1.9 KiB
Power Output vs. Impedance
This note uses plain-text maths and tables so it can render in simple Markdown viewers.
For a sine wave:
RMS voltage = peak voltage / 1.414
RMS current = peak current / 1.414
Power = RMS voltage x RMS current
Power = (RMS voltage x RMS voltage) / resistance
Power = (RMS current x RMS current) x resistance
An amplifier is limited by whichever it reaches first:
- maximum output voltage, or
- maximum output current.
Example: ±16 V supply rails
In an ideal amplifier, ±16 V rails allow a maximum output of 16 V peak.
RMS voltage = 16 / 1.414 = 11.32 V RMS
Into an 8 ohm speaker:
RMS current = 11.32 / 8 = 1.414 A
Power = 11.32 x 11.32 / 8 = 16 W
So the ideal maximum is 16 W into 8 ohms. Extra current capability alone cannot increase this: the amplifier has already reached its voltage limit.
What happens as impedance falls?
If the amplifier can still maintain 11.32 V RMS and its current limit is high enough:
Load 8 ohm 4 ohm 2 ohm
------------------------------------------------
Power 16 W 32 W 64 W
RMS current 1.414 A 2.828 A 5.657 A
Peak current 2 A 4 A 8 A
At a fixed voltage, halving load resistance doubles current. Because power is voltage squared divided by resistance, it also doubles power.
Current-limited example
If an amplifier can supply only 1.414 A RMS into an 8 ohm load:
Power = current x current x resistance
= 1.414 x 1.414 x 8
= about 16 W
That 1.414 A RMS equals 2 A peak. To produce 32 W into 4 ohms at the same voltage, the amplifier would need 4 A peak. To produce 64 W into 2 ohms, it would need 8 A peak.
Important real-world caveat
Real amplifiers cannot usually swing their output all the way to the supply rails, and their supply voltage may sag at high current. Therefore actual power ratings are often lower than these ideal calculations.