Talk:Phase-shift oscillator

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Shouldn't there be another resistor to ground between R1 and C1 ? If not, it is not clear how the phase shift of 180° is obtained. Phds 13:02, 13 December 2006 (UTC)

No - R1 is already "virtually" connected to ground - due to the "virtual short circuit" across the inputs to the op-amp, so the current reaching the negative terminal will be shifted by up to 270 degrees.
Consider the case if R2, C2, R3 & C3 were removed, leaving one end of C1 open circuit and an external stimulus were fed into the open end of C1.
I_{R_1}=\frac{V_{in}}{R_1+\frac{1}{j\omega C_1}} ( due to the virtual short )
I_{R_{fb}}=I_{R_1}
so V_{out}=-I_{R_1} R_{fb}

V_{out}=-\frac{V_{in}R_{fb}}{R_1+\frac{1}{j\omega C_1}}
This expression is shifted according to frequency, without needing another resistor --Ozhiker 14:01, 14 December 2006 (UTC)


[edit] Amplitude

I noticed that there is no expression for amplitude and suppose that it depends on initial conditions. Does that make this circuit impractical as it stands? Should there be a mention of this? Cheers. Gaussmarkov (talk) 01:35, 10 January 2008 (UTC)

[edit] Setup Example

I tried to put together this one, and in the first test I used the following components; 3 1kOhm resistors, 1 29kOhm (Seried 27k, 1k, 1k) resistor for feedback and 3 1µF capacitors. This gave me a freqency of about 500Hz. 82.147.34.250 (talk) 08:52, 20 May 2008 (UTC) Peter Heiberg, Norway