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 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 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
Some typical 12AX7 numbers:
plate resistance: ra = 62.5K
amplification factor: mu = 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 circuit.
Note: the symbol "||" means "in parallel with". Resistors in parallel add in reciprocal, i.e. 1/Rt = 1/R1 + 1/R2.
Rk' = (Rl+ra)/(mu+1)Therefore, the total cathode resistance is the parallel combination of the cathode resistance, Rk', and the cathode resistor, Rk, as below:
= (100K + 62.5K)/(101)
= 1.6K
R = Rk' || Rk
= 1.6K || 820
= 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 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.
- 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)
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.
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)
Design modifications
for a tremolo oscillator
Some typical 12AT7 numbers:
plate resistance: ra = 10.9K
amplification factor: mu = 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 circuit.
Rk' = (Rl+ra)/(mu+1)Therefore, the total cathode resistance is the parallel combination of the cathode resistance, Rk', and the cathode resistor, Rk, as below:
= (47K + 10.9K)/(61)
= 949 ohms
R = Rk' || Rk
= 949 || 1.2K
= 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
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.
- 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.
- 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:
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.
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.
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:2Phase shift network analysis: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)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:where s = jw = j*2*Pi*f
and j = sqrt(-1)Magnitude H(s) = sqrt(A2 + B2)Therefore, the magnitude of the phase lead transfer function is as follows:Phase H(s) = tan-1(B/A)
|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)Therefore, at a frequency of 15.9Hz, the magnitude and phase would be:ø = 90 - tan-1(2*Pi*f*R*C) = 90 - tan-1(2*Pi*f*(1e6)(0.01e-6)) = 90 - tan-1(.06283*f)
|H(jw)| = (0.06283*15.9)/sqrt(1 + .00395*15.92) = 0.707 = -3.01dBIf 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.ø = 90 - tan-1(.06283*15.9) = 45o
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.
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)This can be rearranged and written in terms of the mesh currents as follows:
(2) 0 = Z2(I2-I1) + Z1I2 + Z2(I2-I3)
(3) 0 = Z2(I3-I2) + Z1I3 + Z2I3(1) Vi = (Z1+ Z2)I1 - Z2I2which gives the following matrix equation:
(2) 0 = -Z2I1 + (Z1+2Z2)I2 - Z2I3
(3) 0 = -Z2I2 + (Z1+ 2Z2)I3[Vi] [I1] [(Z1+ Z2) - Z2 0 ]This matrix can be solved for the individual currents by using Cramer's Method as follows:
[0] = [I2] [ -Z2 (Z1+ 2Z2) - Z2 ]
[0] [I3] [ 0 -Z2 (Z1+2Z2) ]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 | Z22I3 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 :
I3 = | -Z2 (Z1+ 2Z2) 0 | = Vi * ____________________
| 0 -Z2 0 | Z13 + 5Z12Z2+ 6Z1Z22 + Z2 3
_________________________
Z13 + 5Z12Z2+ 6Z1Z22 + Z2 3Z2 2
Vo = Vi * _________________________ *Z2
Z13 + 5Z12Z2+ 6Z1Z22 + Z2 3Z23
= 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/Z2,1x3 + 5x2 + 6x + 1
Vo/Vi = _________________________
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 = 0which gives:x2 = -6so:x = + sqrt(-6) = + j*sqrt(6)
If x3 + 6x = 0, the magnitude equation at the frequency of oscillation simplifies to:15x2 + 1
Vo/Vi = __________
Substituting x = - j*sqrt(6) into this equation for x (-j because Z1 is a capacitor) gives:
1 15(- j*sqrt(6))2 + 1 5*(-1*6)+1
Vo/Vi = __________ = __________ = - 1/29
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 3Dividing 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:
Zin = ____________________
Z12 + 4Z1Z2+ 3Z22Z13/Z23+ 5Z12Z2/Z23+ 6Z1Z22/Z23 + Z23/Z23This simplifies to:
Zin = Z2 * _________________________________
Z12Z2/Z23 + 4Z1Z22/Z23+ 3Z23/Z23Z13/Z23+ 5Z12/Z22+ 6Z1/Z2+ 1Substituting the variable 'x' for Z1/Z2,
Zin = Z2 * ___________________________
Z12/Z22 + 4Z1/Z2+ 3x3 + 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)) + 1Simplified, this becomes:
Zin = Z2 * _______________________________________
(- j*sqrt(6))2 + 4(- j*sqrt(6)) + 3- 29 Z2 - 29 Z2 Z2This 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-gain 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.
Zin = ______________ = ______________ = ______________
-3 - 4j*sqrt(6) -3 - j9.798 0.103 + j0.338
Copyright © 1999, 2000 Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.