half and voltage remains constant, power is doubled, hence +3dB), and -3dB in series for the same reason.
Anyway, it's +3dB for parallel (since impedance is cut in half and voltage remains constant, power is doubled, hence +3dB), and -3dB in series for the same reason.
Thanks sr20dem0n-
It that case, lets say I have a 3-way system with two woofers, 4 midrange and one tweeter.
The woofers are 8 ohm @ 90 db. If I wire them in parallel, I get 93 dB @ 4 ohms.
The midrange are 87 db @ 4 ohms each. I wire two pair in parallel getting 8 ohm @ 90 db (each). Then wire the two pair in series and get 87 db @ 4 ohm.
Is this correct logic?
It that case, lets say I have a 3-way system with two woofers, 4 midrange and one tweeter.
The woofers are 8 ohm @ 90 db. If I wire them in parallel, I get 93 dB @ 4 ohms.
The midrange are 87 db @ 4 ohms each. I wire two pair in parallel getting 8 ohm @ 90 db (each). Then wire the two pair in series and get 87 db @ 4 ohm.
Is this correct logic?
Actually for some reason I was thinking of wiring a dual coil speaker in series vs parallel, still caught up on car audio
Anyway, for your application, there's an easier way (IMO) to go about doing it. Take the spl and impedance of one speaker, if you were delivering 1 watt to the speaker at first, figure out how much power it's now getting after rewiring it (if you added a speaker in series, the original speaker has now had its input power cut by a factor of 4, so it's 6dB quieter). Then when you figure out what has happened to that single speaker, do 20*log
where 'n' is the number of speakers to figure out the dB increase you get by having multiple speakers playing at that spl. Then just add that to the efficiency of a single speaker to figure out what the total efficiency is.
So in your case, you have an 8ohm woofer getting 1 watt. If you add another in parallel, the power going to the original is not affected, it's still receiving 1 watt, so it is still playing at 90dB. By adding the second speaker you've doubled the displacement which gives you a 20*log(2)=6dB rise, so the pair of woofers will be playing at 96dB collectively.
The midranges are 87dB each, if you wire them in series/parallel, you've cut the power going to each midrange by a factor of 4, so they're playing at 81dB each. Since you have 4 of them, you have a 20*log(4)=12dB rise, so they'll be playing at 93dB collectively.
Anyway, for your application, there's an easier way (IMO) to go about doing it. Take the spl and impedance of one speaker, if you were delivering 1 watt to the speaker at first, figure out how much power it's now getting after rewiring it (if you added a speaker in series, the original speaker has now had its input power cut by a factor of 4, so it's 6dB quieter). Then when you figure out what has happened to that single speaker, do 20*log
where 'n' is the number of speakers to figure out the dB increase you get by having multiple speakers playing at that spl. Then just add that to the efficiency of a single speaker to figure out what the total efficiency is.So in your case, you have an 8ohm woofer getting 1 watt. If you add another in parallel, the power going to the original is not affected, it's still receiving 1 watt, so it is still playing at 90dB. By adding the second speaker you've doubled the displacement which gives you a 20*log(2)=6dB rise, so the pair of woofers will be playing at 96dB collectively.
The midranges are 87dB each, if you wire them in series/parallel, you've cut the power going to each midrange by a factor of 4, so they're playing at 81dB each. Since you have 4 of them, you have a 20*log(4)=12dB rise, so they'll be playing at 93dB collectively.
No problem
Just remember that in the above example you'll need to adjust for power differences. To send 1w to the 8 ohm woofer you would need to send it 2.83 Vrms, and the final efficiency is still based on sending 2.83 Vrms. To send 1w to the 4 ohm midrange you would need to sent it 2 Vrms, and the final efficiency is still based on sending 2 Vrms. To put them on a common playing field you need to normalize the input voltage, which means cutting it down to 2 Vrms for the woofers. This would cut the power going to the woofers in half and lower the woofers' output by 3dB, so it's 93dB for both the woofers and the midranges if they're on a common amp and receiving 1 watt.
