Formulas for which any equivalent CNF formula has exponential length
Try parity. Parity of $n$ variables has formulas of size $O(n^2)$, but every clause in the CNF must be of width $n$, so there must be at least $2^{n-1}$ of them.

Here are some more details. The $O(n^2)$ formula for parity is obtained by a recursive construction. For simplicity, let's assume that $n = 2^m$ and only count leaves. We use the following formulas:
$$
\begin{align*}
\operatorname{parity}(x_1,\ldots,x_{2n}) = &amp;(\operatorname{parity}(x_1,\ldots,x_n) \land \lnot \operatorname{parity}(x_{n+1},\ldots,x_{2n})) \\ \lor &amp;(\lnot \operatorname{parity}(x_1,\ldots,x_n) \land \operatorname{parity}(x_{n+1},\ldots,x_{2n}))
\end{align*}$$
and $\operatorname{parity}(x_1) = x_1$. Let $P(n)$ be the number of leaves in $\operatorname{parity}(x_1,\ldots,x_n)$. Then
$$ P(2n) = 4P(n), \quad P(1) = 1. $$
The solution to this recurrence is $P(2^n) = 4^n$. Since parity of $n$ variables can be obtained from parity of $2^{\lceil \log n \rceil} &lt; 2n$ variables, in total parity for $n$ variables can be realized using a formula containing at most $4n^2$ leaves. The length of a binary encoding of the formula is $O(n^2\log n)$ (the $\log n$ factor comes from the fact that the indices have length $\log n$).

Regarding CNFs, let's consider DNFs instead; the situation is symmetric. Suppose $\phi = T_1 \lor \cdots \lor T_\ell$ is a DNF for $\operatorname{parity}$. Each $T_i$ must contain $n$ variables: given the values of even $n-1$ variables, you cannot be certain that $\operatorname{parity}$ is true. So each term "catches" one $1$-input of $\operatorname{parity}$, of which there are $2^{n-1}$.