# Wien bridge with smoothed sine wave output

#### Tyropsolos

Joined Jan 16, 2022
3
Hello! So I'm trying to produce a smooth sine wave in a Wien bridge. In order to do so, we can use two zenner diodes in parallel as it is shown in the schematic below:

So the output without the zenner diodes would be:

While using the diodes:

My question is: How can I simulate the same circuit by using a light bulb connected in series with R4 instead of two zenner diodes?
The circuit I used is this:

NOTE: In the last screeshot I'm showcasing the final circuit, with the oscilloscope and the parameters of the light bulb when the VG1 is a pulse of 1V amplitude and 1n width of pulse

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#### LowQCab

Joined Nov 6, 2012
3,579
Building your own Oscillator for frequencies in the Mhz range is going to be extremely difficult.
What are You trying to accomplish ?
Very reliable Chips are available for this.
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#### crutschow

Joined Mar 14, 2008
33,346
I generated the approximate empirical equation below for an incandescent bulb, which may work for you.
You enter the R equation for the resistor Resistance value as shown.
The node Vin must be named in your circuit.
You need to also enter the .param voltage and wattage of the bulb you plan on using (I assume a low volt and watt bulb) for this application.

Last edited:

#### Audioguru again

Joined Oct 21, 2019
6,437
Your zener diodes will never zener because each one has a forward biased diode in parallel with it, ordinary diodes will do the same thing. The diodes do not smooth the waveform, instead they squash the top and bottom but less than if the opamp was clipping.
Your gain is 3.1 minus the diode drop so it will take time for the waveform level to build up when the power is turned on.

A light bulb delay causes the waveform to bounce up and down for a while when the power is turned on and when the frequency is changed.

LvW

#### Ian0

Joined Aug 7, 2020
8,942
You will have trouble simulating the lamp-stabilised Wien-bridge unless you know the thermal characteristics of the lamp.
Whilst the resistance tends to vary as the square-root of the voltage, you also need to know the thermal time constants.
They all tend to oscillate and take a while to settle - even Mr. Hewlett's original one.
The circuit will run at whatever amplitude gives a gain of 3. With the diodes clipping the waveform the gain is <3. With the diodes not conducting the gain is 3.1. It will settle somewhere between the two.

#### crutschow

Joined Mar 14, 2008
33,346
You will have trouble simulating the lamp-stabilised Wien-bridge unless you know the thermal characteristics of the lamp.
You need the thermal characteristics to simulate the settling transient, but you don't need that to simulate the steady-state oscillations.

#### crutschow

Joined Mar 14, 2008
33,346
For low frequency sine waves I like the 4-opamp Bubba oscillator (which can use one quad opamp package).
The oscillations can be softly clamped with two back-back diodes, and the signal from there to the output is filtered by the three low-pass filters that are part of the phase-shift network (giving a 3rd-order rolloff) to minimize output distortion.
Another advantage is that, if the four RC frequency determining component tolerances follow a normal gaussian distribution, then they should average to move the frequency close to what the nominal values predict.

LTspice simulation below:
Note that the clipping distortion observed on the U3 clipping stage output (yellow trace) is not seen on the output stage (green trace).

#### AnalogKid

Joined Aug 1, 2013
10,778
And by making R1, R2, or R6 a combination of a fixed and a variable resistor, you can tune the circuit frequency over a small range if necessary without needing a variable capacitor.

ak

#### Ian0

Joined Aug 7, 2020
8,942
You can achieve similar distortion reduction using the Linsley-Hood Wien bridge circuit, and taking the output from U2, where the distortion from the diodes will be filtered by the Wien network.
The reduction is not as good as three RC networks, but it only needs two op-amps.
There's a compromise - the gain has to be three for steady-state, and more than three to start up. The closer to 3 it is the lower the distortion, but the longer it takes to start up, and, if you aim too close to 3 component tolerances may take the gain lower than 3 and it might not start at all.
The gain of 3 applies to the single op-amp version: in the case of the Linsley-Hood circuit, the gain (R2/R1) should be just more than 2, because of the different topology.

