Confused by decibels

[PeteMcK]

If I wire two speakers that have different Re and efficiencies in series,
how do I work out the resulting difference in output between them?

I'm thinking to start I need to work out the difference in output due to the different voltage across each, but do I use the voltage formula (10 log V1/V2), or the power formula (20 log (P1/P2)?

[pinkmouse]

Frankly, I'd ignore the maths and just measure them.

[5th element]

The trouble here is that the voice coil inductance will alter the impedance at a given frequency too. If both drivers were identical then the impedance of both would be identical at all frequencies, so when wired in series, each driver would see exactly half of the total voltage output by the amplifier.

If the drivers are different though, maybe at 200hz both have an identical impedance of 8 ohms, thus each driver would see an identical voltage across their terminals. However if one driver had a low inductance motor and one didn't, by 1000hz one driver may have an impedance of 10ohms, but one has risen to 15ohms, now the voltage split won't be equal and it will complicate matters.

If you wire them in parallel this isn't a problem, as each driver will then, irrespective of frequency, see the same voltage across their terminals.

[Arty]

suposedly for a given freqvency (in your case every single possible freqvency will yield different results) You have to calculate the voltage they recive. Then You can find the current, and therefore the power they recive. And from the efficiency You can calculate how ould would be eatch driver.

So suposedly voltage will be..
(Vin /(Ra + Rb)) * Ra = Va
Where Vin is the input voltage (or output voltgae of your amplifier)
Ra is unit "a" impedance at a given x freqvency
Rb is unit "b" impedance at a given x freqvency
Va is the voltage across unit "a"

and x is Your choosen freqvency. And results will be different for every freqvency.
and so on.. divide voltgae by impedance, multiply by voltage, you get power, and so on..


Prior You attempt this, i can insure, it is absolute pointles. The chart You are going to get will proove that there is no human imaginable reson to make a setup like this.
even identical drivers in series have.. some problems, as there are no identical drivers.
two absolute different driver will only make things worse.

That is why parelell connection speakers is a more clever solution, at least the voltage across every unit will be exactly the same.

[AllenB]

Originally Posted by PeteMcK

do I use the voltage formula (10 log V1/V2), or the power formula (20 log (P1/P2)?

10 is for power, 20 is for voltage. If these were resistors, I'd look at voltage, but they aren't and I'd agree with Pinkmouse.

[PeteMcK]

you guys are making this too complicated... I've measured the drivers, & am aware that impedance varies with freq, and it's not pointless...

Perhaps I should explain... The idea is to get a rough figure for the relative outputs, to see if different drivers are compatible

Lets just treat them as resistors IN SERIES, what I want to find is the difference in output due to Re.
Then I can add/subtract that figure to the actual efficiency of the drivers to see if I'm getting similar output

I think AllenB is right, the current through both will be the same in series connection, so the difference in output will be due to voltage differences

You guys have unwittingly made a good case for Zobels for series connected drivers though...:-)

[Ron E]

Originally Posted by PeteMcK

you guys are making this too complicated... I've measured the drivers, & am aware that impedance varies with freq, and it's not pointless... Perhaps I should explain... The idea is to get a rough figure for the relative outputs, to see if different drivers are compatible

Amp output is V0
Driver 1 is Z1
Driver 2 is Z2
Zt=Z1+Z2
I=V0/Zt
V1=I*Z1
V2=I*Z2
To calc spl you need to multiply by the voltage sensitivities.
..all vector quantities...

BTW for dB, power ratios are multiplied by 10 and voltage ratios by 20.

[wakibaki]

OP, you've got the power and voltage formulae confused.

OK, if you have a 4R resistor and an 8R resistor and you apply 12VRMS:

4 volts appears across the 4R resistor. The power is V^2/R = (4*4)/4 = 4 Watts.

8 volts appears across the 8R resistor. The power is (8*8)/8 = 8 Watts.

8 Watts to 4 Watts is a factor of 2. A factor of 2 is 10*log10(2) dB (10 because its power) = 10 * 0.30102999566398119521373889472449 = 3dB.

So if it was a 4 ohm speaker with 97dB/W/m and an 8 ohm speaker with 94dB/W/m, the output powers would be identical.

w

Unless I've slipped a cog. Which wouldn't be the first time.

[GM]

Originally Posted by PeteMcK

You guys have unwittingly made a good case for Zobels for series connected drivers though...:-)

Yeah, Le sums, so unless it's very low to start with......... Anyway, if these need to handle anywhere near their rated power, the lower resistance one can turn into an expensive fuse, so trading some efficiency to closely match impedance with parallel resistors is a good plan. Probably a good plan regardless if they are off much more than the variance of drivers within a batch.

GM

[Arty]

Originally Posted by PeteMcK

you guys are making this too complicated... I've measured the drivers, & am aware that impedance varies with freq, and it's not pointless...

Perhaps I should explain... The idea is to get a rough figure for the relative outputs, to see if different drivers are compatible

Lets just treat them as resistors IN SERIES, what I want to find is the difference in output due to Re.
Then I can add/subtract that figure to the actual efficiency of the drivers to see if I'm getting similar output

I think AllenB is right, the current through both will be the same in series connection, so the difference in output will be due to voltage differences

You guys have unwittingly made a good case for Zobels for series connected drivers though...:-)

Yes, impedance correction circuitry is a must have.
I for example just wanted to poit out that speakwers are not ressitors, and the SPL difference between the 2 will be different at any given freqvency. The impedance correctin circuitry will help to reduce it, but will insead the designer will has to deal with the losses on those components, so in no way is this a simple approach.