--- title: "Operating point importance; BJT transconductance" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:q-point-selection-bjt-gm" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/q-point-selection-bjt-gm site: BCT Engineering --- # Operating point importance; BJT transconductance Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2076 Ashwin, 3+4 marks - Video: [Transistor Biasing: What is Q-point? What is Load Line? Fixed Bias Configuration Explained](https://www.youtube.com/watch?v=5T84Jzcgj7M) — ALL ABOUT ELECTRONICS - 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/q-point-selection-bjt-gm - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question For the faithful amplification of signal, selection of operating point is utmost importance. Justify the above statement. Derive transconductance of bipolar junction transistor. ## Short answer The Q-point must be centered in the active region to avoid clipping and ensure faithful amplification. Transconductance gm = IC/VT, derived from iC = IS exp(vBE/VT) giving gm = ∂iC/∂vBE = IC/VT ≈ 40·IC at room temperature. ## Answer #### Selection of the operating point (Q-point) The Q-point (ICQ, VCEQ) is the dc condition when no signal is applied. A signal makes the instantaneous operating point move along the load line around Q. For faithful amplification: - Q must be in the **active region**, in the middle of the load line (VCEQ ≈ VCC/2), so that the positive and negative output swings are equal and not clipped. - If Q is too near saturation (VCE small), the negative half-cycle of the output is clipped (saturation clipping). - If Q is near cut-off (IC small), the positive half-cycle is clipped (cut-off clipping). - If the Q-point is unstable (because of changes in β or temperature), it can drift later into a region where clipping or distortion occurs; hence stable bias circuits are used. - The Q-point also sets gm = ICQ/VT, rπ, the gain and the input resistance. ```text IC IC(sat)|\ | \ Q too high (saturation clipping) | \ ICQ | *Q (best: centre) | \ Q too low (cut-off clipping) +-----\---- VCE VCEQ VCC ``` #### Transconductance of the BJT **gm = ∂iC/∂vBE** at the Q-point: the small-signal change in collector current per unit change in base-emitter voltage. In the active region, iC = IS exp(vBE/VT). With vBE = VBE + vbe and vbe ≪ VT: ```text iC = IC exp(vbe/VT) ≈ IC (1 + vbe/VT) ic = (IC/VT) vbe gm = IC / VT ``` At room temperature VT ≈ 25 mV, so gm = 40 IC. For IC = 1 mA, gm = 40 mA/V. Then rπ = β/gm and re = α/gm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Where should Q be placed for faithful amplification? A. In the active region at about the middle of the load line (VCEQ ≈ VCC/2) B. Near cut-off C. At VCE = 0 D. Near saturation Answer: A. A central Q gives equal positive and negative output swing without clipping. (Key point: Best Q location) 2. What happens if Q is too near saturation? A. Gain doubles B. The positive half-cycle is clipped C. No distortion D. The negative output half-cycle is clipped Answer: D. Small VCE leaves little room for VCE to fall, clipping the negative output peak. (Key point: Q near saturation) 3. What is transconductance of a BJT? A. gm = ∂iC/∂vBE at Q = IC/VT B. gm = IC·VT C. gm = beta/IC D. gm = vBE/iC Answer: A. From iC = IS exp(vBE/VT), the derivative at Q gives IC/VT. (Key point: Definition of gm) 4. What is gm at IC = 1 mA, VT = 25 mV? A. 4 mA/V B. 40 mA/V C. 1 mA/V D. 25 mA/V Answer: B. gm = 1 mA / 25 mV = 40 mA/V. 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