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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):

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

enter image description here

In figure there isn't inverting amplifier.

Thank you for your help.

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  • \$\begingroup\$ Too much confusion. Break feedback loop why? You haven't said what result you want to achieve. You also haven't defined terms, like Vu and Vp. Closing. \$\endgroup\$ – Olin Lathrop Nov 13 '16 at 16:13
  • \$\begingroup\$ Hi @Olin Lathrop, this is the procedure on Sedra-Smith. \$\endgroup\$ – Gennaro Arguzzi Nov 13 '16 at 16:14
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    \$\begingroup\$ We know the method, have been using it in the past 20 years, didn't know how it was called, still used it. It actually does not matter where you break the loop as long as you do it properly (meaning the behaviour of the circuit is unaffected). \$\endgroup\$ – Bimpelrekkie Nov 13 '16 at 16:44
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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:

schematic

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.

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  • \$\begingroup\$ Hi @The Photon, if possible I want to ask you why in the op-amp implementation there isn't the last resistor (see en.wikipedia.org/wiki/Phase-shift_oscillator). Thank you in advance. \$\endgroup\$ – Gennaro Arguzzi Nov 13 '16 at 16:36
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    \$\begingroup\$ Actually there is, in the circuit with the opamp from your link, R1 is the last resistor. The - input of the opamp is a virtual ground so it behaves as a resistor to ground. But anyway, you're only confusing yourself comparing all these schematics. Pick one and fully understand it, then go to the next. \$\endgroup\$ – Bimpelrekkie Nov 13 '16 at 16:42
  • \$\begingroup\$ Yes - this resistor R1 fulfills two tasks: It acts as the "last" resistor for the C-R element; and - at the same time - it is one of the gain-determining resistors of the inverting opamp amplifier. \$\endgroup\$ – LvW Nov 13 '16 at 17:17
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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)

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