class AB audio amp clipping vs. heat

Question

I have built an audio amplifier with a Texas Instruments LM4766 chip amp. It works beautifully. I have performed some torture testing that involved driving a pair of large bass cabinets (rated at 300W), mostly to determine if my proposed heatsink is sufficient.

When I drove the amp as loud as it would go before audible clipping, I noted the maximum temperature of the heatsink tab after it plateaued (59C). Then I pushed it into audible clipping, and the temperature of the heatsink tab dropped to about 56C. In an attempt to keep it somewhat scientific, I used the same audio content for the range of testing.

The question then is this; when the output devices clip, are they entering saturation mode and turning fully on, thereby dissipating less heat? Could it be so simple?

Answer 1

We know from the Maximum Power Transfer Theorem that max power is supplied when the source and load impedances are matched. Since the load is fixed and the source is a voltage, the equivalent source impedance of the NPN high side power transistor inside the LM4766 can be equated to the V/I drop in the Pot below for voltage levels >= 50% V+.

(sorry but the link describes the theory better than I can) another link

Here we forget about the how the circuit amplifies and only focus on how each complementary output stage draws current from each complementary power supply and represent it as an average resistance.

A simple equivalent circuit.

^{simulate this circuit – Schematic created using CircuitLab}

When the average voltage is greater than 50% of the supply voltage towards a clipped output, the less power is dissipated in the driver.

Or in other words when it clips more often, it runs cooler than maximum. ( but only slightly )

Answer 2

As the output voltage increases, the voltage dropped across the output transistors decreases, but the output current increases.

https://www.updatemydynaco.com/documents/Class_B_Amplifier_Dissipation_Calculations.pdf

Gives the equation for sinusoidal output

P = Vcc * 0.45 * Vout/R_L - Vout^2/R_L