The following is derived from material presented by Dr. Cristina Crespo.^{1) 2) 3)}

For the calculations below, the following are assumed: `v_o` is the amplifier output voltage, `i_o` is the output current, and `R_L` is the load resistance. It's also assumed the amplifier has a bipolar power supply with rails `V_(C C)` and `V_(E E)`, where `V_(E E) = -V_(C C)`. We further assume that transistor base currents are negligible (so collector currents equal output currents) and that the zero-conduction dead zone is likewise negligible.

These analyses are based on sinusoidal signals as this is the accepted standard for power and thermal design in audio. The formulae will be valid for other signals with a crest factors of `sqrt(2)`. Alternative analyses based on worst-case signals (i.e., those having a crest factor of one, e.g, square waves), may be instructive but are not presented here.

The average output power `bar(P_L)` into load `R_L` for a sine wave with amplitude `v_(op)` is calculated as:

`bar(P_L) = v_(o RMS) * i_(o RMS) = v_(o RMS) * v_(o RMS)/R_L = (v_(op)/sqrt(2))^2/R_L = (v_(op))^2/(2R_L)`

The maximum power into the load happens when `v_(op)` is at its maximum possible value of `V_(C C)`.

We start by calculating `bar(P_S)_(V+)`, the average power over one cycle delivered by the power supply to the transistor connected to `V_(C C)` (i.e., the “positive” transistor). Letting `i_(C+)` be the collector current flowing through that transistor, this is calculated as:

`bar(P_S)_(V+) = 1/T int_(0)^(T) V_(C C) * i_(C+) dt`

Note the change in the upper integral limit from from `T` to `T/2`. This is because `i_(C+)` from `T/2` to `T` is zero.

Expanding the above yields:

`bar(P_S)_(V+) = 1/T int_(0)^(T/2) V_(C C) * v_(op)/R_L*sin(omega t) dt`

Solving the integral using:

`int sin(ax) dx = -1/a cos(ax)`

yields:

`bar(P_S)_(V+) = 1/T[V_(C C) * v_(op)/R_L * (-T)/(2pi) * cos((2pi)/T t)]_(t=T/2) - 1/T[V_(C C) * v_(op)/R_L * (-T)/(2pi) * cos((2pi)/T t)]_(t=0)`

Owing to symmetry, the negative half of the sinusoidal cycle will consume the same average power as the positive half, therefore the total average power delivered by the power supply is:

`bar(P_S) = bar(P_S)_(V+) + bar(P_S)_(V-) = bar(P_S)_(V+) + bar(P_S)_(V+) = 2 bar(P_S)_(V+)`

Substituting yields:

`bar(P_S) = 2/pi * v_(op)/R_L * V_(C C)`

From the above we can see that `bar(P_S)` increases linearly with `v_(op)`. Therefore, the maximum power delivered by the power supply `bar(P_S)_max` will happen when `v_(op) = V_(C C)`, and therefore:

`bar(P_S)_max = 2/pi * V_(C C)/R_L * V_(C C) = 2/pi * (V_(C C))^2/R_L`

The power dissipated as heat in both transistors `P_D` is the difference between the power consumed and the power delivered to the load:

`P_D = bar(P_S) - bar(P_L)`

The plot of `P_D` vs. `v_(op)` intersects the origin when `v_(op) = 0` and describes an broad curve with a peak somewhat before its final value (when `v_(op) = V_(C C)`). To find the maxima of this curve, we solve for `(d P_d)/(d v_(op)) = 0`. Thus:

`(d P_D)/(d v_(op)) = (2 V_(C C))/(pi R_L) - 2(v_(op)/(2R_L))` ` = (2 V_(C C))/(pi R_L) - v_(op)/(R_L)`

The above is zero when:

`v_(op) = 2/pi V_(C C)`

Substituting this into the relation for `P_D` above yields the worst-case power dissipation `P_(Dmax)`:

`P_(Dmax) = ((2 V_(C C))/(pi R_L) * 2/pi V_(C C)) - [(2/pi V_(C C))^2/(2R_L)]`

Alternatively,

`P_(Dmax) = (2/pi V_(C C))^2 *1/(2R_L)`

Remember that this is the total dissipated power. It will be spread uniformly across all output devices.

The power efficiency `eta` is defined as the ratio of the power delivered to the load to the power consumed:

`eta = bar(P_L) / bar(P_S)`

Substituting values for sinusoidal signals:

`eta = [(v_(op))^2/(2R_L)] / [2/pi * v_(op)/R_L * V_(C C)] = pi/4 * v_(op)/V_(C C)`

The power efficiency when the amplifier is at maximum power dissipation is interesting though not terribly useful:

`eta_("at max power dissipation") = pi/4 * (2/pi V_(C C))/V_(C C)` ` = 1/2 = 50%`

More useful is the maximum efficiency, which happens when `v_(op) = V_(C C)` :

`eta_max = pi/4 * V_(C C)/V_(C C) = pi/4 ~~ 78.5%`