Basic Electrical & Electronics Engineering — Signal Generator, NEC licence examination syllabus (Nepal Engineering Council).
LC Oscillators: Hartley and Colpitts, the radio workhorses
Above about 1 MHz, RC networks stop being practical. Tuned circuits take over.
At radio frequencies an RC oscillator would need impractically tiny resistors and capacitors, and its frequency stability would be poor. An LC tank — an inductor and capacitor exchanging energy back and forth — is naturally frequency-selective and gets better at high frequency, since the components shrink to convenient sizes. Two circuits dominate, distinguished by one thing: whether the tank is tapped on the inductor or the capacitor.
The tank circuit
f = 1 / (2π√(LC)) With L = 100 µH, C = 100 pF.
Quadruple the capacitance and the frequency halves rather than quartering — the square root is doing the work. That is why tuning a radio over a 3:1 frequency range needs a 9:1 capacitor swing.
Energy sloshes between the inductor's magnetic field and the capacitor's electric field, like a pendulum swinging between kinetic and potential energy. Left alone it would decay through resistance; the amplifier's job is to top it up each cycle.
Resonant frequency of any LC tank:
f = 1/(2π√(LC))
This is the same resonance formula from Section 2 — an
oscillator is simply a resonant circuit kept alive by an
amplifier that replaces the energy lost each cycle.
Hartley — tapped inductor
Two inductors (or one tapped coil) with a single capacitor.
f = 1/(2π√(L_eq·C)) where L_eq = L₁ + L₂ + 2M
M = mutual inductance between the coils
(if they are not magnetically coupled, M = 0 and
L_eq = L₁ + L₂)
Feedback fraction: β = L₁/L₂
Gain condition: A ≥ L₂/L₁
Colpitts — tapped capacitor
Two capacitors in series with a single inductor.
f = 1/(2π√(L·C_eq)) where C_eq = C₁C₂/(C₁ + C₂)
(series combination)
Feedback fraction: β = C₁/C₂
Gain condition: A ≥ C₂/C₁
The memory hook: Hartley has two inductors (H for "two Henrys"), Colpitts has two capacitors. And note the mathematical asymmetry — Hartley's inductances add (series inductors), while Colpitts' capacitances combine as a series product-over-sum. Getting that backwards is the most common numerical error in this topic.
⚔️ Hartley vs Colpitts in practice
HartleyEasy to tune over a wide range with a single variable capacitor, so it was standard in old radio receivers' tuning stages. Needs a tapped coil, which is harder to manufacture.
ColpittsBetter frequency stability and a purer waveform, because capacitors are more stable components than tapped inductors. Preferred at higher frequencies and in modern designs. The Clapp oscillator is a Colpitts with an extra series capacitor for even better stability.
Worked numerical 1 — Hartley oscillator
A Hartley oscillator has L₁ = 100 µH, L₂ = 1 mH, mutual inductance M = 20 µH, and C = 200 pF. Find the frequency and the minimum gain.
Equivalent inductance:
L_eq = L₁ + L₂ + 2M
= 100 µH + 1000 µH + 2(20 µH)
= 100 + 1000 + 40 = 1140 µH
= 1.14 mH
Frequency:
f = 1/(2π√(L_eq·C))
= 1/(2π√(1.14 × 10⁻³ × 200 × 10⁻¹²))
= 1/(2π√(2.28 × 10⁻¹³))
= 1/(2π × 4.775 × 10⁻⁷)
= 1/(3.0 × 10⁻⁶)
= 333.3 kHz
Minimum gain:
A ≥ L₂/L₁ = 1000/100 = 10
If the coils were NOT coupled (M = 0):
L_eq = 1100 µH, f = 339.3 kHz
— the mutual inductance shifted f by ~2%. Don't ignore M
when the question provides it.
Worked numerical 2 — Colpitts oscillator
A Colpitts oscillator has C₁ = 100 pF, C₂ = 400 pF and L = 50 µH. Find the frequency and minimum gain.
Equivalent capacitance (SERIES combination):
C_eq = C₁C₂/(C₁ + C₂)
= (100 × 400)/(100 + 400)
= 40 000/500 = 80 pF
Note C_eq (80 pF) is SMALLER than either capacitor —
correct for a series combination. If your C_eq came out
larger than 400 pF you added them by mistake.
Frequency:
f = 1/(2π√(L·C_eq))
= 1/(2π√(50 × 10⁻⁶ × 80 × 10⁻¹²))
= 1/(2π√(4 × 10⁻¹⁵))
= 1/(2π × 6.325 × 10⁻⁸)
= 1/(3.973 × 10⁻⁷)
= 2.517 MHz
Minimum gain:
A ≥ C₂/C₁ = 400/100 = 4
Worked numerical 3 — designing for a target frequency
Design a Colpitts oscillator for 10 MHz using L = 10 µH and a feedback ratio giving gain requirement 5.
From f = 1/(2π√(LC_eq)), find the C_eq needed:
C_eq = 1/(4π²f²L)
= 1/(4π² × (10⁷)² × 10 × 10⁻⁶)
= 1/(39.48 × 10¹⁴ × 10⁻⁵)
= 1/(3.948 × 10¹⁰)
= 2.533 × 10⁻¹¹ F = 25.33 pF
Gain condition A ≥ C₂/C₁ = 5, so C₂ = 5C₁.
Substituting into the series formula:
C_eq = C₁(5C₁)/(C₁ + 5C₁) = 5C₁²/6C₁ = 5C₁/6
25.33 = 5C₁/6
C₁ = 25.33 × 6/5 = 30.4 pF → use 33 pF
C₂ = 5 × 30.4 = 152 pF → use 150 pF
Verify with standard values (33 pF, 150 pF):
C_eq = (33 × 150)/183 = 27.05 pF
f = 1/(2π√(10µ × 27.05p)) = 1/(2π × 1.645 × 10⁻⁸)
= 9.68 MHz (3% low — trim L slightly or accept)
💡 Exam angle: expect a direct numerical. The two things that cost marks are (1) treating Colpitts' capacitors as parallel instead of series, and (2) forgetting the 2M term in Hartley when mutual inductance is given. Always sanity-check: series capacitors give a C_eq smaller than the smallest one. Also be ready for the one-line comparison — Hartley for wide tuning, Colpitts for stability.
Syllabus points
Hartley & Colpitts (frequency numerical)
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