Different Phase Shift Oscillator

The usual phase shift oscillator with 3 CR sections needs a gain of 29 to oscillate. Here is one that only needs a gain of 12:

Is the author correct about the required gain?

How much gain would be needed if there were 4 CR sections?

"The Phantom" "Phil Allison"

** Probably.

But the link does not say what YOU claimed.

** RF will need to be bigger.

Go find a site that has the WHOLE story about such oscillators.

Cos that one is crap.

........ Phil

This last question may seem ambiguous. What I mean is to add one more CR section to the schematic given in the .pdf file.

>

I would recommend that you use Rf >= 12.2*R as a minimum.

By experiment, Rf>=6.27*R sustains oscillation, with perfect parts.

Then let me put it this way. Is the author correct that the criterion for oscillation is RF > 12R? And if another CR section is added, what will be the criterion for oscillation?

I didn't say that it did, Phil. My statements were based on my own point of view about voltage to current conversion followed by current to voltage conversion.

For this case, the exact result I get is Rf > 56*R/9 = 6.222222*R

Are you getting these numbers from simulation?

Compare this circuit to the one in:

Leaving out R1 makes a substantial change in performance.

Consider the passive phase shifting network in the .pdf file. The signal applied to the node labelled Vi is converted to a current Is at the summing junction of the op amp, with a transfer ratio of Is/Vi amps per volt. The passive network is acting as a voltage controlled current source into the summing junction. The op amp is acting as a current controlled voltage source and must supply a transfer ratio of -Vo/Is volts per amp to sustain oscillation. So, the magnitude of the op amp's transfer ratio Vo/Is must be equal to the passive network's transfer ratio Vi/Is, which for this circuit is 12 volts per amp when the R's and C's are all unity, whereas for the one given on the Wikipedia page it's 29.

There was a thread some years ago where somebody wanted to use a bipolar transistor as the active device and get a phase shift oscillator to work at a very low supply voltage.

Usually when the phase shift oscillator is shown with a bipolar (in descriptions found on the web), the load impedance of the transistor and its bias network isn't dealt with properly. With a FET, one could assume the input of the FET is very high impedance, probably negligible with respect to the network.

With a bipolar, I wonder if it might be more suitable to use this topology and just design for a low input impedance, almost a summing junction. Don't have an (un-bypassed) emitter resistor and set the gain with a resistor from the collector to the base, with appropriate biasing arrangements. Could this oscillate with even lower supply voltages than the standard topology?

This same basic topology can also oscillate if the resistors are replaced with capacitors, and the capacitors with resistors. The feedback resistor, Rf, must become a capacitor, because it obviously can't be an oscillator with only two capacitors.

It seems that this topology might be able to oscillate with less "hot" bipolar transistors.

It's well known that the standard 3 section topology (with equal R's and C's), voltage in and voltage out, needs a voltage gain of 29. Sometimes a

4 section version is given, and this version needs a voltage gain of 901/49 = 18.3878. With even more sections, the required gain continues to drop. A 5 section version needs a gain of 15.4354 and a 6 section version needs a gain of 14.1179. Tapering impedances can reduce the required gain even more.

But the topology shown in the referenced .pdf, with equal R's and C's, only needs a "gain" of 12 with 3 sections, and only 6.2 with 4 sections. I wonder what can be done with modest tapering of impedances.

As long as the loading remains small, N cascaded RC sections give

H**N(f) = 1/(1+ (omega RC)**2)**(N/2)

Yes. I threw it into LTspice with an initial condition that bounced it into oscillation and stepped the feedback resistor through a small range to see where the amplitude just barely grew.

Thanks.

I have the feeling that the original circuit could also benefit from differing R and C values in the phase shift network, but had no luck with just guessing changes that would increase the loop gain. I have to get past the mental block about current versus voltage output for the network.

How do you see the ratios of feedback to grounding resistance as gain. It is just an arbitrary resistor ratio.

