Let $E_1$ and $E_2$ be isomorphic elliptic curves. Then the NeronSever group of $S:=E_1 \times E_2$ is generated by 3 elements: $E_1$, $E_2$ and the diagonal. I want to see this in de Rham cohomology (after tensoring by $\mathbb{C}$). Let $dz_1$ and $dz_2$ be the generator of $H^{1,0}$ of $E_1$ and $E_2$ respectively. Then forms like $$ dz_1 \wedge d\bar{z_1}, \ dz_2 \wedge d\bar{z_2}, \ dz_1\wedge dz_2… $$ span a basis of $H^2(S,\mathbb{C}) \cong \mathbb{C}^6$. We see that $dz_1 \wedge d\bar{z_1}$ and $dz_2 \wedge d\bar{z_2}$ represents (certain multiple of) the classes $E_1$ and $E_2$ respectively. What about the diagonal? By considering intersection, we know that them which corresponds to the diagonal contains $$ dz_1 \wedge d\bar{z_1}+dz_2 \wedge d\bar{z_2}, $$ but I don’t know the coefficients of the other terms like $dz_1\wedge dz_2$ and $dz_1\wedge d\bar{z_2}$.

3$\begingroup$ If E has complex multiplication, there is a 4th class the graph of the complex multiplication. $\endgroup$ – aginensky Feb 7 '17 at 16:18
Actually, the sign stuff is much more confusing than I thought, and my scalars were off too, so I've completely rewritten my answer to explain this a little better.
The natural Poincaré pairing is defined by \begin{align*} (,) : H^*(E,\mathbb R) \otimes H^*(E,\mathbb R) &\to \mathbb R\\ \alpha \otimes \beta &\mapsto \int_E \alpha \wedge \beta. \end{align*} The pairing is understood to be $0$ if $\alpha \wedge \beta$ is a $k$form for $k \neq \dim_\mathbb R E = 2$. It can be extended to either a bilinear form or a sesquilinear form $$H^*(E,\mathbb C) \otimes H^*(E,\mathbb C) \to \mathbb C.$$ This is where the confusion starts. Note for example that the sesquilinear form is not Hermitian, because the bilinear form we started with is not symmetric (rather, it's graded symmetric). Thus, it seems more natural to me to pick the bilinear extension: \begin{align*} (,) : H^*(E,\mathbb C) \otimes H^*(E,\mathbb C) &\to \mathbb C\\ \alpha \otimes \beta &\mapsto \int_E \alpha \wedge \beta. \end{align*} Given a basis $x_i$ of $H^*(E)$, we will call a right dual basis $x_i^*$ a basis such that $$(x_i,x_j^*) = \delta_{ij}.$$ We could also work with left dual bases, which differs only up to a suitable sign from this notion. Recall also that the Künneth isomorphism is given by \begin{align*} H^*(E) \otimes H^*(E) &\stackrel\sim\longrightarrow H^*(E \times E)\\ \alpha \otimes \beta &\longmapsto \alpha \times \beta := \pi_1^* \alpha \wedge \pi_2^* \beta. \end{align*} Now if $z$ denotes the coordinate on the universal cover $\mathbb C \twoheadrightarrow E$, then $dz$ is a basis for $H^{1,0}(E)$. Moreover, after possibly rescaling, we get $$\int_E dz \wedge d\bar z = 2i \int_E dx \wedge dy = 2i.$$ In particular, the right dual basis of $(1, dz, d\bar z, dz \wedge d\bar z)$ is $\tfrac{1}{2i}(dz \wedge d\bar z, d\bar z, dz, 1)$. Now, it follows from Poincaré duality and the gradedcommutativity of the cohomology ring that $$[\Delta] = \sum_i (1)^{\deg{x_i}} x_i \times x_i^*,$$ where the $x_i$ are a homogeneous basis for $H^*(E)$, with right dual basis $x_i^*$. See for example Bott–Tu, Lemma 11.22, or Milnor–Stasheff, Theorem 11.11. Thus, in our case, we find \begin{align*} [\Delta] &= \tfrac{1}{2i} \left( 1 \times (dz \wedge d\bar z)  dz \times d\bar z + d\bar z \times dz + (dz \wedge d\bar z) \times 1\right)\\ &= \tfrac{1}{2i} \left( dz_2 \wedge d\bar z_2  dz_1 \wedge d\bar z_2 + d\bar z_1 \wedge dz_2 + dz_1 \wedge d\bar z_1 \right). \end{align*} Note the factor $\frac{1}{2i}$ on the front; this is even necessary to make sure we get a real class! (Compare the class we wrote down with its complex conjugate.)
Remark. More generally, it should be possible to describe the class of a middledimensional subvariety in $E \times E$ in terms of its action on $H^i(E) \to H^i(E)$ defined by Künneth and Poincaré duality. You can start by reading the proof for $\Delta$, and then see how it generalises.