Just remember that in the above example you'll need to adjust for power differences. To send 1w to the 8 ohm woofer you would need to send it 2.83 Vrms, and the final efficiency is still based on sending 2.83 Vrms. To send 1w to the 4 ohm midrange you would need to sent it 2 Vrms, and the final efficiency is still based on sending 2 Vrms. To put them on a common playing field you need to normalize the input voltage, which means cutting it down to 2 Vrms for the woofers. This would cut the power going to the woofers in half and lower the woofers' output by 3dB, so it's 93dB for both the woofers and the midranges if they're on a common amp and receiving 1 watt.
First of all you have to make the assumption which is correct for most solid state amplifiers that output voltage won't vary connecting the speakers in parallel or in series.
For both cases of parallel and series wiring cone area is doubled which doubles SPL or increases it by 3 dB.
For parallel wiring, impedance is reduced by one-half causing wattage consumed by the speaker to double or increase by 3 dB. Therefore for parallel wiring SPL is increased by four times or 6 dB.
For series wiring, impedance is increased by a factor of two and and wattage consumed by the speaker is decreased by one-half or -3 dB. Therefore for series wiring the two effects cancel and SPL is unchanged.
For both cases of parallel and series wiring cone area is doubled which doubles SPL or increases it by 3 dB.
For parallel wiring, impedance is reduced by one-half causing wattage consumed by the speaker to double or increase by 3 dB. Therefore for parallel wiring SPL is increased by four times or 6 dB.
For series wiring, impedance is increased by a factor of two and and wattage consumed by the speaker is decreased by one-half or -3 dB. Therefore for series wiring the two effects cancel and SPL is unchanged.
MCPete said:
Not exactly
Doubling displacement increases output by 6dB, not 3dB, for parallel wiring this takes into account the extra power required to push a second speaker to the same level of displacement as the first. When wiring in series, you cut the power going to the speaker by a factor of 4, but you also double displacement, so the -6dB and +6dB cancel.
Another thing is it depends on if you're talking about voltage sensitivity or power sensitivity. For a 1 watt input, wiring in either series or parallel will increase 1 watt efficiency by 3dB (each speaker gets 1/2 watt, so that's -3dB, and you have 2 speakers which doubles the displacement, so that's +6dB, regardless of how they're wired). For voltage sensitivity, wiring in parallel will increase efficiency by 6dB and wiring in series will leave efficiency unchanged.
surface area * excursion = displacement
surface area is only one part of spl, it also depends on how far the cone is moving, which is proportional to the voltage applied. If you go from one speaker getting 1 watt to 2 speakers getting 1 watt total, the power being delivered to the original speaker has been cut in half, or in other words the voltage has dropped to 70% of the initial, which means the excursion has dropped to 70% of the initial, which means the displacement has dropped to 70% of the initial. Since you have 2 speakers receiving this amount of power each, you end up with 140% of the initial displacement, which is +3dB. Or if you go from one speaker getting 1 watt to 2 speakers getting 1 watt each, the displacement of the original speaker remains the same, so by adding the second you end up with 200% of the initial displacement, which is +6dB.
surface area is only one part of spl, it also depends on how far the cone is moving, which is proportional to the voltage applied. If you go from one speaker getting 1 watt to 2 speakers getting 1 watt total, the power being delivered to the original speaker has been cut in half, or in other words the voltage has dropped to 70% of the initial, which means the excursion has dropped to 70% of the initial, which means the displacement has dropped to 70% of the initial. Since you have 2 speakers receiving this amount of power each, you end up with 140% of the initial displacement, which is +3dB. Or if you go from one speaker getting 1 watt to 2 speakers getting 1 watt each, the displacement of the original speaker remains the same, so by adding the second you end up with 200% of the initial displacement, which is +6dB.
response to post # 8
"Doubling displacement increases output by 6 dB, not 3 dB..."- My post says that doubling cone area increases SPL by +3 dB, not displacement. As you correctly say, as cone excursion remains constant when the pair of speakers are connected in parallel, then SPL relative to that of one of the pair acting alone is increased by +6 dB.
"When wiring in series, you cut the power going to the speaker by a factor of 4, but you also double displacement,..."- Connecting the speakers in series reduces power converted by the series-connected pair by a factor of 2, not 4. Power equals the square of the applied voltage divided by the impedance. The applied voltage remaining constant, connecting in series doubles impedance and therefore also divides wattage by a factor of 2. Ten times the log base 10 of 0.5 equals -3 dB, not -6 dB.
