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--- audio_engineering:class_b_power_calculations [2022/12/13 06:35] – [Power Calculations for Class B Amplifiers] mithat
+++ audio_engineering:class_b_power_calculations [2023/08/01 23:47] (current) – [Maximum power dissipation] mithat
@@ Line 1: / Line 1: @@
 ====== Power Calculations for Class B Amplifiers ======
-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.
-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 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.
+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.)).
 ===== 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:
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-<html><b>Power into the load</b></html>
+**Power into the load**
 `bar(P_L) = (v_(op))^2/(2R_L)`
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 The maximum power into the load happens when `v_(op)` is at its maximum possible value of `V_(C C)`.
-===== Required supply power =====
+===== 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:
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-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:
+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+)`
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-<html><b>Required supply power</b></html>
+**Required supply power for given output
-`bar(P_S) = 2/pi * v_(op)/R_L * V_(C C)`
+**`bar(P_S) = 2/pi * v_(op)/R_L * V_(C C)`
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-<html><b>Maximum required supply power</b></html>
+**Maximum required supply power**
 `bar(P_S)_max = 2/pi * (V_(C C))^2/R_L` \\ or \\
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 `P_D = bar(P_S) - bar(P_L)`
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-`    = (2/pi * v_(op)/R_L * V_(C C)) - [(v_(op))^2/(2R_L)]`
+`    = (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>
-===== Worst-case power dissipation =====
+===== Maximum power dissipation =====
-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:
+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))`
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 `v_(op) = 2/pi V_(C C)`
-Substituting this into the relation for `P_D` above yields the worst-case power dissipation `P_(Dmax)`:
+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)]`
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-<html><b>Worst-case power dissipation</b></html>
+**Maximum power dissipation**
 `P_(Dmax) = 2/pi^2 * (V_(C C))^2/R_L`\\ or\\
 `P_(Dmax) = 4/pi^2 P_(L max)`
-<html><b>occurs when</b></html>
+**occurs when**
 `v_(op) = 2/pi V_(C C)`
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-<html><b>Maximum power efficiency</b></html>
+**Maximum power efficiency**
 `eta_max ~~ 78.5%`
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