The phase-shift oscillator circuit is widely used in laboratories for generating an audio frequency, and for other testing where precision is a must. A special combination of resistors and capacitors are employed to achieve regenerative feedback and thus permit the circuit to generate a continuous alternating current.
Unlike other oscillators, the phase-shift oscillator has no tuned circuit and hence is not as susceptible to detuning by occasional stray capacitances and inductances. Consequently, the output remains fairly stable at the desired frequency. Figs. 1 and 2 show one of the several possible configurations for this special application of the R-C oscillator. The circuit components and their functions are as follows:
V1--Conventional pentode amplifier tube.
C1--Cathode bypass capacitor for preventing degeneration.
R1--Cathode biasing resistor.
C2--Screen-grid bypass or filter capacitor.
R2--Screen-grid voltage-dropping resistor.
RS--Plate-load resistor.
C3, C4, and C5--Feedback, coupling, and phase-shifting capacitors.
R3, R4, and R5--Phase-shifting resistors.
M1--DC power supply.
There are five electron currents at work in this circuit, and each one needs to be analyzed before you can understand how the circuit operates. These currents are:
1. Cathode-to-plate current (conventional tube current) (red).
2. Screen-grid current (solid blue).
3. Current in the first R-C combination (R3 and C3) (dotted green).
4. Current in the second R-C combination (R4 and C4) (dotted blue).
5. Current in the third R-C combination (RS and CS) (solid green). Its associated voltage is applied to the control grid as the feedback voltage.
CURRENT PATHS
Fig. 1 shows the current flow during the first half-cycle, which begins when the grid is most negative and ends when it is most positive. Conditions are just the opposite for the second half-cycle in Fig. 2--it begins when the grid is most positive and ends when it is most negative. In both illustrations, significant voltage polarities are shown as they exist at the end of the half-cycle.
At the start of the first half-cycle, you will recall that the grid voltage is most negative. However, in this circuit (unlike the ones discussed previously), plate current must flow at all times.
Therefore, the most negative grid voltage merely restricts the flow of plate current, instead of cutting it off completely. A low plate current always leads to a high positive plate voltage, particularly when this current must flow through a resistive load such as R6 on its journey to the power supply.
Line 1 of Fig. 3 shows a sine wave of plate voltage. At the start of the first half-cycle this plate voltage has its maximum positive value.
Line 4 of Fig. 3 shows two sine waves of grid voltage. The one in dashed lines is developed across RS by the current indicated in solid green in Figs. 1 and 2, and represents the actual voltage applied at the grid. The solid line is its hypothetical value if the signal could pass through the R-C network unattenuated.
It is of course significant that the grid-voltage waveforms on line 4 and the plate voltage on line 1 of Fig. 3 are 180° out of phase with each other-or more simply, "out of phase," or "of opposite phase." This out-of-phase condition is necessary for the oscillation to continue. Now let us see how this phase shift be tween output and input can be achieved from the combination of resistors and capacitors making up the phase-shifting network in Figs. 1 and 2.
Fig. 1 shows the current conditions at the end of the first half-cycle. Prior to this, the grid voltage has been increasing steadily from negative to positive. Consequently, the plate current, shown in red, has also been steadily increasing, until it reaches maximum. Now it is drawn across the tube by the positive voltage applied to the plate from power supply Ml. (Although shown as a large battery, the power supply can be a vacuum tube rectifier or any other device capable of furnishing DC). Ohm's law tells us the voltage developed, or "dropped," across a resistor by the current flowing through it is proportional to the amount of that current. Consequently, the voltage drop across resistor R6 will be much larger at the end than at the beginning of the first half-cycle. Since the voltage at the plate is always the supply voltage minus the voltage developed across the plate load resistor, we can see why the plate voltage is lowest when the grid voltage is at its highest.
At the end of the first half-cycle, when maximum plate current is flowing, it is convenient to visualize the consequent reduction in plate voltage resulting from the excess of plate-current electrons flowing out of the tube, into load resistor R6, and on to the power supply. Until these electrons can enter the load resistor, they are in a sense "dammed up" at the junction of R6 and coupling capacitor C3. For this reason, during the first half cycle they are shown flowing onto the left plate of C3.
Conversely, during the second half-cycle, when the flow of plate current is restricted by the negative grid voltage, it is convenient to equate the drop in number of plate-current electrons with the corresponding rise in plate voltage. This is done by picturing it as current being drawn out of the left plate of capacitor C3 and into load resistor R6, to the power supply.
Fig. 1 shows current (solid green line) flowing upward through grid-driving resistor RS during the first half-cycle.
