What does series wiring do to sensitivity?

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- Thread starter Rory
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What does series wiring do to sensitivity?

As a result, the total impedance is twice, and the global sensitivity of the system is half (- 3db). If the drivers are identical, the frequency response of the system is close to the individual frequency response of each driver.

If the drivers are not identical, things becomes complicated !

Regards, P.Lacombe

When you wire drivers in series, if they are identical drivers then the measured response is identical to the single drive unit, both in frequency response AND sensitivity. If it was 90dB for 2.83volts for one driver, it will be exactly the same for the two wired in series. However the impedance will be exactly double, so the efficiency will have increased by 3dB.

You are wrong... Il you connect two identical drivers in series, voltage across each is half, intensity is half so each driver is 6 db down. But they are two drivers, so the total sound level is 3 db up. Finally, the resultant sound level -6 +3 = -3 db.

Regards, P.Lacombe.

Hi P.Lacombe, I don't wish to come of as being impolite,so I apologise if needed beforehand as I am writing this quickly before departing for the CEDIA show tomorrow, but I am not wrong at all. Having been a practicing speaker engineer for the last 25 years and worked with the best, I am qualified to speak on this issue. Both theoretically and practically two identical drivers when wired in series will have the same frequency response and sensitivity as the single driver. I have verified this by measurement many many times. I can supply you with the measurement data if you wish. Perhaps you are confusing sensitivity with efficiency and radiated power, but the internationally agreed upon definition of sensitivity for a loudspeaker is farfield sound pressure at a defined distance and axis for an input voltage of 2.83 volts.

regards

Andrew

I beg your leniency because my native language is french, and it's possible that I don't understand exactly what you mean. I apologize for misspelling, syntax errors.

On the other hand, I'm working with amplifiers, loudspeakers, and measuring instruments for over 40 years... but I think that I'm not yet doddering ;-)

Suppose that you connect a driver, nominal impedance 8 Ohms, at the output of an amplifier delivering a 2.83 V standard voltage, the input power in the driver is 2.83 * 2.83 / 8 = 1 Watt

Suppoose that the driver produces 90 dBa in the axis, at 1 m, whith this input power.

If you connect a second driver in series, the voltage across each driver becomes 2.83 / 2 = 1.415 v and the input power in each driver : 1.415 * 1.415 / 8 = 0.25 Watt. Each driver will produce an acoustical level of 90 - 6 = 84 dBa at 1 m.

Since they are two drivers, the total acoustical level will be 84 + 3 = 87 dBa at 1 m. If you actually measure this level with a microphone, the final result can vary between 90 and (theoretical) -infinite dBa, dependig on the phase relationship, according to the position of the measuring microphone, the wavelength of the sound, and the distance between the drivers.

Regards, P.Lacombe.

[Edited by P.Lacombe on 09-06-2001 at 05:06 PM]

Here is an example of typical but basically wrong answer to this question:

“If you connect two identical drivers in parallel, and the driving amplifier has enough current capability to drive, you will get a 6 dB increase in sensitivity because 3 dB comes from the impedance being cut in half, thus causing twice the current to flow. The other 3 dB comes from mutual radiation of driver cones. If you connect the drivers in series, the sensitivity doesn't change compared to one driver. You lose 3 dB because the impedance is doubled, but you gain 3 dB from mutual radiation.”

The final numbers - 6dB gain for parallel connection and 0 dB gain for series connection are right - but the reasoning is wrong.

Sound pressure level (SPL) is only determined by relationship of SPL =20log(delta P)/(delta Po) where (delta P)/(delta Po) is change in supplied power.

As sated, with parallel connection of drivers your amplifier puts out twice the power (P= IU= U²/R) comparing to series connection because of impedance (load) is halved.

So with parallel connection of drivers:

(delta P)/(delta Po)=2 (twice the power supplied)

Hence SPL =20 log 2 =6dB

And with series connection of drivers:

(delta P)/(delta Po)=1(no change in power supplied)

Hence SPL =20 log 1 =0dB

Hope this clarifies it

Argo

I'm sicerely sorry, I cann't agree.

If you consider two drivers of 8 Ohms, series connection of two devices will result in 16 Ohms, parallel connection, 4 ohms. So parallel connection is 4 times lower, not half. And power will be 4 times higher, that is 6 db.

Right formula for power levels is : log A/B in Bels, that is 10 log A/B in deciBels (one deciBel is 1/10 of one Bel). If one calculate with voltages or intensities, its necessary to square the result, that is multiply by 2 the value in deciBels , the formula becomes 20 log A/B.

Sound reinforcement of 3 db caused by mutual radiation is only valid if the distance between drivers is very small in comparison with the wavelength. This is not the case in the whole spectrum of audible frequencies, so one should consider that the actual sound reinforcement is very smaller.

Regards, P.lacombe

[Edited by P.Lacombe on 09-07-2001 at 04:33 PM]

the problem being that many a guitar player with 4x12 marshall style cabs all have the same question/complaint.why do 2 of my 4 identical speakers sound different?

i encountered this also in rebuilding an old pair of column speakers that consisted of four 8" drivers.

i'm not sure if the inductance of the first driver is affecting the second and so on but so far it seems there is no definite research or tech info about this.

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