The way I count it, addition can be done in depth 3.  Assume $a_i$ and $b_i$ are the $i$th bits of the two numbers, where $0$ is the index of the LSB and $n$ of the MSB.  Let us compute the $i$th bit of the sum, $s_i$ in the standard way with carry look ahead:

$s_i = a_i \textrm{ xor } b_i \textrm{ xor } c_i$

where $c_i$ is the carry computed as:

$c_i = \textrm{OR}_{j\mid j < i} (g_j \textrm{ and } p_j)$

and $g_j$ means that the $j$th location "generated" the carry:

$g_j = (a_j \textrm{ and } b_j)$

and $p_j$ means that the carry gets propagated from $j$ to $i$:

$p_j = \textrm{AND}_{k\mid j < k < i} (a_j \textrm{ or } b_j)$

Counting the depth, $p_j$ is depth 2, and $c_i$ is depth 3.  While it would seem that $s_i$ is depth 4 or 5, it really is also just depth 3 since it is a bounded fanin computation of depth 3 circuits so one may push the top two levels down using de-Morgan formulas, while blowing the circuit size by a polynomial amount.

Depth 2 circuits require exponential size to compute $c_n$ since depth two circuits must be DNF or CNF and it is easy to verify that there are exponentially many minterms and maxterms in the function $c_n$.