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### Network # Different Phase Shift Oscillator

T

Jan 1, 1970
0
P

#### Phil Allison

Jan 1, 1970
0
"The Phantom"
"Phil Allison"
Then let me put it this way. Is the author correct that the criterion
for oscillation is RF > 12R?

** Probably.

But the link does not say what YOU claimed.

And if another CR section is added, what will
be the criterion for oscillation?

** RF will need to be bigger.

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

Cos that one is crap.

......... Phil

T

#### The Phantom

Jan 1, 1970
0
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?

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

J

Jan 1, 1970
0
T

#### The Phantom

Jan 1, 1970
0
"The Phantom"

** The link does not say what you claim.

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?

T

#### The Phantom

Jan 1, 1970
0
"The Phantom"
"Phil Allison"

** Probably.

But the link does not say what YOU claimed.

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.

T

#### The Phantom

Jan 1, 1970
0
I would recommend that you use Rf >= 12.2*R as a minimum.

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

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:

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

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.

P

#### Phil Allison

Jan 1, 1970
0
"The Phantom
I didn't say that it did,

** " Here is one that only needs a gain of 12

Is the author correct about the required gain?"

That is a claim that YOU posted that is NOT in the link.

Don't be such a PITA, fucking LIAR.

........... Phil

P

#### Phil Hobbs

Jan 1, 1970
0
The said:
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?

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

H**N(f) = 1/(1+ (omega RC)**2)**(N/2) <) -N atan(omega RC).

For a large number of sections and omega << 1/RC, the phase angle is -N
omega RC. Setting that as pi, we get the gain at f.pi:

H**N(f.pi) = -1/(1+ (pi/N)**2)**N/2 which in the limit N-> infinity is
-1.0.

So if you ignore loading, you can make a phase shift oscillator with any
gain above 1 (see our earlier thread about RC networks with gain).

Cheers,

Phil Hobbs

J

#### John Popelish

Jan 1, 1970
0
The said:
For this case, the exact result I get is Rf > 56*R/9 = 6.222222*R

Are you getting these numbers from simulation?

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.
Compare this circuit to the one in:

http://en.wikipedia.org/wiki/Phase_shift_oscillator
Thanks.

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.

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.
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.

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

T

#### Tam/WB2TT

Jan 1, 1970
0
The Phantom said:
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?
How do you get 29? For a single RC network the loss at 60 degrees is about
2. So, you would need a gain of 2**3=8.

Tam

T

#### The Phantom

Jan 1, 1970
0
How do you get 29? For a single RC network the loss at 60 degrees is about
2. So, you would need a gain of 2**3=8.

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.

T

#### Tam/WB2TT

Jan 1, 1970
0
The Phantom said:
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

T

#### The Phantom

Jan 1, 1970
0
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.

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

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.

T

#### The Phantom

Jan 1, 1970
0
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.

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

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

F

#### Fred Bloggs

Jan 1, 1970
0
The said:
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?

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.

T

#### The Phantom

Jan 1, 1970
0
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,

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".

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