From: "Saved by Internet Explorer 11" Subject: Designing Phase Shift Oscillators Date: Fri, 20 Dec 2013 13:22:36 -0800 MIME-Version: 1.0 Content-Type: multipart/related; type="text/html"; boundary="----=_NextPart_000_0000_01CEFD86.8C53F300" X-MimeOLE: Produced By Microsoft MimeOLE V6.1.7601.17609 This is a multi-part message in MIME format. ------=_NextPart_000_0000_01CEFD86.8C53F300 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Content-Location: http://www.aikenamps.com/PhaseShiftOscillators.html =20 =20 =20 = Designing=20 Phase Shift Oscillators=20
Designing Phase = Shift=20 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=20 tube, transistor, or opamp to generate the required phase shift at a=20 particular frequency to sustain oscillations.  They are = moderately stable=20 in frequency and amplitude, and very easy to design and=20 construct.
Where are they = used?=20
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=20 variations.
How do they = work?=20
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=20 positive feedback).  When used with a single-stage inverting=20 amplification element, such as a tube, transistor, or inverting opamp=20 configuration, the amplifier itself provides 180 degrees of phase = shift (a=20 gain of -A, where A is the gain of the amplification stage). The = remaining 180=20 degrees of phase shift necessary to provide a total of 360 degrees is = provided by an external network of resistors and capacitors.=20

Following is a schematic diagram of a typical phase shift = oscillator: =20
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=20 phase lead1 networks contribute a total of 180 degrees of = phase=20 shift at the oscillation frequency.  Note that a phase shift = oscillator=20 could also be built using four or more phase shift elements, with each = element=20 contributing less overall phase shift at the oscillation = frequency. =20 Normally, there is no need to do this, as it takes extra = components.  A=20 minimum of three phase shift networks is required, however, because = the=20 maximum theoretical phase shift available from any one RC network is = 90=20 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=20 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 =3D = 1/(2*Pi*R*C) =3D=20 1/(2*Pi*1Meg*.01uF) =3D 15.9Hz: =20
Phase Lead = Network           =             &= nbsp;          Phase Lag Network

Following is a plot of the phase shift and attenuation = characteristics of=20 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=20 degrees at 0.1Hz, and shifts through +45 degrees at the -3dB point of = 15.9Hz,=20 continuing on toward 0 degrees above 1kHz. The phase lag network, on = the other=20 hand, starts at 0 degrees, shifts through -45 degrees at the -3dB = point, and=20 continues on towards -90 degrees above 1kHz.  Either one will = provide an=20 effective 0 degrees phase shift when three of them are combined with = the 180=20 degree phase shift of the amplifier as shown in the phase shift = oscillator=20 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=20 the amplifier gain is greater than 29 (which  will bring the = overall loop=20 gain above unity gain, and satisfy the gain criterion for = oscillation).  The oscillations will occur at a frequency given by the following = equation:=20

fo =3D 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=20 purest sine wave, however, it is impractical if tubes of varying gains = may be=20 substituted (this usually requires an adjustment control to trim the = gain), or=20 if the frequency of oscillation must be adjusted in such a manner as = to change=20 the gain of the network.  For these reasons, the gain is usually = made=20 higher, and post-filtering of the waveform is done to remove unwanted = harmonic=20 distortion.

If four phase lead networks are used, the phase shift per section = at the=20 oscillation frequency is lower, therefore, the attenuation of the = network is=20 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=20 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=20 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=20 designed.  This requirement will affect the type of tube=20 chosen.  For example, a 12AU7 will not work in this circuit, = because=20 it has an amplification factor of only 17, which is lower than the = minimum=20 gain of 29.  Note that the actual gain obtained from a tube = is always=20 considerably less than the amplification factor, which is the = theoretical=20 maximum obtainable from the tube, and then only with an infinite = load=20 impedance (a current source load is a good approximation to = this). =20 Typically, a 12AX7, with an amplification factor of 100, will = achieve a=20 gain of around 60 or higher with a resistive plate load, when the = cathode=20 is fully bypassed. The gain of an unbypassed 12AX7 can also be = greater=20 than 29, so you can design the circuit without the bypass = capacitor and it=20 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=20 require gain of 29.   For completeness, this example = shows=20 bypass capacitor calcuation, even though a 12AX7 is=20 used.

