Help me understand this wien bridge oscillator behaviour - Page 1
Forum section: electronics / beginners

author: Dan Moos
author_rank: Frequent Contributor
timestamp: July 03, 2023, 01:18:13 am
content: I built a wein bridge oscillator with R1 and R2 = 10k, C1 and C2 = 10pMath says I should get 1.6 MHzOpamps tried are tl072, ne5332, and rc4558. I Know these are all low freq opamps. I didn't expect a great result. Osc. is lamp stabilized.I do get oscillation. Depending on the opamp, its in the 150 to 300 KHz range.So tell me if I'm understanding things right here...These opamps can't do 1.6 MHz, and so my chosen components along with the capacitances of the opamps have created a different bandpass filter than my components alone?

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2023, 03:25:12 am
content: The circuit will oscillate when the phase around the loop is zero.Since you are pushing the GBW of the op amps, there will be a lot of extra phase at 1.6 MHz.The resistive leg's negative feedback of the Wien bridge (including your lamp) should give a gain of 3 V/V for the positive feedback through the reactive side.The NE5532 has a GBW of 10 MHz (typ), so that gain (including negative feedback) at 1.6 MHz is a bit above 3 V/V (roughly 6 V/V).To verify this, change your caps from 10 pF to 10 nF (1.6 kHz calculated frequency) and see that it works.

author: Dan Moos
author_rank: Frequent Contributor
timestamp: July 03, 2023, 04:20:31 am
content: First, the cap replacement did exactly as you said.Ok, Lets see what I do and don't understandI Know the basics of how audio power amps work, and I know an opamp is basically the same thing. I diff amp, feeding a voltage amp, feeding some sort of push-pull output.And I know an audio amp has a miller cap to kill gain before the phase gets out of hand. I presume opamps do to, and that that cap is the primary bandwidth limit of the circuit. Am I on track?Here's where I'm fuzzier. Tell me if any of this is correct: Since in the oscillator circuit, the opamp is driving a capacitive load, the phase shift is happening on a faster "schedule" for lack of a better term. It gets to 0 degrees sooner. So There is now an interaction between the capacitance in the opamp, and my external filter that causes oscillation at a much lower frequency (the phase shift happened too soon!)Am I close?Also, I haven't seen the use of V/V before. I assume it means Volt per Volt?

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2023, 02:03:34 pm
content: Yes, "V/V" means voltage gain in volts/volt.Phase shift:The op amp's "miller" compensation capacitor produces a 6 dB/octave fall in voltage gain with frequency from a very low frequency up to the GBW for a unity-gain-stable op amp, where the gain is unity.Along with that fall, there is a 90ophase shift in the output voltage.The negative feedback from the two resistors in the negative (resistive) feedback network reduces the closed-loop voltage gain to +3 at reasonable frequencies, and also reduces the phase angle.At very high frequencies, the loop gain (open-loop gain times feedback fraction) reduces the phase angle less, and the phase shift has also increased as you near GBW.Therefore, at high frequencies the closed-loop phase shift becomes rather large: I refer you to the normal textbooks for a quantitative description of the closed-loop phase shift.The small capacitive load from the wien-bridge load has a relatively small effect on the excess phase shift. The open-loop output impedance (maybe 50\$\Omega\$) with 10 pF load has a corner frequency > 300 MHz which gives only a few degrees of phase shift at 1.6 MHz. The actual load impedance from the R-C leg of the bridge is less than 10 pF equivalent capacitance in your original circuit.

author: Kleinstein
author_rank: Super Contributor
timestamp: July 03, 2023, 04:43:31 pm
content: !0 pF is a rather small capacitance and the input capacitance of the OP-amp and parasitic capacitance of the setup can be comparable and contribute to a lower frequency.

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: July 03, 2023, 04:49:38 pm
content: If you need a higher frequency, choose a smaller R and larger C. That way the opamp parasitic capacitance will have no effect on the frequency calculations.

author: Zero999
author_rank: Super Contributor
timestamp: July 03, 2023, 08:25:06 pm
content: Quote from: EEEnthusiast on July 03, 2023, 04:49:38 pmIt's not a parasitic capacitance, but a very real capacitor inside the op-amp, which is responsible for this. The solution is to use a faster amplifier.

author: TimFox
author_rank: Super Contributor
timestamp: July 03, 2023, 09:04:32 pm
content: Quote from: Zero999 on July 03, 2023, 08:25:06 pmYes--the compensation capacitor inside the op amp is required to make a device that is stable with 100% feedback (unity gain).However, there is a small input capacitance that is parasitic to the amplifier's guts.The phase effect of the compensation capacitor on the closed-loop response is a function of the open-loop frequency response and the negative feedback circuit.