The tight-binding parameters and the fixed charge distribution of our model have been calibrated  on ab-initio simulations for a simplified structure consisting of a graphene ribbon with a single substitutional boron atom at different positions along the ribbon transverse section. In Reference , we reported results obtained from ab-initio density functional theory (DFT) simulations, performed with the SIESTA code  with a local density approximation and a double-ζ basis set. From these simulations, it emerged that the quasi-bound states localized around the boron atom introduce resonant backscattering—and thus a conduction reduction—in the hole branch of the transmission spectrum. In the devices we simulated, this effect dominates over the electrostatic effect of the impurity . The energy of the conductance dip depends on the position of the boron impurity with respect to the ribbon edges . From a purely electrostatic point of view, the boron atom, being an electron acceptor, introduces a repulsive potential for electrons; this potential profile was also extracted from the ab-initio simulation. We compared  the transmission spectra and the potential profiles obtained from the DFT calculations with those obtained from the previously described self-consistent tight-binding calculation performed through the ViDES code. A very good agreement was obtained by using the tight-binding parameters typical of undoped graphene and a particular choice of fixed charges. In detail, all the onsite energies were assumed equal to zero. The transfer integral between nearest neighbor atoms was always taken equal to tp = −2.7 eV, except between atoms belonging to the edge dimers of armchair ribbons. In this last case, the transfer integral was set to 1.12tp, in order to account for the reduced interatomic distance at the edges, and to correctly reproduce the energy gap of the ribbon . Regarding the distribution of fixed charges ρfix(r→), the best fitting was obtained considering a fixed charge null at each carbon atom and equal to −e at each boron atom. This can be physically explained observing that the sum of the charge of the nucleus and of all the electrons that are not in the 2pz orbitals is equal to +e for carbon atoms and to 0 for boron atoms. The total charge is the sum of this charge and of the charge of the electrons in the π orbitals (which derive from the 2pz atomic orbitals—i.e., the only orbitals that are considered in the tight-binding calculation). Let us call LDOS the local density of π states for unit area and energy, EF the Fermi energy and Ei the midgap energy. Then, the total charge contained in an area S of the ribbon (containing NC carbon atoms and NB boron atoms) is given by (5)+eNC−e∫SdS∫−∞+∞dELDOS(E)f(E−EF)=+eNC−e∫SdS∫−∞EidELDOS(E)f(E−EF)−e∫SdS∫Ei+∞dELDOS(E)f(E−EF). we can sum and subtract from Equation (5) the charge of the electrons that would fill all the states in the π valence bands; that is, (6)−e∫SdS∫−∞EidELDOS(E). the total number of states in the π valence bands is half the total number of π states, which is 2(NC+NB) because each atom contributes with two 2pz states with different spin. Therefore, the value of Equation (6) is −e(NC+NB). Summing and subtracting this charge from Equation (5), we obtain that the total charge in S is given by (7)+eNC−e(NC+NB)+e∫SdS∫−∞EidELDOS(E)[1−f(E−EF)]+−e∫SdS∫Ei+∞dELDOS(E)f(E−EF)=−eNB+e∫SdS(p−n). therefore, the fixed charge to consider in Equation (2) is given by a charge null at each carbon atom and equal to −e at each boron atom.