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 =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.=20Following is a schematic diagram of a typical phase shift = oscillator:
=20
Phase Shift OscillatorThe 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
Some typical 12AX7 numbers:
plate=20 = resistance: &n= bsp; ra =3D 62.5K
amplification=20 = factor: = mu =3D 100
Where:
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.
Note: the symbol "||" means "in parallel with". = Resistors in=20 parallel add in reciprocal, i.e. 1/Rt =3D 1/R1 + = 1/R2.
Rk' =3D (Rl+ra)/(mu+1) =Therefore, the total cathode resistance is the = parallel=20 combination of the cathode resistance, Rk', and the cathode = resistor,=20 Rk, as below:=20
=3D=20 (100K + 62.5K)/(101)
=3D=20 1.6K
R =3D Rk' || Rk
=3D = 1.6K || 820=20
=3D 542 ohms
The minimum value of bypass capacitor is therefore: =
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.
- 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.=20solving 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)
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.
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)
Design modifications for a tremolo=20
oscillator
- 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.
Some typical 12AT7 numbers:
plate=20 = resistance: &n= bsp; ra =3D 10.9K
amplification=20 = factor: = mu =3D 60
Where:
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.
Rk' =3D (Rl+ra)/(mu+1) =Therefore, the total cathode resistance is the = parallel=20 combination of the cathode resistance, Rk', and the cathode = resistor,=20 Rk, as below:=20
=3D=20 (47K + 10.9K)/(61)
=3D 949=20 ohms
R =3D Rk' || Rk
=3D 949 = || 1.2K=20
=3D 530 ohms
The minimum value of bypass capacitor is therefore: =
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.
- 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.=20solving 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
- 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:
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
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:
1Phase lead network=20 analysis:
A single-section, phase lead network transfer function can = be=20 derived using the voltage divider rule as follows:=202Phase shift=20 network analysis:=20Vo =3D Vi*R/(R+1/sC) =3D = Vi*sRC/(sRC+1)Therefore, the=20 transfer function, H(s), is equal to:=20H(s) =3D Vo/Vi =3D sRC/(sRC+1)=20A 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:=20where s =3D jw =3D j*2*Pi*f
and j = =3D=20 sqrt(-1)Magnitude H(s) =3D sqrt(A2 + B2)=20Therefore, = the=20 magnitude of the phase lead transfer function is as follows:=20Phase H(s) =3D tan-1(B/A)
|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)=20Therefore, at a = frequency of=20 15.9Hz, the magnitude and phase would be:=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)
|H(jw)| =3D (0.06283*15.9)/sqrt(1 + = .00395*15.92) =3D=20 0.707 =3D -3.01dB=20If 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.=F8 =3D 90 - tan-1(.06283*15.9) =3D=20 45o
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.=20A 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.
The transfer function of the phase shift network can be=20 determined as follows:=20Using 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)This can be rearranged and = written in=20 terms of the mesh currents as follows:=20
(2) 0 =3D=20 Z2(I2-I1) + = Z1I2 +=20 Z2(I2-I3)
(3) 0 =3D=20 Z2(I3-I2) + = Z1I3 +=20 Z2I3(1) Vi =3D (Z1+=20 Z2)I1 - Z2I2 =which gives the following = matrix=20 equation:=20
(2) 0 =3D=20 -Z2I1 + = (Z1+2Z2)I2 - Z2I3
(3) 0 =3D=20 -Z2I2 + (Z1+=20 2Z2)I3[Vi] =20 [I1] [(Z1+=20 Z2) -=20 = Z2 = 0 ]This matrix can be solved for the individual currents = by=20 using Cramer's Method as follows:=20
[0] = =3D=20 [I2] [ -Z2 = (Z1+ 2Z2) -=20 Z2 ]=20
[0] = [I3] [ =20 = 0 = -Z2 (Z1+2Z2) ]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; Z22I3 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 =
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&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/Vi = =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 = ; = &= nbsp; x3 + 5x2 + 6x + 1
Vo/Vi = =3D =20 = _________________________
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 0which gives:=20x2 =3D -6so:=20x =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&n= bsp; &nb= sp; &nbs= p; 5x2 + 1
Vo/Vi = =3D =20 = __________
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&n= bsp; &nb= sp; 5(- j*sqrt(6))2 +=20 = 1 = 5*(-1*6)+1
Vo/Vi = =3D =20 = __________ &nb= sp; =3D = __________ =20 =3D -=20 = 1/29
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,=20f =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.=20The 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=20Dividing 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 =
Zin=20 = =3D &nbs= p; ____________________=20 =
&nb= sp; &nbs= p; Z12 + 4Z1Z2+=20 3Z22&= nbsp; &n= bsp; Z13/Z23+=20 5Z12Z2/Z23+ 6Z1Z22/Z23 = +=20 Z23/Z23This=20 simplifies to:=20 =
Zin=20 =3D Z2=20 * _________________________________=20 =
&nb= sp; &nbs= p; Z12Z2/Z23 + = 4Z1Z22/Z23+ = 3Z23/Z23&= nbsp; &n= bsp; Z13/Z23+=20 5Z12/Z22+ 6Z1/Z2+ 1Substituting the = variable 'x'=20 for Z1/Z2,=20 =
Zin=20 =3D Z2=20 * ___________________________=20 =
&nb= sp; &nbs= p; Z12/Z22 +=20 4Z1/Z2+ 3&= 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=20Simplified, this=20 becomes:=20 =
Zin =3D Z2 = * _______________________________________=20 =
&nb= sp; &nbs= p; (- j*sqrt(6))2 + 4(- j*sqrt(6)) + = 3&= nbsp; - 29=20 = Z2 = &= nbsp; - 29=20 = Z2 = &= nbsp; Z2This 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.
Zin =3D = ______________ =20 =3D ______________ =3D ______________=20 =
&nb= sp; -3 -=20 = 4j*sqrt(6) &nb= sp; -3 -=20 = j9.798 &= nbsp; 0.103 + j0.338
Copyright =A9 1999-2012 = Randall=20 Aiken. May not be reproduced in any form without written = approval from=20 Aiken Amplification.