Q.A transistor Colpitts oscillator has L=4mH, C1=10nF and C2=10nF. Determine the frequency of oscillations.
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Concept understanding — Colpitts Oscillator
The Colpitts oscillator is an LC oscillator that takes its feedback from a capacitive voltage divider — hence it uses capacitive feedback. Its tank circuit is a single inductor L in parallel with two capacitors C1 and C2 connected in series, their common junction being grounded. The feedback voltage is developed across C2.
The amplifier and biasing are the same as in other transistor LC oscillators: R1, R2 and RE set the DC bias, CE is the bypass capacitor, CC the coupling capacitor, and the RFC (radio-frequency choke) isolates the AC oscillations from the DC supply.
Working: at switch-on the rising collector current charges C1 and C2. Because their common terminal is earthed, the voltages across the two capacitors are in opposite phase; the voltage across C2 is fed back to the input. When the capacitors are charged they discharge through L, sustaining the oscillation. The tank provides 180∘ and the amplifier 180∘, giving the required 360∘.
Because the two tank capacitors are effectively in series across the inductor, the frequency of oscillation is
f=2πLCeq1,Ceq=C1+C2C1C2
The feedback factor is β=C2C1, and to meet Aβ=1 the amplifier gain must be A=C1C2. Colpitts oscillators produce stable high (radio) frequencies and are used in radio and TV transmitters and receivers and as local oscillators; their main drawback is that they are unsuited to low frequencies, where the required components would be bulky.
In a Colpitts oscillator the two capacitors are effectively in series across the coil, so use Ceq=C1+C2C1C2 with f=2πLCeq1.
[!ANSWER]
With Ceq=5nF and L=4mH, the frequency of oscillations is f≈35.6kHz.
[!TLDR]
The series capacitors give Ceq=5nF, which with L=4mH gives f≈35.6kHz.
This is a standard Karnataka 2nd PUC Electronics Colpitts problem. The tank capacitance is the series combination Ceq=C1+C2C1C2, and the oscillation frequency is f=2πLCeq1.