Note that if there were no 180° shift in phase as the voltage pulse from the plate passes through the three R-C combinations, the current in RS would flow downward instead.
Likewise, during the second half-cycle, the normal downward flow of current through RS would be reversed if it were not for this 180° phase shift.
Suppose this phase shift did not occur? Then, when electrons flowed onto the left plate of capacitor C3, other electrons would flow in unison and in the same direction through all these capacitors, much as they would through a straight wire. Also, when electrons flowed out of the left plate of C3, other electrons would flow in the same direction through all three capacitors.
How, then, does a shift in phase occur in an R-C network? Unfortunately, it is impossible to show the voltage pulse actually changing phase. The dashed diagonal lines in Fig. 3 are meant to strengthen your conviction that such a phase shift does occur, even though not clarifying exactly how. Let us now look more closely at these waveforms, to see what makes them do so.
Fig. 1. Operation of the phase-shift oscillator-positive half-cycle.
PHASE-SHIFT NETWORK
Suppose resistor R3, R4, and RS each have the same resistance, and C3, C4, and CS the same capacitance. Then it is safe to assume that each of the R-C combinations (R3-C3, R4-C4, and RS-CS) will produce equal portions of the total phase shift of 180°; in other words, the applied voltage will be shifted 60° by each R-C combination.
What would happen if no phase shift occurred in any R-C combination during the entire first half-cycle of Fig. 1? Then, the electron current shown by the dotted green line would flow down through R3 and reach its maximum value at the same instant the plate voltage drops to its minimum at the end of the half-cycle. However, the voltage peak across R3 must occur 60°, or one-sixth of a cycle, later. For this reason, the green arrows in Fig. 1 point in both directions through R3. Admittedly this is a crude presentation of what is happening within the resistor and will require help from your imagination.
Fig. 3 shows the voltage waveforms at the plate and at feed back points A, B, and C. Each successive waveform is displaced 60° in phase, and consequently in time, from the preceding one.
As a result, a positive-voltage peak at point C occurs 180°, or half a cycle, after the identical peak at the plate. Since the grid is connected directly to point C, this positive-voltage peak will release maximum plate current through the tube. We already know that in a vacuum tube a high plate current usually coincides with a low plate voltage because of the voltage drop across plate load resistor R6.
Fig. 2. Operation of the phase-shift oscillator-negative half-cycle.
Fig. 3. Voltage waveforms at four key points in the phase-shift oscillator.
Also in Fig. 3, we see that a negative-voltage peak occurs at point C a half-cycle,
or 180°, after the plate voltage reaches its minimum. This peak, applied to
the control grid via the feed back line, drastically reduces the flow of plate
current. Minimum plate current usually coincides with maximum plate voltage,
because the lower current through load resistor R6 reduces the voltage drop
across it.
From the two preceding paragraphs and Fig. 3 we may conclude that the feedback has the appropriate phase to control the current flow in the tube and deliver an alternating current or oscillation at the output point. Several important approximations have been made in arriving at Fig. 3. The signal voltage will of course be attenuated as it passes through the R-C network.
The solid curves of lines 2, 3, and 4 represent the theoretical voltage waveforms at point A, B, and C, respectively, if no attenuation occurred. The waveforms shown in dashed curves indicate the actual reduction in strength at each point. The feed back voltage at point C is only a small fraction of the output voltage which causes it. This is normal for all oscillators-the tube acts as an amplifier; hence, only a small grid-driving voltage is required to control a much larger plate current through the tube and thereby contribute to large swings of plate voltage.
The values of resistors R3, R4, and R5 and capacitors C3, C4, and C5 determine the basic frequency of the voltage delivered to the output point. As an example, if each capacitor has a value of 0.01 microfarad, and each resistor a value of 10,000 ohms, the circuit will oscillate in the region of 700 or 800 cycles per second, well within the range of the human ear.
Let us consider each R-C combination separately (an approximation which the actual currents and voltages refuse to recognize). It is often said of capacitors that the current "leads" the voltage, frequently cited as unassailable evidence that some other related condition exists or will occur. Recall that in inductors the opposite is true-the current lags the voltage. (This latter property has been adequately described in earlier Sections.) Let us give some thought now to the significance of the axiom that the current leads the voltage in capacitors.
This statement is directly related to another widely quoted truth that capacitors will oppose any change in voltage. The existing voltage across the plates of a capacitor can be changed only by adding or withdrawing charged particles (electrons) and these electrons must enter or leave the capacitor first, in order for the voltage to change.
