No, this isn't possible for an undoped semiconductor. The reason it's happening is because the Fermi level is not well defined for your semiconductor at finite temperature, and you're trying to read the valence band maximum (VBM) from the graph of your broadened DOS. I recommend taking the VBM directly from the band energies, instead.
In a DFT program, the states are computed at a discrete set of k-points, and have well-defined eigenvalues ("band energies") at those points. The DOS for this system is a set of spikes (delta-functions) at those energies, but this doesn't give you a smooth, meaningful DOS. The problem is that the DOS should be integrated over the whole Brillouin zone, but all you have are discrete sampling points.
The common solution is to replace each spike with a narrow Gaussian, or similar "smearing function", which has the effect of smoothing the states out in energy. The states are almost unchanged by any reasonable choice of the width of the Gaussian (the "smearing width"). If you increase the smearing width, the DOS gets smoother and the tail of the smearing distribution stretches to higher energies, which naturally pushes some densities-of-states higher in energy. The way you're reading the DOS, this would raise the VBM, even though the states are the same as they were before. Even worse, strictly-speaking the smearing function never goes to zero, so if you "zoom in" on the DOS, you'll never find somewhere that it decays to zero, until you hit the limits of the calculations finite precision.
The actual VBM is the band-energy of the topmost occupied band, so I recommend you just use that instead. If the precise value is important to you, you will need to ensure that you've sampled the Brillouin zone at the correct k-point (or as close to it as possible). You can check this visually by computing and plotting the band-structure (particularly along the high-symmetry directions).