Regarding series wiring (2) 4ohm speakers to get 8ohms
vs
parallel wiring (2) 16ohm speakers to get 8ohms.
When series wiring I get a 3db drop in SPL due to the increased resistance. When parallel wiring I get a 3bd increase in SPL, right?
I'll use 2 drivers (PE's buyout section) for examples.
A 10", 4ohm driver with an SPL of 93db. When wired in series the SPL of the pair would now be 90db.
An 8", 16ohm driver with 83db SPL. When wired in parallel the SPL of this pair would now be 86db.
Now since there would be 4 of each driver (2 per side), would there be any additional increase in SPL because of quantity. If I understand correctly, the (4) 10"'s would be back up to 93db while the (4) 8"s would stay at 86db or would they be 89db (86db + 3db more because of the additional pair?
R/
Jim
vs
parallel wiring (2) 16ohm speakers to get 8ohms.
When series wiring I get a 3db drop in SPL due to the increased resistance. When parallel wiring I get a 3bd increase in SPL, right?
I'll use 2 drivers (PE's buyout section) for examples.
A 10", 4ohm driver with an SPL of 93db. When wired in series the SPL of the pair would now be 90db.
An 8", 16ohm driver with 83db SPL. When wired in parallel the SPL of this pair would now be 86db.
Now since there would be 4 of each driver (2 per side), would there be any additional increase in SPL because of quantity. If I understand correctly, the (4) 10"'s would be back up to 93db while the (4) 8"s would stay at 86db or would they be 89db (86db + 3db more because of the additional pair?
R/
Jim
2litre said:
When you wire 2 drivers in parallel you'll get an increase of 6dB (half the resistance (+3) and double the number of drivers (+3)) compared to one single driver.
When wiring in series there will be a loss of 3 dB (double the resistance (-3), each driver gets half of the applied voltage (-3) but there's a doubling of drivers (+3)) compared to one driver.
When you wire the 4 ohms drivers in series you'll get a SPL of 93-3 = 90 dB. As you mentioned, you're going to use four drivers. The SPL will increase with 3 dB (twice the number of drivers), so finally this turns out to 90+3= 93 dB.
When you wire the 16 ohms drivers in parallel the SPL becomes 83+6 = 89 dB. Because of doubling the number of drivers, the SPL increases with another 3 dB, so you'll end up with 89+3 = 92 dB.
regards
Erwin
Re: Re: series vs parallel SPL?
I believe your second paragraph contradicted your first.
I believe the first paragraph is correct.
When he takes two 4 ohm, 93 dB @ 2.83V/1M speakers and wires them in series, he will have the same sensitivity, 93dB @ 2.83V/1M.
Since the single 4 ohms speaker draws 2 watts when fed 2.83 Volts, and the series pair draws 1 watt at 2.83 Volts, he will be cutting his power in half when he drives the series pair. However, the series pair has twice the cone area of a single driver, and that increases efficiency to compensate.
Same goes for any voltage drive level. The series pair will be the same sensitivity as the single driver for any given voltage level, but will draw only half the power at that voltage level, whatever it is.
e-side said:
I believe your second paragraph contradicted your first.
I believe the first paragraph is correct.
When he takes two 4 ohm, 93 dB @ 2.83V/1M speakers and wires them in series, he will have the same sensitivity, 93dB @ 2.83V/1M.
Since the single 4 ohms speaker draws 2 watts when fed 2.83 Volts, and the series pair draws 1 watt at 2.83 Volts, he will be cutting his power in half when he drives the series pair. However, the series pair has twice the cone area of a single driver, and that increases efficiency to compensate.
Same goes for any voltage drive level. The series pair will be the same sensitivity as the single driver for any given voltage level, but will draw only half the power at that voltage level, whatever it is.
Re: Re: Re: series vs parallel SPL?
I may be wrong, but when wiring resistors in series the applied voltage to the circuit will be divided by the number of resistors.
So when you wire two resistors in series, and you apply 5 Volts to the circuit, each resistor gets 5/2 = 2,5 Volts. The same goes up for loudspeakers (I suppose).
So when you wire two 4 Ohm drivers with a SPL of 93 dB in series, the following happens:
1) - Doubling of the resistance compared to one driver, so 3 dB less;
2) - Because of wiring two drivers in series the voltage applied to the circuit gets divided by the number of drivers, so each driver gets the half of the actual applied voltage to the circuit. This results in another decrease of sound level (I think 3 dB)
3) - Because twice the number of drivers is used, SPL increases with 3 dB
-------------------------
1) 93- 3 = 90 dB;
2) 90- 3(?) = 87 dB;
3) 87+ 3 = 90 dB = final SPL when those two 4 ohms drivers in series.
I'm not exactly sure, but I think that point (2) should be taken into account, causing a different result.
regards
Erwin
kelticwizard said:
I may be wrong, but when wiring resistors in series the applied voltage to the circuit will be divided by the number of resistors.
So when you wire two resistors in series, and you apply 5 Volts to the circuit, each resistor gets 5/2 = 2,5 Volts. The same goes up for loudspeakers (I suppose).
