Multivibrators, like the familiar sinusoidal oscillators, are circuits with regenerative feedback, with the difference that they produce pulsed output. There are three basic types of multivibrator, namely the

**Bistable Multivibrator**, the

**Monostable Multivibrator**and the

**Astable Multivibrator**.

**1. Bistable Multivibrator**

A

*bistable multivibrator*circuit is one in which both LOW and HIGH output states are stable. Irrespective of the logic status of the output, LOW or HIGH, it stays in that state unless a change is induced by applying an appropriate trigger pulse. As we will see in the subsequent pages, the operation of a bistable multivibrator is identical to that of a flip-flop. Figure 1 shows the basic bistable multivibrator circuit. This is the fixed-bias type of bistable multivibrator. Other configurations are the self-bias type and the emitter-coupled type. However, the operational principle of all types is the same. The multivibrator circuit of Fig. 1 functions as follows. In the circuit arrangement of Fig. 1 it can be proved that both transistors Q

_{1}and Q_{2}cannot be simultaneously ON or OFF. If Q_{1}is ON, the regenerative feedback ensures that Q_{2}is OFF, and when Q_{1}is OFF, the feedback drives transistor Q_{2}to the ON state. In order to vindicate this statement, let us assume that both Q_{1}and Q_{2}are conducting simultaneously. Owing to slight circuit imbalance, which is always there, the collector current in one transistor will always be greater than that in the other. Let us assume that I_{C2}> I_{C1}. Lesser I_{C1}means a higher V_{C1}. Since V_{C1}is coupled to the Q_{2}base, a rise in V_{C1}leads to an increase in the Q_{2}base voltage. Increase in the Q_{2}base voltage results in an increase in I_{C2}and an associated reduction in V_{C2}Reduction in V_{C2}leads to a reduction in Q_{1}base voltage and an associated fall in I_{C1}, with the result that V_{C1}increases further. Thus, a slight circuit imbalance has initiated a regenerative action that culminates in transistor Q_{1}going to cut-off and transistor Q_{2}getting driven to saturation. To sum up, whenever there is a tendency of one of the transistors to conduct more than the other, it will end up with that transistor going to saturation and driving the other transistor to cut-off. Now, if we take the output from the Q_{1}collector, it will be LOW (= V_{CE1}sat.) if Q_{1}was initially in saturation. If we apply a negative-going trigger to the Q_{1}base to cause a decrease in its collector current, a regenerative action would set in that would drive Q_{2}to saturation and Q_{1}to cut-off. As a result, the output goes to a HIGH (=+V_{CC}) state. The output will stay HIGH until we apply another appropriate trigger to initiate a transition. Thus, both of the output states, when the output is LOW and also when the output is HIGH, are stable and undergo a change only when a transition is induced by means of an appropriate trigger pulse. That is why it is called a bistable multivibrator.**2. Monostable Multivibrator**

A

*monostable multivibrator*, also known as a*monoshot*, is one in which one of the states is stable and the other is quasi-stable. The circuit is initially in the stable state. It goes to the quasi-stable state when appropriately triggered. It stays in the quasi-stable state for a certain time period, after which it comes back to the stable state. Figure 2 shows the basic monostable multivibrator circuit. The circuit functions as follows. Initially, transistor Q

_{2}is in saturation as it gets its base bias from +V_{CC}through R. Coupling from Q_{2}collector to Q_{1}base ensures that Q_{1}is in cut off. Now, if an appropriate trigger pulse induces a transition in Q_{1}from saturation to cut-off, the output goes to the HIGH state. This HIGH output when coupled to the Q_{1}base turns Q_{1}ON. Since there is no direct coupling from Q_{1}collector to Q_{2}base, which is necessary for a regenerative process to set in, Q_{1}is not necessarily in saturation. However, it conducts some current. The Q_{1}collector voltage falls by I_{C1}R_{C1}and the Q_{2}base voltage falls by the same amount, as the voltage across a capacitor (C in this case) cannot change instantaneously. To sum up, the moment we applied the trigger, Q_{2}went to cut off and Q_{1}started conducting. But now there is a path for capacitor C to charge from V_{CC}through R and the conducting transistor. The polarity of voltage across C is such that the Q_{2}base potential rises. The moment the Q_{2}base voltage exceeds the cut-in voltage, it turns Q_{2}ON, which, owing to coupling through R_{1}, turns Q_{1}OFF. And we are back to the original state, the stable state. Whenever we trigger the circuit into the other state, it does not stay there permanently and returns back after a time period that depends upon R and C. The greater the time constant*RC,*the longer is the time for which it stays in the other state, called the quasi stable state.**3. Astable Multivibrator**

In the case of an astable multivibrator, neither of the two states is stable. Both output states are quasistable. The output switches from one state to the other and the circuit functions like a free running square-wave oscillator.

Figure 3 shows the basic astable multivibrator circuit. It can be proved that, in this type of circuit, neither of the output states is stable. Both states, LOW as well as HIGH, are quasi-stable. The time periods for which the output remains LOW and HIGH depends upon R

_{2}C_{2}and R_{1}C_{1}time constants respectively. For R_{1}C_{1}= R_{2}C_{2}, the output is a symmetrical square waveform. The circuit functions as follows. Let us assume that transistor Q_{2}is initially conducting, that is, the output is LOW. Capacitor C_{2}in this case charges through R_{2}and the conducting transistor from V_{CC}, and, the moment the Q_{1}base potential exceeds its cut in voltage, it is turned ON. A fall in Q_{1}collector potential manifests itself at the Q_{2}base as voltage across a capacitor cannot change instantaneously. The output goes to the HIGH state as Q_{2}is driven to cut-off. However, C_{1}has now started charging through R_{1}and the conducting transistor Q_{1}from VCC. The moment the Q2 base potential exceeds the cut-in voltage, it is again turned ON, with the result that the output goes to the LOW state. This process continues and, owing to both the couplings (Q_{1}collector to Q_{2}base and Q_{2}collector to Q_{1}base) being capacitive, neither of the states is stable. The circuit produces a square-wave output.
## No comments:

## Post a Comment