--- title: "BJT differential amplifier collector waveforms" subject: "Basic Electronics Engineering" chapter: 4 key: "semester_2/basic_electronics_engineering:diff-amp-bjt-waveform" source: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/diff-amp-bjt-waveform site: BCT Engineering --- # BJT differential amplifier collector waveforms Basic Electronics Engineering · Chapter 4: The Operational Amplifier and Oscillator - Asked in: 2072 Magh, 4 marks - Video: [BJT: Differential Amplifier Explained](https://www.youtube.com/watch?v=yXJADXuWtlI) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/basic_electronics_engineering/diff-amp-bjt-waveform - Whole chapter: https://www.bctengineer.com/md/semester_2/basic_electronics_engineering/chapter-4 ## Question Draw the circuit diagram of differential amplifier using BJT and sketch the waveform at the collector terminals for sinusoidal differential input. ## Short answer A BJT differential pair with matched transistors and equal RC. For differential sinusoidal inputs (v1, v2 out of phase), collector voltages vo1 and vo2 are antiphase, each with amplitude Ad·vd/2; vo1 is inverted relative to v1, vo2 inverted to v2. The differential output between collectors has amplitude Ad·vd. ## Answer #### Circuit ```text +VCC +-------+-------+ | | [RC] [RC] | | vo1 o----+ +----o vo2 | | C| |C v1 o----B| Q1 Q2 |B----o v2 E| |E | | +-------+-------+ | [RE] (or constant current source) | -VEE ``` Q1 and Q2 are matched transistors with equal collector resistors RC and a common emitter resistor RE (or constant current source). The inputs v1 and v2 are applied at the bases; the outputs are taken at the collectors. #### Differential input Let v1 = (vd/2) sin ωt and v2 = −(vd/2) sin ωt (equal amplitude, 180° out of phase). - When v1 rises, IC1 rises, so the voltage drop across RC rises and the collector voltage vo1 falls. At the same time v2 falls, IC2 falls and vo2 rises. - The total emitter current stays almost constant (IE1 + IE2 = constant), so the emitter point acts as an ac ground for the differential signal. #### Collector waveforms ```text v1: /\ /\ input at base of Q1 / \ / \ / \/ \ v2: \ /\ / input at base of Q2 \ / \ / \/ \/ vo1: \ /\ / collector of Q1: amplified, 180° to v1 \ / \ / \/ \/ vo2: /\ /\ collector of Q2: amplified, in phase with v1 / \ / \ / \/ \ ``` - vo1 is 180° out of phase with v1; vo2 is 180° out of phase with v2 (that is, in phase with v1). - vo1 and vo2 are therefore in antiphase with each other, and each has amplitude Ad·vd/2. - The output taken between the collectors, vo2 − vo1, has amplitude Ad·vd (twice that of either collector), with differential gain Ad = gm RC. Each single-ended output has a gain of gm RC/2 with respect to vd. - A common-mode input (the same signal on both bases) produces a very small output, because the large RE (or current source) gives strong negative feedback. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When v1 rises, what happens to vo1? A. It falls (180° out of phase with v1) B. It rises in phase C. It stays constant D. It doubles Answer: A. IC1 rises, the drop across RC rises, so the collector voltage falls. (Key point: Collector phase) 2. How are vo1 and vo2 related for differential input? A. In antiphase (180° apart) B. In phase C. 90° apart D. vo2 is zero Answer: A. v2 is opposite to v1, so vo2 is in phase with v1 and opposite to vo1. (Key point: Relative phase of outputs) 3. Why does the emitter point act as ac ground for differential signal? A. IE1 + IE2 stays almost constant, one rises as the other falls B. The emitter is physically grounded C. RE is zero D. The collector is shorted Answer: A. Increase in one emitter current is matched by decrease in the other, so the common emitter voltage barely moves. (Key point: Constant emitter current) 4. Why is the output to a common-mode input very small? A. The large RE or current source gives strong negative feedback B. The transistors are off C. Ad is zero D. RC is zero Answer: A. Both emitter currents rise together, increasing the drop in RE strongly and opposing the change. 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