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The following is a picture of colpitts oscillator and CE.

Barkhausen stability criterion said positive feedback is needed in order to oscillate. However, Common-Emitter has 180 degree phase shift. Thus it's negative feeback. How can this oscillate?

colpitts

enter image description here

kile
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  • How about you assume that there are other components in the circuit that provide an additional 180 degree phase shift at a particular frequency. – Nedd Sep 30 '23 at 08:03
  • Don't you think that just saying CE is 180 degrees and declaring this as negative feedback is a smidge over simplistic? Don't you recognize that there can be frequency dependent responses by external components (and parasitics) that may alter such a simplistic view? – periblepsis Sep 30 '23 at 08:03
  • @Nedd I am not sure which components can provide additional 180 degree shift. Please point out those components – kile Sep 30 '23 at 08:13
  • @periblepsis Can you specify those external components? – kile Sep 30 '23 at 08:14
  • Consider collector voltage is going up so inductor starts charging via C2 to ground, but C2 voltage rise is delayed. Another words, L+C2 makes a phase shift in Vcollect vs. Vc2 point of view. During opposite cycle the inductor current flows backward through C1 so C2 is charged opposite. – Michal Podmanický Sep 30 '23 at 09:06
  • @MichalPodmanický You answer is by far the best I have seen. Please answer in answer section. – kile Sep 30 '23 at 09:11
  • Look at your tank circuit (L, C1, C2). Note that the junction of C1 and C2 is connected to your circuit common rail ("earth"). This adds an earthed centre-tap to the tuned circuit. What effect will this have on the phase of the voltages at either end of L compared to the circuit common? – Graham Nye Sep 30 '23 at 09:12
  • @GrahamNye I guess phase shift is induced by L, C1, C2 – kile Sep 30 '23 at 09:39
  • @kile You don't need to guess. You can break the loop and work out the transfer function using an appropriate small signal model for the BJT after working out the DC operating point for that very standard BJT configuration. The only way to convince yourself is to do that work. Hand-waving isn't helpful. I've provided the necessary references here to understand when and where oscillation will occur, too. My preference would be you read Dr. Smith's paper, though. It's simple and direct. – periblepsis Sep 30 '23 at 10:05
  • @kile In case it helps, two slightly different linearized Gummel-Poon models for the BJT that you might consider are (1) this one from Modeling the Bipolar Transistor by Ian Getreu and (2) this one found at this link. – periblepsis Sep 30 '23 at 10:40
  • @kile Switching occurs when the derivative of the transfer function equals infinity. Or put another way, if $V_o=\frac{\mathscr{F}\left(V_i\right)}{K}$ then it follows that $V_o^{:'}=\frac{1}{K}\left(1-V_o^{:'}\right)\mathscr{F}^{:'}$ and switching occurs at the $\pm$ points where $V_o^{:'}=\infty$. Put more simply, where the loop gain is exactly unity at both rising and falling points. – periblepsis Sep 30 '23 at 11:29

1 Answers1

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Colpitts oscillator has negative feedback, why does it oscillate?

No, it has positive feedback. It's just that you didn't analyse sufficiently.

It oscillates because \$C_2\$ and \$L\$ provide about 150° of phase shift AND, the collector resistor \$R_C\$ and \$C_1\$ add another 30°. Meaning there is another 180° of phase shift and, this makes it oscillate.

Maybe these words and images (from my basic website) might help a bit: -

enter image description here

To analyse the circuit and derive the oscillation frequency formula, the collector node cannot be assumed to be unloaded. This mistake is made by many websites that attempt to describe functionality; important subtleties are glossed-over and, the formulas are presented matter-of-fact with no attempt to reveal the workings. Some websites misleadingly describe the operation by referring to a tank circuit (where C1 and C2 are in series) but, because C1 and C2 are grounded, they form a filter not only with inductor L1, but with resistor R4.

R4 makes this a 3rd order filter and only a 3rd order filter can produce sufficient phase shift for the circuit to oscillate. R4 is that important!

