Q.The frequency of Hartley oscillator is 20kHz. The capacitor used in the feedback network has a value of 576pF. If one of the inductors value is 100mH, calculate the value of the other inductor.
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Concept understanding — Hartley Oscillator
The Hartley oscillator is an LC oscillator that takes its feedback from a tapped (two-section) inductor — hence it is said to use inductive feedback. Its tank circuit is a single capacitor C in parallel with two inductors L1 and L2 connected in series, the junction between them being the tapping point. The output is developed across L1 and the feedback voltage across L2.
A common-emitter transistor amplifier supplies the amplification. Resistors R1, R2 and RE set the DC bias, CE is the emitter bypass capacitor, CC is the coupling capacitor, and a radio-frequency choke (RFC) — a short circuit for DC but an open circuit for AC — isolates the DC supply from the collector oscillations.
Working: at switch-on the collector current charges C, which then discharges through L1 and L2, setting up tank oscillations. The voltages across the two inductor sections are in opposite phase, so the tank contributes 180∘ and the transistor another 180∘, giving the 360∘ loop shift needed for oscillation.
The frequency of oscillation is
f=2πLC1,L=L1+L2
(if the coils are magnetically coupled, L=L1+L2±2M, with +2M for series-aiding and −2M for series-opposition windings). The feedback factor is β=L1L2, and to satisfy Aβ=1 the required amplifier gain is A=L2L1. Hartley oscillators are widely used as local oscillators in radio receivers and as RF signal generators.
The tank formula gives the total inductance L=(2πf)2C1; subtracting the known section L1 leaves the other inductor L2=L−L1.
[!ANSWER]
The other inductor is L2≈9.91×10−3H=9.91mH.
[!TLDR]
The tank needs L≈109.91mH; subtracting L1=100mH gives L2≈9.91mH.
In a Hartley oscillator the two inductor sections in series give L=L1+L2, and the frequency is f=2πLC1. First find the total tank inductance required, then subtract the known section:
L=(2πf)2C1,L2=L−L1.
Step 1 — total inductance: with f=20×103Hz and C=576×10−12F, (2πf)2≈1.579×1010, so L=1.579×1010×576×10−121≈109.91×10−3H.
Step 2 — subtract the known inductor: L2=(109.91−100)×10−3H.