simulate this circuit – Schematic created using CircuitLab

I'm watching Prof. D.C. Dube's lecture (timestamp included) on power amplifiers, and I'm confused about the calculation of DC power \$P_\text{dc}\$ in class A amplifiers, like in one shown in the above schematic. I do understand that we take the following approximations:

$$i_\text{C(min)}=0$$ $$i_\text{C(max)}=2I_\text{CQ}=i_\text{C(sat)}$$ $$v_\text{CE(min)}=0$$ $$v_\text{CE(max)}=2V_\text{CEQ}$$

and that the AC load line characteristics is like this:

enter image description here

DC Power \$P_\text{dc}\$ calculation

$$P_\text{dc} = V_\text{CC}I_\text{CQ}$$

Now, from what I understand \$P_\text{dc}\$ is defined as the average DC power drawn by circuit at the collector (power consumption by biasing resistors is ignored).


Why are we writing the DC power as \$V_\text{CC}I_\text{CQ}\$ rather than \$V_\text{CEQ}I_\text{CQ}\$? \$V_\text{CC}\$ is the maximum value of \$v_\text{CE}\$ when the AC signal is applied. But in the absence of the AC signal (we're calculating DC power after all, so AC signal swing can be neglected), the maximum DC voltage signal at collector should be \$V_\text{CEQ}\$, no? That is, basically I suspect that the DC power should have been

$$P_\text{dc} = V_\text{CEQ}I_\text{CQ} = \frac{1}{2}V_\text{CC}I_\text{CQ}.$$

Do we take \$V_\text{CC}\$ rather than \$V_\text{CEQ}\$ because we include the power consumed by resistors \$R_C\$ and \$R_E\$ in DC power drawn by circuit as well?

Could someone clarify?


1 Answer 1


The power for an amplifier is provided by the supply. The power a supply delivers is its own voltage (that's why we take \$V_\text{CC}\$) times the current drawn from it.

For a Class A power amplifier, with no input signal, the DC current drawn equals only the collector bias current, \$I_\text{CQ}\$, because the current drawn by biasing resistors is ignored.

So, the DC power that the supply delivers equals to: $$ P_\text{dc} = V_\text{CC} \cdot I_\text{CQ} $$


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