Phase Shift Oscillator Question - UPDATE: Circuit is a ring oscillator - Page 1
Forum section: electronics / projects

author: ANTALIFE
author_rank: Frequent Contributor
timestamp: November 22, 2018, 10:58:13 pm
content: HelloI am trying to figure out how thisphase shift based oscillatoris able to well oscillate.From my understanding for oscillations to occur the phase (between input/output) must have a difference of 360deg or a multiple of.Knowing that each RC section gives a phase shift of 60deg and each BJT gives a phase shift 180deg, adding these up I get an expected phase shift of 720deg (60+60+60+180+180+180). However when I simulate the circuit in LTspice and try to manually calculate the phase shift I get a value of ~250deg, am I missing something here?UPDATE: Turns out the circuit is actually a ring oscillator, see orolo's & Hero999's comment

author: xavier60
author_rank: Super Contributor
timestamp: November 23, 2018, 08:38:36 am
content: I can't directly answer the question. I notice that R12 is 2K which will make the phase shift of the first stage much less than the others.

author: spec
author_rank: Frequent Contributor
timestamp: November 23, 2018, 08:47:45 am
content: Hi ANTALIFEI love your question. It is just the sort of thing that I get into a quandary about.I am not going to work it out, because my brain is not functioning yet.But here are some thoughts to consider.The phase angle between a transistor(BJT) base and collector is 180 deg, as you rightly say.But it in not correct to say that the phase angle across a CR network is always 60 degrees: the phase angle varies with frequency.The capacitor connected between the BJT collector and base, as shown on the schematic, are what is physically fitted. But the vittual capacitance seen by the circuit is the value of the physical capacitor multiplied by the voltage gain (A) of the BJT stage.A is worked out practically as follows: The value of a BJT internal emitter resistor (re) is 25/ie where re is in Ohms and ie is in mA. And, providing the BJT has a reasonable current gain (hFE), the A of a BJT is Rc/re, Rc being the external collector resistor. For a typical BJT, like a BC227, with an ie of 1ma and an Rc of 1k, A is 1000/25 = 40. For a lower conductance BJT, like a BC54x, the A may be a touch lower, say 30.

author: bson
author_rank: Unknown
timestamp: November 23, 2018, 11:25:49 am
content: The capacitors bypass base-collector, so are there to limit the bandwidth, which in this circuit effectively sets the oscillation frequency. If they weren't there the circuit would oscillate around the transistors' effective unity frequency. They're in parallel with the transistor gain, not in series. (You'll often see the practice of capacitively coupling input and output of a gain or buffer stage referred to as miller compensation.)

author: spec
author_rank: Frequent Contributor
timestamp: November 24, 2018, 12:38:42 am
content: Quote from: bson on November 23, 2018, 11:25:49 amNot sure what you are saying here. How can a capacitor bypass two separate points on a circuit.Quote from: bson on November 23, 2018, 11:25:49 amAgain not sure what you are saying here. How can limiting the bandwidth set the frequency. The resistor and virtual capacitance set the phase angle which sets the frequency, just like a standard phase shift oscillator. You seem to be thinking about an amplifier where often a miller feedback capacitor is used to roll the gain off to ensure frequency stability. This approach is frequently used in the voltage amplification stage (VAG) of many audio power amplifiers.Quote from: bson on November 23, 2018, 11:25:49 amYes, maybe but cant see the relevance of this statement.Quote from: bson on November 23, 2018, 11:25:49 amAfraid this does not make sense. How can anything be in parallel with gain- the dimensions do not match.Quote from: bson on November 23, 2018, 11:25:49 amYes, and Miller compensation uses the Miller effect as I describe.

author: ANTALIFE
author_rank: Frequent Contributor
timestamp: November 24, 2018, 02:22:43 am
content: Thanks for the input everyone, looks like I will have to do a bit more digging.Also as pointed out I should have said the RC sections give a phase difference of 60deg at a given frequency, and in the LTspice sims this frequency is 59mHz

author: xavier60
author_rank: Super Contributor
timestamp: November 24, 2018, 03:29:12 am
content: Quote from: ANTALIFE on November 24, 2018, 02:22:43 amDid you intend R12 to be 2k?