•
• Assuming a 12AX7 is chosen, following are the calculations for = the=20 gain and output impedance:
Some typical 12AX7 numbers:
plate=20 = resistance:          &n= bsp;     ra =3D 62.5K
amplification=20 = factor:           =   mu =3D 100
• A plate load resistance of 100K and a cathode resistance of = 820 ohms=20 are chosen to provide good gain and linearity. In order to = maximize the=20 gain and minimize the output impedance, the cathode must be = fully-bypassed=20 at the frequency of oscillation. The midband gain for the 12AX7 = with a=20 fully-bypassed cathode is:
•
Av =3D (mu*Rl)/(Rl+ra)
= =3D=20 (100*100K)/(100K+62.5K)
=3D 61.5=20

Where:

Rl =3D the load resistor (100K in this example)
ra =3D = the=20 internal plate resistance (62.5K for a typical 12AX7)
mu = =3D the mu=20 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=20 circuit.

• Next, the output impedance of the stage is calculated. Since = the=20 internal plate resistance is effectively in parallel with the = plate load=20 resistor, the output impedance (for the signal taken off the = plate) will=20 be:
•
R =3D ra || Rl
=3D 62.5K || 100K=20
=3D 38.5K=20

Note: the symbol "||" means "in parallel with".  = Resistors in=20 parallel add in reciprocal, i.e. 1/Rt =3D 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=20 unbypassed) is:
Rk' =3D (Rl+ra)/(mu+1) =
=3D=20 (100K + 62.5K)/(101)
=3D=20 1.6K
Therefore, the total cathode resistance is the = parallel=20 combination of the cathode resistance, Rk', and the cathode = resistor,=20 Rk, as below:=20
R  =3D Rk' || Rk
=3D = 1.6K || 820=20
=3D 542 ohms
• Next, the value of the cathode bypass capacitor must be=20 calculated.  Since the gain of the amplifier must be = maximized at the=20 frequency of oscillation, the minimum capacitor value is chosen = using the=20 following equation for the -3dB point of the amplifier.
•
f =3D 1/(2*PI*R*C)
solving for C:=20

C=3D 1/(2*PI*R*f)

The minimum value of bypass capacitor is therefore: =

C=3D 1/(2*PI*542*7Hz) =3D 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=20 above the required minimum of 29.  However, in order to = maximize gain=20 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=20 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=20 selection:
• Since the impedance is proportional to the shunt element in = the phase=20 shift network2, in this case, the resistor, a suitable=20 impedance value must first be chosen.   The input = impedance of=20 the network  must be large in comparison to the output = impedance of=20 the amplifier, so as to not load the output appreciably, which = would=20 reduce the gain, possibly to a point where it can no longer = sustain=20 oscillations.  A good minimum value is around ten times the = actual=20 output impedance of the amplification stage.  Since the input = impedance is proportional to the shunt element, and is = approximately twice=20 the value of the shunt element at the oscillation frequency, the=20 resistance can be chosen to be around half the required impedance. = This=20 resistance will then determine the value of capacitor necessary to = achieve=20 the desired frequency of oscillation.  Since we have an = output=20 impedance of 38.5K in our example, a good minimum value for the = input=20 impedance is ten times this value, to prevent loading of the = output stage.  Since the resistance value to achieve this impedance = is=20 around half the total impedance, a value of five times the output=20 impedance, or 5*38.5K =3D 193K, will work.

•
• Next, the capacitance value is calculated using the formula = for the=20 frequency:

•
fo =3D 1/(2*Pi*Sqrt(6)*R*C)

solving for C:

C =3D 1/(2*Pi*Sqrt(6)*fo*R) =3D = 1/(2*Pi*2.45*7*193K) =3D  .048uF  (use .047uF as the nearest smaller standard = value)
Since=20 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.=20

solving for R:

R =3D 1/(2*Pi*Sqrt(6)*fo*C) =3D = 1/(2*Pi*2.45*7*.047uF)=20 =3D  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=20 require a larger resistance to achieve the desired = frequency. =20 Likewise, when choosing the resistance, it is best to choose the = next=20 larger size, as this will also increase the input impedance of the = phase=20 shift network.
• The design is then built and tested, and resistor or capacitor = values=20 are trimmed as necessary to provide the exact frequency of = oscillation=20 desired. Following is the output of the completed oscillator using = the=20 values calculated above:

(click on image for larger view)

