Designing Phase Shift
Oscillators for Tremolo Circuits

What is a phase shift oscillator?

"Phase shift oscillator" is the term given to a particular
oscillator circuit topology that uses an RC network in the feedback loop of a
tube, transistor, or opamp to generate the required phase shift at a
particular frequency to sustain oscillations.  They are moderately stable
in frequency and amplitude, and very easy to design and
construct.
Where are they used?

Phase shift oscillators are most commonly used in tremolo circuits
in guitar amplifiers. They are used as the low-frequency oscillator (LFO)
that generates the sinusoidal waveform which amplitude modulates the guitar
signal to produce the characteristic tremolo amplitude
variations.
How do they work?

In order to create and sustain an oscillation at a particular
frequency, a circuit must have a gain higher than unity, and a total phase
shift around the loop of 360 degrees (which is equivalent to 0 degrees, or
positive feedback).  When used with a single-stage inverting
amplification element, such as a tube, transistor, or inverting opamp
configuration, the amplifier itself provides 180 degrees of phase shift (a
gain of -A, where A is the gain of the amplification stage). The remaining 180
degrees of phase shift necessary to provide a total of 360 degrees is provided
by an external network of resistors and capacitors.

Following is a schematic diagram of a typical phase shift oscillator:

Phase Shift Oscillator

The triode is configured as an inverting amplifier to provide the necessary
gain, and the feedback network is connected from the plate to the grid.

The phase shift elements are C1/R1, C2/R2, and C3/R3.  Three of these
phase lead1 networks contribute a total of 180 degrees of phase
shift at the oscillation frequency.  Note that a phase shift oscillator
could also be built using four or more phase shift elements, with each element
contributing less overall phase shift at the oscillation frequency.
Normally, there is no need to do this, as it takes extra components.  A
minimum of three phase shift networks is required, however, because the
maximum theoretical phase shift available from any one RC network is 90
degrees, and the actual phase shift approaches this value asymptotically.

A phase shift oscillator can also be made using three phase lag networks,
which are obtained by swapping the positions of the R and C value components
in the above schematic.  The lag network would require one additional
coupling cap to block the DC on the plate voltage from the grid, and one
additional resistor to provide the grid bias ground reference for V1A, so it
is not normally used.

Following is an example of both a phase lead and a phase lag network,
designed for a 45 degree phase shift at the -3dB point of f = 1/(2*Pi*R*C) =
1/(2*Pi*1Meg*.01uF) = 15.9Hz:

Phase Lead
Network
Phase Lag Network

Following is a plot of the phase shift and attenuation characteristics of
the phase lead and phase lag networks:

(click on image for larger view)

As can be seen from the plot, the phase lead network starts at near +90
degrees at 0.1Hz, and shifts through +45 degrees at the -3dB point of 15.9Hz,
continuing on toward 0 degrees above 1kHz. The phase lag network, on the other
hand, starts at 0 degrees, shifts through -45 degrees at the -3dB point, and
continues on towards -90 degrees above 1kHz.  Either one will provide an
effective 0 degrees phase shift when three of them are combined with the 180
degree phase shift of the amplifier as shown in the phase shift oscillator
schematic.

It can be shown2 that the attenuation of the phase shift
elements in the feedback loop is 1/29, so the oscillator will oscillate if
the amplifier gain is greater than 29 (which  will bring the overall loop
gain above unity gain, and satisfy the gain criterion for oscillation).
The oscillations will occur at a frequency given by the following equation:

fo = 1/(2*Pi*Sqrt(6)*R*C)

In order to obtain the lowest distortion for the best sine wave, the
amplifier should be operated with a gain of exactly 29, which is just the
bare minimum necessary to sustain oscillation.  This will produce the
purest sine wave, however, it is impractical if tubes of varying gains may be
substituted (this usually requires an adjustment control to trim the gain), or
if the frequency of oscillation must be adjusted in such a manner as to change
the gain of the network.  For these reasons, the gain is usually made
higher, and post-filtering of the waveform is done to remove unwanted harmonic
distortion.

If four phase lead networks are used, the phase shift per section at the
oscillation frequency is lower, therefore, the attenuation of the network is
also lower, around 1/18.  This allows use of lower gain tubes if
necessary, since the gain of the amplifier only has to be at least
18.

The design procedure

Following are the steps necessary to design a vacuum tube phase shift
oscillator.  For this example, we will assume that it is desired to make
a three-stage vacuum tube phase shift oscillator with a nominal oscillation
frequency of 7Hz.

