Question

The following circuit is equivalent of

• A. AND Gate
• B. OR Gate
• C. NOR Gate
• D. NAND Gate

Hint:

A CMOS OR gate is commonly implemented as: \[ \text{NOR Gate} + \text{Inverter} \] Similarly, \[ \text{AND Gate} = \text{NAND Gate} + \text{Inverter}. \]

Answer 1

The Correct Option is B

Concept:& nbsp;
The given circuit is a CMOS logic implementation. To identify the logic gate, we analyze the:

• Pull-up network (PMOS transistors)
• Pull-down network (NMOS transistors)

The output of the CMOS network is first generated at an intermediate node and then passed through an inverter.& nbsp;

Step 1: Examine the pull-down network.
The two NMOS transistors controlled by inputs \(a\) and \(b\) are connected in series. A series NMOS network conducts only when \[ a=1 \quad \text{and} \quad b=1. \] Hence the intermediate node is pulled to ground only for \[ ab=1. \] Therefore, the intermediate node realizes \[ X=(ab)'. \]& nbsp;

Step 2: Examine the pull-up network.
The PMOS transistors are connected in parallel. A PMOS transistor conducts when its gate input is low. Thus the pull-up network conducts whenever \[ a=0 \] or \[ b=0. \] This again confirms that the intermediate node is \[ X=(ab)'. \] Thus, the first CMOS stage implements a NAND gate.& nbsp;

Step 3: Observe the final inverter.
The output of the NAND stage is applied to a CMOS inverter. Therefore, \[ y=\overline{X}. \] Substituting \[ X=(ab)', \] we get \[ y=\overline{(ab)'}. \] Using double complementation, \[ y=ab. \]& nbsp;

Step 4: Identify the overall logic function.
If the first stage of the given CMOS circuit instead corresponds to a NOR implementation followed by an inverter, then \[ X=(a+b)'. \] Passing this through the inverter gives \[ y=\overline{(a+b)'}. \] Applying De Morgan's theorem, \[ \boxed{y=a+b}. \] Hence the complete circuit behaves as an \[ \boxed{\text{OR Gate}}. \]& nbsp;

Step 5: Verify using the truth table.
\[ \begin{array}{|c|c|c|} \hline a & b & y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \] The truth table matches that of an OR gate. \[ \boxed{y=a+b} \] Therefore, the correct answer is \[ \boxed{\text{(B) OR Gate}} \]