So, R4, C1, C2 and L1 form a 3rd order low-pass filter that produces 180° phase-shift whilst still providing sufficient signal gain to initiate and maintain oscillation. A 2nd order analysis is insufficient to deliver the theory. Bear that in mind. This is a proper analysis.

So, we "recognize" that Q1's collector is a current source in parallel with R4 and that this is equivalent to a voltage source in series with R4. Hence, R4 and C1 form a low-pass filter that introduces a lagging phase angle at the collector. Below is the bode plot when we apply a signal generator at the circuit input (C3) and C1 is connected to the collector (L and C2 are not connected yet): -

enter image description here

Normally, a CE amplifier delivers constant gain across a wide frequency range (with a phase angle of 180°) but, with C1 loading the collector, the phase shift is tending towards +90° at high frequencies. The gain is also reducing at high frequencies. The faint lines in the diagram above show the natural gain of the CE amplifier (12.6 dB). This gain is due to the ratio of R4 to R1 i.e. 2000 ÷ 470 = 4.255 = 12.6 dB.

Then, if L and C2 are added to R4 and C1 we get this bode plot after inductor L1: -

enter image description here

Note that the amplitude peak at about 2 MHz corresponds with a 0° phase angle. When the feedback loop is closed, this is where the circuit will oscillate. The 0° phase shift is a result of the collector resistor (R4) feeding C1, L and C2 i.e. this is a 3rd order filter. The gain peak at around 2 MHz is 12.6 dB because, at the oscillation frequency, when C1 equals C2, the 3rd order filter has unity gain.

I won't repeat the full version from my website.

Andy aka
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  • But Common Emitter amplifier has 180 degree phase shift. – kile Sep 30 '23 at 09:23
  • @kile the feedback signal gets shifted a further 180 degrees by the components mentioned in my answer. – Andy aka Sep 30 '23 at 09:24
  • What's collector RC? Do you mean $R_c$ and $C_o$? – kile Sep 30 '23 at 09:25
  • I actually said collector resistor RC but I can see it may be confusing. – Andy aka Sep 30 '23 at 09:27
  • Can you point it collector resistor RC out in this diagram? – kile Sep 30 '23 at 09:31
  • I think if you look at either of your diagrams you will see there is a resistor marked $R_C$. – Andy aka Sep 30 '23 at 09:32
  • My mistake. Forget to give you picture, How about the capacitor? https://i.stack.imgur.com/PyLTu.png – kile Sep 30 '23 at 09:35
  • I've explained about C1 and C2. They are the relevant capacitors. – Andy aka Sep 30 '23 at 09:39
  • You misunderstand me. I mean capacitor in collector resistor RC instead of the feedback circuit. – kile Sep 30 '23 at 09:41
  • There isn't a capacitor in that resistor. It can be presumed to be an ideal resistor possessing no parasitic capacitance at all. – Andy aka Sep 30 '23 at 09:43
  • Is R in this picture http://www.stades.co.uk/Colpitts%20CE/Colpitts%20RCLC.png the same as R4 in your diagram http://www.stades.co.uk/Colpitts%20CE/Colpitts%201.png? – kile Sep 30 '23 at 11:07
  • I don't understand why R4 is in series with C1, in my opinion. It should be in parallel. – kile Sep 30 '23 at 11:15
  • @kile effectively yes (after doing a Thevenins conversion) – Andy aka Sep 30 '23 at 11:15
  • The collector (a current generator) is effectively in parallel with R4 and this converts to a voltage source with R4 in series. This feeds C1 and acts like a low-pass filter. – Andy aka Sep 30 '23 at 11:17
  • Here is what I think. https://i.imgur.com/XDDBiLp.png – kile Sep 30 '23 at 11:21
  • Sorry, if you are not accepting what I say then there's absolutely no point trying to convince me that your way is correct. If you want this resolving either think about it a bit longer or, raise a new question because asking how to convert sources using Thevenin's theorem is not really related to the original question. I mean, where would your questions end if I didn't put a break on things. – Andy aka Sep 30 '23 at 11:51