3
\$\begingroup\$

enter image description here

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: enter image description here

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.

\$\endgroup\$
  • \$\begingroup\$ It seems like the calculation steps are at fault here because this is a genuine colpitts oscillator (even if it does use the dinosaur 741 op-amp). You also seem to be specifying the feedback factor (or gain) in volts and this doesn't make sense. Are you aware that it won't produce a reliable sine wave under pretty much any circumstances (just in case you plan to use it)? \$\endgroup\$ – Andy aka Feb 15 '17 at 9:46
  • \$\begingroup\$ I am going to use a high speed Op-Amp for this design. Why do you think it won't produce reliable sine wave? I copied the design from electronics-tutorials.ws/oscillator/colpitts.html. Feedback factor typo corrected, thanks for pointing it out. @Andyaka \$\endgroup\$ – niki_t1 Feb 15 '17 at 9:54
  • 1
    \$\begingroup\$ It is an idealized oscillator. In reality it will need an active gain control circuit to keep the sine wave amplitude at a constant value and away from the power rails. Think about it. If the gain is absolutely perfect you get a perfect sinewave but if that gain increases by 0.000001% the sinewave amplitude drifts upwards until it crashes the power rails. \$\endgroup\$ – Andy aka Feb 15 '17 at 12:02
  • \$\begingroup\$ ok... if I understood it correct ... slight change in gain in upward direction will push the poles towards right and no circuit action pulls it towards frequency axis, and, if gain is reduced by a small fraction, poles go towards left then sine wave dies down. Assuming throughout that the feedback factor is 1 (LC is open circuit at resonance) and will stay so. Please correct me if I am wrong somewhere. @Andyaka \$\endgroup\$ – niki_t1 Feb 15 '17 at 13:38
2
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Please, watch the correction I have introduced (3rd line from the bottom). \$\endgroup\$ – LvW Feb 15 '17 at 11:24
1
\$\begingroup\$

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.
circuit for barely oscillating
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:
phase, amplitude of op-amp output with loop open
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
\$\endgroup\$
  • \$\begingroup\$ sorry but your simulation arrangement is wrong. You must NOT place the ac test source as shown in your circuit. Instead, the ac source must be with one leg either at the opamp output (because it is sufficiently less-resistive, nearly at ground) or directly at ground. This is allowed because the opamp is operated with negative feedback. \$\endgroup\$ – LvW Feb 15 '17 at 21:53
  • \$\begingroup\$ @LvW ? Its a series circuit. Swap R4, V1, so that when you open the loop, the negative end of V1 gets grounded. Same result. \$\endgroup\$ – glen_geek Feb 15 '17 at 22:20
  • \$\begingroup\$ No - perhaps the difference is not too large, however, Middlebrooks method requires subsequent voltage AND current injection, unless the source resistance (in your figure the lower end of V1) is sufficiently small. In your case it is 200 ohms. What was the reason for you to place Vtest not between the opamps output and the 200 ohms resistor? \$\endgroup\$ – LvW Feb 16 '17 at 8:03
  • \$\begingroup\$ @LvW My mistake. Spice "G" model is not voltage-controlled voltage source. It has a 1-ohm output Z, not zero as I had assumed an op-amp model of "E". As you've said, this makes a small error (one ohm out of 200). \$\endgroup\$ – glen_geek Feb 16 '17 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.