--- title: "CE amplifier input resistance and overall gain" subject: "Electronics Devices and Circuit" chapter: 2 key: "semester_3/electronics_devices_and_circuit:ce-input-resistance-gain-680k" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-input-resistance-gain-680k site: BCT Engineering --- # CE amplifier input resistance and overall gain Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor - Asked in: 2082 Baisakh, 8 marks - Video: [#412: Common Emitter Amplifier - Input and Output Impedance](https://www.youtube.com/watch?v=yhuf2_Zu8T8) — w2aew - 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-input-resistance-gain-680k - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2 ## Question Determine input resistance and overall voltage gain of given Common Emitter BJT Amplifier. β=110. [Figure: CE amplifier with +18 V supply; 680 kΩ from +18 V to base; 3 kΩ collector resistor; 0.82 kΩ emitter resistor; source Vs with series Rs = 0.6 kΩ and coupling capacitor C1 at the input; coupling capacitor C2 and load RL = 4.7 kΩ at the output] ## Short answer DC analysis gives IE=2.49mA, re=10Ω. Input resistance is Rin = RB || (β+1)(re+RE) = 680kΩ || 92.1kΩ ≈ 81kΩ. Overall voltage gain (including source) is Avs = -2.19 ≈ -2.2. ## Answer The amplifier has fixed (base-resistor) bias with an unbypassed emitter resistor. Find the dc bias first (for re), then the ac parameters. #### Given VCC = 18 V, RB = 680 kΩ, RC = 3 kΩ, RE = 0.82 kΩ, β = 110, Rs = 0.6 kΩ, RL = 4.7 kΩ, VBE = 0.7 V. #### DC bias ```text IB = (VCC - VBE)/(RB + (β+1) RE) = 17.3/(680 k + 111 × 0.82 k) = 17.3/771.0 k = 22.4 µA IC = β IB = 2.47 mA , IE = 2.49 mA re = VT/IE = 25 mV/2.49 mA = 10.0 Ω ``` #### Input resistance ```text Rib = (β+1)(re + RE) = 111 × (10.0 + 820) = 92.1 kΩ Rin = RB || Rib = 680 k || 92.1 k = 81.1 kΩ ``` #### Voltage gain ```text Av (from base to output) = -(RC || RL)/(re + RE) RC || RL = 3k × 4.7k/7.7k = 1.831 kΩ Av = -1831/830 = -2.21 ``` Overall gain, including the source resistance: ```text Avs = Av × Rin/(Rin + Rs) = -2.21 × 81.1/(81.1 + 0.6) = -2.19 ``` **Answer: Rin ≈ 81 kΩ; overall voltage gain ≈ -2.2** (the gain is small because RE is unbypassed). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the dc bias found first in this problem? A. To find the coupling capacitor value B. To remove RL C. To get IE and hence re = VT/IE for the ac calculations D. To find the load line saturation only Answer: C. re depends on the dc emitter current, which is needed in Rib and Av. (Key point: DC bias first) 2. What is the formula for IB with a base resistor and unbypassed RE? A. IB = (VCC - VBE)/(RB + (beta+1) RE) B. IB = (VCC - VBE)/(RB + RE) C. IB = VCC/(RB + RC) D. IB = (VCC - VBE)/RB Answer: A. The emitter resistor is reflected into the base loop multiplied by (beta+1). (Key point: Base current) 3. How is the input resistance obtained? A. Rin = rpi ≈ 1 kΩ B. Rin = RB || RC ≈ 2.9 kΩ C. Rin = RB + re ≈ 680 kΩ D. Rin = RB || (beta+1)(re + RE) ≈ 81 kΩ Answer: D. Rib = 111 × 830 ≈ 92.1 kΩ in parallel with 680 kΩ gives 81.1 kΩ. (Key point: Input resistance) 4. Why is the overall gain only about -2.2? A. Because VCC is only 18 V B. Because the base resistor is 680 kΩ C. The unbypassed RE (and load RL) make Av = -(RC||RL)/(re+RE) small D. Because beta is low Answer: C. With RE = 820 Ω in the ac path Av = -1831/830 = -2.21; the source resistance reduces it slightly to -2.19. 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