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Power, sensitivity and sound pressure

Watt ratings alone do not tell you how loud a loudspeaker will play. Sensitivity, distance, power compression and excursion are equally important.

Watts are input, not sound at the seat

A speaker labelled 300 W is not automatically louder than an 80 W model. Separate sensitivity, allowable electrical input and maximum clean acoustic output. The finished result also depends on distance, frequency, cabinet alignment and the room. A single power number cannot describe all of these.

Read the sensitivity reference

Sensitivity specifies sound pressure at a defined excitation and distance. A statement such as 88 dB at 2.83 V and 1 m does not mean the speaker has that response at every frequency or that 88 dB is its limit. Match distance, voltage or power reference, acoustic conditions and averaging range before comparing drivers.

2.83 volts is not always one watt

For an ideal 8 Ω resistance, 2.83 V corresponds to about 1 W; at 4 Ω it corresponds to about 2 W. That difference is close to 3 dB of electrical input. A real loudspeaker impedance is complex and frequency dependent, so the simple power equation is a reference comparison rather than a complete broadband power measurement.

Power and level are logarithmic

Doubling power adds ideally about 3 dB. Ten times the power adds 10 dB. Halving distance adds approximately 6 dB for a point source in a free field. These are useful relationships, not universal in-room promises. Directivity, near-field geometry, reflected sound and actual limits modify the result.

The bass may run out of excursion first

At low frequencies, usable volume displacement can limit clean output before the thermal power rating is reached. A larger cone or more linear stroke can help, as can a suitable resonant alignment within its operating region. An amplifier upgrade alone cannot remove a mechanical stop or prevent distortion from excessive movement.

Heat and multiple sources change the picture

As a voice coil heats, resistance typically rises and output may compress. Multiple drivers can share the work, but their level gain depends on spacing, phase, wiring and whether total electrical power or voltage is held constant. Simply quoting an unconditional extra 6 dB leaves those assumptions unresolved.

Make the comparison useful

Write down the frequency range, listening distance and required clean level. Compare sensitivity under matching conditions and then inspect excursion, thermal capability, distortion and the chosen enclosure. Leave practical reserve. A system that stays controlled at the required level is more useful than one with an impressive watt label.

Explore the relationship
Interactive experiment

What does ten times the power buy?

Compare 10 W with 100 W: the ideal gain is 10 dB. Then double distance: the free-field model loses about 6 dB.

Model limits: Ideal point source in a free field with constant sensitivity. No compression, excursion limit or room gain. Not a maximum-SPL prediction.

A closer look · model and equation

L = S + 10·log10(P/1 W) − 20·log10(r/1 m). Reference conditions must match. 2.83 V is approximately 1 W into 8 Ω but 2 W into 4 Ω.

Workshop takeaway

Watts alone do not buy deep bass. Compare sensitivity under matching conditions, then check the actual system limits.

Work it out · answers unfold

Write down your prediction first, then your calculation. Open the answer only when you are ready to compare.

01 · A woofer gives 88 dB and a tweeter 92 dB at the same operating point and reference. What level correction is an initial estimate?

Show worked answer
  1. Difference: 92 - 88 = 4 dB.
  2. Initially attenuate tweeter by 4 dB.
  3. Then measure crossover region.
Calculation
\Delta L=92-88=4\,dB

02 · A 4 Ω driver is measured at 2.83 V. What is electrical power for a purely resistive substitute?

Show worked answer
  1. P = U²/R.
  2. 2.83²/4 = 2.002 W.
  3. Here 2.83 V means about 2 W, not 1 W.
Calculation
P=\frac{2.83^2}{4}\approx2\,W

03 · How do you ideally convert 90 dB at 2.83 V/1 m with 4 Ω to 1 W/1 m?

Show worked answer
  1. Initial power is about 2 W.
  2. Half power reduces level by 3.01 dB.
  3. Result about 87 dB/1 W/1 m.
Calculation
L_{1W}=90-10\log_{10}(2)\approx87\,dB

08 · Two equally strong coherent signals of pressure ratio 1 are in phase. Find sum level relative to one signal.

Show worked answer
  1. Add pressure ratios: 2.
  2. 20 log10(2).
  3. Sum is 6.02 dB above one signal.
Calculation
\Delta L=20\log_{10}2=6.02\,dB

Match drivers together →