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?

Question

Prepp Team · Accepted Answer

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.