A Question on Speaker Driver Efficiency

This is exact quote from cj1965 post:

In theory, half the voltage applied to a driver will result in half the sound pressure in the output (-3db).

Res ipsa loquitur, huh?

Obviously he didn't actually do the math, but based this on belief instead- the exact thing of which he was accusing others...

@timlub +1

Each is not absorbing 1/2 watt when placed in series (with the required 2.83 V applied) and the net acoustical sum of output is approximately 97db - not 94db. Two obvious errors in one sentence. The math doesn't lie and frankly, I'm getting a little tired of the verbal gymnastics at this point. The math is simple and anyone who wants to read the details above can readily see where the error was established. Yeah, you can double the input drive voltage to 5.7v and then the series arrangement will have each woofer dissipating 1/2 watt. You can also multiply the input voltage by 10,000 - it won't tell us anything about the effect placing two identical drivers in series has.

@almarg  Its pretty evident to me that this guy seems to equate '1 watt' with '2.83v' and so is somehow assuming that I meant '2.83 volts' when I was pretty talking about the power (not voltage) being '1 watt'.

Also his intuitive grasp of Kirchoff's Law seems to be weak, and I further find that talking about capacitance when dealing with 2 guitar speakers as opposed to one rather silly, since any such distributed capacitance on the coils (or anywhere else) is going to be quite low and negligible. But even if it was not, in a series connection current would have to flow through both drivers in order for one to work. For all his 'hand waving', he seems to have no grasp of the formulae he presented, how they are used or their significance. 

@kijanki +1

I'm the only one in this thread actually citing equations based on Kirchoff's law while you hand wave generalizations about tube amps.

Really? Hand-waving seems to be what you are doing with this quote...

Somehow you don't seem to make the connection that current has to flow if a circuit is complete. Its Circuit Basics 101 first day stuff.

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

Sounds like you are running a solid state amp. Solid state amps, generally speaking, are usually built to behave as a voltage source. If that is the case, it will make double the power into half the impedance, and thus also half the power if the impedance is doubled. So such an amplifier will make 1/4 of its 4 ohm power into 16 ohms (drivers in series).

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.

One last question for everyone...
- In parallel - each driver would move at EXACTLY the same time (pretty much)
- in series - would the second speaker in the "chain" be moving slightly behind the first speaker due to lag time through the voicecoil of the first speaker ?
- also, would one speaker be affected by the other?
- Would the human ear be able to discern this?

@williewonka , a previous answer was incorrect.

Both speakers will move at EXACTLY the same time regardless of the hookup.

Usually tube amps like higher impedances, so usually its to your advantage to put the speakers in series if you want cleaner sound. In the scenario you described though it will work fine either way if 8 ohms is your goal; two 4 ohm speakers in series or two 16 ohms speakers in parallel. You will not be able to hear any difference other than the differences that might be in the drivers themselves.

To clear up some obvious confusion on this thread, Sensitivity is a voltage measurement and efficiency is a Power measurement. Into 8 ohms both are the same, since sensitivity is 2.83 volts at one meter and that works out to 1 watt.

Into a 4 ohm load, 2.83 volts is 2 watts not 1. So the amp must make twice as much power and thus there is a 3 db increase in sensitivity (but note that the increase in volume is because the amp is making more power, not because the speaker is more efficient). The converse is true into 16 ohms- now the amp makes 1/2 watt to make 2.83 volts so the sensitivity is 3 db less.

So if you have a 97 db 1 watt/1 meter 8 ohm driver, two in series will have the same efficiency (since each is absorbing 1/2 watt) while the sensitivity is 94 db. If you put the two drivers in parallel for a 4 ohm load, the efficiency is the same as 1/2 is absorbed by each driver if 1 watt is applied. However the sensitivity is now 100db.

The trick is that you are using a tube amp and since they can't double power as impedance is cut in half, the efficiency spec is easier to use since it tells you how loud the speaker will play with your amp.