A Question on Speaker Driver Efficiency

@almarg 

Do the math. Take two 8 ohm nominal woofers, connect them in series, and apply 2.83v.

2.83 divided by 16  = .176875 Amps (current "I")

power dissipated into each woofer in this circuit then becomes:

R(I X I) or resistance of each woofer times the square of the current running through it. Thus
8(.176875 X .176875) = .25 Watts or 1/4 watt - not 1/2 watt as atmasphere and you have suggested -

 " two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db "- atmasphere

The above can be expressed another way.
Power dissipated in two woofers in series = net voltage drop across both times the net current flowing through both.
2.83 Volts times .176875  = .5 watts or 1/2 watt total between both drivers in series. If together both woofers are dissipating 1/2 watt, then individually they must dissipate 1/4 watt - thus agreeing with the above calculation. At half the voltage drive level (voltage divided when connected in series), each woofer should produce half the acoustic output and consume 1/4 the power.When their acoustic outputs are summed, the net acoustic result is the same as if for one driver - however the power consumed is cut in half.

This jives with measurements I've made of acoustic output of drivers in series.

" @cj1965 Current in the circuit is still the same. You are confusing voltage across the speaker with current. Imagine simple circuit consisting of voltage source and bunch of inductors and capacitors in series. Now you insert speaker into it. Do you think it will sound differently in different places of insertion? There is only one current in the circuit and two speakers in series have to respond at the same time (unless there is place where "faster current" can escape)." - kijanki

Please read and study the equation I posted above for two inductors (speaker coils) in series with a resistance and capacitance. The equation doesn't lie. Current and voltage are constantly varying and as the equation shows  the voltage representations of each woofer are NOT equivalent to the applied votlage input [V(t)] ,minus the other woofer's voltage. The voltage represented by the capacitance must also be accounted for and it is time dependent. If you disagree with the equation and what it is saying - address your comments/concerns with that equation and its applicability to the subject at hand. I didn't invent Kirchoff's law - I'm merely reciting it in the context of a series connected loudspeaker pair.

You're telling Almarg to do the math?  LOL, Man, you got big mouth.

Incorrect? Really? In theory, they move at EXACTLY the same time only when no capacitance exists in the circuit. Current through the coils is the same at any given time only when the circuit doesn’t have capacitance. We know this is not the case.

@cj1965 It seems that you are forgetting about Kirchoff's Law. Obviously one woofer cannot move if current is not also flowing through the second in a series connection! So that means the other woofer has to move **at the same time** otherwise Kirchoff's Law is violated, which is an impossibility. Further, the current has to also flow the other way (it is an audio waveform after all), so the situation with the woofer's relationship is reversed; it is quite obvious that they will move at exactly the same time and this is true even if the drivers are of different inductance.

Kirhoff's Law was also taught to me in school as the 'Law of energy conservation'. It basically states that there cannot be more energy in a circuit than is put into it, nor can there be any less. So if a watt is put in, the individual parts in the circuit will all dissipate some fraction of that watt in such a way that if you added it all up, it would be exactly 1 watt. Here is the Wikipedia page:
https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

Now if the drivers are dissimilar, they might not *sound* the way you want (one might have a greater voltage drop across it than the other), but if they are the same driver there isn't a downside unless higher impedance is problematic as with a solid state amp (however the solid state amp will be seen to be making less distortion).

If all things are equal, meaning a 4 ohm 8 ohm and 16 ohm driver would all sound equivalent in a given cabinet, Paralleled wired speakers sound better overall than series. I would also recommend that you add a simple impedance correction circuit.   Many amplifiers react better to a consistent impedance. Not all, but No amplifier sounds worse with it. (unless you are dropping impedance too low for a few tubes) still within tubes impedance operating range, even tubes sound better with impedance correction circuits on speakers

This bit is problematic. The drivers are going to sound the same whether in parallel or in series. What **won't** sound the same is the amplifier, which reacts very differently to impedance depending on the amp.

Now in the case of this thread, the amp in question is a tube guitar amp. Tubes, generally speaking, prefer a higher impedance and will make less distortion if the higher impedance is accommodated by a tap on the output transformer. The transformer will run cooler as it is more efficient, so with most tube amps you get a tiny bit more power as well, as well as more extension into the bottom octave of the amp.

Reducing distortion may not be the goal in a guitar amp, where the guitarist's individual 'sound' that he is going for is highly subjective and varies greatly from one guitarist to another, even using the amp amp.

So ultimately, the OP will simply have to try it both ways. The amp won't be damaged by this, and if there are taps for each impedance, they should be employed.

Again, current (and not voltage) is what moves the coil.