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Exercise Problems · Q1

Q.A transistor Colpitts oscillator has L=4 mHL = 4\,\text{mH}, C1=10 nFC_1 = 10\,\text{nF} and C2=10 nFC_2 = 10\,\text{nF}. Determine the frequency of oscillations.

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[!TLDR]

The series capacitors give Ceq=5 nFC_{eq} = 5\,\text{nF}, which with L=4 mHL = 4\,\text{mH} gives f≈35.6 kHzf \approx 35.6\,\text{kHz}.

This is a standard Karnataka 2nd PUC Electronics Colpitts problem. The tank capacitance is the series combination Ceq=C1C2C1+C2C_{eq} = \dfrac{C_1 C_2}{C_1 + C_2}, and the oscillation frequency is f=12πL Ceqf = \dfrac{1}{2\pi\sqrt{L\,C_{eq}}}.

Step 1 — equivalent capacitance: Ceq=(10×10−9)(10×10−9)(10×10−9)+(10×10−9)=10×10−92=5×10−9 FC_{eq} = \dfrac{(10\times10^{-9})(10\times10^{-9})}{(10\times10^{-9}) + (10\times10^{-9})} = \dfrac{10\times10^{-9}}{2} = 5\times10^{-9}\,\text{F}.

Step 2 — frequency: f=12π(4×10−3)(5×10−9)=12π2×10−11≈3.56×104 Hzf = \dfrac{1}{2\pi\sqrt{(4\times10^{-3})(5\times10^{-9})}} = \dfrac{1}{2\pi\sqrt{2\times10^{-11}}} \approx 3.56\times10^{4}\,\text{Hz}.

[!ANSWER]

f≈35.6 kHzf \approx 35.6\,\text{kHz}.

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