The Lindblad form guarantees complete positivity of a Markovian quantum master equation (QME). However, its microscopic derivation for a quantum system weakly interacting with a thermal bath requires several approximations, which may result in inaccuracies in the QME. Here, we explicitly calculate, in a perturbative manner, the equilibrium steady states of various Lindbladian QMEs that were recently derived from the Redfield equation, without resorting to secular approximation. We compare the results with the steady state of the Redfield equation, which coincides with the so-called mean force Gibbs (MFG) state obtained by integrating out the bath degrees of freedom for the Gibbs state of the total Hamiltonian. We explicitly show that the steady states of the Lindbladian QMEs are different from the MFG state. Our results indicate that manipulations of the Redfield equation needed to enforce complete positivity of a QME drives its steady state away from the MFG state. We also find that, in the high-temperature regime, both the steady states of the Lindbladian QMEs and MFG state reduce to the same Gibbs state of a system Hamiltonian under certain conditions.