Complicated impedance of RC poles in a phase shift oscillator - Page 1
Forum section: electronics / beginners

author: Mozee
author_rank: Contributor
timestamp: September 20, 2018, 01:23:15 pm
content: Hello there,While reading Floyd's book I came across something similar and decided to solve it! I just wanted to calculate the total impedance that the op-amp will see when connecting these 3 poles through a resistor of 6.2K to the non-inverting input to form an RC phase shift oscillator, also, I wanted to figure out the value of Rin to find the Rf required for the op-amp so that I can get an Av of 29 or more!In this RC phase shift oscillator, I'm trying to calculate the total impedance of the 3 RC pairs!Now the C3 and R3 have a Z3 of sqrt(Xc3^2 +R3^2) then we will have a parallel R2 with this Z3, can I directly add R2//Z3? as in R2//Z3 = Z2Z2= 1/(1/R2)+(1/Z3)?if that was correct, we will face an Xc of C2 in series with the resulting Z2, can we add Xc2 with Z2 in seriesFinally, should the Rin2 be included in the calculation at all? or should I consider the Rin2 the main Rin and use it to calculate the amp gain and why?Thank you.

author: Benta
author_rank: Super Contributor
timestamp: September 20, 2018, 03:21:12 pm
content: Your inverting input is a virtual ground, so your phase shift network will see R3||Rin. You'll need to select your resistances accordingly.

author: Mozee
author_rank: Contributor
timestamp: September 20, 2018, 08:29:34 pm
content: Thank you, my friend, I Knew that it would act as a virtual ground but I have no deeper knowledge of how would it exactly do so!I calculated the Rin2 // R3 and to get a 6.2K I would have to use Rin2=62K and R3=6.8K and when doing so, the oscillator stops oscillating! No idea why!also, I noticed that even taking out both R3 and Rin2 from the circuit, the oscillator still works and I get an output! What is going on?and not forgetting the main question, Is it possible to add the impedance Z3 of Xc3+R3, to the parallel RESISTOR R2? as in Z(R2&Z3)= R2+Z3and similarly , can I add an impedance to a series reactance as in C2 + Z2

author: bson
author_rank: Unknown
timestamp: September 22, 2018, 01:22:09 am
content: There is no equivalent impedance for a divider. It has 3 ports, not 2, while impedance by its nature is a 2-port property. Hence, the divider has separate input and output impedances. Put differently, can you think of a 2-port equivalent box for a divider?I think what you should be looking at is the transfer function T(s). For unity you'll want the resistors to set an amplifier gain G = |1/T(s)| for some frequency w where s=jw. (Probably your oscillation frequency?) The transfer function for a CR stage is R/(R-1/sC). While series impedances add, series transfer functions multiply, so the resulting transfer function for your cascade is:$$ \frac{C_1 C_2 C_3 R_1 R_2 R_3 s^3}{\left(C_1 R_1 s-1\right) \left(C_2 R_2 s-1\right)\left(C_3 R_3 s-1\right)} $$From this you can see that for non-zero positive values for all Rs and Cs, you have a zero at s=0 and three poles at s=1/CnRn. If you evaluate 1/|T(s)| for s=jw at your resonant frequency w you'll get the gain you need in the amplifier to maintain unity (at that frequency).

author: Audioguru
author_rank: Super Contributor
timestamp: September 22, 2018, 02:43:42 pm
content: It is simple, not complicated. Your very low value opamp resistors are overloading the 3rd RC and your opamp gain is way too low.Here is how an expert does it:

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 22, 2018, 04:20:01 pm
content: Rather embarrassing that TI would publish that. Alas, such are appnotes...If they had increased R_G to, say, 100k or more, and R_F proportionally, it would be about right.Less gain is required if the RC values are staged. Performance is in fact very poor when the R and C are equal values. Instead, the R values should increase, and C decrease proportionally, along the chain. This keeps the poles closer together, increasing phase shift and gain in the transition band. (Of course, this needs an even larger R_G still, which is kind of annoying.)Also, hahahaha, the first one uses a TLV2471 (single), and the second one uses a TLV2471 (single) with a part of a TLV2474 (quad). Nice copypasta work they did there.Anyway, for adding up impedances, no, you cannot take the magnitude, you must work in complex numbers the whole way through. This results in a rather complicated 3rd order equation for the circuit, but that's just the nature of the beast. Plug it into a symbolic math solver and get your results that way.Tim

author: ferdieCX
author_rank: Regular Contributor
timestamp: September 22, 2018, 05:04:12 pm
content: Here is the complete general analysis of the phase-shift oscillator. The text is in Spanish, but you can follow the equations.Resuming it:fo = 1/(2 pi R C sqrt 6) , this way the phase-shift of the networks is 180ºThe gain of the phase-shift network is B = - 1/29, so the gain of the amplifier must be A = - 29The phase-shift network presents to the amplifier a load impedance ZL = 2.83 R /_ -73 º ,so for the amplifier to have a phase'shift of 180º, you need to load it with a resistor much smaller than 2.83 R

author: The Electrician
author_rank: Frequent Contributor
timestamp: September 23, 2018, 09:28:59 am
content: Quote from: bson on September 22, 2018, 01:22:09 amYour transfer function is not correct for the circuit shown in the first post. Your expression would be correct if there were unity gain buffers between the stages of the phase shift network, but in the given circuit, later stages load earlier stages.