# Barkhausen condition on a Wien oscillator

Consider the Wien oscillator:

We can determinate the operational amplifier gain defined as:

$$A(s)=\frac{\bar{V_o}}{\bar{V_a}}$$

and the gain from the feedback network:

$$\beta(s)=\frac{\bar{V_a}}{\bar{V_o}}$$

where $\bar{V_a}$ and $\bar{V_o}$ are the complex amplitudes of $v_a$ and $v_o$, respectively.

For an oscillator we have the Barkhausen condition:

$$A\beta(s)=1$$

To determinate the dimension of the components from the circuit which satisfy the previous condition, we usually calculate $A(s)$ and $\beta(s)$ doing some circuit analysis and then impose $Re\{A\beta(s)\}=1$, $Im\{A\beta(s)\}=0$.

But starting from the definition of $A$ and $\beta$, we can say that:

$$A\beta(s)=\frac{\bar{V_o}}{\bar{V_a}}\frac{\bar{V_a}}{\bar{V_o}}=1$$

which already satisfies the Barkhausen condition for any dimension of the circuit components. So where is the point of finding the dimension of the components that make the circuit an oscillator, when any of them could do it?

This makes me confused...Is this thought correct?

....my only problem is how is possible that Aβ=1 for any dimension of the components

Well, the answer is relatively simple. You have forgotten the most important condition for a circuit to oscillate:

1.) The Barkhausen criterion applies to the LOOP GAIN only. That is the gain of the loop under open conditions. In this case, the product (loop gain) Aβ(s) can assume any values (<1 or >1). This is because in this case Va is NOT identical to the input at the non-inv. opamp terminal. The criterion requires (under ideal conditions) that htis product is unity.

Further conditions:

2.) The Barkhausen criterion requires (and is applicable for) a linear system only. That means: The opamp must be operated in its linear transfer characteristic, and

3.) This criterion must be fulfilled for one single frequency only.

This Barkhausen condition

$$A\beta(s)=1$$

says that the voltage at Vo is equal in amplitude and the phase angle, to the voltage at point Va. But feedback loop must be open.

$$\frac{Vo}{Vin} * \frac{Va}{Vo} = 1$$

You need to consider that the op-amp gain is: -

$\dfrac{V_O}{V_A} = 1 + \dfrac{R2}{R1}$

Then you work out the attenuation due to the positive feedback elements Zs and Zp. Then you should be on the right path.

• I think you interpreted wrongly my doubt... Could you read the colored part of my question? – Élio Pereira Dec 30 '15 at 18:56
• Well yes, the b-condition for oscillation is unity overall gain for a perfect self-starting oscillator. – Andy aka Dec 30 '15 at 19:41

Your colored part of the Q only applies if the circuit is operating -- specifically if the opamp can drive so that its inverting and non-inverting inputs are at the same voltage (same AC signal).

• But that satisfies Barkhausen condition for any dimension of the circuit components. So, where is the point of finding the dimension of the components that make the circuit an oscillator, when all of them do it? I know how to proceed to find them and I know that is the needed way.. my only problem is how is possible that $A\beta$=1 for any dimension of the components if the property of oscillator depends on them? – Élio Pereira Dec 30 '15 at 19:16