Amplifier design:

An amplifier with a gain of greater than 29 must first be
designed.  This requirement will affect the type of tube
chosen.  For example, a 12AU7 will not work in this circuit, because
it has an amplification factor of only 17, which is lower than the minimum
gain of 29.  Note that the actual gain obtained from a tube is always
considerably less than the amplification factor, which is the theoretical
maximum obtainable from the tube, and then only with an infinite load
impedance (a current source load is a good approximation to this).
Typically, a 12AX7, with an amplification factor of 100, will achieve a
gain of around 60 or higher with a resistive plate load, when the cathode
is fully bypassed. The gain of an unbypassed 12AX7 can also be greater
than 29, so you can design the circuit without the bypass capacitor and it
will work fine. as long as plate and cathode resistors are calculated to
provide the required gain.  This will not the case with lower-gain
tubes, where you will still need the bypass capacitor to achieve the
require gain of 29.   For completeness, this example shows
bypass capacitor calcuation, even though a 12AX7 is
used.

Assuming a 12AX7 is chosen, following are the calculations for the
gain and output impedance:

Some typical 12AX7 numbers:

plate
resistance:
ra = 62.5K
amplification
factor:
mu = 100

A plate load resistance of 100K and a cathode resistance of 820 ohms
are chosen to provide good gain and linearity. In order to maximize the
gain and minimize the output impedance, the cathode must be fully-bypassed
at the frequency of oscillation. The midband gain for the 12AX7 with a
fully-bypassed cathode is:

Av = (mu*Rl)/(Rl+ra)
=
(100*100K)/(100K+62.5K)
= 61.5

Where:

Rl = the load resistor (100K in this example)
ra = the
internal plate resistance (62.5K for a typical 12AX7)
mu = the mu
of the tube (100 for a typical 12AX7)

The calculated gain of 61.5 is higher than the required minimum
gain of 29, so this amplifier will work in the phase shift oscillator
circuit.

Next, the output impedance of the stage is calculated. Since the
internal plate resistance is effectively in parallel with the plate load
resistor, the output impedance (for the signal taken off the plate) will
be:

R = ra || Rl
= 62.5K || 100K

= 38.5K

Note: the symbol "||" means "in parallel with".  Resistors in
parallel add in reciprocal, i.e. 1/Rt = 1/R1 + 1/R2.

Next, the resistance seen looking into the cathode must be calculated,
as it is needed to determine the value of bypass capacitor needed.

The resistance seen looking into the cathode ( with Rk
unbypassed) is:

Rk' = (Rl+ra)/(mu+1)
=
(100K + 62.5K)/(101)
=
1.6K
Therefore, the total cathode resistance is the parallel
combination of the cathode resistance, Rk', and the cathode resistor,
Rk, as below:

R  = Rk' || Rk
= 1.6K || 820

= 542 ohms

Next, the value of the cathode bypass capacitor must be
calculated.  Since the gain of the amplifier must be maximized at the
frequency of oscillation, the minimum capacitor value is chosen using the
following equation for the -3dB point of the amplifier.

f = 1/(2*PI*R*C)
solving for C:

C= 1/(2*PI*R*f)

The minimum value of bypass capacitor is therefore:

C= 1/(2*PI*542*7Hz) = 42uF

The amplifier gain will be down -3dB at this point, corresponding
to a gain decrease of 0.707 times 61.5, or 43.5, which is still well
above the required minimum of 29.  However, in order to maximize gain
and keep the phase shift associated with the cathode bypass capacitor to a
minimum, the capacitor value should be increased to around five to ten
times the minimum calculated value, such as 470uF, giving a -3dB point of
0.62Hz.

The bypass capacitor to achieve high gain at very low frequencies can
get quite large if with small values of  the cathode resistor. With
the 12AX7, there is enough excess gain that the circuit will still work,
even at very low frequencies, but it may be a problem with other lower
gain tubes.

Phase shift network component
selection:

Since the impedance is proportional to the shunt element in the phase
shift network2, in this case, the resistor, a suitable
impedance value must first be chosen.   The input impedance of
the network  must be large in comparison to the output impedance of
the amplifier, so as to not load the output appreciably, which would
reduce the gain, possibly to a point where it can no longer sustain
oscillations.  A good minimum value is around ten times the actual
output impedance of the amplification stage.  Since the input
impedance is proportional to the shunt element, and is approximately twice
the value of the shunt element at the oscillation frequency, the
resistance can be chosen to be around half the required impedance. This
resistance will then determine the value of capacitor necessary to achieve
the desired frequency of oscillation.  Since we have an output
impedance of 38.5K in our example, a good minimum value for the input
impedance is ten times this value, to prevent loading of the output
stage.  Since the resistance value to achieve this impedance is
around half the total impedance, a value of five times the output
impedance, or 5*38.5K = 193K, will work.

