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S Dec 31, 2021 at 18:52 history suggested Uwe CC BY-SA 4.0
Wien bridge, not Wein
Dec 31, 2021 at 17:44 review Suggested edits
S Dec 31, 2021 at 18:52
Dec 31, 2021 at 17:08 comment added Benjamin Crawford Ctrl-Alt-Tut I have added an edit to my question with a more traditional version of your LTSpice circuit. Could you shed light on the utility of A1? Is it needed to enable variable frequency? Thanks.
Dec 31, 2021 at 14:50 comment added G36 The output frequency is \$ \large Fo = \frac{1}{2 \pi R_6 C_2} * \sqrt{a}\$ where a = R6/R1. but still it is Wien bridge oscillator.
Dec 31, 2021 at 14:00 comment added Neil_UK R1 is connected to U1's -ve input. U1's +ve input is grounded. U1 has feedback, and is active. Therefore U1 forces its -ve input to be the same voltage as its +ve input, ie 0 V. This configuration is also known as a 'virtual ground'. What's a 'charging impedance'? C1 is driven from the output by C2 and R6, and loaded by R1 to (virtual) ground. You can calculate the impedance of those components if you feel like it.
Dec 31, 2021 at 13:45 comment added Benjamin Crawford Ctrl-Alt-Tut Thanks for the analysis. I would say I expect a sqrt dependence in a Wien Bridge (f=1/2\pi\sqrt{R_1R_2C_1C_2}). Could you explain the charging impedance for C1? How does R1 see ground through the op amp?
Dec 31, 2021 at 12:31 comment added G36 But this is a Wein bridge (R6+C2 and R1||C1). R1 is a part of a Wien bridge too. Notice that for R1 = R6 the A2 , A1 gains together is (1 + R4/R3) + R4/R3*R2/R1 = 3 V/V . And this is exactly what we need in Wein Bridge when R6 = R1; C1 = C2. But when we change the R1 we also change the bridge gain, But at the same time, A1 will compensate for this (increases the loop gain)
Dec 31, 2021 at 11:40 history edited Neil_UK CC BY-SA 4.0
added 273 characters in body
Dec 31, 2021 at 11:33 history edited Neil_UK CC BY-SA 4.0
added 273 characters in body
Dec 31, 2021 at 11:26 history answered Neil_UK CC BY-SA 4.0