quantum mechanics - Is the conserved current probability or charge in Klein-Gordon and Dirac equations?

The conserved currents in KG and Dirac are:

K-G : $j^{\mu}=i(\phi^*\partial^{\mu}\phi-\phi\partial^{\mu}\phi^*)$

Dirac: $j^{\mu}=\bar{\psi}\gamma^{\mu}\psi$

$j^0$ is positive definite in Dirac's but not on KG's. One simple way to see this is noticing that, in KG, the conserved current can be written us $j^{\mu}=2p^{\mu}|N|^2$ where $N$ is a normalization factor. When energy is negative, so is $j^0$.

This problem is solved thinking of $j^{\mu}$ as a charge current and not probability. This makes sense because solutions with positive and negative energy (particles and antiparticles) have opposite charge densities.

How does this interpretation apply to the Dirac current? it is positive definite so it can't change sign for negative energy solutions...