Feedback factor of Op-Amp based Colpitts Oscillator

Question

Above is the picture of the oscillator for which I am trying to calculate the feedback factor of the oscillator at resonant frequency. The forward path gain is $$A=R_2/R_1$$.

I want to calculate the feedback factor to calculate the loop gain, so that I can derive the condition for oscillation, and set appropriate values of A.

So for analysis purpose I redrew like following:

So the feedback factor can be: $$\beta=V_{out} / V_{test}$$.

But given the values of capacitors, and inductor in the first diagram, and setting frequency at resonant frequency, I am getting an insanely low value of feedback factor (~10E-30). Clearly I am doing something wrong in the analysis phase.

Can someone be kind enough to guide me here. Is this the right way to calculate the feedback factor? For clarity purpose, I have not put the calculation steps here.

Also I will replace 741 with a very high speed Op-Amp during implementation level.

Answer 1

niki - of course, your test circuit cannot work because you have connected an ideal ac source to the low-resistive output of the opamp (shorting the opamp output). For finding the loop gain the feedback loop must be opened at a suitable node for injecting the test signal at this node.

The problem in your circuit is that we have two feedback ways (forget the term "forward" in your figure - it`s nonsense). Hence, opening both feedback loops (which is, theoretically necessary) will cause problems with the very large open-loop gain Aol of the opamp. Hence, we need a method which would work also for the idealized case with Aol approaching infinite.

Therefore, we must treat the opamp with feedback as a finit gain amplifier with an unknown gain which is frequency-dependent and determined by ALL elements.

Method: Open the loop at the left (correction: right) node of R3 and inject the test signal Vin. The ratio Vout/Vin gives you the loop gain which must have a magnitude of slightly above unity at the zero-phase frequency. Tune this gain with R2.

Answer 2

Where to open the loop, and apply a test signal? You must recognize that op-amps like 741 have less-than-ideal characteristics at 12 kHz where this circuit tends toward oscillation. Choosing a correct model influences the final loop gain, and phase shift. For example, at 12 kHz, a 741 is pretty much an integrator, so nearly 90 degree phase shift is added. And its open-loop gain is near 80, because GBW is near unity at 1 MHz.
Even so, a 741 has excess open-loop gain for this circuit to oscillate (a voltage gain of about three should be enough). A simple op-amp model whose open-loop gain at DC is 100,000, and whose gain=1 @ 1 MHz is used in a SPICE simulation. This model is likely good enough to get reasonably close results, but be aware that its infinite input resistance, and zero output resistance is optimistic. With a voltage source inserted, the loop can be excited. Resistor R1 is adjusted until the resonant peak becomes extremely narrow, indicating that loop gain is very near the critical oscillating point. This occurs with R2=266918 ohms.

Now if you open the loop with this circuit, by disconnecting R4 (200 ohms) from 741 output, and connecting to ground, you get op-amp output voltage of U1 looking like this:

You can see that with one millivolt applied, op-amp output voltage is very nearly 1 mV with zero phase at a frequency near 11.7 kHz. Why open the loop at this point? Because the op-amp model has an ideal voltage source at its output, just like the ideal source V1. A real 741 will not have an ideal voltage source, but will have some small resistance. And it will likely have some additional phase shift too. You could add this resistance to R4.

XU1 N004 0 N005 opamp Aol=100K GBW=1Meg
R1 N005 N004 2.66918k
R2 N004 N002 1k
R4 0 N003 200
L1 N001 N002 10m
C1 N002 0 .024µ
C2 N001 0 .24µ
V1 N001 N003 SINE(0 1m 11.69898k 0) AC 1m Rser=0
.ac lin 10000 11680 11720
.IC i(l1)=10u
.backanno
.subckt opamp 1 2 3
G1 0 3 2 1 {Aol}
R3 3 0 1.
C3 3 0 {Aol/GBW/6.28318530717959}
.ends opamp

.end