Bistable Multivibratoron 24/2/2012 & Updated on 7/9/2018
Initially, let us consider that the SPDT switch is position 1 which inturn grounds the base of the transistor Q1. As a result, Q1 will be OFF (cutoff region) while its collector will be held at VCC, due to which the output at O1 will go high. This inturn forward biases the BE junction of transistor Q2, switching it ON (into saturation mode of operation). Due to this, the collector current flows through the collector resistor RC2, shorting the collector terminal of Q2 to ground. Thus, for this case, the output at O2 terminal goes low.
This state of the circuit remains unchanged for an indefinite period of time, unless triggered externally. In this case, the act of changing the switch position from 1 to 2 acts like an external trigger for the circuit. When done so, the base of transistor Q2 will be grounded, switching it OFF (cutoff region). This also causes the VCC to appear at the collector terminal of Q2, which inturn results in a high output at O2 terminal. Further, at this state, Q1 will switch ON (gets into saturation mode of operation) as it has its base connected to the collector terminal of Q2 via R2. Due to this, the collector terminal of Q1 will be shorted to ground, causing the output at the terminal O1 to go low. This state of the circuit is again maintained until triggered once again.
From the explanation presented, the following two points can be concluded on the nature of the bistable circuits.
- Bistable circuits are not self-triggered as they rely on the user-provided trigger inputs so as to change their state.
- In these circuits, the output wave-forms obtained at the terminal O1 and O2 are complementary to each other, always.
Rating = 0 & Total votes = 0