Electronics > Beginners

Complicated impedance of RC poles in a phase shift oscillator

(1/2) > >>

Mozee:
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 = Z2
Z2= 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 series
Finally, 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.

Benta:
Your inverting input is a virtual ground, so your phase shift network will see R3||Rin. You'll need to select your resistances accordingly.

Mozee:
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+Z3 ???
and similarly , can I add an impedance to a series reactance as in C2 + Z2

bson:
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).

Audioguru:
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:

Navigation

[0] Message Index

[#] Next page

There was an error while thanking
Thanking...
Go to full version
Powered by SMFPacks Advanced Attachments Uploader Mod