--- title: "AC analysis of differential amplifier" subject: "Advanced Electronics" chapter: 1 key: "semester_4/advanced_electronics:differential-amplifier-ac-analysis" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/differential-amplifier-ac-analysis site: BCT Engineering --- # AC analysis of differential amplifier Advanced Electronics · Chapter 1: Operational Amplifier Circuits - Asked in: 2070 Chaitra, 5 marks - Video: [AC Analysis | Differential Amplifier | Analog Electronics| op-amp | IC-741](https://www.youtube.com/watch?v=ggoN__bvOXQ) — Graph Guru - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/differential-amplifier-ac-analysis - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1 ## Question Discuss an ac analysis of differential amplifier with necessary diagrams and mathematical expressions. ## Short answer For differential mode, emitter is ac ground; half‑circuit yields Ad = gm·RC (double‑ended) or gm·RC/2 (single‑ended), Rid = 2rπ. Common‑mode half sees 2RE, Acm ≈ –RC/(2RE), CMRR ≈ gm·RE. A tail current source greatly increases CMRR. ## Answer A differential amplifier amplifies the difference vd = v1 − v2 and rejects the common-mode signal vcm = (v1 + v2)/2. Its ac analysis is done by splitting any input into these two parts and using the half-circuit method. ```text +VCC +VCC | | RC RC +-- vo1 vo2 --+ | | v1 ---B Q1 Q2 B--- v2 | | +-------+-------+ | E RE (tail) | -VEE ``` #### Differential-mode analysis For v1 = +vd/2 and v2 = −vd/2, the change in IE1 equals minus the change in IE2, so the current through RE stays constant and the emitter node is **ac ground**. ```text Half circuit (CE stage, emitter grounded): ic1 = gm·vd/2 , ic2 = -gm·vd/2 vo1 = -gm·RC·vd/2 , vo2 = +gm·RC·vd/2 Double-ended: Ad = (vo2 - vo1)/vd = gm·RC Single-ended: Ad = gm·RC/2 Differential input resistance: Rid = 2rπ ``` #### Common-mode analysis For v1 = v2 = vcm, both emitter currents change together, so the current in RE is doubled. Each half sees an emitter resistance of 2RE. ```text Half circuit: CE stage with 2RE in the emitter Acm = vo1/vcm = -gm·RC / (1 + 2gm·RE) ≈ -RC/(2RE) Common-mode input resistance ≈ β·(2RE) per side Double-ended output: vo2 - vo1 = 0 (ideally Acm = 0) ``` #### CMRR and total output ```text Single-ended: CMRR = |Ad/Acm| = (gm·RC/2)/(RC/2RE) ≈ gm·RE CMRR(dB) = 20 log10 |Ad/Acm| vo = Ad·vd + Acm·vcm ``` For a high CMRR, RE should be very large; this is why the tail resistor is replaced by a constant current source (current mirror), whose output resistance is hundreds of kΩ. Using double-ended output with matched halves also cancels the common-mode output. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can a differential-mode half circuit be drawn with emitter grounded? A. The emitter current in RE is constant so the node is ac ground B. The collector is grounded C. Q1 and Q2 are off D. RE is replaced by a short Answer: A. Equal and opposite emitter-current changes cancel in RE. (Key point: Half circuit) 2. In common-mode analysis, what emitter resistance does each half see? A. 2RE B. RE/2 C. 0 D. RE Answer: A. Both emitter currents flow through RE, doubling its current, so each half sees 2RE. (Key point: Common mode) 3. What is CMRR (single-ended) of the BJT pair? A. About RC/RE B. About rπ C. About gm RE D. About 1/gm Answer: C. (gm RC/2)/(RC/2RE) = gm RE. (Key point: CMRR) 4. Why is the tail resistor replaced by a current source? A. To get a very large RE and hence high CMRR B. To raise Acm C. To lower Ad D. To reduce gm Answer: A. A mirror has very high output resistance, making Acm very small. 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