Common Source Amplifier

Every audio preamplifier and RF stage you encounter likely uses a common source amplifier at its heart. The configuration gives voltage gain that can exceed 100, making it the MOSFET equivalent of the BJT common emitter amplifier.

Core Concept

In the common source (CS) configuration, the source terminal is the common node between input and output. The AC signal is applied to the gate through a coupling capacitor, and the amplified output is taken from the drain through another coupling capacitor.

The voltage gain arises because a small change in VGS causes a large change in drain current through the transconductance gm. That current change develops a voltage across the drain resistor RD. The output is inverted: when Vin rises, VGS rises, ID rises, and the voltage drop across RD increases, pulling Vout down. This is the 180-degree phase inversion that defines the CS stage.

Adding a source resistor RS improves DC bias stability but reduces AC gain unless RS is bypassed by a capacitor CS. With CS in place, RS is shorted for AC signals, and the full gain gm*(RD||rd) is recovered. The output resistance of the CS stage is approximately RD in parallel with the MOSFET drain resistance rd.

Key Equations

Voltage gain (with RS bypassed): Av = -gm * (RD || rd)

Voltage gain (without bypass, RS unbypassed): Av = -gm*RD / (1 + gm*RS)

Transconductance: gm = 2*sqrt(K*IDQ) where K is the MOSFET process parameter and IDQ is the DC drain current.

Input resistance: Rin = R1 || R2 (MOSFET gate draws no DC current)

Output resistance: Rout = RD || rd where rd is the MOSFET drain resistance (1/lambda*IDQ, often 10s of kΩ).

Given:
  VDD = 15 V
  RD  = 4.7 kΩ
  RS  = 1 kΩ  (bypassed by CS = 47 μF)
  IDQ = 2 mA
  K   = 1 mA/V^2
  rd  = 50 kΩ  (from datasheet or given)

Why this formula:
  Source is bypassed so RS is AC-shorted. Gain = -gm*(RD||rd).

Step 1 - Find gm:
  gm = 2 * sqrt(K * IDQ)
     = 2 * sqrt(1e-3 * 2e-3)
     = 2 * sqrt(2e-6)
     = 2 * 1.414e-3
     = 2.83 mA/V

Step 2 - Find RD || rd:
  RD || rd = (4700 * 50000) / (4700 + 50000)
           = 235,000,000 / 54700
           = 4295 Ω  ≈ 4.3 kΩ

Step 3 - Voltage gain:
  Av = -gm * (RD || rd)
     = -2.83e-3 * 4295
     = -12.15

Final Answer:
  Av = -12.15  (gain magnitude 12.15, 180° phase inversion)

Exam Tip: GATE questions on CS amplifier gain often omit rd or state it is infinite. When rd is infinite, Av = -gm*RD. If RS is not bypassed, use Av = -gm*RD/(1+gm*RS). Forgetting the (1+gm*RS) denominator when RS is unbypassed is the most common mistake. Also note that Rin = R1||R2, not infinity, because the bias resistors appear in parallel at the input.

Key Properties

• The CS amplifier produces 180-degree phase inversion between input and output, identical to the BJT common emitter.
• Voltage gain magnitude is gm*RD (when rd is infinite and source is bypassed), typically 10 to 50 for practical circuits.
• Input resistance equals R1||R2 from the bias network, which can be 10 MΩ or more if large resistors are used.
• Output resistance is RD||rd, approximately equal to RD when rd >> RD.
• Bypassing RS with a capacitor CS restores full AC gain without degrading DC bias stability.
• The bandwidth is limited by parasitic capacitances Cgs and Cgd; Cgd causes the Miller effect that reduces high-frequency gain.
• Used in RF amplifiers, audio stages, and sensor interface circuits where high input impedance is needed.

Quick Revision

• Input: gate. Output: drain. Common terminal: source.
• Phase: 180-degree inversion (inverting amplifier).
• Voltage gain (bypassed): Av = -gm*(RD||rd).
• Voltage gain (unbypassed RS): Av = -gm*RD / (1 + gm*RS).
• gm = 2*sqrt(K*IDQ); higher IDQ gives higher gm and higher gain.
• High input impedance: gate draws no current; Rin set only by bias resistors.
• Cgd causes Miller effect, reducing bandwidth at high frequencies.
• Exam trap: Students apply the gain formula Av = -gm*RD even when RS is not bypassed, ignoring the (1+gm*RS) factor that reduces gain significantly.