Colpitts oscillator not oscillating

Question

I have tried breadboarding a simple Colpitts oscillator, just to see how it works (and to get to use my 'scope for something more interesting than measuring static voltages).

I have been following this example, specifically the second circuit design:

_{(source: play-hookey.com)}

I'm feeding it 5 V, the resistors are all at 1k, L is a .22 µH fixed inductor, Q is a 2N4401, C1 and C2 are .001 µF ceramics and the unlabeled cap at the base is a 220 pF ceramic (maybe way too low?), and I'm probing between emitter and ground.

Now, admittedly, these values are more or less randomly chosen from my component drawers. In this case I am not interested in obtaining a specific frequency as long as it's low enough for me to measure it (50 MHz), so I figured I could just throw in any values for the caps and the inductor, as long as they were high enough - I've read that this can actually be a pretty accurate method of measuring capacitance and inductance respectively, based on the frequency you get.

Questions:

1. Why is the circuit not oscillating? I'm measuring a DC voltage of 1.87 V at my probe point.
2. How do you calculate the proper resistor values (or ratios)?
3. What's the base cap used for? Just power decoupling?
4. Am I probing in the right place?

Answer 1

My answer:

1.) I do not know because I didn`t recalculate the circuit. However, it is YOUR task to find a suitable design (not using random parts values), see point 2).

2.) At first, you must understand the circuit (why it can oscillate). There is a frequency-selective feedback network with a bandpass characteristic (L in parallel with C1 and C2). Do not overlook that the supply voltage is identical with signal ground.

Hence, at the midband frequency the phase shift will be zero. A part of this signal (depending on the C1-C2 ratio) is fed back to the emitter establishing the required positive feedback (loop gain).

3.) It is the task of the base capacitor to keep the base at signal ground (transistor in common base configuration and positive gain, see 2).

4.) The classical (normal) output for common base stages is at the collector .

Answer 2

1) A series LC circuit, such as you have, cannot be a resonator, because there is no COMPLETE PATH for the currents to circulate. Add a 0.01UF from Top_of_inductor to Ground; that completes the circulation path.

2) The common-base amplifier has a gain of TWO, set by the capacitor ratio. Reduce the upper capacitor value, by a factor of 5?

3) The Rout of the transistor (early effect) is loading the tuned circuit. Rout is strong function of Vce, if you have only a few volts across the transistor. This Rout may be so low the energy can never build up. Try 9 volts.

4) Only parallel resonant circuits, those with complete circulating path, can store energy. But series resonant are great for nulling-energy in filters.

5) assuming 15MHz is the resonant frequency (using sqrt(25,330/Luh*CpF)), the capacitive impedance on Emitter is 50 ohms (two caps in parallel). The emitter Rin (input to the CommonBase amplifier) is approximately 13 Ohms, from 26 ohms at 1mA/1.9mA actual. Thus energy transfer 50ohm/13ohm might be improved, should energy transfer turn out to be important (the maths might reveal this)

EDIT Found this Colpitts oscillator schematic, which solves the circulating-current problem;

^{simulate this circuit – Schematic created using CircuitLab}