Short note: ac equivalent of common source MOSFET amplifier

In a common-source (CS) MOSFET amplifier the input is applied to the gate, the output is taken from the drain, and the source is common (usually bypassed to ground). The ac equivalent circuit is obtained by short-circuiting the dc supply and the large capacitors and replacing the MOSFET with its small-signal model.

Small-signal model of the MOSFET

• The gate current is zero, so the gate-source path is an open circuit.
• The drain current is a voltage-controlled current source gm·vgs.
• ro is the output (drain-source) resistance.

gm = 2k (VGSQ − Vt) = 2 √(k IDQ), ro = 1/(λ IDQ)

AC equivalent circuit

 vi o---+----o G          D o----+--------+--------+--------+----o vo
        |    +                   |        |        |        |
      [RG]  vgs        gm·vgs (↓)       [ro]     [RD]     [RL]
        |    -                   |        |        |        |
 GND ---+----o S          S o----+--------+--------+--------+---- GND

Here RG = R1 ∥ R2 is the gate bias resistance, and RD and RL are the drain and load resistors, which are in parallel with ro for ac. The gate (input) and drain (output) are separate nodes; the source is at ac ground.

Results

• Input resistance: Ri = RG (very large, in the MΩ range)
• Output resistance: Ro = RD ∥ ro ≈ RD
• Voltage gain: Av = vo/vi = −gm (ro ∥ RD ∥ RL) ≈ −gm (RD ∥ RL)

The negative sign shows a 180° phase shift between input and output. If the source resistor RS is not bypassed, the gain falls to Av = −gm (RD ∥ RL)/(1 + gm RS).