#### LvW

Joined Jun 13, 2013
1,683
You can achieve similar distortion reduction using the Linsley-Hood Wien bridge circuit, and taking the output from U2, where the distortion from the diodes will be filtered by the Wien network.
The reduction is not as good as three RC networks, but it only needs two op-amps.
The shown circuit belongs to the large class of bandpass-oscillators.
If you want to exploit the filtering properties of the last stage - I would not recommend the shown circuit (because of the poor bandwidth of the filter).
With one more resistor you can use a Sallen-Key bandpass which can be designed with much better selectivity.

#### Ian0

Joined Aug 7, 2020
8,942
The shown circuit belongs to the large class of bandpass-oscillators.
If you want to exploit the filtering properties of the last stage - I would not recommend the shown circuit (because of the poor bandwidth of the filter).
With one more resistor you can use a Sallen-Key bandpass which can be designed with much better selectivity.
In which case, it's no longer a Wien bridge, which is what the TS wanted.
A Sallen-and-Key bandpass is non-inverting topology, so it wouldn't oscillate with an inverting feedback loop.

#### LvW

Joined Jun 13, 2013
1,683
In which case, it's no longer a Wien bridge, which is what the TS wanted.
A Sallen-and-Key bandpass is non-inverting topology, so it wouldn't oscillate with an inverting feedback loop.
As I have mentioned already - it is a classical bandpass oscillator.
It is NOT a "Wien bridge" oscillator . There is no bridge at all!
Of course, we need a gain-oft-two amplifier (non-invereting).

#### AnalogKid

Joined Aug 1, 2013
10,778
There is no bridge at all!
Ah, here's a side question. Will that circuit oscillate if the two bandpass poles are not equal. For example, what happens if they are two octaves apart?

ak

#### Ian0

Joined Aug 7, 2020
8,942
Ah, here's a side question. Will that circuit oscillate if the two bandpass poles are not equal. For example, what happens if they are two octaves apart?

ak
There will be a point (presumably one octave from each pole at √(f1.f2) where the "Wien" network has zero phase shift. The attenuation will be considerably more than 3 at that point. If the amplifier had enough gain then it should oscillate. I think that the point with zero phase shift should be unique.
But there aren't two bandpass poles - the Wien network has one pole and one zero.

#### crutschow

Joined Mar 14, 2008
33,346
I would not recommend the shown circuit (because of the poor bandwidth of the filter).
What is "poor bandwidth".
The filter has a slower rolloff than a Salen-Key filter but it has the same rolloff slope and high frequency attenuation.

#### LvW

Joined Jun 13, 2013
1,683
Ah, here's a side question. Will that circuit oscillate if the two bandpass poles are not equal. For example, what happens if they are two octaves apart?

ak
The only consequence would be that the pole-Q is reduced (larger bandwidth).

#### LvW

Joined Jun 13, 2013
1,683
I think, the name is not correct. On the other hand - it is just a name...so what?

#### LvW

Joined Jun 13, 2013
1,683
But there aren't two bandpass poles - the Wien network has one pole and one zero.
The Wien network is a second-order passive bandpass - hence, it has two poles (look at the 2nd-order denominator)

#### Ian0

Joined Aug 7, 2020
8,942
I think, the name is not correct. On the other hand - it is just a name...so what?
Wien Bridge - a crossing of the river Danube (Donau)?
If you think of the original meaning of "bridge" as a measurement circuit, as in Wheatstone bridge, then if one element is a thermistor or a filament lamp, or a pair of diodes, then no oscillator fits the original definition. You can't measure anything when your reference is variable!
If redrawn then the JLLH circuit can be just as much a bridge as all the others. There are now two op-amps, but both have the non-inverting input grounded, so it is essentially amplifying the difference between the two sides of the bridge.
You are, of course, correct about the poles - now that I have drawn the phase plot. I got confused by the zero at the origin (knew there was a zero somewhere).
Probably we should just call it a Wien oscillator?