Next, the capacitance value is calculated using the formula for the
frequency:

fo = 1/(2*Pi*Sqrt(6)*R*C)

solving for C:

C = 1/(2*Pi*Sqrt(6)*fo*R) = 1/(2*Pi*2.45*7*193K) =
.048uF  (use .047uF as the nearest smaller standard value)
Since
capacitor values are more commonly available in 10% values, and resistors
are more commonly available in 5% or even 1% values, the resistor value
should be recalculated based on the standard capacitor value chosen.

solving for R:

R = 1/(2*Pi*Sqrt(6)*fo*C) = 1/(2*Pi*2.45*7*.047uF)
=  198K  (use 200K as the nearest larger standard value)

When choosing the capacitor, it is best to choose the next smaller size
because this will make the input impedance larger, because it will
require a larger resistance to achieve the desired frequency.
Likewise, when choosing the resistance, it is best to choose the next
larger size, as this will also increase the input impedance of the phase
shift network.

The design is then built and tested, and resistor or capacitor values
are trimmed as necessary to provide the exact frequency of oscillation
desired. Following is the output of the completed oscillator using the
values calculated above:

(click on image for larger view)

A small amount of harmonic distortion is present, as
evidenced by the "kink" at the bottom edges of the sine wave.  This
can be eliminated by filtering the output slightly, either with a post RC
filter, or by adding a capacitor across the plate load resistor to act as
a first-order lowpass filter to reduce a bit of the distortion.  The
filter cutoff frequency cannot be made too low at the plate resistor, or
the gain will be reduced below the level necessary to sustain
oscillation.   A good first choice is to select a capacitor that
will produce a -3dB point around 3 times the oscillation frequency.
This can be calculated using the output impedance of the stage as follows:

C = 1/(2*Pi*3*fo*R) = 1/(2*Pi*3*7*38.5K) =
0.197uF  (use 0.2uF as the nearest standard value)

If the starting gain of the amplifier is too low, this extra filter
may lower the gain too much to sustain oscillations, in which case it
should be increased, or an RC post filter should be used after the
oscillator.

The final schematic and output plot are shown below:

Completed Phase Shift Oscillator Design

(click on image for larger view)

Note the
harmonic distortion has been reduced significantly in the above output, at
the expense of some amplitude reduction.  The filter capacitor value
could be reduced to the point of just cleaning up the distortion to a
satisfactory level for more gain, if desired.

Design modifications for a tremolo
oscillator

There are several factors that must be taken into consideration when
designing a phase shift oscillator for use as a tremolo oscillator in a
guitar amplifier. They are:

A tremolo circuit must have a variable frequency oscillator, which is
adjustable by a single potentiometer.  The oscillator needs to have a
relatively wide frequency variation (typically from around 2Hz to 8Hz), the
need to maintain a relatively constant amplitude over the entire adjustment
range of the pot, and the need to keep the potentiometer value low enough to
be able to use standard, off-the-shelf parts.

The ideal method of adjusting the frequency is to use a triple pot, to
control all three phase shift sections. This is not always practical, so
only one section is usually adjusted.  It is usually best to adjust
the last phase shift section, rather than the first one after the
amplifier, as it will usually afford a wider range of control.

The requirement for lower value resistors forces the design to use
larger capacitors, which have lower reactance at the frequency of
oscillation. Because of this,  it is better to design for a lower
output impedance by using a tube with a lower internal plate resistance, or
sacrifice some gain by using lower value plate resistances.  This may,
however, require a larger cathode bypass capacitor to maintain gain at lower
frequencies.

When using the phase lead network, the impedance of the phase shift
network is proportional to the value of R, so it should be made as large as
practical, relative to the output impedance of the amplifier. This will
minimize the gain variations as the frequency is adjusted, however, this is
in contrast to the need to keep the potentiometer and resistance values
low,  so the best overall solution is to use the lowest practical
output impedance.

The design procedure

As in the above fixed-frequency design example, the first step is to
design a suitable amplifier.  For this example, we will assume a
frequency adjustment range of 2Hz to 8Hz is desired. Note that the difference
between the minimum frequency selected is very important at these low
frequencies, because it determines the adjustment range of the
oscillator.  For example, a range from 1Hz to 7Hz requires an oscillator
that can be adjusted over a 7:1 range, while a range of 2Hz to 8Hz only
requires an oscillator that can be adjusted over a 4:1 range, even though the
total frequency difference is the same.  This makes a big difference in
the design of the frequency adjustment portion of the oscillator.

Amplifier design:

An amplifier with a gain of greater than 29 must first be
designed.  The requirement for wide range adjustability with low
value potentiometers requires a low plate resistance tube. The 12AT7 is
chosen for this example.

