So here, I've redrawn the circuit schematic for the common source amplifier.

And we're going to use this schematic to determine the reason for

the negative sign in the gain expression of the previous slide.

Now remember, for DC currents and voltages, the currents and

voltages that's at the operating point of the transistor, the capacitors have

impedances that are large enough that we can consider them to be open circuits.

So, any DC currents flowing through RD, must flow in to the drain of the mosfet,

it can't flow through C2 to the output because of this large impedance.

Now, if we apply an AC signal here, a time-bearing sinusoidal voltage, when

this circuit is operating in its mid-band region, at that frequency, the impedance

of these capacitors is small enough that we can consider them to be short circuits.

So, an AC voltage here is connected by a short circuit to the output.

But DC currents here, this C2 presents an open circuit.

Now, we know, because of the relationship between the gate to source voltage and

the drain current, that as this AC voltage increases,

the drain current will also increase.