A small amount of harmonic distortion is = present, as=20 evidenced by the "kink" at the bottom edges of the sine = wave.  This=20 can be eliminated by filtering the output slightly, either with a = post RC=20 filter, or by adding a capacitor across the plate load resistor to = act as=20 a first-order lowpass filter to reduce a bit of the = distortion.  The=20 filter cutoff frequency cannot be made too low at the plate = resistor, or=20 the gain will be reduced below the level necessary to sustain=20 oscillation.   A good first choice is to select a = capacitor that=20 will produce a -3dB point around 3 times the oscillation = frequency. =20 This can be calculated using the output impedance of the stage as = follows:=20

C =3D 1/(2*Pi*3*fo*R) =3D 1/(2*Pi*3*7*38.5K) = =3D =20 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=20 should be increased, or an RC post filter should be used after the = oscillator.

The final schematic and output plot are shown below: =20
Completed Phase Shift Oscillator Design

(click on image for larger view)

Note = the=20 harmonic distortion has been reduced significantly in the above = output, at=20 the expense of some amplitude reduction.  The filter capacitor = value=20 could be reduced to the point of just cleaning up the distortion to = a=20 satisfactory level for more gain, if desired.

Design modifications for a tremolo=20 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:=20
• A tremolo circuit must have a variable frequency oscillator, = which is=20 adjustable by a single potentiometer.  The oscillator needs to = have a=20 relatively wide frequency variation (typically from around 2Hz to = 8Hz), the=20 need to maintain a relatively constant amplitude over the entire = adjustment=20 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=20 amplifier, as it will usually afford a wider range of control.
• The requirement for lower value resistors forces the design to = use=20 larger capacitors, which have lower reactance at the frequency of=20 oscillation. Because of this,  it is better to design for a = lower=20 output impedance by using a tube with a lower internal plate = resistance, or=20 sacrifice some gain by using lower value plate resistances.  = This may,=20 however, require a larger cathode bypass capacitor to maintain gain = at lower=20 frequencies.
• When using the phase lead network, the impedance of the phase = shift=20 network is proportional to the value of R, so it should be made as = large as=20 practical, relative to the output impedance of the amplifier. This = will=20 minimize the gain variations as the frequency is adjusted, however, = this is=20 in contrast to the need to keep the potentiometer and resistance = values=20 low,  so the best overall solution is to use the lowest = practical=20 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=20 between the minimum frequency selected is very important at these low=20 frequencies, because it determines the adjustment range of the=20 oscillator.  For example, a range from 1Hz to 7Hz requires an = oscillator=20 that can be adjusted over a 7:1 range, while a range of 2Hz to 8Hz = only=20 requires an oscillator that can be adjusted over a 4:1 range, even = though the=20 total frequency difference is the same.  This makes a big = difference in=20 the design of the frequency adjustment portion of the oscillator. =

• Amplifier = design:

•
• An amplifier with a gain of greater than 29 must first be=20 designed.  The requirement for wide range adjustability with = low=20 value potentiometers requires a low plate resistance tube. The = 12AT7 is=20 chosen for this example.
Some typical 12AT7 numbers:
plate=20 = resistance:          &n= bsp;       ra =3D 10.9K
amplification=20 = factor:           =   mu =3D 60
• A plate load resistance of 47K is chosen.  Plotting this = load=20 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=3D=20 1.2K is chosen.  In order to maximize the gain, the cathode = must be=20 fully-bypassed at the frequency of oscillation. The midband gain = for the=20 12AT7 with a fully-bypassed cathode is:
•
Av =3D (mu*Rl)/(Rl+ra)
= =3D=20 (60*47K)/(47K+10.9K)
=3D 48.7=20

Where:

Rl =3D the load resistor (47K in this example)
ra =3D = the internal=20 plate resistance (10.9K for a typical 12AT7)
mu =3D the mu = of the=20 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=20 circuit.

• Next, the output impedance of the stage is calculated. Since = the=20 internal plate resistance is effectively in parallel with the = plate load=20 resistor, the output impedance (for the signal taken off the = plate) will=20 be:
•
R =3D ra || Rl
=3D 10.9K || 47K=20
=3D 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=20 unbypassed) is:
Rk' =3D (Rl+ra)/(mu+1) =
=3D=20 (47K + 10.9K)/(61)
=3D 949=20 ohms
Therefore, the total cathode resistance is the = parallel=20 combination of the cathode resistance, Rk', and the cathode = resistor,=20 Rk, as below:=20
R  =3D Rk' || Rk
=3D 949 = || 1.2K=20
=3D 530 ohms
• Next, the value of the cathode bypass capacitor must be=20 calculated.  Since the gain of the amplifier must be = maximized at the=20 lowest frequency of oscillation, the minimum capacitor value is = chosen=20 using the following equation for the -3dB point of the = amplifier.
•
f =3D 1/(2*PI*R*C)
solving for C:=20

