MOSFET Amplifier Configurations

MOSFET amplifier stages (common-source, common-drain, common-gate) are the building blocks of analog and RF ICs. GATE EC tests their gain and impedance characteristics (1–2 marks).

Key results

• Small-signal: g_m = 2I_D/V_ov = √(2μC_ox(W/L)I_D); r_o = 1/(λI_D).
• Common-source (CS): A_v = −g_m(R_D ∥ r_o); high gain, 180° phase, high R_in (gate). With source degeneration R_S: A_v ≈ −g_m R_D/(1+g_m R_S).
• Common-drain (source follower): A_v = g_m R_S/(1+g_m R_S) ≈ 1; buffer, high R_in, low R_out ≈ 1/g_m.
• Common-gate (CG): non-inverting, low R_in ≈ 1/g_m, current buffer; used in cascode/RF.

Stage	Gain	R_in	R_out	Phase
Common-source	−g_m R_D (high)	∞ (gate)	R_D	180°
Source follower	≈ 1	∞	1/g_m	0°
Common-gate	high	1/g_m	R_D	0°

Exam Tricks & Tips

• 🎯 g_m = 2I_D/V_ov — the fastest way to relate transconductance to bias.
• 🎯 CS = high inverting gain; source follower = unity-gain buffer (low output impedance 1/g_m); CG = low-input-impedance current buffer.
• 🎯 Cascode (CS + CG) boosts output resistance and bandwidth (reduces Miller effect).
• 🎯 Source degeneration lowers gain to −g_m R_D/(1+g_m R_S) but improves linearity/bandwidth.
• 🎯 MOSFET gate draws no DC current → essentially infinite DC input resistance.
• ❌ Common mistake: using BJT r_π for a MOSFET — the MOSFET gate is open (infinite R_in), no r_π.

Expected exam pattern

1–2 marks: compute g_m, r_o, CS/follower gain, output resistance, or identify the configuration from its impedance/gain signature.

Quick recap

g_m=2I_D/V_ov, r_o=1/(λI_D). CS: A_v=−g_m(R_D∥r_o), high R_in, 180°. Follower: A_v≈1, R_out≈1/g_m. CG: low R_in≈1/g_m, non-inverting. Cascode boosts R_out & bandwidth.

Cover the answer, recall, then check. 11 cards on MOSFET amplifier configurations for GATE EC.

Q1. Write the MOSFET transconductance in terms of I_D and V_ov.
A1. g_m = 2I_D/V_ov = √(2μC_ox(W/L)·I_D).

Q2. Write the MOSFET output resistance.
A2. r_o = 1/(λI_D) (from channel-length modulation).

Q3. Give the common-source voltage gain.
A3. A_v = −g_m(R_D ∥ r_o); it is inverting (180°).

Q4. Characterize the source follower (common-drain).
A4. A_v ≈ 1 (non-inverting), very high input resistance, low output resistance ≈ 1/g_m — a buffer.

Q5. Characterize the common-gate stage.
A5. Non-inverting, low input resistance ≈ 1/g_m, acts as a current buffer; used in cascodes/RF.

Q6. What is the DC input resistance at a MOSFET gate?
A6. Essentially infinite (no gate current).

Q7. Effect of source degeneration R_S on CS gain?
A7. A_v ≈ −g_m R_D/(1 + g_m R_S): reduces gain but improves linearity and bandwidth.

Q8. What does a cascode configuration achieve?
A8. Higher output resistance and bandwidth (reduced Miller effect) by stacking a CG stage on a CS stage.

Q9. Output resistance of a source follower?
A9. Approximately 1/g_m.

Q10. Which single stage gives high inverting voltage gain?
A10. The common-source stage.

Q11. Why does the MOSFET model lack an r_π (unlike the BJT)?
A11. The gate draws no current, so there is no finite input resistance between gate and source.

MOSFET amplifiers are the analog backbone of CMOS chips, and their three single-transistor configurations mirror the BJT's but with a purely voltage-controlled, infinite-input-resistance device. GATE tests small-signal gain, input/output resistance, and configuration selection — usually as a quick gm-based gain calculation.

Core concept: Bias a MOSFET in saturation, then use its small-signal model (gm, ro) to find the gain and impedances of common-source, common-drain, and common-gate stages.

