Question: Let's consider the following class A power amplifier:

[![enter image description here](https://i.stack.imgur.com/eYeqC.png)](https://i.stack.imgur.com/eYeqC.png)

And here is the time-domain simulation of the power dissipated across the transistor:

[![enter image description here](https://i.stack.imgur.com/aBQqY.png)](https://i.stack.imgur.com/aBQqY.png)

As we can see, the power varies from approx \$0.1W\$ to \$1W\$. How can we interpret this?

Answer: 
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.
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)