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schematic

simulate this circuit – Schematic created using CircuitLab

I would check that in the Wien bridge sinusoidal oscillator the gain A_V=1+(R5+R4)/R3) is actually A_V=3 as the Barkhausen stability criterion requires. So I build this circuit whit a potentiometer of 10k instead of R4 and I obtain that for R4=6.38K (so R3 becomes 13.7k) the oscillation starts. With this configuration the circuit should have a gain A_V=2.2 (with the formula mentioned before)but I observe a gain of 2.9-3. How can I consider the contribution of the diods to the gain formula?

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  • \$\begingroup\$ What you have shown is not an oscillator. There is no frequency-dependent feedback. Your "R3 becomes 13.7k" makes no sense. Try a better explanation. \$\endgroup\$
    – WhatRoughBeast
    Dec 5, 2015 at 21:24
  • \$\begingroup\$ Of course it is not an oscillator. In order to obtain an oscillator you have to connect point A to non-inverting input V+ through a short circuit (without the sinusoidal voltage source). Now, according to the Barkhausen stability criterion the circuit sustains steady-state oscillations only for frequency for which the loop gain is=1. If you compute this condition in this case you obtain A_V=3. So I build a new circuit in order to verify that actually A_V=3 when the oscillation starts. \$\endgroup\$
    – gilgamesht
    Dec 5, 2015 at 21:55
  • \$\begingroup\$ gilgameshd, why not measuring the loop gain directly? Connect node A to the non-inv. opamp input and disconnect the positive feedback path R7...C1..... Now - excite this path (via R7) with a testsignal and measure the opamp output voltage. \$\endgroup\$
    – LvW
    Dec 6, 2015 at 10:17

1 Answer 1

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The gain you measure depends on the amplitude of the signal in this circuit. This is the function of the diodes.

For example, if the output signal amplitude is small (say less than 0.6 V), then neither of the diodes across the resistor (R6 ? -- it's hidden) will conduct, and your gain will be 4x (with the 10k resistors). When the amplitude exceeds about 2.4 V peak, then at the extremes of the sinusoidal output, the diodes will conduct, and effectively limit the voltage across R6 to about 0.6 V. This reduces the gain beyond this amplitude to 3x (the other resistors control it).

If you were to draw a graph, the gain (slope of VOUT vs. VIN) would be 4x until the peaks reached 2.4 V, then would reduce to 3x for the portion of signal beyond that. Because gain changes with signal amplitude, you would get distortion.

This is the exact purpose of the diodes in this circuit -- they allow the gain to change, and if you get the resistors (and other components) right, for small signal amplitudes the overall lop gain will be just greater than 1, and oscillations will build up. However, when the amplitude reaches about 2.4 V, the gain will fall, and the amplitude will stabilize around where the 'average' gain is 1.0

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  • \$\begingroup\$ When the oscillation is already started (and the threshold of 2.4V is exceeded) , by increasing the input voltage the gain A_V=1+(R5+R4)/R3) goes down quickly...this effect should be connected with diods. How can I consider the contribution of the diods to the gain formula in order to include this effect? \$\endgroup\$
    – gilgamesht
    Dec 6, 2015 at 0:00
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    \$\begingroup\$ there isn't a way to do this analytically precisely. For a rough approximation, consider that the output voltage peak is limited to about 0.6 V across each diode. At that point the gain begins to fall. If (before that point) the amplifier gain is just slightly > 3, it will fall to < 3 when the amplitude is such that there is 0.6 V on the diode. This means the peak V is 3x this, or about 1.8 V. \$\endgroup\$
    – jp314
    Dec 6, 2015 at 0:37
  • \$\begingroup\$ In principle, I agree to the jp314 comments. But one should consider that the output amplitude depends on the amount of "excess loop gain". That means: If the small-signal loop gain at w=wo is LG=1.3 we need more resistor reduction (diode opening) than in case of LG=1.1. In the latter case, the diodes open just a little and the corresponding voltage drop perhaps is (0.3...0.4) volts only (instead of 0.6V) \$\endgroup\$
    – LvW
    Dec 6, 2015 at 10:11

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