Also for connecting in series, compared to displacement produced by only one of the pair connected to the amplifier, displacement remains the same which is what it means to say that connecting in series produces no net change of SPL (I think that you know better & you just made a slip.). Cone area has been doubled, but excursion has been reduced by one-half. Twice the area of air is moved, but that doubled mass of air is moved one-half the distance that a single speaker would move it driven by the amp's full voltage.
Your discussion of "power sensitivity" (I'm not sure exactly what you are getting at with that phrase) is interesting. I agree that the same result is obtained for both series & parallel connecting where amplifier gain is adjusted so that output power stays constant for all three conditions (single speaker and parallel or series connected pair). However I would say that SPL stays the same for all three conditions.
With the voltage drop across each of the pair of either series or parallel connected speakers 0.7 times that which would be applied across the single speaker- arrangement, displacement by each of the pair is one-half of that of the speaker singly driven. That is, SPL of each of the pair is reduced by 20 times log base 10 of 0.7 or -3 dB. Therefore, as the acoustic outputs of the pair are additive, there is no gain or loss of SPL.
"Doubling displacement increases output by 6 dB, not 3 dB..."- My post says that doubling cone area increases SPL by +3 dB, not displacement. As you correctly say, as cone excursion remains constant when the pair of speakers are connected in parallel, then SPL relative to that of one of the pair acting alone is increased by +6 dB.
"When wiring in series, you cut the power going to the speaker by a factor of 4, but you also double displacement,..."- Connecting the speakers in series reduces power converted by the series-connected pair by a factor of 2, not 4. Power equals the square of the applied voltage divided by the impedance. The applied voltage remaining constant, connecting in series doubles impedance and therefore also divides wattage by a factor of 2. Ten times the log base 10 of 0.5 equals -3 dB, not -6 dB.
Also for connecting in series, compared to displacement produced by only one of the pair connected to the amplifier, displacement remains the same which is what it means to say that connecting in series produces no net change of SPL (I think that you know better & you just made a slip.). Cone area has been doubled, but excursion has been reduced by one-half. Twice the area of air is moved, but that doubled mass of air is moved one-half the distance that a single speaker would move it driven by the amp's full voltage.
Your discussion of "power sensitivity" (I'm not sure exactly what you are getting at with that phrase) is interesting. I agree that the same result is obtained for both series & parallel connecting where amplifier gain is adjusted so that output power stays constant for all three conditions (single speaker and parallel or series connected pair). However I would say that SPL stays the same for all three conditions.
With the voltage drop across each of the pair of either series or parallel connected speakers 0.7 times that which would be applied across the single speaker- arrangement, displacement by each of the pair is one-half of that of the speaker singly driven. That is, SPL of each of the pair is reduced by 20 times log base 10 of 0.7 or -3 dB. Therefore, as the acoustic outputs of the pair are additive, there is no gain or loss of SPL.
Hopefully this will explain it better to those confused:
http://www.linkwitzlab.com/faq.htm#Q21
Cheers,
Davey.
http://www.linkwitzlab.com/faq.htm#Q21
Cheers,
Davey.
Re: response to post # 8
My post says that doubling cone area increases SPL by +3 dB, not displacement. As you correctly say, as cone excursion remains constant when the pair of speakers are connected in parallel, then SPL relative to that of one of the pair acting alone is increased by +6 dB.
You're right, but you can't compare dB changes while only looking at cone area, because output is proportional to the product of cone area and excursion. I always look at the whole picture, it gives a much clearer idea of what's going on IMO. Your results weren't wrong, but it's not a very easy-to-understand way of looking at it, especially for someone that's new at this kind of thing.
Connecting the speakers in series reduces power converted by the series-connected pair by a factor of 2, not 4. Power equals the square of the applied voltage divided by the impedance. The applied voltage remaining constant, connecting in series doubles impedance and therefore also divides wattage by a factor of 2. Ten times the log base 10 of 0.5 equals -3 dB, not -6 dB.
As I said before, I look at what happens to a single speaker, and then adjust for how many speakers you now have playing. When you're going from 1 speaker to 2 speakers wired in series, the power being delivered to the original speaker will be cut by a factor of 4, -6dB. Then since you have two speakers playing at this level, the total displacement is twice that of the one speaker, +6dB. Net change is 0dB, as we both said.