C=3D 1/(2*PI*R*f)

The minimum value of bypass capacitor is therefore: =

C=3D 1/(2*PI*530*2Hz) =3D 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=20 times the minimum calculated value, such as 820uF, giving a -3dB = point of=20 0.62Hz.  Note that the required capacitor value can get = rather large=20 if low frequency oscillators are designed with tubes that have low = gains=20 and low internal plate resistances.

• Phase shift network component=20 selection:
• Since the impedance is proportional to the shunt element in = the phase=20 shift network2, in this case, the resistor, a suitable=20 impedance value must first be chosen.   The input = impedance of=20 the network  must be large in comparison to the output = impedance of=20 the amplifier, so as to not load the output appreciably, which = would=20 reduce the gain, possibly to a point where it can no longer = sustain=20 oscillations.  A good minimum value is around ten times the = actual=20 output impedance of the amplification stage.  Since the input = impedance is proportional to the shunt element, and is = approximately twice=20 the value of the shunt element at the oscillation frequency, the=20 resistance can be chosen to be around half the required impedance. = This=20 resistance will then determine the value of capacitor necessary to = achieve=20 the desired frequency of oscillation.  Since we have an = output=20 impedance of 8.85K in our example, a good minimum value for the = input=20 impedance is ten times this value, to prevent loading of the = output stage.  Since the resistance value to achieve this impedance = is=20 around half the total impedance, a value of five times the output=20 impedance, or 5*8.85K =3D 44.3K, will work.

•
• Next, the capacitance value is calculated using the formula = for the=20 frequency:

•
fo =3D 1/(2*Pi*Sqrt(6)*R*C)

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

C =3D 1/(2*Pi*Sqrt(6)*fo*R) =3D = 1/(2*Pi*2.45*8*44.3K) =3D  .183uF  (use 0.18uF as the nearest smaller standard = value)
Since=20 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.=20

solving for R:

R =3D 1/(2*Pi*Sqrt(6)*fo*C) =3D = 1/(2*Pi*2.45*8*.18uF) =3D  45K  (use 47K as the nearest larger standard = value)
When=20 choosing the capacitor, it is best to choose the next smaller size = because=20 this will make the input impedance larger, because it will require = a=20 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.=20

• Next, in order to vary the frequency, the resistance must be = varied=20 over a range equivalent to the frequency range required.  In = this=20 example, the range must be 4:1, so a triple pot of 4*47K, or 188K = is=20 required, so a 200K is chosen as the the nearest standard = value.  The=20 triple pot is chosen in this example because it provides the best = range of=20 adjustment.

Following is the schematic of the = finished=20 circuit:=20 =20

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=20 capacitor value from 0.18uF to 0.15uF to achieve the original=20 specification of 2Hz to 8Hz. If desired, a lowpass filter can be = made=20 using a capacitor across the 47K plate resistor as demonstrated in = the=20 fixed frequency design, but it will only be effective at the = higher=20 frequencies, as it would require a variable cutoff frequency = filter to=20 properly filter over the entire adjustment range of the = oscillator.=20

Design=20 considerations for using a single pot to control frequency=20
It is desirable from a parts cost and availability standpoint to be = able=20 to use a single pot to control frequency instead of a triple = pot.  While=20 this will cause a moderate amount of amplitude variation with = frequency, this=20 is not too much of a problem with a guitar amplifier because it is = natural to=20 adjust the speed and intensity interactively until the desired effect = is=20 achieved, as long as the variation isn't too noticeable.  The = main design=20 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=20 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=20 resistance value, otherwise the oscillations may stop at higher=20 settings.  The amount of frequency variation is limited to = around 3:1=20 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=20 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=20 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=20 selection:
• The impedance of the phase shift network will remain the same = as=20 previously calculated, so the originally calculated resistor value = of 200K=20 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=20 increased from 470uF.

