Skip to content
Solved Examples · Example 6

Q.The frequency of Colpitts oscillator is 18 MHz18\,\text{MHz}. Design the value of inductor to be used if C1=100 pFC_1 = 100\,\text{pF} and C2=10 pFC_2 = 10\,\text{pF}.

Karnataka PUCTextbookNumericImportance★★★★★est
36% · 31/86 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

[!TLDR]

The series capacitors give Ceq=9.091 pFC_{eq} = 9.091\,\text{pF}; rearranging the tank formula for the given 18 MHz18\,\text{MHz} yields L≈8.597 μHL \approx 8.597\,\mu\text{H}.

The Colpitts tank capacitance is the series combination Ceq=C1C2C1+C2C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}, and the frequency is f=12πL Ceqf = \dfrac{1}{2\pi\sqrt{L\,C_{eq}}}. Squaring and rearranging for the unknown inductor,

L=1(2πf)2 Ceq.L = \frac{1}{(2\pi f)^2\,C_{eq}}.

Step 1 — equivalent capacitance: Ceq=(100×10−12)(10×10−12)(100×10−12)+(10×10−12)=1000×10−12110≈9.091×10−12 FC_{eq} = \dfrac{(100\times10^{-12})(10\times10^{-12})}{(100\times10^{-12}) + (10\times10^{-12})} = \dfrac{1000\times10^{-12}}{110} \approx 9.091\times10^{-12}\,\text{F}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.