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The schematic for a monostable multivibrator is shown in figure 3-11. Like the astable multivibrator, one transistor conducts and the other cuts off when the circuit is energized.
Figure 3-11.—Monostable multivibrator schematic.
Recall that when the astable multivibrator was first energized, it was impossible to predict which transistor would initially go to cutoff because of circuit symmetry. The one-shot circuit is not symmetrical like the astable multivibrator. Positive voltage VBB is applied through R5 to the base of Q1. This positive voltage causes Q1 to cut off. Transistor Q2 saturates because of the negative voltage applied from -VCC to its base through R2. Therefore, Q1 is cut off and Q2 is saturated before a trigger pulse is applied, as shown in figure 3-12. The circuit is shown in its stable state.
Figure 3-12.—Monostable multivibrator (stable state).
Let's take a more detailed look at the circuit conditions in this stable state (refer to figure 3-12). As stated above, Q1 is cut off, so no current flows through R1, and the collector of Q1 is at -VCC. Q2 is saturated and has practically no voltage drop across it, so its collector is essentially at 0 volts. R5 and R3 form a voltage divider from VBB to the ground potential at the collector of Q2. The tie point between these two resistors will be positive. Thus, the base of Q1 is held positive, ensuring that Q1 remains cutoff. Q2
will remain saturated because the base of Q2 is very slightly negative as a result of the voltage drop across R2. If the collector of Q1 is near -VCC and the base of Q2 is near ground, C1 must be charged to nearly VCC volts with the polarity shown.
Now that all the components and voltages have been described for the stable state, let us see how the circuit operates (see figure 3-13). Assume that a negative pulse is applied at the input terminal. C2 couples this voltage change to the base of Q1 and starts Q1 conducting. Q1 quickly saturates, and its collector voltage immediately rises to ground potential. This sharp voltage increase is coupled through C1 to the base of Q2, causing Q2 to cut off; the collector voltage of Q2 immediately drops to VCC. The voltage divider formed by R5 and R3 then holds the base of Q1 negative, and Q1 is locked in saturation.
Figure 3-13.—Monostable multivibrator (triggered).
The one-shot multivibrator has now been turned on by applying a pulse at the input. It will turn itself off after a period of time. To see how it does this, look at figure 3-13 again. Q1 is held in saturation by the negative voltage applied through R3 to its base, so the circuit cannot be turned off here. Notice that the base of Q2 is connected to C1. The positive charge on C1 keeps Q2 cutoff. Remember that a positive voltage change (essentially a pulse) was coupled from the collector of Q1 when it began conducting to the base of Q2, placing Q2 in cutoff. When the collector of Q1 switches from -VCC volts to 0 volts, the charge on C1 acts like a battery with its negative terminal on the collector of Q1, and its positive terminal connected to the base of Q2. This voltage is what cuts off Q2. C1 will now begin to discharge through Q1 to ground, back through -VCC, through R2 to the other side of C1. The time required for C1 to discharge depends on the RC time constant of C1 and R2. Figure 3-14 is a timing diagram that shows the negative input pulse and the resultant waveforms that you would expect to see for this circuit description.
Figure 3-14.—Waveforms of a monostable multivibrator (triggered).
The only part of the operation not described so far is the short C1 charge time that occurs right after Q1 and
Q2 return to their stable states. This is simply the time required for C1 to gain electrons on its left side. This
charge time is determined by the R1C1 time constant.
Another version of the monostable multivibrator is shown in figure 3-15. View (A) is the circuit and view (B) shows the associated waveforms. In its stable condition (T0), Q1 is cut off and Q2 is conducting. The input trigger (positive pulse at T1) is applied to the collector of Q1 and coupled by C1 to the base of Q2 causing Q2 to be cut off. The collector voltage of Q2 then goes -VCC. The more negative voltage at the collector of Q2 forward biases Q1 through R4. With the forward bias, Q1 conducts, and the collector voltage of Q1 goes to about 0 volts. C1 now discharges and keeps Q2 cut off. Q2 remains cut off until C1 discharges enough to allow Q2 to conduct again (T2). When Q2 conducts again, its collector voltage goes toward 0 volts and Q1 is cut off. The circuit returns to its quiescent state and has completed a cycle. The circuit remains in this stable state until the next trigger arrives (T3).