• The minimum value of bypass capacitor is therefore: =

C=3D 1/(2*PI*542*2Hz) =3D 147uF

As indicated in the original design, this value of = capacitor will result in the amplifier response being down -3dB at the lowest=20 oscillation frequency of 2Hz.  It is best to increase the = capacitor=20 value by a factor of five to ten to avoid the gain loss at the = lower=20 frequency range.  A good compromise value between size and = frequency=20 response is 820uF, giving a -3dB point of 0.4 Hz. =

• Next, the capacitance value is calculated using the formula = for the=20 frequency:

•
fo =3D 1/(2*Pi*Sqrt(6)*R*C)

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

C =3D 1/(2*Pi*Sqrt(6)*fo*R) =3D = 1/(2*Pi*2.45*4*200K) =3D  .081uF  (use 0.082uF as the nearest standard = value)
• Next, in order to vary the frequency, the resistance must be = varied=20 over a broader range than needed with the triple potentiometer=20 version.  In order to achieve a 3:1 ratio,  the = potentiometer=20 range must be around  5:1, so a pot of 5*200K, or 1Meg,  = is=20 required.

•
• The potentiometer is connected in series between the first = resistor=20 and ground. In order to adjust the frequency both above and below = the=20 nominal value of 4Hz, the first resistor value is lowered from = 200K to=20 around 1/5 the value, or 40K (39K is chosen as the nearest = standard=20 value).

Following is the schematic of the = finished=20 circuit: =20
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=20 performance for a guitar amplifier tremolo oscillator.=20

Design considerations: = footswitch=20 and startup issues for tremolo circuits=20
The only way to get an oscillator like this = to start=20 rapidly and reliably is to introduce an outside interference, in the = way of a=20 voltage or current transient. In general, the larger the transient, = the=20 quicker the start.  The typical method of turning the tremolo on = and off=20 is to use a footswitch that kills the biasing of the amplifier section = of the=20 oscillator.=20

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=20 footswitch provides the shock start in the form of a transient as the = cathode=20 voltage rises from the short to the 1.6V nominal stable point, but the = bypass=20 cap, which must be there in order to attain enough gain to oscillate, = now=20 works against things by slowing down and limiting the magnitude of the = startup=20 transient.  As can be seen, the previously mentioned requirement = for a=20 large value bypass capacitor in order to get enough gain at low = frequencies is=20 now at odds with the requirement for a fast transient to start the = oscillator=20 reliably.  In these cases, the capacitor needs to be designed for = a value=20 just above that required for reliable oscillation.  In some = cases, the=20 oscillator will still be slow to start, or may not start at = all.

It is usually easier to get these type = oscillators to=20 start if the amplifier stays on, biased to the proper point of = operation, and=20 an outside AC transient is introduced somewhere in the circuit. = However, this=20 is not so easy to do when all you have is a footswitch that must be = grounded=20 on one side, and have a safe voltage on the other terminal, and no = circuit to=20 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=20 the power supply, and connect the footswitch to the junction of the = two=20 resistors.  Typically the resistor should be around ten times the = value=20 of the center resistor. When the switch is grounded, the oscillator is = off,=20 because the AC feedback path is broken, but the DC bias on the tube = remains=20 the same, because the coupling caps on either side block the DC = voltage. =20 This leaves the amplifier biased properly for normal operation. When = the=20 switch is opened, there is a fast, relatively high voltage AC = transient that=20 is coupled into the grid circuit, which starts the oscillations = rapidly. The=20 drawback is that the junction of the two resistors is at about 1/11 = the supply=20 voltage normal operation, which means this voltage appears on the = center=20 terminal of the footswitch, which may not be safe, particularly if the = center=20 to ground resistor fails. Also, this circuit may not speed up the = initial=20 power-on startup delay, only the footswitch startup delay, but this is = usually=20 acceptable.

Another method is to connect the "startup" resistor to the = cathode of=20 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=20 will start up a bit slower than the higher voltage "kickstart", = because the=20 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=20 footswitch to the bottom of R2.  The power supply node will = usually have=20 several volts of sawtooth 120Hz ripple riding on it,  which will = provide=20 the necessary "noise" to kick-start the oscillator.