Some typical 12AT7 numbers:

plate
resistance:
ra = 10.9K
amplification
factor:
mu = 60

A plate load resistance of 47K is chosen.  Plotting this load
line on the characteristic curves of a 12AT7 at 300V, results in a
required grid bias point of around -3V/2.5mA, so a resistor of 3V/2.5mA=
1.2K is chosen.  In order to maximize the gain, the cathode must be
fully-bypassed at the frequency of oscillation. The midband gain for the
12AT7 with a fully-bypassed cathode is:

Av = (mu*Rl)/(Rl+ra)
=
(60*47K)/(47K+10.9K)
= 48.7

Where:

Rl = the load resistor (47K in this example)
ra = the internal
plate resistance (10.9K for a typical 12AT7)
mu = the mu of the
tube (60 for a typical 12AT7)

The calculated gain of 48.7 is higher than the required minimum
gain of 29, so this amplifier will work in the phase shift oscillator
circuit.

Next, the output impedance of the stage is calculated. Since the
internal plate resistance is effectively in parallel with the plate load
resistor, the output impedance (for the signal taken off the plate) will
be:

R = ra || Rl
= 10.9K || 47K

= 8.85K

Next, the resistance seen looking into the cathode must be calculated,
as it is needed to determine the value of bypass capacitor needed.

The resistance seen looking into the cathode ( with Rk
unbypassed) is:

Rk' = (Rl+ra)/(mu+1)
=
(47K + 10.9K)/(61)
= 949
ohms
Therefore, the total cathode resistance is the parallel
combination of the cathode resistance, Rk', and the cathode resistor,
Rk, as below:

R  = Rk' || Rk
= 949 || 1.2K

= 530 ohms

Next, the value of the cathode bypass capacitor must be
calculated.  Since the gain of the amplifier must be maximized at the
lowest frequency of oscillation, the minimum capacitor value is chosen
using the following equation for the -3dB point of the amplifier.

f = 1/(2*PI*R*C)
solving for C:

C= 1/(2*PI*R*f)

The minimum value of bypass capacitor is therefore:

C= 1/(2*PI*530*2Hz) = 150uF

The amplifier gain will be down -3dB at this point, corresponding
to a gain decrease of 0.707 times 48.7, or 34, which is still above the
required minimum of 29.  However, in order to maximize gain and keep
the phase shift associated with the cathode bypass capacitor to a
minimum, the capacitor value should be increased to around five to ten
times the minimum calculated value, such as 820uF, giving a -3dB point of
0.62Hz.  Note that the required capacitor value can get rather large
if low frequency oscillators are designed with tubes that have low gains
and low internal plate resistances.

Phase shift network component
selection:

Since the impedance is proportional to the shunt element in the phase
shift network2, in this case, the resistor, a suitable
impedance value must first be chosen.   The input impedance of
the network  must be large in comparison to the output impedance of
the amplifier, so as to not load the output appreciably, which would
reduce the gain, possibly to a point where it can no longer sustain
oscillations.  A good minimum value is around ten times the actual
output impedance of the amplification stage.  Since the input
impedance is proportional to the shunt element, and is approximately twice
the value of the shunt element at the oscillation frequency, the
resistance can be chosen to be around half the required impedance. This
resistance will then determine the value of capacitor necessary to achieve
the desired frequency of oscillation.  Since we have an output
impedance of 8.85K in our example, a good minimum value for the input
impedance is ten times this value, to prevent loading of the output
stage.  Since the resistance value to achieve this impedance is
around half the total impedance, a value of five times the output
impedance, or 5*8.85K = 44.3K, will work.

Next, the capacitance value is calculated using the formula for the
frequency:

fo = 1/(2*Pi*Sqrt(6)*R*C)

solving for C, and using the value of the highest frequency (8Hz)
in the oscillator range:

C = 1/(2*Pi*Sqrt(6)*fo*R) = 1/(2*Pi*2.45*8*44.3K) =
.183uF  (use 0.18uF as the nearest smaller standard value)
Since
capacitor values are more commonly available in 10% values, and resistors
are more commonly available in 5% or even 1% values, the resistor value
should be recalculated based on the standard capacitor value chosen.

solving for R:

R = 1/(2*Pi*Sqrt(6)*fo*C) = 1/(2*Pi*2.45*8*.18uF) =
45K  (use 47K as the nearest larger standard value)
When
choosing the capacitor, it is best to choose the next smaller size because
this will make the input impedance larger, because it will require a
larger resistance to achieve the desired frequency.  Likewise, when
choosing the resistance, it is best to choose the next larger size, as
this will also increase the input impedance of the phase shift network.