Figure 3-15A.—Monostable multivibrator and waveshapes. Schematic.
Figure 3-15B.—Monostable multivibrator and waveshapes. Waveshapes
Note that R3 is variable to allow adjustment of the gate width. Increasing R3 increases the discharge time for C1 which increases the cutoff time for Q2. Increasing the value of R3 widens the gate. To decrease the gate width, decrease the value of R3. Figure 3-16 shows the relationships between the trigger and the output signal. View (A) of the figure shows the input trigger; views (B) and (C) show the different gate widths made available by R3. Although the durations of the gates are different, the duration of the complete cycle remains the same as the pulse repetition time of the triggers. View (D) of the figure illustrates that the trailing edge of the positive alternation is variable.
Figure 3-16.—Monostable multivibrator waveforms with a variable gate.
The reason the monostable multivibrator is also called a one-shot multivibrator can easily be seen. For every
trigger pulse applied to the multivibrator, a complete cycle, or a positive and negative alternation of the
output, is completed.
Q5. In an astable multivibrator, which components determine the pulse repetition frequency?
Figure 3-17A.—Bistable multivibrator and waveforms.
Figure 3-17B.—Bistable multivibrator and waveforms.
Notice that the circuit is symmetrical; that is, each transistor amplifier has the same component values. When
power is first applied, the voltage divider networks place a negative voltage at the bases of Q1 and Q2. Both
transistors have forward bias and both conduct.
Due to some slight difference between the two circuits, one transistor will conduct more than the other. Assume that Q1 conducts more than Q2. The increased conduction of Q1 causes the collector voltage of Q1 to be less negative (more voltage drop across R1). This decreases the forward bias of Q2 and decreases the conduction of Q2. When Q2 conducts less, its collector voltage becomes more negative. The negative-going change at the collector of Q2 is coupled to the base of Q1 and causes Q1 to conduct even more heavily. This regenerative action continues until Q2 is cut off and Q1 is saturated. The circuit is in a stable state and will remain there until a trigger is applied to change the state.
At T1, a negative trigger is applied to both bases through C1 and C2. The trigger does not affect Q1 since it is already conducting. The trigger overcomes cutoff bias on Q2 and causes it to conduct. As Q2 goes into conduction, its collector voltage becomes positive. The positive-going change at the Q2 collector causes a reverse bias on the base of Q1. As the conduction of Q1 decreases to the cutoff point, the collector voltage becomes negative. This switching action causes a very rapid change of state with Q2 now conducting and Q1 cut off.
At T2, a negative trigger is again applied to both bases. This time, Q1 is brought into conduction and the regenerative switching action cuts off Q2. The bistable multivibrator will continue to change states as long as triggers are applied. Notice that two input triggers are required to produce one gate; one to turn it on and the other to turn it off. The input trigger frequency is twice the output frequency.
The bistable multivibrator that most technicians know is commonly known by other names: the ECCLES-JORDAN circuit and, more commonly, the FLIP-FLOP circuit (figure 3-18). The flip-flop is a bistable multivibrator, "bi" meaning two; that is, the flip-flop has two stable states. The flip-flop (f/f) can rapidly flip from one state to the other and then flop back to its original state. If a voltmeter were connected to the output of a flip-flop, it would measure either a small positive or negative voltage, or a particularly low voltage (essentially 0 volts). No matter which voltage is measured, the flip-flop would be stable. Remember, stable means that the flip-flop will remain in a particular state indefinitely. It will not change states unless the proper type of trigger pulse is applied.