Appendix A:  The math behind the = phase shift=20 oscillator:

A single-section, phase lead network transfer function can = be=20 derived using the voltage divider rule as follows:=20
Vo =3D Vi*R/(R+1/sC) =3D = Vi*sRC/(sRC+1)
Therefore, the=20 transfer function, H(s),  is equal to:=20
H(s) =3D Vo/Vi =3D sRC/(sRC+1)=20

where s =3D jw =3D j*2*Pi*f
and     j = =3D=20 sqrt(-1)

A complex number of the form C =3D A + jB = has both a=20 magnitude and a phase.  The magnitude is equal to the square = root of=20 the sum of the squares of the real and imaginary parts, and the = phase is=20 equal to the arctangent of the imaginary part divided by the real = part, as=20 shown below:=20
Magnitude H(s) =3D sqrt(A2 + B2)=20

Phase H(s) =3D tan-1(B/A)

Therefore, = the=20 magnitude of the phase lead transfer function is as follows:=20
|H(jw)| =3D sqrt[(wRC)2]/(sqrt(12 = +=20 (wRC)2)) =3D (wRC)/sqrt(1+=20 w2R2C2)
And the phase = of the=20 phase lead transfer function is:=20
=F8 =3D (phase of numerator - phase of denominator) =3D=20 tan-1(wRC/0) - tan-1( wRC/1) =3D = 90o -=20 tan-1( wRC)
Using the example given, R =3D = 1Meg and C =3D=20 0.01uF:=20
|H(jw)| =3D = (2*Pi*f)*(1e6)(0.01e-6)/sqrt(1 + (2*Pi*f)2*(1e6)2*(0.01e-6)2) =3D=20 (0.06283*f)/sqrt(1 + .00395*f2)=20

=F8 =3D 90 - tan-1(2*Pi*f*R*C) =3D 90 -=20 tan-1(2*Pi*f*(1e6)(0.01e-6)) =3D = 90 -=20 tan-1(.06283*f)

Therefore, at a = frequency of=20 15.9Hz, the magnitude and phase would be:=20
|H(jw)| =3D (0.06283*15.9)/sqrt(1 + = .00395*15.92) =3D=20 0.707 =3D -3.01dB=20

=F8 =3D  90 - tan-1(.06283*15.9) =3D=20 45o

If f =3D 0 is substituted into the = equations,=20 the resultant magnitude goes to zero (as it should, since the = capacitor=20 blocks DC) and the resultant phase goes to 90 degrees.
If f =3D = infinity=20 is substituted into the equations, the resultant magnitude goes to = 1, or=20 0dB, and the resultant phase shift goes to zero.=20

A plot of the magnitude and phase of the transfer function can be = made by=20 substituting in values of f and solving for the resulting magnitude = and=20 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=20 frequency, and plug in a value of 60 degrees to find the resultant = frequency=20 at which oscillation will occur for given values of R and C.  = The=20 procedure for determining frequency of oscillation for a = three-section phase=20 lead network is described in the next section on phase shift network = analysis.

2Phase shift=20 network analysis:=20
The transfer function of the phase shift network can be=20 determined as follows:=20

Using mesh analysis, and substituting general impedance variables = Z1 for C and Z2 for R, the following three = equations=20 can be derived:

(1)  Vi =3D Z1I1 + = Z2(I1-I2)
(2)  0 =3D=20 Z2(I2-I1) + = Z1I2 +=20 Z2(I2-I3)
(3)  0 =3D=20 Z2(I3-I2) + = Z1I3 +=20 Z2I3
This can be rearranged and = written in=20 terms of the mesh currents as follows:=20
(1)  Vi =3D (Z1+=20 Z2)I1 - Z2I2 =
(2)  0 =3D=20 -Z2I1 + = (Z1+2Z2)I2 - Z2I3
(3)  0 =3D=20 -Z2I2 + (Z1+=20 2Z2)I3
which gives the following = matrix=20 equation:=20
[Vi]    =20 [I1]    [(Z1+=20 Z2)     -=20 = Z2          =     0        ]
   = =3D=20 [I2]    [    -Z2        = (Z1+ 2Z2)     -=20 Z2     ]=20
      = [I3]    [     =20 = 0            = -Z2       (Z1+2Z2) ]
This matrix can be solved for the individual currents = by=20 using Cramer's Method as follows:=20

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

=
|= =20 (Z1+ Z2)     -=20 = Z2          = 0         |
Det = =3D =20 |    -Z2     = (Z1+ 2Z2)     -=20 Z2       |  = =3D =20 Z13 + = 5Z12Z2+ 6Z1Z2 2 + Z2 = 3=20
=20 |      = 0           =20 -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=20 determinant as follows:

