--- title: "Crystal oscillator parallel and series resonance" subject: "Electronics Devices and Circuit" chapter: 5 key: "semester_3/electronics_devices_and_circuit:crystal-oscillator" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/crystal-oscillator site: BCT Engineering --- # Crystal oscillator parallel and series resonance Electronics Devices and Circuit · Chapter 5: Signal Generator and Waveform-Shaping Circuits - Asked in: 2079 Baisakh, 4+2+2 marks; 2068 Baisakh, 2 marks (asked 2 times) - Video: [Crystal Oscillator Fully Explained](https://www.youtube.com/watch?v=aOelJw5bREE) — ElectronIshq - 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/crystal-oscillator - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-5 ## Question Explain the working principle of crystal oscillator with diagrams operating in both parallel and series resonance mode. ## Short answer A quartz crystal oscillator relies on piezoelectric resonance. In series mode the crystal in the feedback path oscillates at fs = 1/(2π√(LC)). In parallel mode it acts as an inductor in a Colpitts circuit, oscillating between fs and fp determined by load capacitance. ## Answer A crystal oscillator uses a piezoelectric quartz crystal as the frequency-determining element in place of an LC tank. When an ac voltage is applied, the crystal vibrates mechanically at its natural frequency, and the vibration generates an ac voltage (piezoelectric effect). Because the crystal has a very high Q (10⁴ to 10⁶) and its frequency hardly changes with temperature, the oscillator is very stable (a few ppm). #### Equivalent circuit and two resonances ```text +-----[ C0 ]-----+ C0 = holder (shunt) capacitance | | o------+--[L]--[C]--[R]-+------o ``` L, C and R represent the mass, stiffness and losses of the crystal (L large, C very small, R small). - **Series resonance** (L with C): fs = 1 / (2π √(LC)). The crystal impedance is at its minimum and purely resistive (R). - **Parallel resonance** (L with the series combination of C and C0): fp = 1 / (2π √(L Ceq)), with Ceq = C C0/(C + C0). It is slightly higher than fs, and the impedance is at its maximum. Between fs and fp the crystal is inductive. #### (a) Series resonance mode ```text +VCC | [RC] | +---------B-C NPN ----+----o Output | E | [R1] | | | [RE]||[CE] | | | | [R2] GND | | | +-------[Crystal]------+ ``` The crystal is in the feedback path from the collector to the base. At fs its impedance is minimum, so the feedback is maximum; the circuit oscillates at fs. Away from fs the impedance rises and the feedback drops, so oscillation is sustained only at fs. #### (b) Parallel resonance mode (Pierce / Colpitts type) ```text Collector o----+--------+ | | [C1] [Crystal] | | Ground o--+ | | | [C2] | | | Base o--+--------+ ``` Between fs and fp the crystal behaves as a high-Q inductor, and it replaces the inductor L of the Colpitts oscillator. Together with C1 and C2 it forms the tank, and oscillation occurs at a frequency between fs and fp: f = 1/(2π √(L Ceq)), where Ceq = C·CL/(C + CL) and CL = C0 + C1C2/(C1 + C2) is the load capacitance seen by the crystal Parallel mode needs a specified load capacitance (given by the manufacturer) for the correct frequency. Applications: clocks in microprocessors, watches, communication transmitters and frequency counters. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does a quartz crystal act as a frequency-determining element? A. It vibrates at its natural frequency under ac voltage due to the piezoelectric effect and has very high Q B. It stores charge like a large capacitor C. It amplifies the signal D. It has very low Q so it is broadband Answer: A. Piezoelectric vibration at a fixed natural frequency with Q of 10^4 to 10^6 gives very stable frequency. (Key point: Piezoelectric principle) 2. What is the crystal impedance at series resonance fs? A. Maximum B. Minimum and purely resistive C. Capacitive D. Infinite Answer: B. At fs, L and C cancel leaving only R, so the impedance is minimum. (Key point: Series resonance impedance) 3. In the parallel-resonance (Pierce/Colpitts) mode, how does the crystal behave? A. As a capacitor replacing C1 B. As a high-Q inductor between fs and fp, replacing L C. As a short circuit D. As a resistor Answer: B. Between fs and fp the crystal is inductive, so it replaces the inductor of the Colpitts tank. 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