Common Gate Amplifier

RF front-end circuits and current buffer stages rely on the common gate amplifier for its unique property: low input impedance and excellent high-frequency behavior. Unlike the common source, it passes signal from source to drain without phase inversion.

Core Concept

In the common gate (CG) configuration, the gate is the common terminal. The gate is connected to AC ground through a bypass capacitor CG, even though DC bias sets a finite gate voltage. The input signal enters at the source terminal, and the output is taken from the drain.

The most important property of the CG amplifier is its low input impedance, approximately 1/gm. This makes it ideal for matching to low-impedance sources like 50-ohm antennas in RF systems. Current entering the source passes directly through the channel and out the drain, so the signal is not inverted. The CG stage is a non-inverting amplifier.

Because the gate is grounded for AC signals, there is no Miller effect from the gate-drain capacitance Cgd. This gives the CG configuration superior high-frequency performance compared to the CS stage. The CG amplifier is commonly used in cascode configurations paired with a CS stage to extend bandwidth while maintaining gain.

Key Equations

Voltage gain: Av = +gm * (RD || rd) (positive, non-inverting)

Input resistance: Rin = 1/gm || RS_bias (approximately 1/gm, which is low, typically 100 to 500 Ω)

Output resistance: Rout = RD || rd (approximately RD when rd >> RD)

Transconductance: gm = 2*sqrt(K*IDQ)

Current gain: unity (Iout/Iin = 1), since drain current equals source current in a MOSFET.

Given:
  VDD  = 12 V
  RD   = 3.3 kΩ
  IDQ  = 2 mA
  K    = 0.8 mA/V^2
  rd   = 40 kΩ

Why this formula:
  CG configuration with source bypass; non-inverting gain = +gm*(RD||rd).

Step 1 - Find gm:
  gm = 2 * sqrt(K * IDQ)
     = 2 * sqrt(0.8e-3 * 2e-3)
     = 2 * sqrt(1.6e-6)
     = 2 * 1.265e-3
     = 2.53 mA/V

Step 2 - Find RD || rd:
  RD || rd = (3300 * 40000) / (3300 + 40000)
           = 132,000,000 / 43300
           = 3049 Ω ≈ 3.05 kΩ

Step 3 - Voltage gain:
  Av = +gm * (RD || rd)
     = 2.53e-3 * 3049
     = +7.71

Step 4 - Input resistance:
  Rin = 1/gm = 1 / 2.53e-3 = 395 Ω

Final Answer:
  Av = +7.71 (non-inverting), Rin = 395 Ω

Exam Tip: GATE often compares CG and CS in the same problem. Remember: CG has no phase inversion (Av is positive), and input resistance is 1/gm (low). CS has phase inversion (Av is negative) and high input resistance. The CG's low Rin is its defining feature; mixing this up with CS gives the wrong input resistance by orders of magnitude.

Key Properties

• The CG amplifier is non-inverting: output at drain is in phase with input at source.
• Input impedance is low, approximately 1/gm, typically 100 to 500 Ω for practical bias conditions.
• No Miller effect because gate is AC-grounded; Cgd does not create a feedback path.
• Excellent high-frequency performance; used in RF cascode stages for bandwidths exceeding 1 GHz.
• Current gain is approximately unity; the stage is primarily a voltage amplifier with impedance transformation.
• Output resistance is high, approximately RD, making it a good current source driving a load.
• In cascode circuits, the CG stage sits above a CS stage, combining high gain with wide bandwidth.

Quick Revision

• Input: source. Output: drain. Common terminal: gate (AC-grounded).
• Phase: non-inverting (Av is positive).
• Voltage gain: Av = +gm*(RD||rd).
• Input resistance: 1/gm (low, 100 to 500 Ω typically).
• No Miller effect; superior high-frequency performance vs CS.
• Used in RF cascode amplifiers and low-impedance source matching.
• Current gain is unity; output current equals input current.
• Exam trap: Students assign phase inversion to the CG stage because they confuse it with the CS amplifier. CG does not invert; only CS and CD produce a phase relationship change (CS inverts, CD does not, CG does not).