Topic: Digital Circuits
The voltage V in the circuit given below is :
- A. $Y=(AB+CD+CEB)$
- B. $Y=(AB+CD+AED+CEB)$
- C. $Y=(AB+AED+CEB)$
- D. $Y=(AB+CD)$
Answer: B. $Y=(AB+CD+AED+CEB)$
Explanation: The given circuit consists of an NMOS pull-down network configured as a bridge (A, B, C, D, and cross-connecting transistor E). The output is pulled low when there is a conducting path to ground. The four possible paths to ground are through transistors:
1. A and B ($AB$)
2. C and D ($CD$)
3. A, E, and D ($AED$)
4. C, E, and B ($CEB$)
Therefore, the logic function evaluating the ON condition of the pull-down network is $Y = AB + CD + AED + CEB$.