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?


enter image description here

  • \$\begingroup\$ How about you assume that there are other components in the circuit that provide an additional 180 degree phase shift at a particular frequency. \$\endgroup\$
    – Nedd
    Commented Sep 30, 2023 at 8:03
  • \$\begingroup\$ 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? \$\endgroup\$ Commented Sep 30, 2023 at 8:03
  • \$\begingroup\$ @Nedd I am not sure which components can provide additional 180 degree shift. Please point out those components \$\endgroup\$
    – kile
    Commented Sep 30, 2023 at 8:13
  • \$\begingroup\$ @periblepsis Can you specify those external components? \$\endgroup\$
    – kile
    Commented Sep 30, 2023 at 8:14
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Sep 30, 2023 at 9:06

1 Answer 1


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.

  • \$\begingroup\$ But Common Emitter amplifier has 180 degree phase shift. \$\endgroup\$
    – kile
    Commented Sep 30, 2023 at 9:23
  • \$\begingroup\$ @kile the feedback signal gets shifted a further 180 degrees by the components mentioned in my answer. \$\endgroup\$
    – Andy aka
    Commented Sep 30, 2023 at 9:24
  • \$\begingroup\$ What's collector RC? Do you mean \$R_c\$ and \$C_o\$? \$\endgroup\$
    – kile
    Commented Sep 30, 2023 at 9:25
  • \$\begingroup\$ I actually said collector resistor RC but I can see it may be confusing. \$\endgroup\$
    – Andy aka
    Commented Sep 30, 2023 at 9:27
  • \$\begingroup\$ Can you point it collector resistor RC out in this diagram? \$\endgroup\$
    – kile
    Commented Sep 30, 2023 at 9:31

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