Posted on Feb 6, 2014

This circuit is easy to recognise by the three equal-value capacitors and two equal-value resistors connected to the base of the transistor. We have already mentioned the signal at the output of a transistor connected in COMMON EMITTER mode is FALLING when the signal at the input is RISING. This is called OUT OF PHASE signals and you cannot connec

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t the output directly to the input to get the transistor to oscillate. If you do, the rising output will be fed into the base to turn the transistor ON and this will reduce the output. The signal has to be delayed by a short period of time to allow the output voltage to rise and then the transistor can be turned ON to reduce the voltage on the output. This delay is created by a set of capacitors and resistors on the base. We have already mentioned the fact that a capacitor takes a period of time to charge and this feature is utilized in the PHASE-SHIFT OSCILLATOR circuit. The value of the components create a time-delay and this sets the frequency at which the circuit operates. But it`s a bit more complex than a single "time-delay" arrangement. Each capacitor is doing something different at each part of the cycle. The interesting point to note is the feedback signal only has to be about 1/50th of the collector signal for the circuit to operate as the gain of the transistor is about 50 to 100. This means there can be a lot of ATTENUATION (reduction) in the signal and the circuit will still operate. The signal through the 1M finds it difficult to pull the base up or down "quickly" because the 22n capacitor has a very large "holding effect" on the signal. This voltage is passed to the three uncharged capacitors and they pull the base up very quickly to turn the transistor ON. The secret to the operation of the circuit is the...

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