Next, in order to vary the frequency, the resistance must be varied
over a range equivalent to the frequency range required.  In this
example, the range must be 4:1, so a triple pot of 4*47K, or 188K is
required, so a 200K is chosen as the the nearest standard value.  The
triple pot is chosen in this example because it provides the best range of
adjustment.

Following is the schematic of the finished
circuit:

Completed Variable Phase Shift Oscillator Design

This design, using standard values, ended up with a frequency adjust
range from 1.5 Hz to 7.3Hz.  It can be trimmed by decreasing the
capacitor value from 0.18uF to 0.15uF to achieve the original
specification of 2Hz to 8Hz. If desired, a lowpass filter can be made
using a capacitor across the 47K plate resistor as demonstrated in the
fixed frequency design, but it will only be effective at the higher
frequencies, as it would require a variable cutoff frequency filter to
properly filter over the entire adjustment range of the oscillator.

Design
considerations for using a single pot to control frequency

It is desirable from a parts cost and availability standpoint to be able
to use a single pot to control frequency instead of a triple pot.  While
this will cause a moderate amount of amplitude variation with frequency, this
is not too much of a problem with a guitar amplifier because it is natural to
adjust the speed and intensity interactively until the desired effect is
achieved, as long as the variation isn't too noticeable.  The main design
considerations are:

The amplifier needs to have higher gain than the nominal gain of
29.  This is to account for gain variations as the impedance variation
and gain of the network is greater with the single pot adjustment.

The phase shift network should be designed for a frequency around half
the maximum, and the pot should be used to set gains above and below the
nominal frequency by using a smaller series frequency limit resistor than
the calculated nominal value for the half frequency.

The pot should have a range of no more than five times the nominal
resistance value, otherwise the oscillations may stop at higher
settings.  The amount of frequency variation is limited to around 3:1
or so.

The design procedure

As in the above fixed-frequency design example, the first step is to
design a suitable amplifier.  For this example, we will assume a
frequency adjustment range of 2Hz to 6Hz is desired. This is in keeping with
the approximate 3:1 range limit imposed by the single-pot design.

Amplifier design:

Since we need an amplifier with excess gain, we will use the
12AX7 amplifier design from the first fixed-frequency example. However,
since we want a nominal frequency of around 4Hz, the phase shift components
will be recalculated as follows:

Phase shift network component
selection:

The impedance of the phase shift network will remain the same as
previously calculated, so the originally calculated resistor value of 200K
will be used.

Since the frequency will be varied over a range lower than the
original design spec of 7Hz, the cathode bypass capacitor value should be
increased from 470uF.

The minimum value of bypass capacitor is therefore:

C= 1/(2*PI*542*2Hz) = 147uF

As indicated in the original design, this value of capacitor will
result in the amplifier response being down -3dB at the lowest
oscillation frequency of 2Hz.  It is best to increase the capacitor
value by a factor of five to ten to avoid the gain loss at the lower
frequency range.  A good compromise value between size and frequency
response is 820uF, giving a -3dB point of 0.4 Hz.

Next, the capacitance value is calculated using the formula for the
frequency:

fo = 1/(2*Pi*Sqrt(6)*R*C)

solving for C, and using the value of the nominal frequency (4Hz)
in the oscillator range:

C = 1/(2*Pi*Sqrt(6)*fo*R) = 1/(2*Pi*2.45*4*200K) =
.081uF  (use 0.082uF as the nearest standard value)

Next, in order to vary the frequency, the resistance must be varied
over a broader range than needed with the triple potentiometer
version.  In order to achieve a 3:1 ratio,  the potentiometer
range must be around  5:1, so a pot of 5*200K, or 1Meg,  is
required.

The potentiometer is connected in series between the first resistor
and ground. In order to adjust the frequency both above and below the
nominal value of 4Hz, the first resistor value is lowered from 200K to
around 1/5 the value, or 40K (39K is chosen as the nearest standard
value).

Following is the schematic of the finished
circuit:

Completed Single-Pot Variable Phase Shift Oscillator Design

This oscillator has a range of 2Hz to 6.5Hz, with an amplitude variation
from 204V at 6.5Hz to 163V at 2Hz.  As expected, the amplitude variation
is quite large, compared to the triple pot version, and the frequency
adjustment range is smaller.  However, this should be acceptable
performance for a guitar amplifier tremolo oscillator.

Design considerations: footswitch
and startup issues for tremolo circuits

The only way to get an oscillator like this to start
rapidly and reliably is to introduce an outside interference, in the way of a
voltage or current transient. In general, the larger the transient, the
quicker the start.  The typical method of turning the tremolo on and off
is to use a footswitch that kills the biasing of the amplifier section of the
oscillator.

