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Exercise Problems · Q2

Q.A Colpitts oscillator circuit is to generate a frequency of 24 kHz24\,\text{kHz}. The capacitors used are C1=0.2 μFC_1 = 0.2\,\mu\text{F} and C2=0.22 μFC_2 = 0.22\,\mu\text{F}. Find the value of an inductor used.

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[!TLDR]

With Ceq≈0.1048 μFC_{eq} \approx 0.1048\,\mu\text{F} and f=24 kHzf = 24\,\text{kHz}, the design gives L≈0.42 mHL \approx 0.42\,\text{mH}.

The Colpitts tank uses the series capacitance Ceq=C1C2C1+C2C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}; the frequency f=12πL Ceqf = \dfrac{1}{2\pi\sqrt{L\,C_{eq}}} is rearranged for the unknown inductor:

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

Step 1 — equivalent capacitance: Ceq=(0.2×10−6)(0.22×10−6)(0.2+0.22)×10−6=0.044×10−60.42≈1.048×10−7 FC_{eq} = \dfrac{(0.2\times10^{-6})(0.22\times10^{-6})}{(0.2 + 0.22)\times10^{-6}} = \dfrac{0.044\times10^{-6}}{0.42} \approx 1.048\times10^{-7}\,\text{F}.

Step 2 — inductor: with f=24×103 Hzf = 24\times10^{3}\,\text{Hz}, (2πf)2≈2.274×1010(2\pi f)^2 \approx 2.274\times10^{10}, so L=12.274×1010×1.048×10−7≈4.2×10−4 HL = \dfrac{1}{2.274\times10^{10}\times1.048\times10^{-7}} \approx 4.2\times10^{-4}\,\text{H}.

[!ANSWER]

L≈0.42 mHL \approx 0.42\,\text{mH}.

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