--- title: "CE with unbypassed RE; application of CB amplifier" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-unbypassed-output-impedance-cb-application" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-unbypassed-output-impedance-cb-application site: BCT Engineering --- # CE with unbypassed RE; application of CB amplifier Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2076 Chaitra, 4+3+1 marks - Video: [CE Amplifier with Unbypassed RE - Small Signal Analysis using Hybrid Equivalent Model](https://www.youtube.com/watch?v=XcDXJ1S2Sfk) — Padmasri Naban - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-unbypassed-output-impedance-cb-application - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Draw common emitter transistor amplifier circuit (emitter bias with unbypassed emitter capacitor) and find its output impedance and voltage gain. Write application of common base amplifier. ## Short answer With unbypassed RE, output impedance is approximately RC. Voltage gain is Av = -(RC||RL)/(re+RE). Common base amplifiers are used in high-frequency/RF circuits, current buffers, and cascode stages due to low input impedance and no Miller effect. ## Answer #### Circuit (voltage divider bias, RE unbypassed) ```text +VCC | +-----+-----+ | | R1 RC | | C2 vs--||--+------B C----+----||----+---- vo C1 | \ / | | R2 NPN (vo node) RL | | | GND E GND | RE (no bypass capacitor) | GND ``` C1 and C2 are coupling capacitors; RE is not bypassed, so it appears in the ac path. #### Output impedance Set vs = 0 and look into the collector. The unbypassed RE (series feedback) raises the resistance looking into the collector: ```text Zout(collector) = ro [1 + gm (RE || rπ)] → ∞ (ro = ∞) Zo = RC || Zout(collector) ≈ RC ``` So the output impedance of the stage is approximately **RC** (RE only raises the collector impedance, which is already very large). #### Voltage gain With the T-model: vb = ie (re + RE) and vo = -ie (RC || RL), so ```text Av = vo/vb = -(RC || RL)/(re + RE) ``` (If RE ≫ re: Av ≈ -(RC||RL)/RE, which is set by the resistors alone.) The input impedance is Zin = R1 || R2 || (β+1)(re + RE). #### Applications of the common base amplifier - High-frequency and RF amplifiers (VHF/UHF), because there is no Miller effect and the bandwidth is wide. - Current buffers, and stages that need a low input impedance (e.g. matching to low-impedance sources such as microphones or antennas). - The upper stage of a cascode (CE-CB) amplifier. - Constant current sources and current mirrors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the output impedance of the CE stage with unbypassed RE? A. About RE B. About RC, since the collector impedance looking in is very large C. About rpi D. About re Answer: B. Looking into the collector gives ro[1+gm(RE||rpi)], very large, so Zo ≈ RC. (Key point: Output impedance) 2. What is Av with unbypassed RE? A. -(RC||RL)/(re + RE) B. +RE/(re+RE) C. -gm RC D. -(RC||RL)/RE only Answer: A. vb = ie(re+RE) and vo = -ie(RC||RL). (Key point: Voltage gain) 3. If RE >> re, what determines the gain? A. VT/IE B. VCC C. beta only D. Resistor ratio -(RC||RL)/RE, nearly independent of transistor parameters Answer: D. The gain becomes stable and set by resistors. (Key point: Gain when RE >> re) 4. Which is a typical application of the CB amplifier? A. Voltage regulator pass element as buffer B. Digital NOT gate C. Audio power output stage D. High-frequency/RF amplifier or cascode top stage Answer: D. CB has no Miller effect and a wide bandwidth, suited to RF and the upper stage of a cascode. 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