--- title: "Colpitts oscillator frequency" subject: "Electronics Devices and Circuit" chapter: 5 key: "semester_3/electronics_devices_and_circuit:colpitts-oscillator" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/colpitts-oscillator site: BCT Engineering --- # Colpitts oscillator frequency Electronics Devices and Circuit · Chapter 5: Signal Generator and Waveform-Shaping Circuits - Asked in: 2080 Baisakh, 2+4 marks; 2075 Chaitra, 6 marks; 2074 Ashwin, 5 marks (asked 3 times) - Video: [Colpitts Oscillator Explained](https://www.youtube.com/watch?v=1fgw-ONlAcc) — 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/colpitts-oscillator - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5 ## Question Draw the circuit diagram of Colpitts oscillator. Derive its frequency of oscillation. ## Short answer The Colpitts oscillator uses a capacitive divider (C1, C2) and an inductor L in the tank. Feedback is taken across C2; the CE stage and divider give 360° phase. Frequency of oscillation: f = 1/[2π√(L C1C2/(C1+C2))]. ## Answer The Colpitts oscillator is an LC oscillator whose tank uses a tapped **capacitor** pair (C1, C2) with a single inductor L. The feedback voltage is taken across one capacitor. #### Circuit diagram (BJT, common emitter) ```text +VCC | +---------+ | | [R1] [RFC] | | | +----||---+---------+ | | Cc2 | | | C [C1] | +-----+------B--| Q1 | [L] | | E GND | | [R2] | | | | | [RE]||[CE] [C2] | | GND | | | | GND | | | | | +---------||--------------+---------+ Cc1 ``` The collector is coupled through Cc2 to the top of the tank. C1 runs from the top of the tank to ground, C2 from ground to the bottom of the tank, and L is across the whole C1-C2 series combination. The bottom of the tank is coupled back to the base through Cc1. The output is taken from the collector (or the tank). Tank connections in a simple form: ```text Collector o----+------+ | | [C1] | | [L] Emitter/GND o--+ | | | [C2] | | | Base o--+------+ ``` C1 is between collector and ground, C2 between base and ground (emitter), and L is across the series combination. Bias: voltage divider R1, R2; emitter resistor RE with bypass capacitor CE; RF choke in the collector supply. #### Working The CE stage gives a 180° phase shift. The capacitor divider (with its junction at ground) gives another 180° between the collector and base voltages. So the total is 360° and the feedback is positive. The feedback fraction is β = C1/C2. #### Derivation of frequency Neglecting loading by the transistor (hie, hoe), the tank is L in parallel with the series combination of C1 and C2: Ceq = C1 C2 / (C1 + C2) At resonance the inductive and capacitive reactances are equal: ωL = 1/(ω Ceq) **f = 1 / (2π √(L Ceq)) = 1 / (2π √( L C1 C2 / (C1 + C2) ))** (A more exact derivation using the h-parameter model gives the same frequency when the transistor loading is negligible.) **Condition for oscillation:** with Aβ ≥ 1 and β = C1/C2, the transistor gain must satisfy hfe ≥ C2/C1 **Example:** L = 100 µH, C1 = 1 nF, C2 = 1 nF: Ceq = 0.5 nF; f = 1/(2π√(100×10⁻⁶ × 0.5×10⁻⁹)) = 712 kHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the tank of a Colpitts oscillator? A. Two inductors and one capacitor B. A tapped capacitor pair (C1, C2) with a single inductor L C. A quartz crystal only D. Three RC sections Answer: B. Colpitts uses a capacitor divider C1, C2 across an inductor L, in contrast to Hartley. (Key point: Colpitts tank) 2. In the frequency derivation, what is the effective capacitance Ceq of the tank? A. C1 + C2 B. C1 C2/(C1 + C2) C. C1 - C2 D. sqrt(C1 C2) Answer: B. C1 and C2 are in series across L, so Ceq = C1C2/(C1 + C2). (Key point: Equivalent capacitance) 3. How is the frequency obtained? A. Setting loop gain to 29 B. Equating inductive and capacitive reactances: ωL = 1/(ωCeq) C. Setting hfe = 1 D. Setting C1 = 2 C2 Answer: B. At resonance XL = XC, giving f = 1/(2π√(L Ceq)). (Key point: Resonance condition) 4. With β = C1/C2, what must the transistor gain satisfy? A. hfe ≥ C1/C2 B. hfe ≥ C2/C1 C. hfe ≤ C2/C1 D. hfe ≥ 29 Answer: B. Aβ ≥ 1 with β = C1/C2 gives hfe ≥ C2/C1. 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