# Wien-bridge oscillator - what is happening in the first seconds?

#### Hung N.

Joined Dec 2, 2019
3
Hello,
I can't devise the initial behaviour of Wien-bridge oscillator.

I thought that the output "randomly" switches to one polarity (due to noise, asymmetry or initial conditions) and then the V(-) is immediately V(out)/3. V(+) stays zero and because of higher voltage (for positive V(out)) on the negative input of the op-amp, the output changes to a negative voltage, this behaviour is fast repeated and the output should alternate.
But no simulation showed similar results. So, I am certainly wrong.

Can someone give me an explanation of what is happening before it starts to oscillate, please?
I am also interested in why the oscillations are delayed in LTSpice simulation and delay depends on the time of transient analysis.

Joined Mar 10, 2018
4,057
Recommend you post LTC sim file so that forum can take a look at it.

Regards, Dana.

#### crutschow

Joined Mar 14, 2008
25,133
Can someone give me an explanation of what is happening before it starts to oscillate, please?
Typically, an oscillator has a small amount of positive feedback to start and maintain the oscillation.
When starting, the signal has to supply energy to the reactive components as they increase the signal increases in amplitude.
The time it takes for this to build, is determined by the circuit design and the positive feedback gain.
Higher gain would make the oscillator start faster, but could generate more distortion in the signal, as the signal reaches its maximum amplitude.

To help the oscillator start in LTspice, skip the initial operating point calculation, by selecting the top or bottom options in the
Transient analysis (below). • Hung N. and Delta prime

#### Hung N.

Joined Dec 2, 2019
3
Thanks, but I still don't understand.

If the op-amp set the output to positive saturation voltage, e.g. +15 V, the capacitors in negative feedback will be charging. But in the beginning, the V(+) is 0 and V(-) = 5 V (1/3 of V(out)). Therefore, the op-amp tries to set negative output, and it changes V(-) to -5 V. V(+) is probably still less, so the output will change polarity again and then alternates as fast as op-amp allows.

Why doesn't it work in the way I write?

#### LvW

Joined Jun 13, 2013
1,056
Hung N.....start your analysis with a parallel L-C combination (tank circuit). When you apply a short current or voltage puls into the circuit it will react with an oscillation process - however, the amplitude will continuously decrease because the circuit always has losses (parasitic resistive parts) .

It is the working principle of an oscillator to continuously compensate these losses and maintain a constant amplitude. For this purpose, an amplifier is needed which amplifies the oscillation signal with a fixed gain value G.
The output of this amplifier is connected again to the oscillating block (feedback loop) and will supply it with a kind of "kick" in the correct time slot - and thus enhances the amplitude again. More correct: the amplifier delivers energy into the oscillating part of the circuit which exactly can compensate the energy loss due to the ohmic effects.

This is identical to the following requirement: When the gain G of the amplifier compensates the loss factor k the product G*k must be unity. This is the well-known Barkhausen condition for oscillation.

The mentioned "kick" must meet the correct time slot (remember the pendulum clock) - and this is identical to the requirement of zero phase shift within the feedback loop. For a LC tank circuit, this can be fulfilled at the resonant frequency wo=1/SQRT(LC).

The same applies to another bandpass consisting of RC elements only (WIEN oscillator). The WIEN network has a resonant frequency (zero phase shift) at wo=1/RC with a damping factor k=1/3. Hence, the amplifier gain must be G=3.
(In reality, slightly larger in order to safely start the oscillation).

#### Analog Ground

Joined Apr 24, 2019
411
Thanks, but I still don't understand.

If the op-amp set the output to positive saturation voltage, e.g. +15 V, the capacitors in negative feedback will be charging. But in the beginning, the V(+) is 0 and V(-) = 5 V (1/3 of V(out)). Therefore, the op-amp tries to set negative output, and it changes V(-) to -5 V. V(+) is probably still less, so the output will change polarity again and then alternates as fast as op-amp allows.

Why doesn't it work in the way I write?
Can you post your schematic? Your comment about "output to positive saturation" and "capacitors in negative feedback" makes me think there might be something wrong in the design. Have you switched the positive and negative inputs?

#### Hung N.

Joined Dec 2, 2019
3
I'm sorry for the confusion. The capacitors are in positive feedback.

I don't know how the oscillator works. I wrote my idea of its function, not the results of a simulation. But I know that it's incorrect and I want to know why.

I imagine that the operation amplifiers should compare voltages on inputs and amplify a difference. If the non-inverting input voltage is higher, there will be positive saturation in the output.

In the case of the oscillator, it would mean that the output changes voltage on inverting input (its polarity) and then this voltage would change the polarity of output voltage, and it would happen again over and over.

It's not probably about oscillators, maybe I misunderstood principles of op-amps.

#### LvW

Joined Jun 13, 2013
1,056
I imagine that the operation amplifiers should compare voltages on inputs and amplify a difference. If the non-inverting input voltage is higher, there will be positive saturation in the output.
Yes - in principle, there are two ways for explaining the circuit.

1.) Both RC elements (series and parallel) form together with the two resistors (as a voltage divider) a frequency-dependent bridge (called WIEN bridge). When the bridge is tuned (oscillation condition) the voltage difference between the two nodes which are connected to the opamp is (ideally) zero...and we need an amplifier with infinite gain for closing the loop. In reality, the bridge is not fully tuned and the remaining voltage (some µVolts) is amplified with a gain in the range of 1E5.
However, for my opinion, this explanation cannot give a good and understandable insight how the circuit really works.
More than that, it is not so easy to verify the oscillation condition (Barkhausen).
Therefore - I prefer another explanation:

2.) Both RC elements (series and parallel) form a bandpass with a center frequency wo=1/RC and with a damping factor k=1/3 at this frequency. The two remaining resistors are allocated to the opamp and constitute a negative feedback path. Hence, now we have a fixed-gain amplifier which must have a gain of G=3 for compensating the losses of the WIEN bandpass (G*k=1). This condition fulfills Barkhausens oscillation condition.

• Hung N.