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I am trying to understand the following tuned collector feedback oscillator extracted from here using LTSpice. I understand that L1 and C1 creates the resonance at the oscillation frequency and L2 provides the feedback.

enter image description here

However, The voltage at the output (V(C3)) is not sinusoidal as expected. Please refer to the following waveform.

enter image description here

If I understand it correctly, transistor goes to saturation (edit: because of high gain). When capacitors C2 and C4 are removed (as suggested in comment), output become close to sinusoidal (refer the following). enter image description here

I am not sure why the circuit behave this way, and what is the need of C2 and C4?

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  • \$\begingroup\$ Get rid of C4 and see what happens - I think you have too much circuit gain. \$\endgroup\$ – Andy aka Mar 1 '18 at 10:45
  • \$\begingroup\$ @Andy, question is modified by adding results when C4 is removed \$\endgroup\$ – Pojj Mar 1 '18 at 11:00
  • \$\begingroup\$ It looks like you still have too much gain. Try lowering L2 incrementally to get a better looking sinewave. Then take it even lower to find the point at which it won't start oscillating. No need to show pictures at each step. \$\endgroup\$ – Andy aka Mar 1 '18 at 11:23
  • \$\begingroup\$ @Andy, thanks, but changing L2 didn't help. but removing both C1 and C4 helped to obtain an output close to sinusoidal. But I am not sure the operation here. I edited the question accordingly. \$\endgroup\$ – Pojj Mar 1 '18 at 11:46
  • \$\begingroup\$ You can't get rid of C1 because that sets the tank tuning. Try L2 = 0.02 uH i.e. significantly smaller then take it smaller again but remember it will take longer to start oscillating because the positive feedback is less so make sure you take a longer time to wait for it to begin oscillating (up to 1 ms). \$\endgroup\$ – Andy aka Mar 1 '18 at 11:58
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The type of oscillator you are simulating needs careful consideration to get anywhere near decent sinusoidal performance. In its simple form you will never get a great sinewave purity because the output waveform has nothing other than the power rails to control amplitude. Yes you have feedback to make it "sing" but there is no active control element that can make the amplitude stable AND keep its output sinusoidal.

So currently, the output "hits" the power rails (one or the other or both) and this controls the output amplitude by limiting/clipping.

However, your simplified circuit has too much positive feedback for "adequate" performance. Look what I've done below; L2 has reduced to 0.01 uH and I've added 10 ohm in series with the main collector inductor (for realism): -

enter image description here

But still, the output is "hitting" the bottom limit and clipping because.... it needs something that can stabilize the output amplitude.

This can be achieved with a JFET in series with the feedback to the base. The standard way is to rectify the output level to get a "measure" of the output amplitude then use this "measure" to control the JFET so that it starts to lower gain as amplitude rises above a certain threshold.

It can also be done with diodes and here is my attempt: -

enter image description here

Now you have about 10 volts peak to peak and much better sine wave purity: -

enter image description here

Diodes used were 1N4148 but any fast recovery signal diode should be OK.

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  • \$\begingroup\$ thanks, it worked with your circuit. And it also gave a nice sine wave when I remove C2 and make coupling lower (to around 0.5). Here, I still didn't get the role of C2. \$\endgroup\$ – Pojj Mar 1 '18 at 14:06
  • \$\begingroup\$ C2 is not necessarily needed when operating the circuit with the diodes (as I have done). However I would keep it because it does provide a ripple free voltage at the junction of the 47k and 22k. It also makes the AC impedance at that node close to zero so it's clearer what the base driving impedance is when analysing. \$\endgroup\$ – Andy aka Mar 1 '18 at 14:10
  • \$\begingroup\$ Thank you for your valuable inputs, I am now trying to understand the operation of feedback stabilizing network. I can see that feedback has reduced in value, and a phase difference (close to 90 deg.) has introduced by the diode-resistor network. Then how the zero phase condition (for oscillation) is generated? Also, what is the role of R4 here? \$\endgroup\$ – Pojj Mar 5 '18 at 15:16
  • \$\begingroup\$ I'm not sure where the phase shift is relative to. Sure there must be an added 180 degrees to oscillate so that confused me too. R4 is typical of a 100 uH coil and it was placed there because some resistance needs to be put there in real life circuits. The diodes clip the signal on the way back to the base hence they stabilize the fed back sine wave into a heavily clipped sine wave. But that's no big deal because the tank restores good sinewave purity as seen on the spectral result in my answer; 2nd harmonic down over 30 dB. \$\endgroup\$ – Andy aka Mar 5 '18 at 16:35
  • \$\begingroup\$ Thank you, For phase, I mean the voltage of base relative to L2- and Vout are about 90 degrees out of phase, whereas, that phase difference was 180 degrees for the initial circuit (without feedback stabilizing network). I got your other points. \$\endgroup\$ – Pojj Mar 6 '18 at 6:39

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