One of the problems with this kind of footswitch
circuit is that the amplifier is cut off by removing the bias on the cathode
with the shorting switch. In order to oscillate, it has to first amplify. The
footswitch provides the shock start in the form of a transient as the cathode
voltage rises from the short to the 1.6V nominal stable point, but the bypass
cap, which must be there in order to attain enough gain to oscillate, now
works against things by slowing down and limiting the magnitude of the startup
transient.  As can be seen, the previously mentioned requirement for a
large value bypass capacitor in order to get enough gain at low frequencies is
now at odds with the requirement for a fast transient to start the oscillator
reliably.  In these cases, the capacitor needs to be designed for a value
just above that required for reliable oscillation.  In some cases, the
oscillator will still be slow to start, or may not start at all.

It is usually easier to get these type oscillators to
start if the amplifier stays on, biased to the proper point of operation, and
an outside AC transient is introduced somewhere in the circuit. However, this
is not so easy to do when all you have is a footswitch that must be grounded
on one side, and have a safe voltage on the other terminal, and no circuit to
generate a startup pulse to inject into the circuit.

One way to accomplish this is to add a large value
resistor from the "center resistor", or R2 as shown in the above examples, to
the power supply, and connect the footswitch to the junction of the two
resistors.  Typically the resistor should be around ten times the value
of the center resistor. When the switch is grounded, the oscillator is off,
because the AC feedback path is broken, but the DC bias on the tube remains
the same, because the coupling caps on either side block the DC voltage.
This leaves the amplifier biased properly for normal operation. When the
switch is opened, there is a fast, relatively high voltage AC transient that
is coupled into the grid circuit, which starts the oscillations rapidly. The
drawback is that the junction of the two resistors is at about 1/11 the supply
voltage normal operation, which means this voltage appears on the center
terminal of the footswitch, which may not be safe, particularly if the center
to ground resistor fails. Also, this circuit may not speed up the initial
power-on startup delay, only the footswitch startup delay, but this is usually
acceptable.

Another method is to connect the "startup" resistor to the cathode of
the oscillator tube, as used in many Fender circuits, such as the Vibrolux
6G11.  This will provide a smaller, safer DC voltage for startup. but
will start up a bit slower than the higher voltage "kickstart", because the
cathode voltage is only a volt or two.

A third method, as used in some DeArmond amps, is to connect the "startup"
resistor between R1 and  B+ node at the screen or plate, and connect the
footswitch to the bottom of R2.  The power supply node will usually have
several volts of sawtooth 120Hz ripple riding on it,  which will provide
the necessary "noise" to kick-start the oscillator.

Appendix A:  The math behind the phase shift
oscillator:

1Phase lead network
analysis:

A single-section, phase lead network transfer function can be
derived using the voltage divider rule as follows:

Vo = Vi*R/(R+1/sC) = Vi*sRC/(sRC+1)
Therefore, the
transfer function, H(s),  is equal to:

H(s) = Vo/Vi = sRC/(sRC+1)

where s = jw = j*2*Pi*f
and     j =
sqrt(-1)

A complex number of the form C = A + jB has both a
magnitude and a phase.  The magnitude is equal to the square root of
the sum of the squares of the real and imaginary parts, and the phase is
equal to the arctangent of the imaginary part divided by the real part, as
shown below:

Magnitude H(s) = sqrt(A2 + B2)

Phase H(s) = tan-1(B/A)

Therefore, the
magnitude of the phase lead transfer function is as follows:

|H(jw)| = sqrt[(wRC)2]/(sqrt(12 +
(wRC)2)) = (wRC)/sqrt(1+
w2R2C2)
And the phase of the
phase lead transfer function is:

ø = (phase of numerator - phase of denominator) =
tan-1(wRC/0) - tan-1( wRC/1) = 90o -
tan-1( wRC)
Using the example given, R = 1Meg and C =
0.01uF:

|H(jw)| = (2*Pi*f)*(1e6)(0.01e-6)/sqrt(1
+ (2*Pi*f)2*(1e6)2*(0.01e-6)2) =
(0.06283*f)/sqrt(1 + .00395*f2)

ø = 90 - tan-1(2*Pi*f*R*C) = 90 -
tan-1(2*Pi*f*(1e6)(0.01e-6)) = 90 -
tan-1(.06283*f)

Therefore, at a frequency of
15.9Hz, the magnitude and phase would be:

|H(jw)| = (0.06283*15.9)/sqrt(1 + .00395*15.92) =
0.707 = -3.01dB

ø =  90 - tan-1(.06283*15.9) =
45o

If f = 0 is substituted into the equations,
the resultant magnitude goes to zero (as it should, since the capacitor
blocks DC) and the resultant phase goes to 90 degrees.
If f = infinity
is substituted into the equations, the resultant magnitude goes to 1, or
0dB, and the resultant phase shift goes to zero.

