Monostable Multivibrator

Posted on Feb 5, 2014

The schematic for a monostable multivibrator is shown below. Like the astable multivibrator, one transistor conducts and the other cuts off when the circuit is energized. 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-sh

Monostable Multivibrator
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ot 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. The circuit is shown in its stable state. Let`s take a more detailed look at the circuit conditions in this stable state. 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. 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...

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