author: ANTALIFE
author_rank: Frequent Contributor
timestamp: November 24, 2018, 03:33:48 am
content: Quote from: xavier60 on November 24, 2018, 03:29:12 amSorry forgot to reply, yup having it as 2K (instead of 100K) gives me a nicer waveform plus gets the oscillation going faster (see pic)

author: bson
author_rank: Unknown
timestamp: November 24, 2018, 07:57:37 am
content: Quote from: spec on November 24, 2018, 12:38:42 amBy placing it parallel with whatever else is between the two nodes.

author: awallin
author_rank: Frequent Contributor
timestamp: November 24, 2018, 12:06:56 pm
content: maybe try the simulation "open loop" i.e. by removing the feedback wire completely.compute bode-plots around the frequency where you expect it to oscillate and figure out where Barkhausen criteria are (are not) fulfilled:https://en.wikipedia.org/wiki/Barkhausen_stability_criterion

author: orolo
author_rank: Frequent Contributor
timestamp: November 24, 2018, 01:29:31 pm
content: This is not a phase shift oscillator. It's more like the classic LED chaser circut, a ring of switches that trigger each other. When Q4 is saturated, Q6 must be off to keep Q4's base high. Since Q4 is saturated, its collector is low, which will turn Q5 off. As Q5 turns off, its collector goes up, which then turns Q6 on. As Q6 turns on and goes into saturation, Q4 is turning off.So the thing goes:Q4(sat), Q5(turn off), Q6(turn on) -> Q4(turn off) Q5(turn on) Q6(sat) -> Q4(turn on) Q5(sat) Q6(turn off) -> Q4(sat) Q5(turn off) Q6(turn on) -> etc.If your circuit is symmetric, you should see identical waveforms in the transistor collectors, 120 degrees out of phase.

author: spec
author_rank: Frequent Contributor
timestamp: November 24, 2018, 05:25:51 pm
content: Quote from: orolo on November 24, 2018, 01:29:31 pmWow!

author: spec
author_rank: Frequent Contributor
timestamp: November 24, 2018, 05:50:03 pm
content: @ ANTALIFEWhat frequency is the circuit actually oscillating at? I can't make out the legend on the scope screen. Also what supply voltage does the circuit have.My back of cigarette packet calculation shows around 500Hz.

author: Zero999
author_rank: Super Contributor
timestamp: November 24, 2018, 08:02:24 pm
content: Quote from: orolo on November 24, 2018, 01:29:31 pmYes, it's a ring oscillator.

author: ANTALIFE
author_rank: Frequent Contributor
timestamp: November 24, 2018, 08:30:37 pm
content: Quote from: spec on November 24, 2018, 05:50:03 pmCircuit oscillates at 59mHz (yup milli) also supply voltage is 4V

author: ANTALIFE
author_rank: Frequent Contributor
timestamp: November 24, 2018, 08:39:27 pm
content: Quote from: orolo on November 24, 2018, 01:29:31 pmOh wow, the steps are quite cool. Also you are totally right, making all elements symmetrical gives "identical waveforms in the transistor collectors, 120 degrees out of phase" as you said. Thanks for solving the mystery

author: xavier60
author_rank: Super Contributor
timestamp: November 24, 2018, 10:52:10 pm
content: You could have realized that earlier when it was pointed out that "R12 is 2K which will make the phase shift of the first stage much less than the others"

author: ANTALIFE
author_rank: Frequent Contributor
timestamp: November 24, 2018, 10:56:44 pm
content: Quote from: xavier60 on November 24, 2018, 10:52:10 pmBut the thing is that I was comparing the circuit to a wrong model (phase shift oscillator), I had no idea that I should have been looking at it as a ring oscillator

author: spec
author_rank: Frequent Contributor
timestamp: November 25, 2018, 09:11:07 am
content: Anybody like to give the formula for determining the frequency of that oscillator? 59miliHertz is hellishly low.