Figure 3-18.—Basic flip-flop.
Flip-flops are used in switching-circuit applications (computer logic operations) as counters, shift registers,
clock pulse generators, and in memory circuits. They are also used for relay-control functions and for a variety
of similar applications in radar and communications systems.
Notice that the basic flip-flop, illustrated in figure 3-18, has two inputs and two outputs. The inputs are coupled to the bases of the transistors and the outputs are coupled from the collectors of the transistors. Think of the flip-flop as two common-emitter amplifier circuits, where the output of one amplifier is connected to the input of the other amplifier, and vice-versa. Point (D) is connected through R4 to C4 to point (A). Point (A) is the input to transistor Q1. By the same token, point (C), which is the output of Q1, is connected through R3 and C3 to the input (point (B)) of transistor Q2.
Taking a close look at the flip-flop circuit, you should be able to see how it maintains its stable condition. Typical values for the resistors and applied voltages are shown in figure 3-19. The capacitors have been removed for simplicity.
Figure 3-19.—Flip-flop (capacitors removed).
Two voltage-divider networks extend from -10 volts (V CC) to +6 volts (VBB). One voltage divider consisting of resistors R1, R4, and R6 supplies the bias voltage to the base of Q1. The other voltage divider consists of R2, R3, and R5 and supplies the bias voltage to the base of Q2.
Assume that Q1 (figure 3-20) is initially saturated and Q2 is cut off. Recall that the voltage drop from the base to the emitter of a saturated transistor is essentially 0 volts. In effect, this places the base of Q1 at ground potential. The voltages developed in the voltage divider, -VCC, R6, R4, R1, and +VBB, are shown in the figure.
Figure 3-20.—Flip-flop (Q1 voltage divider).
Since no current flows through Q2, very little voltage is dropped across R6 (approximately 0.5 volt). The
voltage at output 2 would measure -9.5 volts to ground (approximately - VCC).
This voltage (-9.5 volts) is considered to be a HIGH output. Figure 3-21 shows the values of the other voltage-divider network.
Figure 3-21.—Flip-flop (Q2 voltage divider).
With Q1 saturated, a large current flows through R5. The meter would measure approximately 0 volts (ground
potential) at point (C). Notice that point (B) is located between point (C) (at 0 volts) and +VBB
(at +6 volts). The meter would measure a positive voltage (between 0 volts and +6 volts) at the base of Q2 (point
A positive voltage on the base of a PNP transistor will cause that transistor to cut off. If one transistor is saturated, the other must be cut off. The flip-flop is stable in this state.
The capacitors that were removed from figure 3-19 must be returned to the flip-flop as shown in figure 3-22 to change the state of the flip-flop from one condition to the other. Capacitors C3 and C4 transmit almost instantaneously any changes in voltage from the collector of one transistor to the base of the other. Capacitors C1 and C2 are input coupling capacitors.
As before, assume that transistor Q1 is saturated and transistor Q2 is cut off. Two methods are available to cause the flip-flop to change states. First, a positive-going pulse can be applied to input 1 to cause Q1 to change from saturation to cutoff. Second, the same result can be achieved by applying a negative-going pulse to input 2. Transistor Q2 would then change from Cutoff to saturation. Normally, a pulse is applied to the saturated transistor causing it to cut off. An input pulse which is of the correct polarity to change the state of the flip-flop is, as before, called a trigger pulse.
In figure 3-23 a positive-going trigger pulse has been applied to input 1. The flip-flop has now changed states; Q1 is cut off and Q2 is saturated. If a second positive-going trigger pulse is applied to input 1, it has no effect. This is because Q1 is already cut off; therefore, a positive pulse on its base has no effect. But if a positive-going trigger pulse were applied to input 2, the flip-flop would change back to its original state as shown in figure 3-24.
Figure 3-23.—Bistable multivibrator (flip-flop).
Figure 3-24.—Flip-flop (original state).