# Class A power amplifier: power dissipation

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?

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)