Driver efficiency vs. overall speaker efficiency


If a speaker uses a driver that has an efficiency of, let's say 87.5db/2.83V/m, then theoretically the speaker can never have an efficiency of greater than 97.5db?
andy2
High-efficiency speakers-

Are they just going to give more sound-per watt, or will they also communiate more texture (micro-dynamics?) of the sound, along with the extra dB?
Here we go again. One thing we modern folk oft do is to use the WRONG terminology. 'Efficiency' is different than 'sensitivity'. Speaker-system efficiency normally is measured 1 meter from the speaker/driver when driven with exactly 1 watt of power measured into the speaker's rated impedance. IOW, a '4-ohm' speaker system's output is measured while being driven by TWO volts into 4 ohms (which equals ONE WATT). Decades ago, with the increasing popularity of solidstate amps*, the audio industry started publishing sensitivity ratings, and not all of them do it correctly.

Sensitivity is measured with a signal created by driving an 8-Ohm load resistor with 2.83 volts. That does happen to be 1 watt into 8 ohms, but it's not 1 watt into 6 or 4 or 3 or 2 ohms. The correct way to state this is 'sensitivity at 1M using 2.83v. into 8 Ohms' or something like that. And BTW, it's not PER meter, it's at 1 meter.

*--Solidstate amplifiers are constant-voltage devices and not constant-current devices are as most tubed amps. That is, because a SS amp's output impedance is so low, it will maintain its output voltage into decreasing loads until it reaches its maximum current output. A tubed amp generally will NOT increase its current output much as the load impedance decreases.
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"Efficiency"???...The only thing I see quoted in specs is SPL in units of dB at 1 meter for 2.83 volts. This is 1 watt for an 8 ohm load.

Power = Volts squared divided by ohms.

(2.83 * 2.83) / 8

For 4 ohms it's 2 watts.
""Efficiency"???...The only thing I see quoted in specs is SPL in units of dB at 1 meter for 2.83 volts. This is 1 watt for an 8 ohm load.

Power = Volts squared divided by ohms.

(2.83 * 2.83) / 8

For 4 ohms it's 2 watts."

What's your point? I already did that math.

My point is that there's a difference between the 2 terms, and those of us who aren't aware and careful, including several members who posted above, tend to mix them.
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Jeffreybehr....My points are:

1. The term "Efficiency" is just confusing. I don't see it quoted by manufacturers.

2. You gave the answer, but did not "do the math" so that others could understand where you are coming from. I am not disagreeing with you. Just making a clarification.