The theory

Beginner — the small-signal model

With the MOSFET in saturation at drain current ID:

• Transconductance gm = √(2·kn·ID) = 2ID/VOV = kn·VOV (VOV = VGS − VT).
• Output resistance ro = 1/(λID) (channel-length modulation).
• Gate current = 0, so input resistance is essentially infinite — the key MOSFET advantage over the BJT.

Intermediate — the three configurations

Config	Small-signal gain	Zin	Zout	Role
Common-source (CS)	−gm(RD‖ro), inverting	∞ (gate)	RD‖ro	voltage amp
Common-drain (source follower)	gm·RL/(1+gm·RL) ≈ 1	∞	≈ 1/gm (low)	buffer
Common-gate (CG)	+gm(RD‖ro), non-inverting	≈ 1/gm (low)	high	high-freq/current buffer
CS is the workhorse voltage amplifier. A source degeneration resistor RS reduces gain to ≈ −gm·RD/(1 + gm·RS) but linearises and stabilises.

Advanced — active loads and cascode

• Active (current-source) load: replacing RD with a transistor current source gives a very high load resistance, so gain approaches the intrinsic gain gm·ro = 2/(λ·VOV) — the maximum a single device can give.
• Cascode (CS + CG stacked): boosts output resistance to ≈ gm·ro² and gain to (gm·ro)², while improving high-frequency response by killing the Miller effect on the input device.
• Gain is fundamentally limited by gm·ro, which shrinks as devices scale — a driver of modern analog design.

Worked example

A common-source amplifier: ID = 0.5 mA, VOV = 0.5 V, RD = 10 kΩ, λ ≈ 0 (ignore ro). Find gm and voltage gain.
gm = 2ID/VOV = 2(0.5 mA)/0.5 V = 2 mA/V.
Av = −gm·RD = −2 mA/V × 10 kΩ = −20 (inverting).
If an active load raised the effective load to 100 kΩ, the gain would approach −gm·ro ≈ −200 — the value of an active load.

GATE relevance

1–2 mark NAT/MCQ: gm from bias, CS/CD/CG small-signal gain and impedances, source degeneration, active-load/cascode gain (gm·ro, (gm·ro)²), and configuration selection. Parallels and often contrasts with the BJT versions.

Exam tricks and shortcuts

• gm = 2ID/VOV = √(2·kn·ID); intrinsic gain = gm·ro.
• CS = inverting voltage amp; CD (source follower) = buffer (gain ≈ 1); CG = non-inverting, low Zin.
• MOSFET Zin = ∞ (no gate current) — a defining difference from the BJT.
• Mnemonic: "CS/CD/CG ↔ CE/CC/CB" for the BJT analogy.

Key formulas

• Common source: A_v = −g_m(R_D ∥ r_o); high gain, inverting.
• Common drain (source follower): A_v = g_m R_S/(1 + g_m R_S) ≈ 1; high input, low output impedance.
• Common gate: A_v = g_m(R_D ∥ r_o); low input impedance, non-inverting.
• Transconductance: g_m = √(2μ_nC_ox(W/L)I_D); output resistance r_o = 1/(λI_D).

Common-source gives voltage gain, source-follower buffers, and common-gate provides current buffering.

Worked Example

Problem: A common-source MOSFET amplifier has g_m = 2 mA/V, output resistance r_o = 50 kΩ and drain resistor R_D = 10 kΩ. (a) Find the midband voltage gain. (b) A source-degeneration resistor R_S = 500 Ω is now added; find the new gain (neglect r_o for this part).

Solution:
(a) Common-source gain (with r_o in parallel with R_D):
A_v = −g_m·(R_D ∥ r_o).
R_D ∥ r_o = (10k × 50k)/(10k + 50k) = 500k/60 ≈ 8.33 kΩ.
A_v = −(2 × 10⁻³)(8.33 × 10³) = −16.7.

(b) With source degeneration R_S (neglecting r_o), the gain is reduced by the factor (1 + g_m R_S):
A_v = −g_m·R_D / (1 + g_m·R_S).
g_m·R_S = (2 × 10⁻³)(500) = 1.
A_v = −(2 × 10⁻³ × 10 × 10³) / (1 + 1) = −20/2 = −10.

Degeneration trades gain (from −16.7 to −10) for improved linearity and bandwidth.

Answer: (a) A_v ≈ −16.7; (b) with R_S = 500 Ω, A_v = −10.