Your discussion of "power sensitivity" (I'm not sure exactly what you are getting at with that phrase) is interesting. I agree that the same result is obtained for both series & parallel connecting where amplifier gain is adjusted so that output power stays constant for all three conditions (single speaker and parallel or series connected pair). However I would say that SPL stays the same for all three conditions.
With the voltage drop across each of the pair of either series or parallel connected speakers 0.7 times that which would be applied across the single speaker- arrangement, displacement by each of the pair is one-half of that of the speaker singly driven. That is, SPL of each of the pair is reduced by 20 times log base 10 of 0.7 or -3 dB. Therefore, as the acoustic outputs of the pair are additive, there is no gain or loss of SPL.
What I was getting at is the fact that how the efficiency changes when wiring in series or parallel depends on which efficiency you're looking at. Most data sheets will give a 1 watt sensitivity and a 2.83V sensitivity. For an 8ohm speaker these will be the same, but for a 4ohm speaker the 2.83V rating will be 3dB higher. I was just pointing out that the whole +6dB in parallel, +0dB in series only applies to the voltage sensitivity. It really didn't have much to do with the thread, I was just rambling.
Anyway, yes it will go up 3dB in either series or parallel, it won't stay the same. You're right, if you wire in series or parallel and readjust the input to stay at 1w, you will have cut the voltage being applied to the original speaker by 30% and it will be playing at -3dB. But you have two speakers playing at this level, which gives +6dB. The net change is +3dB. Or just take the examples from before which we both agree upon. If you wire 2 speakers in parallel you get +6dB, but power has doubled. If you cut the power in half to go back to what it was, you're down to +3dB. Or in series you get +0dB, but power has been cut in half. If you double the power to go back to what it was, you're up to +3dB.
My post says that doubling cone area increases SPL by +3 dB, not displacement. As you correctly say, as cone excursion remains constant when the pair of speakers are connected in parallel, then SPL relative to that of one of the pair acting alone is increased by +6 dB.
You're right, but you can't compare dB changes while only looking at cone area, because output is proportional to the product of cone area and excursion. I always look at the whole picture, it gives a much clearer idea of what's going on IMO. Your results weren't wrong, but it's not a very easy-to-understand way of looking at it, especially for someone that's new at this kind of thing.
Connecting the speakers in series reduces power converted by the series-connected pair by a factor of 2, not 4. Power equals the square of the applied voltage divided by the impedance. The applied voltage remaining constant, connecting in series doubles impedance and therefore also divides wattage by a factor of 2. Ten times the log base 10 of 0.5 equals -3 dB, not -6 dB.
As I said before, I look at what happens to a single speaker, and then adjust for how many speakers you now have playing. When you're going from 1 speaker to 2 speakers wired in series, the power being delivered to the original speaker will be cut by a factor of 4, -6dB. Then since you have two speakers playing at this level, the total displacement is twice that of the one speaker, +6dB. Net change is 0dB, as we both said.
Your discussion of "power sensitivity" (I'm not sure exactly what you are getting at with that phrase) is interesting. I agree that the same result is obtained for both series & parallel connecting where amplifier gain is adjusted so that output power stays constant for all three conditions (single speaker and parallel or series connected pair). However I would say that SPL stays the same for all three conditions.
With the voltage drop across each of the pair of either series or parallel connected speakers 0.7 times that which would be applied across the single speaker- arrangement, displacement by each of the pair is one-half of that of the speaker singly driven. That is, SPL of each of the pair is reduced by 20 times log base 10 of 0.7 or -3 dB. Therefore, as the acoustic outputs of the pair are additive, there is no gain or loss of SPL.
What I was getting at is the fact that how the efficiency changes when wiring in series or parallel depends on which efficiency you're looking at. Most data sheets will give a 1 watt sensitivity and a 2.83V sensitivity. For an 8ohm speaker these will be the same, but for a 4ohm speaker the 2.83V rating will be 3dB higher. I was just pointing out that the whole +6dB in parallel, +0dB in series only applies to the voltage sensitivity. It really didn't have much to do with the thread, I was just rambling.
