6
\$\begingroup\$

I am trying to understand oscillator basics. I thought I already understood how a capacitor works, but I still get confused about it.

This circuit is an RC phase shift oscillator. I understand that we need 3 capacitors to get a 180° phase shift in the feedback circuit (3×60°, but the exact phase shift depends on the capacitor and resistor values, right?).

phase shift oscillator From here

After I thought I knew why a capacitor causes phase shift I realized that at the output there is a bypass capacitor.

So we have here a 180° phase shift because of the transistor, and 180° phase shift because of the feedback RC circuit. But when we look at the output voltage there should be additional phase shift because of the output bypass capacitor, right? So the output voltage of the circuit has a different phase than the emitter voltage because there is an additional bypass capacitor. And what about the phase shift of the capacitor which is parallel with Re?

Can somebody explain to me how this phase shift thing really works? I am really confused right now.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Do you understand this formula: \$I_C = C\cdot\frac{dv}{dt}\$? Have you seen it before? Have you used it before? \$\endgroup\$
    – Andy aka
    Sep 27, 2022 at 8:05
  • \$\begingroup\$ NB: since the network is a high pass filter, one should be aware that the "oscillation" should not be "sinusoidal". \$\endgroup\$
    – Antonio51
    Sep 27, 2022 at 8:56
  • \$\begingroup\$ @Antonio51 why wouldn't it be sinusoidal ? wouldn't the higher frequencies just disappear since their overall feedback phase would be different than 2k*Pi ? \$\endgroup\$
    – Rahmany
    Sep 27, 2022 at 9:17
  • 1
    \$\begingroup\$ Have you simulated this circuit? Will add in an answer later. The higher frequencies do not dissapear, they are 'amplified'. \$\endgroup\$
    – Antonio51
    Sep 27, 2022 at 9:38

5 Answers 5

1
\$\begingroup\$

If the output cap is an open circuit, it doesn't have any effect.

If there's a grounded load resistor on the other side (or another circuit stage), the capacitor tends to be very large so it's practically a short circuit to AC current, with little to no phase effect.

The same about the cap. in parallel with Re - it's very large so it blocks DC, but is basically a short circuit with hardly any phase shift to AC current.

If there's a phase shift after all, say 2 degrees, and the transistor shifts 180 degrees, then the oscillation frequency would be the frequency where the 3 oscillator capacitors shift by 178 degrees. (Slightly different than the 180 degree frequency).

But once more, probably those large 10uF+ caps won't cause even a 2 degrees shift.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ thank you its clear now! \$\endgroup\$
    – Darko
    Sep 29, 2022 at 13:10
  • \$\begingroup\$ Great - thanks for picking my answer. To be honest, I feel like others put a bit more effort in their answers, but glad my way of explaining worked for you. \$\endgroup\$
    – ee_student
    Sep 30, 2022 at 0:26
7
\$\begingroup\$

As far as the oscillation condition is concerned, it is the phase shift within the feedback loop that only matters! Therefore, the coupling capacitor at the collector node has practically no influence on the oscillation condition ("practically" means: There will be a only pretty small influence due to the finite resistance at the collector node).

The same applies to the emitter bypass capacitor. The value of this capacitor is selected to be much larger than the capacitors in the feedback path. For this reason, it nearly works like a short for the frequency of oscillation. This is necessary because the gain of the transistor amplifier should have a fixed value for the frequency range of interest (no frequency-dependent negative feedback caused by the emitter circuitry).

Additional comment (just for your understanding): It is correct that a 3rd-order C-R (or R-C) ladder structure can provide a 180deg phase shift at one specific frequency. However, it is not true that each stage will contribute 60deg to this phase shift. This would be the only case when all three sections would be decoupled. The distribution of the particular phase shifts is more involved due the loading effects. In particular, this is true because the last (most right) capacitor is connected to R in parallel to r_in of the transistor stage (in the lower kOhm range).

\$\endgroup\$
4
\$\begingroup\$

This circuit is an RC phase shift oscillator. I understand that we need 3 capacitors to get a 180° phase shift in the feedback circuit (3×60°, but the exact phase shift depends on the capacitor and resistor values, right?).

