Design a Colpitts Oscillator in a common collector configuration using a single BJT. Provide base current to get the transistor operating properly, and you can do this using a potential divider (between Vcc and gnd): use resistor values as high as possible (bearing in mind the loading effect of the base current) in order not to load the LC circuit too much, and aim for a base voltage of about half the supply.
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Once you`ve done this you`ll need to isolate the LC circuit from the base, dc-wise, and you`d do this using a capacitor (whose value is much higher than the tuning capacitors). No, the supply must connect directly to the collector, as well as the divider - there needs to be a path for standing current in the transistor. So delete the top capacitor. The divider should connect directly to the base, as you`ve shown, but both the L and the Cs need to be coupled to the base via a capacitor. If you don`t have this capacitor, the L will short out the bottom resistor in the divider chain, and you`ll not be biasing the transistor correctly. No, remove the leftmost capacitor. Disconnect the L and "top" C from the base. Connect them both to a capacitor, and connect the other side of the capacitor to the base. Hope that`s clear! The frequency is the resonant frequency of the inductor in parallel with the two series capacitors. The formula for capacitors in series and for a parallel LC resonance is in Google. Don`t go too wild with the "large enough": think microfarads, not millifarads. Capacitors suitable for resonating can be inconvenient in large values: electrolytics won`t suit this job. If you get your oscillator to work, but with a rotten waveform, a small resistor in the lead between the emitter and the junction of the two capacitors forming Ct may help. Do not make this too big, or the oscillator will not run any more. from the...
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