This is a general question: What is the power consumption of an Op Amp with supply voltage Vcc, output voltage Vrms and a resistive load R?

Ignoring the amp's quiescent current I would say:

Power = Vcc * Vrms / R

But this sounds strange: If the load is 16 Ohm, Vrms=0.1V and Vcc = 1.8V
then the amp draws 11mW to supply only 0.6mW of power to the load, that's 5% efficiency. Aren't op amps nearly ideal devices?

Any thoughts are welcome.

 

The power to the load is (Vrms^2) / R where Vrms is the rms voltage across the load. The power drawn from the supply is Vcc * Iave where Iave is the average (not RMS) load current. I leave you to do the math.

Op amps are nearly ideal for many characteristics but their output is a Class B type stage and the efficiency of such a stage is always much less than 100%.

 

Thanks for the feedback.

 

An opamp cannot drive a 16 ohm load. The minimum load for most opamps is 2k ohms but a few can drive 600 ohms.

The output of an amplifier is a variable resistor in series with the supply voltage and the load. So of course the amplifier dissipates power making heat.