Electronic – MOSFET common source amplifier output impedance calculation

It's called negative feedback and the biasing attempts to keep the gate at "just the right amount of voltage" - should the drain voltage droop a bit low, the DC voltage applied to the gate reduces and, this in turn, turns the fet off slightly and, in turn, this makes the drain voltage rise a bit. Thus there is a measure of stability! It's proper negative feedback.

R6 (the pot) in the source connection is only 100 ohm max and this will alter things a little bit but it's not a showstopper at one end of its travel or at the other. Overall, AC gain will be a tad limited when R6 is at 100 ohms but it won't affect things too much.

So, in short, feeding the gate half the supply doesn't necessarily keep the FET biased as well as the "negative feedback" method AND, a source resistor of 0 to 100 ohms isn't going to change things that much. Think about how much voltage the gate needs (with respect to source) to adequately "turn on" the FET - if it is (say) 2V then the drain has to be at 4V (due to R3 and R4) - if it needs 3V then the drain has to be at 6V. Because we can't be as sure about a FET (compared to a BJT) this is a sensible way of biasing it. Having said that I've seen plenty of BJT circuits that use EXACTLY the same method for biasing the base i.e. collector feedback.

On a slightly off-question point - opamps work precisely this way.

The true answer to your question unfortunately involves some bits of advanced calculus. Small signal models are derived from a first-order multi-variable Taylor expansion of the true non-linear equations describing the actual circuit behavior. This process is called circuit linearization.

Let's consider a very simple example with only one independent variable. Assume you have a non-linear V-I relationship for a two-terminal component that can be expressed in some mathematical way, for example \$i = i(v) \$, where \$i(v)\$ represents the math relationship (a function). Regular (i.e. one-dimension) Taylor expansion of that relation around an arbitrary point \$V_0\$, gives:

$$ i = i(V_0) + \dfrac{di}{dv}\bigg|_{V_0} \cdot (v-V_0) + R = i(V_0) + \dfrac{di}{dv}\bigg|_{V_0} \cdot \Delta v + R $$

where \$R\$ is an error term which depends on all the higher powers of \$\Delta v = v - V_0\$. The linearization consists in neglecting the higher order terms (R) and describe the component with the linearized equation:

$$ i = i(V_0) + \dfrac{di}{dv}\bigg|_{V_0} \cdot \Delta v $$

This is useful, i.e. gives small errors, only if the variations are small (for a given definition of small). That's where the small signal hypothesis is used.

Keep well in mind that the linearization is done around a point, i.e. around some arbitrarily chosen value of the independent variable V (that would be your quiescent point, in practice, i.e. your DC component). As you can see, the Taylor expansion requires to compute the derivative of \$i\$ and compute it at the same quiescent point \$V_0\$, giving rise to what in EE term is a differential circuit parameter \$\frac{di}{dv}\big|_{V_0}\$. Let's call it \$g\$ (it is a conductance and it is differential, so the lowercase g). Moreover, \$g\$ depends on the specific quiescent point chosen, so if we are really picky we should write \$g(V_0)\$.

Note, also, that \$i(V_0)\$ is the quiescent current, i.e. the current corresponding to the quiescent voltage. Hence we can call it simply \$I_0\$. Then we can rewrite the above linearized equation like this:

$$ i = I_0 + g \cdot \Delta v \qquad\Leftrightarrow\qquad i - I_0 = g \cdot \Delta v \qquad\Leftrightarrow\qquad \Delta i = g \cdot \Delta v $$

where I defined \$\Delta i = i - I_0\$.

This latter equation describes how variations in the current relate to the corresponding variations in the voltage across the component. It is a simple linear relationships, where DC components are "embedded" in the variations and in the computation of the differential parameter g. If you translate this equation in a circuit element you'll find a simple resistor with a conductance g.

To answer your question directly: there is no trace of DC components in the linearized (i.e. small signals) equation, that's why they don't appear in the circuit.

The same procedure can be carried out with components with more terminals, but this requires handling more quantities and the Taylor expansion becomes unwieldy (it is multi-variable and partial derivatives pop out). The concept is the same, though.

Small signal models are nothing more than the circuit equivalent of the differential parameters obtained by linearizing the multi-variable non-linear model (equations) of the components you're dealing with.

To summarize:

• You choose a quiescent point (DC operating point): that's \$V_0\$
• You compute the dependent quantities at DC (DC analysis): you find \$I_0\$
• You linearize your circuit around that point using the DC OP data: you find \$g\$
• You solve the circuit for small variations (aka AC analysis) using only the differential (i.e. small-signal) model \$g\$.