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Answered 2012-04-03 13:12:01

The circuit has two stable states that change alternatively with maximum transition rate because of the "accelerating" positive feedback. It is implemented by the coupling capacitors that instantly transfer voltage changes because the voltage across a capacitor cannot suddenly change. In each state, one transistor is switched on and the other is switched off. Accordingly, one fully charged capacitor discharges (reverse charges) slowly thus converting the time into an exponentially changing voltage. At the same time, the other empty capacitor quickly charges thus restoring its charge (the first capacitor acts as a time-setting capacitor and the second prepares to play this role in the next state). The circuit operation is based on the fact that the forward-biased base-emitter junction of the switched-on bipolar transistor can provide a path for the capacitor restoration.

State 1 (Q1 is switched on, Q2 is switched off):

In the beginning, the capacitor C1 is fully charged (in the previous State 2) to the power supply voltage V with the polarity shown in Figure 1. Q1 is on and connects the left-hand positive plate of C1 to ground. As its right-hand negative plate is connected to Q2 base, a maximum negative voltage (-V) is applied to Q2 base that keeps Q2 firmly off. C1 begins discharging (reverse charging) via the high-resistive base resistor R2, so that the voltage of its right-hand plate (and at the base of Q2) is rising from below ground (-V) toward +V. As Q2 base-emitter junction is backward-biased, it does not impact on the exponential process (R2-C1 integrating network is unloaded). Simultaneously, C2 that is fully discharged and even slightly charged to 0.6 V (in the previous State 2) quickly charges via the low-resistive collector resistor R4 and Q1 forward-biased base-emitter junction (because R4 is less than R2, C2 charges faster than C1). Thus C2 restores its charge and prepares for the next State 2 when it will act as a time-setting capacitor. Q1 is firmly saturated in the beginning by the "forcing" C2 charging current added to R3 current; in the end, only R3 provides the needed input base current. The resistance R3 is chosen small enough to keep Q1 (not deeply) saturated after C2 is fully charged.

When the voltage of C1 right-hand plate (Q2 base voltage) becomes positive and reaches 0.6 V, Q2 base-emitter junction begins diverting a part of R2 charging current. Q2 begins conducting and this starts the avalanche-like positive feedback process as follows. Q2 collector voltage begins falling; this change transfers through the fully charged C2 to Q1 base and Q1 begins cutting off. Its collector voltage begins rising; this change transfers back through the almost empty C1 to Q2 base and makes Q2 conduct more thus sustaining the initial input impact on Q2 base. Thus the initial input change circulates along the feedback loop and grows in an avalanche-like manner until finally Q1 switches off and Q2 switches on. The forward-biased Q2 base-emitter junction fixes the voltage of C1 right-hand plate at 0.6 V and does not allow it to continue rising toward +V.

State 2 (Q1 is switched off, Q2 is switched on):

Now, the capacitor C2 is fully charged (in the previous State 1) to the power supply voltage V with the polarity shown in Figure 1. Q2 is on and connects the right-hand positive plate of C2 to ground. As its left-hand negative plate is connected to Q1 base, a maximum negative voltage (-V) is applied to Q1 base that keeps Q1 firmly off. C2 begins discharging (reverse charging) via the high-resistive base resistor R3, so that the voltage of its left-hand plate (and at the base of Q1) is rising from below ground (-V) toward +V. Simultaneously, C1 that is fully discharged and even slightly charged to 0.6 V (in the previous State 1) quickly charges via the low-resistive collector resistor R1 and Q2 forward-biased base-emitter junction (because R1 is less than R3, C1 charges faster than C2). Thus C1 restores its charge and prepares for the next State 1 when it will act again as a time-setting capacitor...and so on... (the next explanations are a mirror copy of the second part of Step 1).

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No triggering is used on an astable multivibrator because it can cause it to become unstable.

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the astable multivibrator has two quasi stable states whereas a monostable multivibrator has one stable and one quasi stable state. Also in astable multivibrator, the output of the capacitor is used as the input at the non inverting terminal of the comparator, whereas in monostable multivibrator the state transition is caused by thge application of a negative trigger pulse on yhe inverting input of the comparator.

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