A plot of the magnitude and phase of the transfer function can be made by
substituting in values of f and solving for the resulting magnitude and
phase numbers.

It is important to note that the frequency at which a 60 degree phase
shift occurs in a single phase lead section is not the same frequency at
which a 180 degree phase shift occurs in a three section phase lead
network, so you can't just solve the single-section phase equation for
frequency, and plug in a value of 60 degrees to find the resultant frequency
at which oscillation will occur for given values of R and C.  The
procedure for determining frequency of oscillation for a three-section phase
lead network is described in the next section on phase shift network
analysis.

2Phase shift
network analysis:

The transfer function of the phase shift network can be
determined as follows:

Using mesh analysis, and substituting general impedance variables
Z1 for C and Z2 for R, the following three equations
can be derived:

(1)  Vi = Z1I1 +
Z2(I1-I2)
(2)  0 =
Z2(I2-I1) + Z1I2 +
Z2(I2-I3)
(3)  0 =
Z2(I3-I2) + Z1I3 +
Z2I3
This can be rearranged and written in
terms of the mesh currents as follows:

(1)  Vi = (Z1+
Z2)I1 - Z2I2
(2)  0 =
-Z2I1 + (Z1+2Z2)I2
- Z2I3
(3)  0 =
-Z2I2 + (Z1+
2Z2)I3
which gives the following matrix
equation:

[Vi]
[I1]    [(Z1+
Z2)     -
Z2
0        ]
[0]   =
[I2]    [
-Z2        (Z1+
2Z2)     -
Z2     ]

[0]      [I3]
[
0
-Z2       (Z1+2Z2)
]
This matrix can be solved for the individual currents by
using Cramer's Method as follows:

First, the characteristic determinant of the matrix is calculated as
follows:

|
(Z1+ Z2)     -
Z2
0         |
Det =
|    -Z2     (Z1+
2Z2)     -
Z2       |  =
Z13 + 5Z12Z2+
6Z1Z2 2 + Z2 3

|
0
-Z2
(Z1+2Z2) |

Next, the individual currents can be solved by substituting the
voltage matrix into the appropriate position in the numerator matrix and
solving the resulting determinant, and dividing by the characteristic
determinant as follows:

|
(Z1+ Z2)     -
Z2
Vi
|
Z22
I3  =
|
-Z2       (Z1+
2Z2)
0        |
=         Vi
*     ____________________

|
0
-Z2
0
|
Z13 + 5Z12Z2+
6Z1Z22 + Z2 3

_________________________

Z13 + 5Z12Z2+
6Z1Z22 + Z2
3
I3 was chosen first so the output
voltage could be determined in order to derive the transfer function.
Since I3 is now known, the output voltage is simply I3
multiplied by the last shunt impedance, Z2 :

Z2 2
Vo =
Vi *  _________________________  *Z2

Z13 + 5Z12Z2+
6Z1Z22 + Z2 3

Z23

=      Vi *
_________________________

Z13 + 5Z12Z2+
6Z1Z22 + Z2 3

Therefore, the transfer function is:

Z23

Vo/Vi    =
_________________________

Z13 + 5Z12Z2+
6Z1Z22 + Z2
3

This can be rewritten by dividing the numerator and denominator
by  Z23 as follows:

1
Vo/Vi    =
_________________________

Z13/Z23 +
5Z12/Z22+
6Z1/Z2 + 1

Now, substituting the variable 'x' for Z1/Z2,

1
Vo/Vi    =
_________________________

x3 + 5x2 + 6x + 1

The criteria for oscillation is positive feedback, i.e., the total phase
shift around the loop have to be equivalently zero (a multiple of 360
degrees) and a gain of unity or greater. This criterion can be met by a
total phase shift through the network of 180 degrees, since the amplifier
contributes 180 degrees to the phase shift through it's inversion.  In
order for the phase shift to be 180 degrees, the imaginary parts of the
transfer function must be zero.  The squared term, 5x2 and
the constant 1, are purely real, since the square of j is equal to -1, so
the following equation must be met for a zero imaginary part:

x3 + 6x = 0
which gives:

x2 = -6
so:

x = + sqrt(-6) = + j*sqrt(6)

If  x3 + 6x = 0, the magnitude equation at the
frequency of oscillation simplifies to:

1
Vo/Vi    =
__________

5x2 + 1

Substituting x = - j*sqrt(6) into this equation for x (-j because
Z1 is a capacitor) gives:

1
1
Vo/Vi    =
__________
=        __________
=     -
1/29

5(- j*sqrt(6))2 +
1
5*(-1*6)+1

This means that the transfer function of the phase shift network has a
gain of 1/29, and the negative sign indicates a phase inversion of 180
degrees.  Therefore, in order to satisfy the gain criterion for
oscillation, the amplifier must have a gain of -29 (once again, the negative
sign indicating phase inversion of 180 degrees). The gain can be higher than
this, but the distortion will be lowest at the minimum gain necessary to
sustain oscillations.