It doesn't have to be C-R. It can also be a R-C. Although one brings phase lead and the other brings phase lag the overall effect doesn't change. And yes, the RC (or CR) determines the amount of phase lead (or lag).

But when we look at the output voltage there should be additional phase shift because of the output bypass capacitor, right?

True, the bypass cap parallel to Re and the coupling cap at the output (with the load resistance or simply the following block's input resistance) do bring a phase shift but they are selected large enough so the phase shifts caused by them are negligible.

As for how it works, it's quite simple: The first requirement of oscillation is positive feedback. Without the C-R feedback network the amplifier itself is an inverting amplifier: It outputs an amplified and 180° shifted copy of the input signal.

If the output signal (signal at the collector) was fed to the input (i.e. base) with a resistor only then the feedback would be negative feedback.

Now the feedback network comprises a number of C-Rs and the entire network brings a total phase shift of 180°, so the transistor's base will see 180° shifted (or in other words, inverted) copy of the output signal. So the output signal is inverted and fed back to the input. And we know that the amplifier configuration bring another 180° by its nature (because it's an inverting amplifier) and therefore makes the total phase shift 360°. So the negative feedback turns into a positive feedback.

\$\endgroup\$
4
  • \$\begingroup\$ "the RC (or CR) determines the amount of phase lead (or lag)." I don't think that's true : R & C define the cutoff frequency of the filter. And the number of stages define the frequency of oscillation f_osc(N) (where N is the number of CR/RC filters stages). The 60° phase lag is only mentioned because it's a 3 stage feedback network and is only true for the oscillation frequency f_osc(3). If it was a 4-stage feedback network, then f_osc(4) would have 45° phase lag at each stage. \$\endgroup\$
    – Rahmany
    Sep 27, 2022 at 9:38
  • 1
    \$\begingroup\$ @Rahmany simulate it, then: Build a C-R network, apply a sine wave input and observe the output's phase lead by adjusting the component values. \$\endgroup\$ Sep 27, 2022 at 9:50
  • \$\begingroup\$ yes, in a single RC filter the output's phase is given by arctan(Xc/R). But in this specific oscillator circuit, the values of R and C are used to set f_osc. Each stage will always have 180°/N of phase shift at f_osc regardless of the values of R and C (f_osc will change of course). But because phase shift is a function of frequency, then the only phase shift value that's pertinent is the one at f_osc which will be always the same for the same number of RC stages. \$\endgroup\$
    – Rahmany
    Sep 27, 2022 at 10:07
  • 1
    \$\begingroup\$ Rahmany - your last sentence is not true. It would be true only if the three sections would be decoupled (in this case, the oscillation frequency would also be shifted). But due to coupling/loading effects the calculation of the individual phase shifts is more involved. \$\endgroup\$
    – LvW
    Sep 27, 2022 at 13:36
2
\$\begingroup\$

First, I would add the transient analysis of that circuit.
It oscillates, but 'phases' could be difficult to be measured here.

enter image description here

\$\endgroup\$
2
\$\begingroup\$

For stable amplitude oscillation to occur there needs to be a gain of 1 around the total loop but also there needs to be a phase lag of 360 degrees or 0 degrees around the complete loop (Barkhausen Criteria). For your configuration of phase shift oscillator the 3 high pass filters will create about 180 degrees of lead followed by the amplifiers (approximate) inversion.

The oscillator will oscillate at a frequency where the complete loop phase is 0 degrees. If the amplifier's emitter by-pass capacitor adds a little phase lead such that the phase lag across the amplifier is a little less than -180 degrees then the oscillator will oscillate at the frequency where the phase lead across the 3 RC networks is a little less than +180 degrees.

To give an example, with made up numbers. The oscillator may oscillate at a frequency where the phase lag across the amplifier is -175 degrees and where the phase lead across the 3 RC networks is +175 degrees. The oscillator will find the oscillation frequency where the total loop phase is 0 degrees.

The output capacitor is outside the loop and so has no bearing on the phase shift around the loop or the oscillation frequency. However there will be some phase shift across that capacitor and the amount of phase shift will depend upon the size of the capacitor and the size of the amplifier's load resistor. The load resistance is not shown.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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