The Boolean relation at the output stage Y fo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The circuit consists of two switches $A$ and $B$ connected to diodes,which act as an $OR$ gate input stage. When either switch $A$ or $B$ is close

Vedclass · Accepted Answer

$\overline{A} \cdot \overline{B}$

Vedclass · Answer

(C) The circuit consists of two switches $A$ and $B$ connected to diodes,which act as an $OR$ gate input stage. When either switch $A$ or $B$ is closed (logic $1$),the base of the $NPN$ transistor receives a high voltage,causing it to conduct. When the transistor conducts,the collector voltage drops to near $0 \ V$ (logic $0$). If both switches are open (logic $0$),the base is grounded through the resistor,the transistor is in cutoff $(OFF)$,and the output $Y$ is pulled up to $+5 \ V$ (logic $1$). This behavior corresponds to a $NOR$ gate,where $Y = \overline{A+B}$. According to De Morgan's theorem,$\overline{A+B} = \overline{A} \cdot \overline{B}$.$A$$B$$Y$$0$$0$$1$$1$$0$$0$$0$$1$$0$$1$$1$$0$

The Boolean relation at the output stage $Y$ for the following circuit is:

Similar Questions

Identify the logic operation carried out by the given circuit :-

In which of the logic gate the following statement is true? "The output is high when either of the inputs is high but not if both inputs are high"

The following configuration of gates is equivalent to $:-$

The output $(Y)$ of the logic circuit given below is $0$ only when:

What will be the input of $A$ and $B$ for the Boolean expression $\overline{(A+B) \cdot(A \cdot B)}=1$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module