Common-source amplifier

The common-source amplifier is the workhorse single-transistor MOSFET amplifier: input at the gate, output at the drain, source held at AC ground. It is the MOSFET analogue of the Common-emitter amplifier, and it is the default voltage amplifier — high gain, inverting, near-infinite input resistance.

Why it amplifies

Bias the MOSFET in saturation at some operating point (see Operating point), then add a small signal $v_{g s}$ on top of the DC gate–source bias. In saturation the device is a voltage-controlled current source: the small signal produces a drain-current change

$i_{d} = g_{m} v_{g s}$

where $g_{m}$ is the MOSFET transconductance. That current flows through the drain resistor $R_{D}$ , and the output is the drain voltage. Because $i_{d}$ flowing into the drain pulls the drain voltage down,

$v_{o} = - i_{d} R_{D} = - g_{m} R_{D} v_{g s} \Rightarrow A_{v} = \frac{v _{o}}{v _{g s}} = - g_{m} R_{D}$

The minus sign means the stage is inverting (180° phase shift). Intuitively: the transistor turns an input voltage into a current via $g_{m}$ , and the resistor turns that current back into a much larger output voltage.

A small input $v_{g s}$ on top of the DC bias produces a current $i_{d} = g_{m} v_{g s}$ which drops across $R_{D}$ to give a much larger output $v_{d}$ ; the negative sign makes the amplifier inverting.

Input at gate, output at drain — the most popular MOS amplifier configuration. Inverting, high gain.

Including the transistor’s own output resistance $r_{o}$ (from Channel-length modulation), the load on the drain is $R_{D} ∥ r_{o}$ , so

$A_{v} = - g_{m} (R_{D} ∥ r_{o})$

Input resistance: essentially infinite — the gate is insulated by the Gate oxide and draws no current.
Output resistance: $R_{D} ∥ r_{o}$ , looking back into the drain.

See Input and output resistance (amplifier) for why these matter when stages are cascaded.

Maximum gain trade-off

Using $g_{m} = 2 I_{D} / V_{O V}$ and the fact that the DC drop across $R_{D}$ is $I_{D} R_{D} = V_{DD} - V_{D S}$ ,

$∣ A_{v} ∣ = g_{m} R_{D} = \frac{2 ( V _{DD} - V _{D S} )}{V _{O V}}$

Large gain wants small $V_{D S}$ (more room for the drop across $R_{D}$ ) and small Overdrive voltage $V_{O V}$ (larger $g_{m}$ ). But $V_{D S}$ cannot fall below $V_{O V}$ without leaving saturation, and very small $V_{O V}$ leaves almost no signal swing before clipping. Trading this excess gain away deliberately is exactly what Source degeneration does.

Maximum achievable gain $∣ A_{v} ∣_{ma x} = (V_{DD} - V_{D S} ∣_{B}) / (V_{O V} /2)$ : large gain wants low $V_{D S}$ at the bias point and small $V_{O V}$ .

Common-source with source degeneration

Adding an un-bypassed resistor $R_{S}$ in the source introduces Negative feedback: a rise in $i_{d}$ raises the source voltage by $i_{d} R_{S}$ , which eats into the effective $v_{g s}$ . The gain drops to

$A_{v} = - \frac{g _{m} R _{D}}{1 + g _{m} R _{S}}$

When $g_{m} R_{S} ≫ 1$ this collapses to $A_{v} \approx - R_{D} / R_{S}$ — a gain set purely by a resistor ratio, independent of the process-dependent transistor parameters. That robustness is the whole point of Source degeneration. To get the AC gain back while keeping $R_{S}$ for DC bias stability, a Bypass capacitor is placed across $R_{S}$ . The full board-level circuit is the Discrete-circuit MOSFET amplifier.

Idriss Rami — Notes

Explorer

Common-source amplifier

Why it amplifies

Maximum gain trade-off

Common-source with source degeneration

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