--- title: "BJT as amplifier: gm, rπ, re" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:bjt-amplifier-gm-rpi-re" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bjt-amplifier-gm-rpi-re site: BCT Engineering --- # BJT as amplifier: gm, rπ, re Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2072 Kartik, 6+2 marks - Video: [Hybrid-π Model](https://www.youtube.com/watch?v=yUEss0Dl6ww) — Neso Academy - 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/bjt-amplifier-gm-rpi-re - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Describe functions of BJT as amplifier with the help of transfer characteristics (ic - vBE graph), and find expressions for gm, rπ and re. Also show that β = gm rπ and rπ = (β+1) re. ## Short answer The BJT amplifies because a small change in vBE moves the operating point along the exponential iC–vBE curve, producing a collector current change ic = gm vbe. gm = IC/VT, rπ = β/gm = β VT/IC, and re = α/gm ≈ VT/IE. Since rπ = β/gm, β = gm rπ, and rπ = (β+1) re because ie/ib = β+1. ## Answer #### BJT as an amplifier (transfer characteristic) The transfer characteristic of the BJT is the plot of iC against vBE (exponential): iC = IS exp(vBE/VT). ```text iC | / | _/ | _.- <- slope = gm at Q | _.* Q |__.-' +--+-----+-------- vBE 0.5 0.7 ``` The transistor is biased in the active region at Q (VBE, IC). A small signal vbe added to VBE moves the operating point along the curve and gives a collector current change ic = gm vbe. This current through RC gives an output voltage vo = -ic RC = -gm RC vbe, which is larger than the input, so the BJT amplifies. For small signals (vbe < 10 mV) the curve at Q is nearly a straight line, and the distortion is low. #### gm, rπ and re **Transconductance:** ```text gm = ∂iC/∂vBE |Q = (IS/VT) exp(VBE/VT) = IC / VT ``` **Input resistance seen at the base:** since ib = ic/β = gm vbe/β ```text rπ = vbe/ib = β/gm = β VT/IC ``` **Resistance seen at the emitter:** since ie = ic/α = gm vbe/α ```text re = vbe/ie = α/gm = VT/IE ``` #### Proofs 1. β = gm rπ: from rπ = β/gm, gm rπ = β. ✓. 2. rπ = (β+1) re: since ie = (β+1) ib and the same vbe, ```text rπ/re = (vbe/ib)/(vbe/ie) = ie/ib = β + 1 rπ = (β+1) re ``` ✓ (also α/gm × (β+1) = β/gm since α = β/(β+1)). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the BJT amplify, using the iC-vBE characteristic? A. A small vbe on the Q-point changes iC by gm vbe, and iC through RC gives a larger output voltage B. The curve is linear so no change occurs C. vbe is amplified directly without current D. The base current flows through RC Answer: A. ic = gm vbe, and vo = -ic RC = -gm RC vbe, which is larger than vbe. (Key point: How BJT amplifies) 2. What is gm? A. IC/VT, the slope of the iC-vBE curve at Q B. VT/IC C. β/IC D. IC VT Answer: A. Differentiating iC = IS exp(vBE/VT) gives gm = IC/VT. (Key point: Transconductance) 3. What is rπ in terms of β and gm? A. gm/β B. β gm C. β/gm D. 1/(β gm) Answer: C. ib = ic/β = gm vbe/β, so rπ = vbe/ib = β/gm. (Key point: rπ expression) 4. Why is rπ = (β+1) re? A. Because ie = (β+1) ib for the same vbe, so rπ/re = ie/ib B. Because ic = (β+1) ib C. Because re = β rπ D. Because α = 1 Answer: A. rπ = vbe/ib and re = vbe/ie, so their ratio is ie/ib = β+1. 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