Oscillator/filter frequency. Can someone offer a formula?

[brett]

Hi.
I'm using a simple oscillator based on a combination of RG Keen's and Bill Bowden's designs.  They use filters to produce positive feedback at a particular frequency, which quickly becomes the oscilliation frequency.

Here's a graphic of it (on the right) from Bill's website:? 

Funny thing is that when I built it, the frequency was quite different to what I expected.  It was lower by about 4 times.  With 0.1uF caps and 2k2 resistors the frequency was about 170 Hz, which is close to 1/(8.pi.R.C), but nowhere near the 1/(4.pi.R.C) that I expected.  ??

Anybody want to explain this or suggest a formula for the resonant frequency ?

PS I'll post my take on this circuit in a couple of days.

[gez]

tolerances, hidden capacitance of semiconductors, loading?

http://www.aikenamps.com/PhaseShiftOscillators.html

PS  Wrong value parts? 

[brett]

Thanks Gez
The link shows this oscillator:

and gives the formula f = 1/(2.pi.R.C.squareroot(6))
I presume that the squareroot 6 is because there are 3 phase shifts for 180 degrees in this oscillator? (my circuit has 2 shifts).
I'll either fiddle with my circuit and either work out the answer empirically or dig into the math at the link to work out the formula for a 2 stage oscillator.

cheers

[bioroids]

I think you're supposed to use all three stages to get a proper 180º  phase shift.
I'm not sure exactly why, something had to do with reaching the 180º asymptotically.

What is the schem you are using?

Miguel

[brett]

Doh!
Of course!  The filter I'm using is the equivalent of the one top, right.  It has THREE stages, not two, because the input resistance of transistor forms a third filter.  That's why my results looked strange.  I'll go back and calculate the base resistance of my BJT and make the other resistances roughly equal to that.

PS I now realise that you can't form an oscillator with two single pole filters because the maximum phase shift possible in each stage is 90 degrees and that occurs at 0 Hz.

[gez]

PS I now realise that you can't form an oscillator with two single pole filters because the maximum phase shift possible in each stage is 90 degrees and that occurs at 0 Hz.

You can, sort of.  I'll answer you later (got to go out soon) and post a schemo...it's related to this type of circuit.

[gez]

I'll go back and calculate the base resistance of my BJT and make the other resistances roughly equal to that

The schematic on the right, the way it's biased it has low input impedance.  It's more usual (well, in text books) to use divider bias for higher input impedance then calculate its equivalent resistance and use this for the phase shift network.

I have a full analysis of this type of filter, but it doesn't explain intuitively what's going on (though PM me if you want it).  All you need to know is the amp provides a 180 degree shift and the network a further 180 degrees (60 degrees per leg).  Amp needs a min gain of 29.

The other oscillator I mentioned comes form a National (if I recall) app note.  Outlined in this thread (read through it and you'll get all the info).

http://www.diystompboxes.com/smfforum/index.php?topic=26740.0

Formula:

1/(2.pi.C.square root R1.R2) 

where R1 is resistor to ground and R2 the resistor in the feedback loop.

The circuit can easily be run at 9V without the esoteric chip, but the component values seem picky for some reason (stick in your own values and it doesn't always work...weird).  The app note (full title mentioned in the thread) gives more details.

[A.S.P.]

http://www.national.com/ms/LB/LB-16.pdf

[brett]

Thanks everyone.
Yes, for most BJTs (hFE 250 to 500) the input impedance is low because of the very low emitter current (controlled by that 470k bias resistor).

[gez]

Thanks everyone.
Yes, for most BJTs (hFE 250 to 500) the input impedance is low because of the very low emitter current (controlled by that 470k bias resistor).

Not what I meant Brett, it's the Miller effect on the 470k bias resistor that creates the low input impedance (shunts current past the input).  Think of it as the feedback resistor in an op-amp and you'll see why it's low.  BJTs don't have the 'infinite gain' of op-amps, so the resistor has some effective resistance, but it's still low and needs to be taken into consideration.

[George Giblet]

Here's the problems:

The phase shift ideally consists of three time constants formed by three RC networks.

If you take the second circuit.  Two of the R's which form the RC networks are the 7.5k resistors.  The corresponding C's are the C's preceeding the R's.  When you get to the last stahe you have a C but there is no corresponding R.  The R is formed by the input impedance of the transistor stage.   That's where gez is up to. 

The gain of that stage is about 30.  So if you use the Miller effect concept the input impedance caused by the 470k resistor is 470/30 ~ 16kohm.  The input impedance isn't just Miller effect, there's also the input impedance of the transistor itself.  From my rough calculations the  emitter resistor needs to be around 400ohms.  The transistors internal emitter resistance is about 85 ohms   If you assume the gain of the transistor is say 200 the input impedance will be around 100khm.  So 100kohm//16kohm = 13.8kohm.

So the third stage has an effective RC network of 50nF and 13.8k.   

The second problem is the transistor stage has an output impedance (It will be in the order of ~2k).  This resistance appear in series with the first capacitor and the effect is it modifies the behaviour of the phase shift network.  Obviously adding series resistance to the circuit will increase the time constants and decrease the frequency.

The fact the third stage is 13.8k and not 7.5k also causes the frequency to drop ie. the third R is different.

The transistor output impedance and the non-equal R's means the frequency will be lower than usual frequency equation (f = 1/(2 pi sqrt(6) RC) implies.

These non ideal conditions are neither good or bad.  All it means is the equation doesn't work because it's or the right equation forthe job.  Nonetheless the equation will give you a ball park answer.

The third problem is you have changed the R's to 2.2k.  This will cause some issues.   The 2k or so output impedance of the transistor stage is now loaded quite heavily buy the 2k2 resistors.  Also the effective series 2k will have a bigger effect on the network time constants since it's much closer to the R's in the circuit - this means the frequency will even further deviate from the equation.  When you deviate the circuit from the ideal this much you can run into trouble.

You probably don't want to change the R's too much from 7.5k.

As a side issue there are many deviations on the phase shift oscillator.  If you take the first circuit with the opamp it uses a different network (C's to ground instead of R's to ground).  This has a different formula.   Also with that particular circuit the last stage is heavily loaded by the 7.5k input impedance to the opamp, which modifies the cirucit time constants, and that  means even the correct formula won't give the right answer.