Abstract

Cu is usually considered as an effective dopant to introduce shallow acceptors in Zn chalcogenides because it is on the left-hand side of Zn in the Periodic Table. Here, using first-principles calculations based on the hybrid functional with spin polarization, we show that contrary to the common expectation, Cu substituting Zn ($\mathrm{C}{\mathrm{u}}_{\mathrm{Zn}}$) in bulk Zn chalcogenides actually generates rather deep acceptor levels in ZnO, ZnS, and ZnSe, i.e., 2.91, 1.03, and 0.53 eV above the valence-band maximum (VBM), respectively, except in ZnTe (0.13 eV). More interestingly, the absolute Cu impurity energy level does not follow the variation of the VBM, decreasing from ZnTe to ZnSe to ZnS to ZnO, instead, it is the highest in ZnO. The abnormal behavior of $\mathrm{C}{\mathrm{u}}_{\mathrm{Zn}}$ in ZnO is attributed to the fact that, due to the very low O $2p$-orbital energy, the $\mathrm{C}{\mathrm{u}}_{\mathrm{Zn}}$ defect wave function has dominantly localized the Cu $3d$-orbital component, whereas in other Zn chalcogenides, anion $p$ states are dominant. The localized Cu $3d$ state leads to the enhanced exchange energy that elevates the acceptor level, which explains why the Cu impurity level is abnormally deep in ZnO. This finding provides insight in designing shallow acceptor levels in II--VI semiconductors.

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