How the Colpitts Op-Amp Oscillator circuit provides sinusoidal waveforms?

Actually, the circuits as shown is NOT sinusoidal. Like any positive feedback system, the amplitude increases until it is limited by circuit factors, such as saturation of the transistor or opamp. This results in a distorted waveform, usually squashed at the top or bottom of the sine wave.

This is countered by the sharpness, the Q, of the tuned circuit, and by high frequency rool-off of the gain of the opamp.

Reducing the feedback can also reduce the distortion.

Bottom line, the distortion may be very noticeable or it may be small, depending on circumstances.

Taking the party contained in the hyperlink decrease than: The sensitivity is round 26mV per g, it fairly is gravity acceleration (9.81m/s/s). this actual sensor covers from 1kHz to 20kHz. The minimum capacitance is round 800pf and the minimum frequency 1kHz so the amplifier desires to have an enter resistance a minimum of one million megohm so it would not attenuate frequencies round 1kHz. a significantly better resistance may enable operation at decrease frequencies than diverse. this is a non inverting op-amp as contained in the 2d hyperlink. i do not recognize what acceleration aspect you pick, this sensor covers from 0.1g - 400g, and with 1g the signal is 26mV RMS, and about 70mV p-p (top to top). At 400G it should be 400 circumstances this, round 28V p-p. If the amplifier has +9V and -9V substances the utmost enter and output is about 12V p-p. making use of an LF411 op-amp with twin +/- 15V substances it will be round 22V p-p which I anticipate genuinely matches your maximum want. It is smart to have a voltage divider making use of two * a million megohm resistors on the enter if extreme g is anticipated. The earnings may be adjusted to in fantastic condition an information measuring equipment, perhaps 1V p-p for 1g acceleration. The voltage earnings for that case ought to hence be 100mV/70mV = 14.28 circumstances. this may be 100K for R2 and a couple of * 15K in parallel for R1 contained in the hyperlink decrease than. word there are 2 separate substances, which will be 2 small batteries to provide +9V and -9V. Notes: Piezo sensors in many circumstances use "value amplifiers" because they're capacitive in nature. it type of feels this one makes use of a classic voltage amplifier, so already has a equipped in a million megohm resistor. If the signal seems irrelevant, attempt including a load resistor of one million megohm or better by potential of the sensor, and if nevertheless no longer acceptable, attempt a cost amplifier, that you will be able to seem up. Use an op-amp with jfet inputs like the LF411. this may provide some safe practices hostile to over-voltage if the sensor is dropped and so on.

The L C1 leg supplies positive feedback. The series reactance of the inductor and capacitor is inductive, but the feedback signal from capacitor C1 supplies resistor R1 an inverted signal. The op amp amplifies the signal by the ratio -R2/R1, resulting in positive feedback

X lc1 = X l - X c1 so I lags V in this leg of the oscillator

V out/ (jX lc1) = I lc1 where I lc1 is the current in the LC1 leg, I in the LC1 leg lags V out by 90 degrees

V c1 = I lc1(1/jX c1) where V c1 is the voltage across C1, V c1 lags the current in the LC1 leg by 90 degrees. So V c1 is 180 degrees out of phase with Vout.

Capacitor C2 resonants with the inductance of the LC1 leg. The frequency out is 1/(2π√(LCeq))

Ceq = C1||C2 = C1(C2)/(C1 + C2)