====== Power Calculations for Class B Amplifiers ======

A majority of the following is derived from material presented by [[https://www.oit.edu/directory/cristina-crespo|Dr. Cristina Crespo]]((Cristina Crespo. “Class B Output Stage: Push-Pull Network.” Video. //YouTube//, May 23, 2017. https://www.youtube.com/watch?v=xXIZOg0BKag.)) ((———. “Class B Push-Pull Stage: Power Calculations.” Video. //YouTube//, May 23, 2017. https://www.youtube.com/watch?v=YltVJ3KzaR4.)) ((———. “Example: Class B Output Stage.” Video. //YouTube//, May 23, 2017. https://www.youtube.com/watch?v=et5wzj2Xh7E.)).

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 likewise is 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 factor 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.

===== Power into the load =====
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)`

<WRAP center tip round box 60%>
**Power into the load**

`bar(P_L) = (v_(op))^2/(2R_L)`
</WRAP>

The maximum power into the load happens when `v_(op)` is at its maximum possible value of `V_(C C)`.
===== Required supply power for given output =====

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`
<WRAP indent>
`              = 1/T int_(0)^(T/2) V_(C C) * v_o/R_L dt` \\
`              = 1/T int_(0)^(T/2) V_(C C) * v_(op)/R_L*sin(omega t) dt`
</WRAP>

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`
<WRAP indent>
`               = 1/T int_(0)^(T/2) V_(C C) * v_(op)/R_L*sin(2pi f t) dt`\\ 
`               = 1/T int_(0)^(T/2) V_(C C) * v_(op)/R_L*sin((2pi)/T t) dt`
</WRAP>

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)`
<WRAP indent>
` = 1/T * V_(C C) * v_(op)/R_L * (-T)/(2pi) * [cos((2pi)/T t)_(t=T/2) - cos((2pi)/T t)_(t=0)]`\\ 
` = -(V_(C C) * v_(op))/(2pi R_L) * [cos((2pi)/T t)_(t=T/2) - cos((2pi)/T t)_(t=0)]`\\ 
` = -(V_(C C) * v_(op))/(2pi R_L) * [cos((2pi)/T T/2) - cos((2pi)/T 0)]`\\ 
` = -(V_(C C) * v_(op))/(2pi R_L) * [cos(pi) - cos(0)]`\\ 
` = -(V_(C C) * v_(op))/(2pi R_L) * (-1 - 1)`\\ 
` = -(V_(C C) * v_(op))/(2pi R_L) * (-2) = (V_(C C) * v_(op))/(pi R_L) = 1/pi * v_(op)/R_L * V_(C C)`
</WRAP>

Owing to symmetry, the transistor connected to `V_(E E)` will consume the same average power over one cycle as the one connected to `V_(C C)`, 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)`

<WRAP center round tip 60%>
**Required supply power for given output

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

===== Maximum required supply power =====

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)`. Thus:

`bar(P_S)_max = 2/pi * V_(C C)/R_L * V_(C C) = 2/pi * (V_(C C))^2/R_L`
<WRAP indent>
`             = 2/pi * 2bar(P_L)_max = 4/pi * bar(P_L)_max`
</WRAP>

<WRAP center round tip 60%>
**Maximum required supply power**

`bar(P_S)_max = 2/pi * (V_(C C))^2/R_L` \\ or \\
`bar(P_S)_max = 4/pi * bar(P_L)_max`
</WRAP>

===== Power dissipation =====

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)`
<WRAP indent>
`    = (2/pi * v_(op)/R_L * V_(C C)) - [(v_(op))^2/(2R_L)]` \\ 
`    = ((2 V_(C C))/(pi R_L) * v_(op)) - [(v_(op))^2/(2R_L)]` 
</WRAP>

===== Maximum power dissipation =====

The plot of `P_D` vs. `v_(op)` describes a broad curve that intersects the origin when `v_(op) = 0` and has 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 maximum (i.e., 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)]`
<WRAP indent>
`         = (2/pi V_(C C))^2 *1/R_L - [(2/pi V_(C C))^2/(2R_L)]` \\ 
`         = (2/pi V_(C C))^2 *1/(2R_L)` \\ 
`         = 2/pi^2 * (V_(C C))^2/R_L`
</WRAP>

Alternatively,

`P_(Dmax) = (2/pi V_(C C))^2 *1/(2R_L)`
<WRAP indent>
`         = (2/pi)^2 * (V_(C C))^2/(2R_L)`\\
`         = (2/pi)^2 P_(L max)`
`         = 4/pi^2 P_(L max)`
</WRAP>

<WRAP center round tip 60%>
**Maximum power dissipation**

`P_(Dmax) = 2/pi^2 * (V_(C C))^2/R_L`\\ or\\ 
`P_(Dmax) = 4/pi^2 P_(L max)`

**occurs when**

`v_(op) = 2/pi V_(C C)`
</WRAP>

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

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 yields:

`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%`

<WRAP center round tip 60%>
**Maximum power efficiency**

`eta_max ~~ 78.5%`
</WRAP>