=
|= =20 (Z1+ Z2)     -=20 = Z2           =      Vi        = |            =             &= nbsp;           &n= bsp;     Z22
I3  = =3D   =20 |   =20 = -Z2       (Z1+=20 2Z2)      =20 0        |  =20 =3D         Vi=20 *     ____________________=20
=20 |      = 0           =20 = -Z2          = ;    0        = |            =              Z13 + = 5Z12Z2+=20 6Z1Z22 + Z2 = 3=20
=20 _________________________=20 =
&nb= sp;  Z13 + = 5Z12Z2+=20 6Z1Z22 + Z2=20 3
I3 was chosen first so the = output=20 voltage could be determined in order to derive the transfer = function. =20 Since I3 is now known, the output voltage is simply = I3=20 multiplied by the last shunt impedance, Z2 :=20 =

&nbs= p;            = ;            =        Z2 2
Vo = =3D     =20 Vi *  _________________________  *Z2 = =
&nb= sp;            Z13 + 5Z12Z2+ = 6Z1Z22 + Z2 3 =

=

&nbs= p;            = ;            =         Z23
=20 =3D      Vi * =20 _________________________

=

&nbs= p;            = ; Z13 + 5Z12Z2+ = 6Z1Z22 + Z2 3 =

Therefore, the transfer function is:

=
&= nbsp;           &n= bsp;           &nb= sp;     Z23=20
Vo/V   = =3D   =20 _________________________=20 =
&nb= sp;            Z13 + = 5Z12Z2+=20 6Z1Z22 + Z2=20 3

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

=
&= nbsp;           &n= bsp;           &nb= sp;     1
Vo/Vi    = =3D   =20 _________________________=20 =
&nb= sp;        Z13/Z23 +=20 5Z12/Z22+ 6Z1/Z2 + 1

Now, substituting the variable 'x' for = Z1/Z2,=20

=
&= nbsp;           &n= bsp;           &nb= sp;     1
Vo/Vi    = =3D   =20 = _________________________
= ;            =             &= nbsp;        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=20 degrees) and a gain of unity or greater. This criterion can be met = by a=20 total phase shift through the network of 180 degrees, since the = amplifier=20 contributes 180 degrees to the phase shift through it's = inversion.  In=20 order for the phase shift to be 180 degrees, the imaginary parts of = the=20 transfer function must be zero.  The squared term, = 5x2 and=20 the constant 1, are purely real, since the square of j is equal to = -1, so=20 the following equation must be met for a zero imaginary part:

x3 + 6x =3D 0
which gives:=20
x2 =3D -6
so:=20
x =3D + sqrt(-6) =3D + = j*sqrt(6)

If  x3 + 6x =3D 0, the magnitude equation at = the=20 frequency of oscillation simplifies to:

=
&= nbsp;           &n= bsp;     1
Vo/Vi    = =3D   =20 = __________
&n= bsp;           &nb= sp;           &nbs= p;  5x2 + 1

Substituting x =3D - j*sqrt(6) into this equation for x (-j = because=20 Z1 is a capacitor) gives:

=
&= nbsp;           &n= bsp;     = 1            =             &= nbsp;           &n= bsp;   1
Vo/Vi    = =3D   =20 = __________          &nb= sp; =3D        = __________  =20 =3D     -=20 = 1/29
&n= bsp;           &nb= sp;     5(- j*sqrt(6))2 +=20 = 1            =     5*(-1*6)+1

This means that the transfer function of the phase shift network = has a=20 gain of 1/29, and the negative sign indicates a phase inversion of = 180 degrees.  Therefore, in order to satisfy the gain criterion = for=20 oscillation, the amplifier must have a gain of -29 (once again, the = negative=20 sign indicating phase inversion of 180 degrees). The gain can be = higher than=20 this, but the distortion will be lowest at the minimum gain = necessary to=20 sustain oscillations.

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

Since x =3D Z1/Z2 =3D (1/(jwC))/R =3D = 1/(jwRC) =3D=20 -j*sqrt(6),

w =3D 1/(sqrt(6)*R*C)
Since w =3D 2*Pi*f,=20
f =3D 1/(2*Pi*sqrt(6)*R*C)
This is the = frequency which=20 will give a total phase shift through the network of -180 degrees, = which=20 will result in oscillation.=20

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=20 is done as follows:

