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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?

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    \$\begingroup\$ yes............. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 16 '18 at 1:13
  • \$\begingroup\$ @TonyStewart.EEsince'75 if you posted a scribbled picture of that as the answer, i would upvote it \$\endgroup\$ – Tyler Stone Jan 16 '18 at 1:44
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    \$\begingroup\$ Though it is a totem-pole output, its tendency to have a thermal runaway event is negated by special protection circuits. Not just any given amplifier can survive saturated outputs. \$\endgroup\$ – Sparky256 Jan 16 '18 at 2:22
  • \$\begingroup\$ this chip amp does indeed have some protection features built in, but after a good hour of thumping the big speakers, i tripped the 1/2A circuit breaker on the power transformer. Amp seems to be fine. \$\endgroup\$ – Tyler Stone Jan 16 '18 at 2:33
  • \$\begingroup\$ That's essentially what clipping is. \$\endgroup\$ – Brian Drummond Jan 16 '18 at 11:29
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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.

schematic

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 )

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    \$\begingroup\$ "max power is supplied when the source and load impedances are matched. Since the load is fixed and the source is a voltage..." - I think what you are trying to say is that maximum power is transferred to (and dissipated by) the transistor when its resistance matches the speaker. But then the transistor is the load, and the speaker is the (fixed) source impedance. \$\endgroup\$ – Bruce Abbott Jan 16 '18 at 3:48
  • \$\begingroup\$ Actually I was referring to impedance and the fixed load resistance is the speaker. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 16 '18 at 4:30
  • \$\begingroup\$ Thus my equivalent circuit was the equivalent NPN DC resistance that would absorb the loss between the total power supplied and delivered, meaning highest power absorbed is at max power transfer condition, meaning at 50% V+ (even though equivalent Rce is not a design figure) \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 16 '18 at 4:41
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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

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