Two oscillators use the same inductor (L) and the same capacitor (C) values. One is Colpitts with $C_1$ = $C_2$ and the other is Hartley with $L_1$ = $L_2$ with negligible mutual inductance (M). Which of the following is true about their resonant frequencies?
Colpitts frequency is higher
Both oscillators resonate at \(f = \frac{1}{2\pi\sqrt{L_{eq}C_{eq}}}\), so the frequency depends on the effective inductance and capacitance of the tank.
The Colpitts tank splits the capacitance into two series capacitors, giving a smaller effective value \(C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C}{2}\) when \(C_1 = C_2 = C\), with a single inductor L.
The Hartley tank splits the inductance into two series inductors, giving a larger effective value \(L_{eq} = L_1 + L_2 = 2L\) (mutual inductance negligible), with a single capacitor C.
Comparing the L·C products: Colpitts gives \(L\cdot\frac{C}{2} = \frac{LC}{2}\) whereas Hartley gives \(2L\cdot C = 2LC\); the Colpitts product is four times smaller, so its \(\sqrt{L_{eq}C_{eq}}\) is smaller and its frequency is higher.
Hence, the answer is Colpitts frequency is higher.
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