Anyway, yes it will go up 3dB in either series or parallel, it won't stay the same. You're right, if you wire in series or parallel and readjust the input to stay at 1w, you will have cut the voltage being applied to the original speaker by 30% and it will be playing at -3dB. But you have two speakers playing at this level, which gives +6dB. The net change is +3dB. Or just take the examples from before which we both agree upon. If you wire 2 speakers in parallel you get +6dB, but power has doubled. If you cut the power in half to go back to what it was, you're down to +3dB. Or in series you get +0dB, but power has been cut in half. If you double the power to go back to what it was, you're up to +3dB.
Oops! It's not possible that SPL of series connecting is the same maintaining constant amplifier voltage or wattage output. I agree with sr20dem0n that where constant wattage is maintained, then SPL relative to that of the single speaker is +3 dB for both parallel and series connecting. That is, current through the voice coils of the pair of speakers in both cases is reduced by the factor 0.7. Therefore displacement is increased by the factor 0.7 times 2 equals 1.4 and
20 times log(base 10) of 1.4 = +3 dB.
My apologies to all participants for my at least initially hazy analysis.
20 times log(base 10) of 1.4 = +3 dB.
My apologies to all participants for my at least initially hazy analysis.
my post #7
I've noticed that my method of solving SPL for the four different cases we've considered given by my post #7 works. That is, if you assign +3 dB to two identical speakers both connected to the output voltage corresponding to a doubling of cone area and then add to that, in decibels, the net change of TOTAL electrical power that is converted to an acoustic output, you arrive at the correct result.
This can be explained as follows. SPL is directly proportional to the rate of work of pushing the air around. Doubling cone area doubles the rate of doing work (or wattage) as for some unit of excursion twice the number of air molecules are pushed. Calculating a change of wattage in decibels is 10 times the log of the ratio of wattages of the two states being compared.
10 times log 2 = +3 dB.
Next add the the change of the TOTAL amount of electrical power that undergoes electroacoustical conversion and you arrive at what we've agreed actually occurs.
Maintaining constant voltage, for parallel and series connecting, electrical power respectively increases (+) by 3 dB and decreases (-) by 3 dB. Therefore SPL respectively becomes +6 dB and 0 dB.
Maintaining constant wattage, for parallel and series connecting, there is no change of electrical wattage being converted and the change of SPL in both cases is +3 dB.
I've noticed that my method of solving SPL for the four different cases we've considered given by my post #7 works. That is, if you assign +3 dB to two identical speakers both connected to the output voltage corresponding to a doubling of cone area and then add to that, in decibels, the net change of TOTAL electrical power that is converted to an acoustic output, you arrive at the correct result.
This can be explained as follows. SPL is directly proportional to the rate of work of pushing the air around. Doubling cone area doubles the rate of doing work (or wattage) as for some unit of excursion twice the number of air molecules are pushed. Calculating a change of wattage in decibels is 10 times the log of the ratio of wattages of the two states being compared.
10 times log 2 = +3 dB.
Next add the the change of the TOTAL amount of electrical power that undergoes electroacoustical conversion and you arrive at what we've agreed actually occurs.
Maintaining constant voltage, for parallel and series connecting, electrical power respectively increases (+) by 3 dB and decreases (-) by 3 dB. Therefore SPL respectively becomes +6 dB and 0 dB.
Maintaining constant wattage, for parallel and series connecting, there is no change of electrical wattage being converted and the change of SPL in both cases is +3 dB.
need help
hii all
i have sony amplifier xm-504z 4 channel amp. that is stable at 2 ohm
and i have 4 speaker pioneer TS-A6971e 6x9 350w 4 ohm impedance
so, i'm planing to hook up 2 speaker at one channel
each speaker 4 ohm impedance when hook it in parallel it'll be 2 ohm total impedance
i was asking that the speaker support 2 ohm ?????
or it will blow up?
hii all
i have sony amplifier xm-504z 4 channel amp. that is stable at 2 ohm
and i have 4 speaker pioneer TS-A6971e 6x9 350w 4 ohm impedance
so, i'm planing to hook up 2 speaker at one channel
each speaker 4 ohm impedance when hook it in parallel it'll be 2 ohm total impedance
i was asking that the speaker support 2 ohm ?????
or it will blow up?
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