Let's consider the following class A power amplifier:
And here is the time-domain simulation of the power dissipated across the transistor:
As we can see, the power varies from approx \$0.1W\$ to \$1W\$. How can we interpret this?
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2 Answers
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When the base is driven at the positive peak, the emitter-follower's emitter current is at a maximum because it must supply both the current sink current as well as the current through \$R_1\$. But this is also at the same point where the \$V_\text{CE}\$ is at a minimum. So there is a power minima here.
When the base is driven at the negative peak, the emitter-follower's emitter current is at a minimum because it now only has to supply the difference between what's required by the current sink and what is being supplied through \$R_1\$. But this is also at the same point where the \$V_\text{CE}\$ is at a maximum. Again, there is a power minima here.
So there are two power-minima in each cycle. Their minima values do not have to be equal to each other, because of the \$V_\text{BE}\$ voltage.
The two power maxima should be about the same value and will occur right when the signal voltage is near \$0\:\text{V}\$. You can work out this out simply enough. Ignoring \$V_\text{BE}\$ entirely for now (assuming the signal appears directly as the emitter voltage), the power in the BJT is roughly speaking: \$P=\left(V_+-V_\text{IN}\right)\cdot\left(I_1+\frac{V_\text{IN}}{R_1}\right)\$. Taking the derivative it works out that the maximum power occurs when \$V_\text{IN}\approx \frac{1}{2}\left(V_+-I_1\cdot R_1\right)\$. For your current setup, this occurs at about \$V_\text{IN}\approx 0\:\text{V}\$
You can test out this idea by changing your positive supply rail voltage to \$14\:\text{V}\$, for example. Then you should find that the peak power occurs at about \$V_\text{IN}\approx +2\:\text{V}\$.
That's really all there is to it.
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The difference is only due to the Vbe DC offset on the resistor.
Since Vin=Vb is 0V dc , Ve = Vin - Vbe (neglecting Rbe)
If you add Vbe= 0.7V dc offset to sig.gen. input, then the output will be 804mW for every peak and Voltage will be symmetrical about 0V.
Some voltage swing is lost due to \$r_{be}\$ at a base emitter current Ib= (Isink/hFE) ( in theory 9^2/100 = 810 mWpeak)
