One can perform a simple rotation of the neutral bidoublet fields to obtain two new scalar fields $$h_1^0 = \frac{v_1 \phi_1^0 + v_2 \phi_2^0}{\sqrt{v_1^2 + v_2^2}},~~~~ h_2^0 = \frac{v_2 \phi_1^0 - v_1 \phi_2^0}{\sqrt{v_1^2 + v_2^2}}.~~~~$$ In this rotated basis, only one of these new fields ($h_1^0$) gets a non-zero VEV. This along with a redefinition of the couplings gives $$\hspace*{-0.3cm}
{M_{u} = Y^q v'_1,~~M_d = \wt{Y}^q v'_1,~~M_l = \wt{Y}^l v'_1,~~M_\nu^D = Y^l v'_1,}
\label{eq:fmass}$$ where $\left< h_1^0 \right> = v'_1 $ is the VEV in the redefined basis and $$\begin{aligned}
Y^q &=& \frac{1}{v'_1}\left( Y^{q1} v_1+Y^{q2} v_2 \right),~~\wt Y^q = \frac{1}{v'_1}\left( Y^{q1} v_2+Y^{q2} v_1 \right),\notag \\ \wt{Y}^l &=& \frac{1}{v'_1}\left(Y^{l1} v_2+Y^{l2}v_1\right),~~Y^l = \frac{1}{v'_1}\left(Y^{l1} v_1+Y^{l2}v_2\right).~~\end{aligned}$$ The $\delta^+$ field is responsible for producing the Majorana mass terms in the neutrino mass matrix which are given as [@pavel1]: $$\begin{aligned}
\displaystyle {(M_\nu^L)}^{\alpha \gamma}&&=\frac{1}{4\pi^2}\lambda'_L{^{\alpha \beta}}m_{e_\beta}\sum_{i=1}^3 \text{Log}\left(\frac{M_{h_i}^2}{m_{e_\beta}^2}\right) \times V_{5i}\left [ (Y_l^\dagger)^{\beta \gamma}V_{2i}^*-(\wt Y_l^\dagger)^{\beta \gamma}V_{1i}^* \right ] \ + \ \alpha \leftrightarrow \gamma 
\,,\notag \\
\displaystyle {(M^R_\nu)}^{\alpha \gamma}&&=\frac{1}{4\pi^2}\lambda'_R{^{\alpha \beta}}m_{e_\beta}\sum_{i=1}^3\text{Log}\left(\frac{M_{h_i}^2}{m_{e_\beta}^2}\right) \times V_{5i}\left[(Y_l)^{\beta \gamma}V_{1i}^*-(\wt Y_l)^{\beta \gamma}V_{2i}^*\right] \ + \ \alpha \leftrightarrow \gamma \,. 
\label{RHM}\end{aligned}$$ Here $\alpha$, $\beta$ and $\gamma$ each run from $1-3$, $V_{ij}$ corresponds to the $ij$-th element of the charged Higgs boson mixing matrix $V$ defined in Eq. \[eq:hdiag\], 