--- title: "AC equivalent: Rin and Rout of divider-bias amplifier" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-ac-equivalent-rin-rout-100k-20k" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-ac-equivalent-rin-rout-100k-20k site: BCT Engineering --- # AC equivalent: Rin and Rout of divider-bias amplifier Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2069 Chaitra, 8 marks - Video: [BJT Small Signal Analysis: Common Emitter Fixed Bias and Voltage Divider Bias](https://www.youtube.com/watch?v=wbDUDRlmUuM) — 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/ce-ac-equivalent-rin-rout-100k-20k - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Draw the ac equivalent circuit for given circuit and find its input and output resistances. Assume β = 100 for the BJT. [Figure: BJT amplifier with Vcc = +15 V; base divider R1 = 100 kΩ from Vcc and R2 = 20 kΩ to ground; collector resistor RC = 5 kΩ with output Vo at the collector; emitter resistor RE = 1 kΩ to ground; input vi at the base] ## Short answer DC bias gives re ≈ 16.2 Ω. With unbypassed RE, the base sees (β+1)(re+RE) ≈ 102.6 kΩ. Rin = R1||R2||102.6 kΩ ≈ 14.3 kΩ; Rout is simply RC = 5 kΩ. ## Answer The circuit has a base divider (R1 = 100 kΩ, R2 = 20 kΩ), RC = 5 kΩ, and RE = 1 kΩ in the emitter, which is assumed unbypassed (no bypass capacitor is shown). VCC = 15 V, β = 100. #### DC bias (to find re) ```text VBB = 15 × 20/120 = 2.5 V , RB = 100 k || 20 k = 16.67 kΩ IB = (2.5 - 0.7)/(16.67 k + 101 × 1 k) = 1.8/117.7 k = 15.3 µA IC = 1.53 mA , IE = 1.545 mA re = 25 mV/1.545 mA = 16.2 Ω ``` #### AC equivalent circuit VCC is replaced by ground: R1 and R2 are in parallel from the base to ground; the transistor T-model (re in the emitter path, with current source α ie to the collector); RE in series from the emitter to ground; RC from the collector to ground; the output is at the collector. ```text vi--+----B--------C---+--- vo | \ re (α ie)| R1||R2 E RC | | | GND RE GND | GND ``` #### Input resistance ```text Rib = (β+1)(re + RE) = 101 × (16.2 + 1000) = 102.6 kΩ Rin = R1 || R2 || Rib = 16.67 k || 102.6 k = 14.3 kΩ ``` #### Output resistance Looking into the collector, the transistor is a current source (ro = ∞), so ```text Rout = RC = 5 kΩ ``` (Voltage gain, if needed: Av = -RC/(re + RE) = -4.9.) **Answer: Rin ≈ 14.3 kΩ, Rout ≈ 5 kΩ.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the ac equivalent drawn? A. RC removed B. VCC kept as a source C. RE shorted though unbypassed D. VCC replaced by ground so R1 and R2 are in parallel at the base Answer: D. The dc supply is an ac ground. (Key point: AC equivalent rules) 2. What are VBB and RB? A. 1.8 V and 20 kΩ B. 15 V and 120 kΩ C. 2.5 V and 16.67 kΩ D. 2.5 V and 120 kΩ Answer: C. VBB = 15×20/120 = 2.5 V; RB = 100k||20k = 16.67 kΩ. (Key point: DC thevenin) 3. How is Rin found? A. R1 || R2 || rpi B. R1 || R2 || (beta+1)(re + RE) ≈ 14.3 kΩ C. RE || re D. R1 + R2 Answer: B. RE is unbypassed, so (beta+1)(re+RE) = 102.6 kΩ in parallel with 16.67 kΩ. (Key point: Rin) 4. What is Rout? A. R1||R2 B. RE = 1 kΩ C. re D. RC = 5 kΩ Answer: D. Looking into the collector sees a current source, leaving RC. 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