Design and Explain a 4-Bit Carry-Lookahead Adder (CLA)

A carry-lookahead adder (CLA) speeds up addition by reducing the time to determine carries. Instead of waiting for each carry to ripple through bit positions, a CLA computes carries from generate and propagate terms using combinational logic. The core idea is:

• $g_i = a_i b_i$
• $p_i = a_i \oplus b_i$
• Carries are expressed as functions of $c_0$ and the $g$/$p$ terms.

This approach forms the standard carry lookahead structure used in multi-bit adders.

Key terms for this course section:

• generate
• propagate
• carry-in
• carry-out
• critical path

4-bit CLA design goal

Design a 4-bit combinational circuit that:

1. Computes $p_0..p_3$ and $g_0..g_3$.
2. Computes internal carries $c_1..c_4$ in parallel from $c_0$.
3. Computes sums $s_0..s_3$.

We will use the standard CLA recurrence to derive each carry: $c_{i+1} = g_i + p_i c_i$

Then expand recursively so that each $c_{i}$ becomes a Boolean expression in terms of $c_0$ and the $g$/$p$ signals.

Boolean carry network (final expressions)

For a 4-bit CLA:

Carries

$$\begin{aligned} c_1 &= g_0 + p_0 c_0 \\ c_2 &= g_1 + p_1 g_0 + p_1 p_0 c_0 \\ c_3 &= g_2 + p_2 g_1 + p_2 p_1 g_0 + p_2 p_1 p_0 c_0 \\ c_4 &= g_3 + p_3 g_2 + p_3 p_2 g_1 + p_3 p_2 p_1 g_0 + p_3 p_2 p_1 p_0 c_0 \end{aligned}$$

Sums

$$s_i = p_i \oplus c_i \quad \text{for } i=0,1,2,3$$

Example: compute a 4-bit CLA addition

Let:

• $a = 1011_2$ ($a_3a_2a_1a_0 = 1,0,1,1$)
• $b = 0110_2$ ($b_3b_2b_1b_0 = 0,1,1,0$)
• $c_0 = 1$

We compute $g_i=a_i b_i$ and $p_i=a_i\\oplus b_i$.

Step 1: table of g and p

So:

• $g_0=0, g_1=1, g_2=0, g_3=0$
• $p_0=1, p_1=0, p_2=1, p_3=1$

Step 2: compute carries

• $c_1 = g_0 + p_0 c_0 = 0 + 1\\cdot 1 = 1$
• $c_2 = g_1 + p_1 g_0 + p_1 p_0 c_0 = 1 + 0\\cdot 0 + 0\\cdot 1\\cdot 1 = 1$
• $c_3 = g_2 + p_2 g_1 + p_2 p_1 g_0 + p_2 p_1 p_0 c_0$ $= 0 + 1\\cdot 1 + 1\\cdot 0\\cdot 0 + 1\\cdot 0\\cdot 1\\cdot 1 = 1$
• $c_4 = g_3 + p_3 g_2 + p_3 p_2 g_1 + p_3 p_2 p_1 g_0 + p_3 p_2 p_1 p_0 c_0$ $= 0 + 1\\cdot 0 + 1\\cdot 1\\cdot 1 + 1\\cdot 1\\cdot 0\\cdot 0 + 1\\cdot 1\\cdot 0\\cdot 1\\cdot 1 = 1$

So carries are: $c_0=1, c_1=1, c_2=1, c_3=1, c_4=1$.

Step 3: compute sums

Use $s_i = p_i \\oplus c_i$:

• $s_0 = p_0 \\oplus c_0 = 1 \\oplus 1 = 0$
• $s_1 = p_1 \\oplus c_1 = 0 \\oplus 1 = 1$
• $s_2 = p_2 \\oplus c_2 = 1 \\oplus 1 = 0$
• $s_3 = p_3 \\oplus c_3 = 1 \\oplus 1 = 0$

Therefore:

• $s = 0010_2$ with final carry $c_4=1$. Interpreting the full result: $c_4 s_3 s_2 s_1 s_0 = 1\,0\,0\,1\,0_2 = 18_{10}$.

Quick check by arithmetic:

• $a=11$, $b=6$, $c_0=1$ ⇒ $11+6+1=18$ ✅

Visual interpretation of operation

A CLA first classifies each bit pair as:

• Generate ($g_i=1$): forces a carry regardless of previous carry.
• Propagate ($p_i=1$): carries the previous carry forward.

Then the carry network computes $c_1..c_4$ with AND/OR combinations of $g$ and $p$, and the sum stage uses XOR with the corresponding carry.