How does complex loading affect an op amp's power dissipation

I'm trying to calculate power dissipation in a complex load. I know that the power loss due to the resistive component is dissipated, and that the power loss due to the reactive component is canceled out due to an equal amount of power being generated at the load and sent back. My question is how does the op amp handle this, is the power dissipation of the op amp equal to P_resistive+P_reactive? How would I interpret this on an oscilloscope?

Thanks in advanced

EDIT: I should clarify, when I ask how this affects the op amp power dissipation, I really mean how does this affect the junction temperature.

• No. If it was supplying 1V and 10mA to the load and powered by 15V, it would drop 14V at that 10mA current. Your best bet is to measure supply currents, multiply by supply voltages (for + and - rails separately, and subtract power delivered to load. – user_1818839 Oct 16 '20 at 22:09
• See if this helps, it's related. – a concerned citizen Oct 17 '20 at 11:34

1 Answer

An op-amps internal power dissipation is due to its output being a class AB stage (usually) and they are about 60% power efficient in delivering load power. In other words, if the load power is 60 mW then the op-amp will be dissipating 40 mW.

But this is at maximum peak-to-peak output voltage. If if the op-amp is required to deliver only half the maximum peak-to-peak output voltage then efficiency drops significantly. The graph below compares an op-amp output against a class D driver but, the important thing to note is how a class AB stage drops efficiency when it is required to deliver less signal to the load (x-axis): -

Regarding reactive load currents versus resistive load currents, all the op-amp knows is that it is being asked to deliver current and that current passes through output transistors that will have voltage across them. That makes those output transistors dissipate power.

It's the same for a lossy power transmission line feeding a resistive/reactive load. The generator at the sending end only produces the resistive power required but the cable wiring resistance losses means that the cable dissipates power based on current flow ($$\I^2R\$$ losses) and that current flow is both reactive and resistive if the load is reactive and resistive.