In calculating the electron dispersion you probably obtained the diagonalized Hamiltonian in the momentum space
$$ H=\sum_\mathbf{k}\left[c^{\dagger}_{\mathbf{k}A},c^{\dagger}_{\mathbf{k}B}\right]\left[\begin{array}{cc}0 & \Delta(\mathbf{k})\\ \Delta^{\dagger}(\mathbf{k}) &0\end{array}\right]\left[\begin{array}{c}c_{\mathbf{k}A} \\ c_{\mathbf{k}B}\end{array}\right]. $$
If you you chose your $x$ axis along the zigzag direction (arXiv:1004.3396), the two nonequivalent Dirac valleys are $\mathbf{K}_\kappa=\left(\kappa\frac{4\pi}{3\sqrt{3}a},0\right)$, $\kappa=\pm1$ and $\mathbf{K}_{-1}=\mathbf{K}^{\prime}$, andwhere $a$ is the C-C distance. Then $\Delta(\mathbf{k})=-t\left(1+e^{-i\mathbf{k}\cdot\mathbf{a}_1}+e^{-i\mathbf{k}\cdot\mathbf{a}_2}\right)$, where $t$ is the hopping term, and $\mathbf{a}_1$$\mathbf{a}_1=\left(\sqrt{3}a/2,3a/2\right)$ and $\mathbf{a}_2$$\mathbf{a}_2=\left(-\sqrt{3}a/2,3a/2\right)$ are the lattice vectors.
Taylor expanding $\Delta(\mathbf{k})$ up to linear terms around those two points you obtain
$$ \Delta(\mathbf{k})=\kappa\frac{3ta}{2}q_x-i\frac{3ta}{2}q_y $$
where $\mathbf{q}$ is the displacement momenta from the $\mathbf{K}_\kappa$ point. Promoting these displacement momenta to operators you obtain the Hamiltonian
$$ H=\hbar v_F\left[\begin{array}{cccc}0 & q_x-iq_y & 0 & 0\\q_x+iq_y & 0 & 0 & 0\\0 & 0 & 0 & -q_x-iq_y\\0 & 0 & -q_x+iq_y & 0\end{array}\right] $$
where $v_F=\frac{3ta}{2\hbar}$ is the Fermi velocity. This is in $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{A\mathbf{K}^{\prime}},\Psi_{B\mathbf{K}^{\prime}}\right]^T$ basis, if you rearrange your basis as $\left[\Psi_{A\mathbf{K}},\Psi_{B\mathbf{K}},\Psi_{B\mathbf{K}^{\prime}},\Psi_{A\mathbf{K}^{\prime}}\right]^T$ you get the compact form
$$ H=\hbar v_F\tau_z\otimes\boldsymbol{\sigma}\cdot\mathbf{k} $$
where $\tau_z$ acts in the valley space. This is similar to the Dirac-Weyl equation for relativistic massless particles, where instead of $v_F$ you get the speed of light
$$ H=\pm\hbar c\boldsymbol{\sigma}\cdot\mathbf{k} $$
where $+$ denotes right-handed antineutrions, and $-$ denotes left-handed neutrions. The differences are that $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y\right)$ for graphene acts in pseudospin space and $\boldsymbol{\sigma}=\left(\sigma_x,\sigma_y,\sigma_z\right)$ for neutrinos acts in real spin space.