# Oscillator frequency

1. Oct 15, 2014

### tuttyfruitty

I'm trying to determine the frequency of this oscillator . The traditional resonant frequency equation for the tank circuit doesn't give the correct answer as the feed back capacitor effects the circuit .

i would also like to know what the feedback capacitor does? any help would be great.

2. Oct 15, 2014

### davenn

hi there

welcome to PF :)

strange place for the output to be coming from ... I would have normally expected it to be off the output of the Op-amp. Why are you taking the output from the input to the op-amp ?

The feedback cap is making the op-amp a lowpass filter ( ie an integrator)

Dave

3. Oct 16, 2014

### tuttyfruitty

Hi Dave,

The circuit is used at work and the design is a few years old, so I'm unsure why the output is on the input pin. I have found 2 versions of the circuit, one with a variable resistor and one with the capacitor. I understand that the variable resistor is to "tune" the oscillator so it maintains its oscillations and the resistance of the variable resistor must match the losses of the tank circuit.

I will do some research on the integrator, do you know what the low pass filter is doing regards to maintaining oscillations?

4. Oct 16, 2014

### zoki85

I would try to see here what is min and max impedance of the network seen from the terminals of the assigned output. This is a way one usually gets resonant frequencies.

5. Oct 16, 2014

### sophiecentaur

The OpAmp output terminal will be fairly low impedance and so the feedback capacitor will appear as is it was in parallel with the resonator capacitor. That might suggest the operating frequency might be a bit lower than you expected. Around f/√2, perhaps, if the capacitances are the same?

6. Oct 16, 2014

### Staff: Mentor

For some working circuit you've seen, what are the values of feedback capacitor and tuned-circuit capacitor?

7. Oct 16, 2014

### tuttyfruitty

The cap in the tank circuit is 10 uF and the feedback cap is 470uF, the oscillation frequency is 2174 Hz, for completeness, the inductor is 470 uH

8. Oct 16, 2014

### The Electrician

It looks to me as though the impedance seen at pin 3 of the opamp (with the LC tank disconnected) will be a negative capacitance:

http://en.wikipedia.org/wiki/Negative_impedance_converter

Adjusting one of the resistors would certainly change the apparent value of the negative capacitance.

With a variable resistor in place of the capacitor (from opamp output to pin 3), the opamp circuit would present a negative resistance. That could be adjusted to compensate for the losses in the tank circuit, keeping the tank on the edge of oscillation.

9. Oct 16, 2014

### tuttyfruitty

Ive just done some research on negative impedance, that's some interesting stuff! but how does the negative capacitor effect the circuit?is it sourcing current into the tuned circuit?

10. Oct 16, 2014

### The Electrician

The frequency of the LC tank is determined by the resonance frequency of L and C. If some negative capacitance is connected in parallel with the tank's C, the effective value of capacitance in parallel with the L will be changed.

11. Oct 16, 2014

### Mike_In_Plano

Looks like a negative resistance shunted across an LC. I've designed these in before, though normally I used a gyrator instead of the inductor.
Normally, the positive feedback would be a resistor rather than cap. Also, the values are wacky. I get 2.32 kHz as the tank frequency, but a z of only 6.9 Ohms for the reactive parts! That comes to .146 Ohms for the 470uF cap, so it's ESR is probably a fair percentage of it's Z (explaining it even working) and the remaining negative capacitance (inductance) will drive the frequency up.

12. Oct 17, 2014

### Staff: Mentor

Are you able to connect an oscilloscope to the output of the OP-AMP and describe the waveform you see?

13. Oct 17, 2014

### zoki85

Output is actually input. Normally, point "6 to ground" should be considered output, no?

14. Oct 17, 2014

### tuttyfruitty

Once I get the chance I will connect a scope up to the output.

Mike,

Just trying to get my head around your answer, from what I have read about negative capacitors, a decrease in voltage across the cap will create a increase in current . As you have worked out the resonant frequency of the tank circuit is 2.321K Hz, as soon as the feedback cap is in place the output (pin 3) changes frequency to 2.174 K Hz . My understanding is that the circuit needs to "make up" for the loses of the tank circuit which as you have calculated is 6.85 Ohms where as the reactance of the feedback cap is only 0.146 Ohms.

when I replace the feedback cap with a variable resistor the circuit oscillates when the feedback resistor is about 3.4 Ohms From my understanding, as the negative feedback part of the system causes a gain of 2 the feedback resistor needs to be half the impedance of the tank circuit.

Using that as an example I would say that the reactance of the feedback cap needs to be 3.4 Ohms. Is this saying that impedance of the cap Is made up of 0.146 Ohms of reactance and 3.2542 Ohms of ESR.

I am usual "bulk standard" electrolytic caps.

15. Oct 17, 2014

### zoki85

As far as I can see the connection diagram, you don't need to introduce "negative capacitor" to explore resonant frequencies.
Op-amp has huge input impedance, and the network can be treated as passive one for that.

16. Oct 17, 2014

### Mike_In_Plano

Cannot say what the loss impedance is from the given info. The Z numbers correspond to loss-less inductor and capacitor. I would need the Q of the inductor at that frequency, and the ESR of the 10uF at that frequency to know the loss.

17. Oct 26, 2014

### Staff: Mentor

Awaiting the result with eager anticipation....