=
|    =20 1          -=20 = Z2          =         0        = |            =          Z12 + 4Z1Z2+=20 3Z22
I1  = =3D   =20 |     0       (Z1+ = 2Z2)        = -Z2           |   = =3D         =20 ____________________=20
=20 |      0         =20 = -Z2          (Z1+ 2Z2) =20 = |            =    Z13 + = 5Z12Z2+=20 6Z1Z22 + Z2 = 3=20
=20 _________________________=20 =
&nb= sp;  Z13 + = 5Z12Z2+=20 6Z1Z22 + Z2=20 3

Taking the reciprocal to get the input impedance gives:

=
&= nbsp;         Z13 + = 5Z12Z2+=20 6Z1Z22 + Z2 = 3=20
Zin=20 = =3D          &nbs= p; ____________________=20 =
&nb= sp;           &nbs= p; Z12 + 4Z1Z2+=20 3Z22
Dividing both numerator and = denominator by Z23, after multiplying the = denominator=20 and the entire equation by Z2/Z2, (to make the = function proper in terms of degree of exponents in the numerator and = denominator) gives:=20 =
&= nbsp;           &n= bsp;   Z13/Z23+=20 5Z12Z2/Z23+ 6Z1Z22/Z23 = +=20 Z23/Z23
Zin=20 =3D      Z2=20 *     _________________________________=20 =
&nb= sp;           &nbs= p;           Z12Z2/Z23 + = 4Z1Z22/Z23+ = 3Z23/Z23
This=20 simplifies to:=20 =
&= nbsp;           &n= bsp;   Z13/Z23+=20 5Z12/Z22+ 6Z1/Z2+ 1
Zin=20 =3D      Z2=20 *     ___________________________=20 =
&nb= sp;           &nbs= p;           Z12/Z22 +=20 4Z1/Z2+ 3
Substituting the = variable 'x'=20 for Z1/Z2,=20 =
&= nbsp;           &n= bsp;   x3 + 5x2 + 6x + 1
Zin=20 =3D      Z2 = *   _______________=20 =
&nb= sp;           &nbs= p;      x2 + 4x + 3

At the frequency of oscillation, Zin  = becomes:

=
&= nbsp;           &n= bsp;   (- j*sqrt(6))3 + 5(-j*sqrt(6))2 + = 6(-j*sqrt(6)) + 1=20
Zin =3D      Z2 = *   _______________________________________=20 =
&nb= sp;           &nbs= p;            (- j*sqrt(6))2 + 4(- j*sqrt(6)) + = 3
Simplified, this=20 becomes:=20 =
&= nbsp;        - 29=20 = Z2          =             &= nbsp; - 29=20 = Z2          =             &= nbsp;   Z2
Zin =3D   = ______________ =20 =3D   ______________  =3D  ______________=20 =
&nb= sp; -3 -=20 = 4j*sqrt(6)          &nb= sp;      -3 -=20 = j9.798           &= nbsp; 0.103 + j0.338
This impedance equation indicates that = the input=20 impedance at the oscillation frequency is proportional to = Z2, but=20 not Z1.    If the frequency is to be = varied, this=20 impedance must remain constant, or the amplitude of the oscillations = will=20 vary.  If the gain of the amplifier is set to the critical = value of 29,=20 and the impedance decreases, the gain will drop and the oscillations = will=20 decay to zero. Likewise, if the impedance increases, the gain will = increase,=20 and the oscillation amplitude will increase, and there will also be = an=20 increase in the distortion in the output.  Therefore, the best = method=20 of varying the frequency of oscillation is to vary all three Z1 = impedances simultaneously.  This will require a = triple-gang=20 variable capacitor, which is not practical at the low frequencies = involved=20 in audio oscillators used in tremolo circuits.  A better = approach would=20 be to use the phase lag network version, and vary the three Z1=20 impedances simultaneously with a triple-gang potentiometer, = since they=20 are resistances in the phase lag configuration.  This requires = an extra=20 isolation capacitor as well as an extra bias resistor, which must be = large=20 in relation to the series resistance, to avoid loading the = network.  In=20 addition, the phase lag version has much higher distortion of the = sine wave=20 at the plate output.  Alternately, the gain of the amplifier = could be=20 made much larger than 29, and the output will have amplitude = variations as=20 well as distortion of the sine wave as the amplitude is = adjusted.  The=20 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=20 desired level.  Also, the amplitude variations can be minimized = by=20 making the phase shift network impedance much larger than the output impedance of the amplifier stage.

Copyright =A9 1999-2012 = Randall=20 Aiken.  May not be reproduced in any form without written = approval from=20 Aiken Amplification.

Revised=20 02/04/12