Question

The MOSFET switches shown in the circuit are ideal.

Which of the following is the correct option for Boolean logical expression of the output (OUT), and the maximum possible power (P) consumed by the circuit?

• A. OUT = \(\overline{AB + \bar C}\), P = 5 mW
• B. OUT = \(\overline{(A + B)\bar C}\), P = 5 mW
• C. OUT = \(\overline{AB\bar C} + \bar C\), P = 7.5 mW
• D. OUT = \(\overline{ABC}\), P = 7.5 mW

Hint:

OUT is pulled low only when every pull-down switch conducts together; find which input combination makes both resistor branches draw current at once.

Answer 1

The Correct Option is D

Step 1: Identify the pull-up and pull-down structure.
The OUT node is pulled up to \(5\) V through the \(5\) k\(\Omega\) resistor. It is pulled down to ground through the NMOS switches gated by \(A\) and \(B\), together with the switching action set up by the \(10\) k\(\Omega\) resistor and the transistor gated by \(C\), so OUT is pulled low only when a complete conducting path exists that needs all three of \(A\), \(B\), and \(C\) satisfied together.

Step 2: Recall how an ideal NMOS switch behaves.
An NMOS switch conducts (behaves like a short) when its gate is HIGH (logic \(1\)) and is open (like a break) when its gate is LOW (logic \(0\)). With the switches ideal, when the pull-down path conducts, OUT is pulled all the way to \(0\) V; when no pull-down path conducts, OUT is pulled all the way up to \(5\) V through the resistor (no current flows, so there is no drop across the \(5\) k\(\Omega\)).

Step 3: Combine the conditions for a conducting pull-down path.
The pull-down path conducts, forcing \(OUT=0\), only when \(A=1\), \(B=1\), and \(C=1\) all hold together; if even one of \(A\), \(B\), \(C\) is \(0\), the path is broken and no current can flow to ground, so OUT is pulled high.

Step 4: Write the Boolean expression.
\[ OUT = \overline{A\cdot B\cdot C} \]
This is a 3-input NAND function; it is \(1\) for every input combination except \(A=B=C=1\), where it is \(0\).

Step 5: Find the current through the \(5\) k\(\Omega\) branch.
This branch carries current only when \(A=B=C=1\) (the only case where OUT is pulled to \(0\) V):
\[ I_{5k} = \frac{5\text{ V}}{5\text{ k}\Omega} = 1\text{ mA} \implies P_{5k} = 5\text{ V}\times1\text{ mA} = 5\text{ mW} \]
For every other input combination this branch draws no current.

Step 6: Find the current through the \(10\) k\(\Omega\) branch.
This branch, together with the transistor gated by \(C\), forms a simple inverter for the internal node. It draws current whenever \(C=1\) (the transistor gated by \(C\) is then ON, pulling its node to \(0\) V and putting the full \(5\) V across the \(10\) k\(\Omega\)):
\[ I_{10k} = \frac{5\text{ V}}{10\text{ k}\Omega} = 0.5\text{ mA} \implies P_{10k} = 5\text{ V}\times0.5\text{ mA} = 2.5\text{ mW} \]

Step 7: Find when both branches draw current at the same time.
Both branches conduct together only when \(A=B=C=1\) (this satisfies \(A=B=C=1\) for the \(5\) k\(\Omega\) branch and \(C=1\) for the \(10\) k\(\Omega\) branch at once). This is exactly the condition that gives the maximum total current drawn from the \(5\) V supply.

Step 8: Add the two powers for the maximum case.
\[ P_{max} = P_{5k}+P_{10k} = 5\text{ mW}+2.5\text{ mW} = 7.5\text{ mW} \]

Step 9: Rule out the other options.
Options (A) and (B) both state \(P=5\) mW, which only accounts for the \(5\) k\(\Omega\) branch and misses the extra \(2.5\) mW drawn through the \(10\) k\(\Omega\) branch when \(C=1\). Option (C), on simplification, reduces to \(\bar A+\bar B+C+\bar C=1\) (a constant \(1\)), since \(C+\bar C\) is always true, so it cannot describe a circuit whose output changes with the inputs.

Final Answer:
\[ \boxed{OUT=\overline{ABC},\quad P_{max}=7.5\text{ mW}} \]