By using trigonometry it is possible to calculate the exact number of degrees the voltage developed across the capacitor will lag the current which produces it. The resultant is called the vector relationship between reactances and resistances, or be tween reactive and resistive voltages.
The maximum voltage developed across R3 should occur one sixth of a cycle after the maximum current flow into C3. Hence, the reactive voltage across the capacitor will be 1.73 times the resistive voltage. The only frequency which satisfies these conditions turns out to be 920 cycles per second. The reactance and resistance at the given frequency are arrived at from the trigonometric relationship existing between two legs of a right triangle with an included angle of 60°, as follows: C = .01 microfarad (10^-8 farads), R = 10,000 ohms (104 ohms), Xc=l.73R
We know the formula for capacitive reactance is: 1 Xe= 2.,,.fc
Therefore, substituting this formula for Xe we have: 1 2 pi fC = 1. 73R
Rearranging for f: 1 f = -2 pi r_X __ C_x_1 __ 7_3_x_R_
Substituting the known values and solving: 1 f = 6.28 X 10^-8 X 1.73 X 10^4 1 6.28 X 1.73 X 10^-4
1 10.86 X 10^-4 = 920 cycles.
If each of the three R-C combinations could be considered separately, a frequency of 920 cycles per second would satisfy the requirement that the reactive voltage lead the resistive volt age by 60°. This is another way of saying that the capacitor current should lead the resistor current by 60°, since the current through a resistor is always in phase with the voltage developed across the resistor. The three R-C combinations would then cause a 180° phase shift between the input current and feedback voltage. This solution is not 100% correct because all the current driven through capacitor C3 does not flow downward through R3, but divides between C4 and R3 in proportion to the impedance offered by each path. Likewise, when there is a negative-voltage peak at point A, the current it drives into C4 does not all flow down ward through R4 (the current shown by the dotted blue line), but again divides between R4 and C5 in proportion to the impedance offered by each path. The significance of these multiple-current paths is that the amplitudes and phase shifts of the voltages achieved at points A, B, and C cannot be calculated separately as was done earlier.
Also, these amplitudes and phase shifts will be modified by the type of circuit which follows.
Summary
1. Current must first flow into or out of a capacitor before the voltage across it can be changed. Therefore, the current leads the voltage in a capacitor.
2. In an R-C combination, maximum current will flow through the capacitor before it will through the resistor.
3. The voltage developed across a resistor by a current flowing through it is in phase with that current. Consequently, the voltage across a resistor lags the capacitor current.
4. A capacitor opposes the flow of electron current, the amount varying inversely with the frequency of the applied current, in accordance with the standard reactance formula.
5. The current in a capacitor leads the resulting voltage and the resistor current by the same number of degrees, which can be calculated by triangulation. The capacitance and resistance are considered two legs of a right triangle. When these two values are known, the third, or phase, angle can then be determined.
SCREEN-GRID AND FILTER CURRENTS
Little has been said so far about the remaining currents in the circuit. The screen-grid current, shown in solid blue, follows a closed path within the tube, from cathode to screen grid. Here it exits and flows through screen-dropping resistor R2 and the power supply, then through common ground and cathode biasing resistor R1, and back to the cathode.
During the first half-cycle, the control-grid voltage becomes more and more positive, and the plate- and screen-grid currents also steadily increase. This demand is satisfied by the electrons on the upper plate of cathode filter capacitor C1, and their departure from the top plate draws an equal number (labeled "cathode filter current") from ground to the bottom plate.
The increase in screen-grid current during the first half-cycle also drives an excess of electrons onto the upper plate of screen grid filter capacitor C2. Coincidentally, an equal number of electrons are driven from the lower plate to ground. This is the screen-grid filter current, also shown in solid blue.
In Fig. 2 we see that during the second half-cycle, when the control-grid voltage goes negative and reduces the plate and screen currents, both filter currents flow in the opposite direction.
The upper plate of C1 becomes positively charged and draws electrons to it through R1, permitting an equal number to flow back to ground from the lower plate. Therefore, the cathode filter current will always have the same frequency as the basic oscillator frequency. The cathode filter capacitor acts as a "shock absorber"--it keeps the voltage at the cathode from changing as the tube current changes.
The reduction in screen current during the second half-cycle permits the excess of electrons on the upper plate of C2 to be drawn off through R2 and the power supply. An equal number will then be drawn upward from ground to the lower plate of C2. Thus this capacitor also acts as a shock absorber by keeping the screen-grid voltage from changing. It does this by maintaining the flow of electrons constant through R2 so that the voltage drop (or rise) across it will likewise be steady. Thus, the screen-grid filter current will also remain at the basic oscillator frequency.