# Wien Bridge Oscillator: Why does equating the real part to 0 give the gain equation?

I am learning about Wien bridge oscillators. Following Experiment No. 9: WIEN BRIDGE OSCILLATOR USING OPAMP, they use the following schematic: They then arrive at the following equation half-way down page 3: They then equate the imaginary part to 0 to find the resonant frequency, i.e: This makes sense, because at resonance the current is in phase with the voltage, hence the imaginary part goes to 0. I don't understand the next part where they:

To obtain the condition for gain at the frequency of oscillation, equate the imaginary part to zero.

which expressed as an equation gives you: Why can you just equate the imaginary part to zero (i.e. the real part goes to 0) to find the gain needed at resonance?

EDIT: After @TimWescott's great answer I created a re-written version of the equation which shows the "missing link", i.e. Im() + Re() = 0: • ** the imaginary part to zero (i.e. the real part goes to 0)** Real part is not the imaginary part. What does resonance mean? To find the frequency you equate the imaginary part to 0. They have made R & C leading = R & C lagging. May 1, 2021 at 23:45
• @StainlessSteelRat sorry I see how that is a tad confusing, what I meant was that when they use the terminology "equate the imaginary part to zero", I was imaging Im() = 0 and they had expressed the equation as Im() = Re(), so that you could view them as replacing Re() in the Im() = Re() equation with 0. May 2, 2021 at 0:57

Because they're starting with the Barkhausen Criterion; loop gain = 1. The loop gain is $$\left( 1 + \frac{R_3}{R_4} \right)\left(\frac{RC s}{(RC s)^2 + 3RCs + 1} \right)$$.

Everything else comes from setting that to one, and then doing some math. Basically, they set that to one: $$\left( 1 + \frac{R_3}{R_4} \right)\left(\frac{RC s}{(RC s)^2 + 3RCs + 1} \right)$$

Then they make an equation where "stuff = 0". In order for "stuff" to be zero, then both it's imaginary and real parts have to be zero (not just one or the other).

It happens, conveniently, that in the circuit as shown, the real part depends on $$\R\$$, $$\C\$$, and the frequency of oscillation, $$\\omega\$$. So setting it to zero finds you $$\\omega\$$ as a function of $$\R\$$ and $$\C\$$.

Again, conveniently, once you know the frequency of oscillation (because you set the real part to zero), $$\R\$$, $$\C\$$, and $$\\omega\$$ drop out of the equation and you can calculate the gain.

But it is not "why can you just set the imaginary part to zero". You have to set both the imaginary and the real parts to zero, then solve that system of two equations to get two unknowns.

• Thanks! I have updated my question with a re-written version of the equation helping show what you just explained. May 2, 2021 at 1:44
• Is the same equation twice in the post intended? May 2, 2021 at 14:37
• Yes............. May 2, 2021 at 17:32

gbmhunter, you have explicitely shown the "WIEN-Robinson" bridge (heart of the WIEN oscillator).

Another view of the same circuit sometimes helps to understand the principle better (in conjunction with Barkhausens oscillation criterion):

If you redraw the circuit allocating the resistors R3 and R4 to a separate negative feedback loop, you have a fixed gain amplifier (gain: 1+R3/R4). Now - this fixed gain amplifier has an additional positive feedback loop (RC-series and RC-parallel) which resembles a passive RC bandpass having zero phase shift at w=wo=1/RC (with a damping factor d=1/3).

Hence, for fulfilling Barkhausens condition the fixed-gain amplifier block must exhibit a gain of 1/d=3 (unity loop gain for one frequency w=wo only). For all other frequencies the oscillation condition cannot be met because of the phase shift for w not equal to wo (zero phase shift within the loop required).

Additional note: Because the unity loop gain condition cannot be met exactly (parts tolerances) and for ensuring a safe start of oscillation, in reality the loop gain is made somewhat larger than unity (fixed gain stage app with a gain of 3.2 - 3.5). In such a case we accept a sort of clipping of the oscillation amplitude (power rail) or we add a kind of "soft" NON-linearity (diodes, FET-regulation) to softly reduce the gain for rising amplitudes.

There is a much simpler way to compute this. ( coincidentally with @LvW)

We know that:

• there is virtual ground such that Vin- = Vin+ ....(1)
• Av- = -Rfb/Rin = -R3/R4
• Av+ = 1 + |Av-|
• at resonance $$\|X_C(f)|= R \$$ forms a right-angle impedance
• in series $$\|Z(f)|= \sqrt{2}*R\$$ to parallel $$\|Z(f)|=\dfrac{R}{\sqrt{2}} \$$ yields the ratio = 2
• so to satisfy (1) the ratios must match so Av-=2.00 to make perfect sine bridge oscillator - the pot doesn't need to be grossly out by 4k7/1k but it speeds up oscillator into square wave, so one might choose 10% higher than 2 or reduce 10% with precision tolerance parts or use a soft limiter at 100x the impedance
• thus making Av-= - 2.0 and Av+=3.0 the net gain = 1.00 to satisfy the Barkhausen unity-gain net-positive-feedback criterion