How do you get 29? For a single RC network the loss at 60 degrees is about

1. So, you would need a gain of 2**3=8.

Tam

That's only if the successive CR sections are separated by unity gain buffers. Otherwise, the later sections load the earlier ones and that changes the attenuation and phase shift.

Yeah, figured that out. The gain of 29 only holds if the three sections are identical. If the succeeding sections are higher impedance then you need less gain. (per SWCAD). I have seen the number 8 quoted in the literature, must have used 2 emitter followers.

Tam

From left to right, label the first section as C2-R2, the second as C1-R1 and the final cap as C. Define a non-existent resistor as R (to be used in the math). Select a value for R and C, and set C2 = 9*C, R2 = R/9; set C1 = 3*C, R1 = R/3. Then the condition for oscillation is Rf > 3.852*R. This is with scaling of 9:3:1.

Similarly, with scaling of 100:10:1 the condition for oscillation is Rf >

2.442*R.

If you have a voltage controlled voltage source (just a voltage amplifier), with input Vi and output Vo, the ratio Vo/Vi volts per volt is called the voltage gain, or just "the gain". Given a current controlled current source, the ratio Io/Ii amps per amp is called the current gain, or just "the gain" when the parties to the discussion understand that it's a current amplifier.

What if you have a current controlled voltage source with a transfer ratio of Io/Vi amps per volt? Is not the transfer ratio a kind of "gain"? Anyway, that's what I mean by "gain".

And, in fact, the simple inverting amplifier that you would have if the phase shifting network in the referenced circuit were replaced by a single resistor, Ri, between Vi and the minus input of the opamp would have a voltage gain given by -Rf/Ri, an arbitrary resistor ratio.

The standard voltage in, voltage out, equal R and C, 3-section phase shifting network has a "gain" (voltage transfer ratio) at the frequency of oscillation of 1/29. This "gain" doesn't depend on the impedance level of the network; the output voltage of the (unloaded) network is always 1/29 of the input voltage at the frequency of oscillation.

The network under discussion:

has the property that the current delivered to the minus input of the opamp, at the frequency of oscillation, varies with impedance level. If the current delivered to that node is called Is, then the ratio Is/Vi is

1/(12*R), and the "attenuation", if you will, of the network increases with impedance level. This "attenuation" must be made up by the opamp, and we see that as R gets larger, then Rf also becomes larger.

If you consider the "canonical" network, with R and C equal to unity, then that number 12 in the 1/(12*R) expression becomes the current to voltage transfer ratio of the (opamp + feedback resistor), which is acting as a current controlled voltage source.

The dependence on impedance vanishes if you take the ratio Rf/R, so this is a fundamental property of the circuit. To repeat, it's the "gain" (current to voltage transfer ratio) of the (opamp + feedback resistor) circuit with unity R's and C's.

I probably should have explained this point of view in more detail in the first post, but I had been playing with the circuit for a while and it had become familiar to me.

Have a look at the expression for the feedback resistor at:

That's a nice well-written little tutorial. There's nothing new about that particular topology, you can find it many linear op amp application notes dating back 35 years, and it is the one most used because the DC output level is easily set and common mode response is eliminated. In fitting this particular arrangement to the ideal Barkhausen block diagram, the op amp portion is broken into two parts, the differentiator with gain is included in the Barkhausen feedback block along with the RC's, and the low output impedance op amp drive is accounted for by an ideal virtual unity gain non-inverting buffer. You might look at sensitivity of wo with respect to component values before you add additional sections, and you would not the op amp GBW to figure into the oscillation frequency with any significance.

Can you cite a manufacturer's AN #? In particular, do you know of any that provide analyses of the factors you mention? Usually, they just give a circuit and expressions for required gain and frequency of oscillation.

I've dug a couple of interesting old papers on the topic and posted them over on ABSE, under the subject "PSO paper #x".