Break feedback loop of oscillator

Question

How could I choise the break point on the feedback loop of an oscillator (goal: evaluate the loop gain \$\beta \space A\$)? The Sedra-Smith said that it is convenient to break the feedback loop at the aplifier output. I found a case in which this not works: a phase-shift oscillator with an inverting amplifier. In this case, if a break the feedback loop at the amplifier output Vu, i obtain Vu=0 because the "+" terminal of op-amp is at 0V and hence the loop gain \$\beta \space A\$ is 0 (Vp is the voltage source on the opposite side of Vu):

$$\beta \space A=\frac{Vu}{Vp}=0$$

In figure there isn't inverting amplifier.

Thank you for your help.

Answer 1

When you "break the loop", you're not expecting the new (hypothetial) circuit to still be an oscillator. You're just making clear what the loop path is so that you can see what \$\beta\$ and \$A\$ are.

In order to see these, you typically inject a new source at the point where you broke the loop.

So your new circuit with the loop broken should look like this:

^{simulate this circuit – Schematic created using CircuitLab}

Now you can calculate the transfer function from V1 to Vfb to get \$\beta\$, and of course the forward gain is just -K.

In cases where you are modeling the circuit at the level of transistors, you will likely also have to add additional components to maintain the bias point of the circuit, while breaking the feedback loop at the signal frequency.

Answer 2

The feedback loop of an oscillator must be opened for measuring or simulating the loop gain. This is necessary to show resp. prove if the circuit meets the oscillation criterion (Barkhausen: Loop gain equal resp. larger than unity).

For this purpose, the loop must be opened at a point where this opening does not change the loading conditions during closed-loop conditions. For the present case, it is the best choice to open the loop at the opamp output.

The loop gain will be, of course, NOT be zero. However, you must connect, of course, an AC signal voltage at the opening (and NOT a DC source as shown in the figure) and measure the opamp output. The frequency of this AC source must be varied within a range that, of course, contains the expected oscillation frequency.

(The Photon has "beaten" me by 3 minutes)