The frequency of oscillation is also determined by the above oscillation
criteria, as shown below:

Since x = Z1/Z2 = (1/(jwC))/R = 1/(jwRC) =
-j*sqrt(6),

w = 1/(sqrt(6)*R*C)
Since w = 2*Pi*f,

f = 1/(2*Pi*sqrt(6)*R*C)
This is the frequency which
will give a total phase shift through the network of -180 degrees, which
will result in oscillation.

The input impedance of the network can be determined by first calculating
I1 with a Vi of 1V, then taking the reciprocal of it.  This
is done as follows:

|
1          -
Z2
0
|
Z12 + 4Z1Z2+
3Z22
I1  =
|     0
(Z1+ 2Z2)
-Z2
|   =
____________________

|
0
-Z2
(Z1+ 2Z2)
|
Z13 + 5Z12Z2+
6Z1Z22 + Z2 3

_________________________

Z13 + 5Z12Z2+
6Z1Z22 + Z2
3

Taking the reciprocal to get the input impedance gives:

Z13 + 5Z12Z2+
6Z1Z22 + Z2 3

Zin
=
____________________

Z12 + 4Z1Z2+
3Z22
Dividing both numerator and
denominator by Z23, after multiplying the denominator
and the entire equation by Z2/Z2, (to make the
function proper in terms of degree of exponents in the numerator and
denominator) gives:

Z13/Z23+
5Z12Z2/Z23+
6Z1Z22/Z23 +
Z23/Z23
Zin
=      Z2
*     _________________________________

Z12Z2/Z23 +
4Z1Z22/Z23+
3Z23/Z23
This
simplifies to:

Z13/Z23+
5Z12/Z22+
6Z1/Z2+ 1
Zin
=      Z2
*     ___________________________

Z12/Z22 +
4Z1/Z2+ 3
Substituting the variable 'x'
for Z1/Z2,

x3 + 5x2 + 6x + 1
Zin
=      Z2 *
_______________

x2 + 4x + 3

At the frequency of oscillation, Zin  becomes:

(- j*sqrt(6))3 + 5(-j*sqrt(6))2 + 6(-j*sqrt(6)) + 1

Zin =      Z2
*   _______________________________________

(- j*sqrt(6))2 + 4(- j*sqrt(6)) + 3
Simplified, this
becomes:

- 29
Z2
- 29
Z2
Z2
Zin =   ______________
=   ______________  =  ______________

-3 -
4j*sqrt(6)
-3 -
j9.798
0.103 + j0.338
This impedance equation indicates that the input
impedance at the oscillation frequency is proportional to Z2, but
not Z1.    If the frequency is to be varied, this
impedance must remain constant, or the amplitude of the oscillations will
vary.  If the gain of the amplifier is set to the critical value of 29,
and the impedance decreases, the gain will drop and the oscillations will
decay to zero. Likewise, if the impedance increases, the gain will increase,
and the oscillation amplitude will increase, and there will also be an
increase in the distortion in the output.  Therefore, the best method
of varying the frequency of oscillation is to vary all three Z1
impedances simultaneously.  This will require a triple-gang
variable capacitor, which is not practical at the low frequencies involved
in audio oscillators used in tremolo circuits.  A better approach would
be to use the phase lag network version, and vary the three Z1
impedances simultaneously with a triple-gang potentiometer, since they
are resistances in the phase lag configuration.  This requires an extra
isolation capacitor as well as an extra bias resistor, which must be large
in relation to the series resistance, to avoid loading the network.  In
addition, the phase lag version has much higher distortion of the sine wave
at the plate output.  Alternately, the gain of the amplifier could be
made much larger than 29, and the output will have amplitude variations as
well as distortion of the sine wave as the amplitude is adjusted.  The
amplitude variation is not too much of an issue with a guitar amplifier,
because it is normal to set the speed and then adjust the intensity to the
desired level.  Also, the amplitude variations can be minimized by
making the phase shift network impedance much larger than the output
impedance of the amplifier stage.

Copyright © 1999-2012 Randall
Aiken.  May not be reproduced in any form without written approval from
Aiken Amplification.

Revised
02/04/12