So when you wire two 4 Ohm drivers with a SPL of 93 dB in series, the following happens:
1) - Doubling of the resistance compared to one driver, so 3 dB less;
2) - Because of wiring two drivers in series the voltage applied to the circuit gets divided by the number of drivers, so each driver gets the half of the actual applied voltage to the circuit. This results in another decrease of sound level (I think 3 dB)
3) - Because twice the number of drivers is used, SPL increases with 3 dB
-------------------------
1) 93- 3 = 90 dB;
2) 90- 3(?) = 87 dB;
3) 87+ 3 = 90 dB = final SPL when those two 4 ohms drivers in series.
I'm not exactly sure, but I think that point (2) should be taken into account, causing a different result.
regards
Erwin
Erwin:
Let's try this a different way.
Let's compare a ten inch single unit speaker to a series unit speaker, as if the two speakers in series are one speaker.
To do this, we use Bullock and White's Dos BoxModel, where the user inputs certain speaker parameters in the dark blue area illustrated below, and the aqua green area shows the calculated values. One of these calculated values is sensitivity per watt-not per 2.83 volts.
Forget the box section for now-we are just dealing with the driver.
Bullock and White's program, available here is very useful in this respect. Note: Ignore the TL program-they admit that it was an experiment which failed. But the Dos BoxModel for closed, vented, and PR is just fine.
We take a typical 10 inch speaker with a sensitivity of 93 dB per watt, and fill in typical Thiele-Small values for it below:
Let's try this a different way.
Let's compare a ten inch single unit speaker to a series unit speaker, as if the two speakers in series are one speaker.
To do this, we use Bullock and White's Dos BoxModel, where the user inputs certain speaker parameters in the dark blue area illustrated below, and the aqua green area shows the calculated values. One of these calculated values is sensitivity per watt-not per 2.83 volts.
Forget the box section for now-we are just dealing with the driver.
Bullock and White's program, available here is very useful in this respect. Note: Ignore the TL program-they admit that it was an experiment which failed. But the Dos BoxModel for closed, vented, and PR is just fine.
We take a typical 10 inch speaker with a sensitivity of 93 dB per watt, and fill in typical Thiele-Small values for it below:
Attachments
Now, in order to combine two speakers into one, we change three things:
We double the Re - It goes from 3.0 ohms to 6.0 ohms.
We double Sd, the Cone Area. It goes from 350 sq cm, (56 sq inches) to 700 sq cm, (112 sq inches).
We double the Vas. Everyone knows you must double the box volume when you add another speaker to get the same response. So Vas goes from 80 L, (2.8 cu ft), to 160 L, (5.7 cu ft).
What happens when we do this? We find our sensitivity per watt jumps from 93 dB to 96 dB.
Of course, since our 8 ohm somposite spekaer is an 8 ohmer, it will get only hald the power delivered to it that the 4 ohm single speaker will, so the SPL [per 2.83 Volts will equal the 4 ohm speaker. Even though it is only drawing half the power. Because the the 8 ohm composite speaker is 3 db more sensitive per watt.
We double the Re - It goes from 3.0 ohms to 6.0 ohms.
We double Sd, the Cone Area. It goes from 350 sq cm, (56 sq inches) to 700 sq cm, (112 sq inches).
We double the Vas. Everyone knows you must double the box volume when you add another speaker to get the same response. So Vas goes from 80 L, (2.8 cu ft), to 160 L, (5.7 cu ft).
What happens when we do this? We find our sensitivity per watt jumps from 93 dB to 96 dB.
Of course, since our 8 ohm somposite spekaer is an 8 ohmer, it will get only hald the power delivered to it that the 4 ohm single speaker will, so the SPL [per 2.83 Volts will equal the 4 ohm speaker. Even though it is only drawing half the power. Because the the 8 ohm composite speaker is 3 db more sensitive per watt.
Attachments
One interesting thing I never realized until now. When we combine the two speakers together, and therefore double Re, Sd (Cone Area), and Vas, the the calculated BL rating also doubles from 7.99 to 15.99.
Makes sense. When we combine two speakers, we are also combining two voice coil and magnet systems as well.
Makes sense. When we combine two speakers, we are also combining two voice coil and magnet systems as well.
Attached is a visual of (8) 16ohm drivers that are 100dB in two configurations to end up with 8ohm. The figures in the visual represent power conversion only (i.e 3db), but even if acoustical conversions are entered, you'll still arrive at 112dB for both.
Both configurations end up at 8ohm, but one group has 6 parallel groups, and one has 3 parallel groups.
Which configuration is more efficient?
Both configurations end up at 8ohm, but one group has 6 parallel groups, and one has 3 parallel groups.
Which configuration is more efficient?
vadi said:
Probably yes, but adding a resistor in series with the woofer will change the parameters of the woofer (Qes and Qts).
When your woofer has, for example, the following parameters:
Re = 8
Qms=4
Qes=0,4
Qts = 0,36
When you wire a 4 ohm resistor in series with the woofer, I believe Qes and Qts change as follow:
Qes' = Qes(1+(resistor/Re)) = 0,4 (1+(4/8)) = 0,6
Qts' = (Qms . Qes') / (Qms + Qes') = (4 . 0,6) / (4 + 0,6) = 0,52
I'm not sure the above will happen when you wire a resistor in parallel with the woofer, but the impedance changes because of the resistor.
regards
Erwin
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