- Asked 3/28
- 2080 Baisakh · 2+4 marks
- 2075 Chaitra · 6 marks
- 2074 Ashwin · 5 marks
Colpitts oscillator frequency
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))].
Draw the circuit diagram of Colpitts oscillator. Derive its frequency of oscillation.
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)
+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:
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Applications of LC oscillators; Colpitts oscillator · Electronics Devices and Circuit, chapter 5
- Hartley oscillator frequency · Electronics Devices and Circuit, chapter 5 · asked 4×
- Hartley vs Colpitts for FM carrier · Electronics Devices and Circuit, chapter 5
- RC phase shift oscillator · Electronics Devices and Circuit, chapter 5 · asked 6×
- Barkhausen criteria and RC phase shift oscillator · Electronics Devices and Circuit, chapter 5 · asked 3×
- Voltage controlled oscillator using 555 · Electronics Devices and Circuit, chapter 5 · asked 4×
- Damped LCR oscillation, damped frequency · Engineering Physics, chapter 1 · asked 2×
- R-H range of k and oscillation frequency · Control Systems, chapter 4 · asked 2×
- Wien bridge oscillator circuit and operation · Basic Electronics Engineering, chapter 4 · asked 4×
- Phase and frequency error in local oscillator (DSB and SSB) · Communication Systems, chapter 2 · asked 2×