1
\$\begingroup\$

enter image description here

According to the solution, it says it's because the phase shift is wrong. The gain network is non-inverting, so I assume that means that the LC feedback network was not 0 degrees for that particular frequency.

So I tried to work it out by hand but I am stuck here. Is my knowledge that the LC feedback not being 0 degrees for f = 1/root(2LC) the reason why it doesn't work? If so, what is the shift exactly, and how can I calculate the phase shift from the equations below

enter image description here

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

Consider the Wien bridge oscillator circuit: -

enter image description here

It is very similar to the oscillator in the question except that the lower capacitor is is an inductor in the question. At this point alarm bells should possibly start ringing to inform you that replacing the cap with an inductor will not work.

However, lets look at the frequency and phase response of the RC Wien bridge network. This is the filter section of the Wien bridge oscillator circuit (C2 is replaced by L in the question): -

enter image description here

Here's the phase response: -

enter image description here

As can be seen, at resonance, the phase shift is zero as one expects - this produces oscillation at Fr. With C2 changing to an inductor, the phase shift will be 90 degrees and will therefore not oscillate. This last point presumes you understand what happens when L and C are combined at resonance. If not then look on this page for a simulation.

\$\endgroup\$
16
  • 1
    \$\begingroup\$ Nice, (+1) It's most likely in your links, but @Raaj also has to adjust the gain resistors to over come the 1/3 drop in amplitude. \$\endgroup\$ Nov 20 '14 at 13:21
  • \$\begingroup\$ For my opinion, it is problematic to compare both circuits, because replacing C2 by an inductor gives a HIGHPASS which hardly can be compared with a classical bandpass response.(By the way: Not WEIN but WIEN!). \$\endgroup\$
    – LvW
    Nov 20 '14 at 13:38
  • 1
    \$\begingroup\$ Ah I get it..its a high pass filter, hence the 90 degree phase shift at fr \$\endgroup\$
    – Raaj
    Nov 20 '14 at 14:04
  • \$\begingroup\$ I wish however, I could understand the math it takes to derive that 90 degrees. \$\endgroup\$
    – Raaj
    Nov 20 '14 at 14:09
  • 2
    \$\begingroup\$ @Raaj, look into my answer (transfer function) and verify what happens at the frequency w=wo: Only the middle part of the denominator with "s=jw" is there (pure imaginary). This is identical to 90 deg phase shift. \$\endgroup\$
    – LvW
    Nov 20 '14 at 15:39
1
\$\begingroup\$

Because your handwriting is not easy to read I didn`t follow your calculation in detail - however, the denominator of the transfer function for the RLC feedback circuit cannot be real for finite frequencies (the imaginary part cannot be zero for s=jw). However, this is required to have a zero phase at a finite frequency - as a precondition for producing oscillations. Hence, the circuit cannot work as a n oscillator. (The imag. part of the transfer function always must contain a difference in order to become zero for a finite frequency).

EDIT 1: The transfer functionof the RLC feedback path is H(s)=N(s)/D(s) with

N(s)=LCs^2 and D(s)=1+s(L/R+RC)+2LCs^2

EDIT 2: OK - I have tried to follow your calculation. It is correct, and - as expected - it is the transfer function of a second-order highpass (phase starts at +180 deg and approaches zero for infinite frequencies).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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