Power Output vs. Impedance
For a sine wave:
\[ V_{\mathrm{RMS}} = \frac{V_{\mathrm{peak}}}{\sqrt{2}}, \qquad I_{\mathrm{RMS}} = \frac{I_{\mathrm{peak}}}{\sqrt{2}} \] \[ P = V_{\mathrm{RMS}} I_{\mathrm{RMS}} = \frac{V_{\mathrm{RMS}}^2}{R} = I_{\mathrm{RMS}}^2 R \]An amplifier is limited by whichever it reaches first: its maximum output voltage or its maximum output current.
Example: ±16 V supply rails
Ideally, ±16 V rails allow 16 V peak at the output:
\[ V_{\mathrm{RMS}} = \frac{16}{\sqrt{2}} = 11.32\ \mathrm{V} \]Into an 8 Ω speaker:
\[ I_{\mathrm{RMS}} = \frac{11.32}{8} = 1.414\ \mathrm{A} \qquad P = \frac{11.32^2}{8} = 16\ \mathrm{W} \]So the ideal maximum is 16 W into 8 Ω. More available current alone cannot increase that figure because the amplifier is already at its voltage limit.
As impedance falls
If the amplifier can maintain 11.32 V RMS and has enough current capacity:
| Load | Power | RMS current | Peak current |
|---|---|---|---|
| 8 Ω | 16 W | 1.414 A | 2 A |
| 4 Ω | 32 W | 2.828 A | 4 A |
| 2 Ω | 64 W | 5.657 A | 8 A |
At a fixed voltage, halving load resistance doubles current and doubles power.
Current-limited example
If the amplifier can provide only 1.414 A RMS into 8 Ω:
\[ P = I_{\mathrm{RMS}}^2 R = (1.414)^2 \times 8 \approx 16\ \mathrm{W} \]That is 2 A peak. Maintaining the same voltage would require 4 A peak for 32 W into 4 Ω, and 8 A peak for 64 W into 2 Ω.
Real-world caveat
Real amplifiers need some voltage headroom below the supply rails and their supply voltage can sag under heavy current draw. Their actual output is therefore often below these ideal figures.