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| Phase Shift Oscillator Question - UPDATE: Circuit is a ring oscillator |
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| ANTALIFE:
--- Quote from: orolo on November 24, 2018, 01:29:31 pm ---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. --- End quote --- Oh 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 :D |
| xavier60:
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" |
| ANTALIFE:
--- Quote from: xavier60 on November 24, 2018, 10:52:10 pm ---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" --- End quote --- But 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 |
| spec:
Anybody like to give the formula for determining the frequency of that oscillator? 59miliHertz is hellishly low. |
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