Are non-standard models always not well-founded? Are non-standard models of ZF set theory by definition always not well-founded? And it seems it is, because it must be. 
But then, Wikipedia says that when there is a set that is a standard model of ZF, there exists smallest set $L_k$ that is standard model. So, this seems to confuse me a lot.
 A: By a standard model of ZF we mean a model $( M , E )$ of ZF where the relation $E$ is real membership on $M$, i.e., $x \mathrel{E} y \; \Leftrightarrow \; x \in y$.  The relation $E$ is usually denoted either simply by $\in$, or by $\in_M$ to emphasise that we've restricted the domain.
Given any standard model $( M , \in_M )$ of ZF we can construct an isomorphic non-standard model $( M^* , E^* )$ quite easily by taking $M^* = \{ \{ x \} : x \in M \}$, and defining $\{ x \} \mathrel{E^*} \{ y \} \; \Leftrightarrow \; x \in y$.  This model will be well-founded.
A: There are several different senses in which a set model of ZF set theory could be "standard". I have marked this answer as community wiki, so please feel free to add any other definitions that you think of.


*

*The model could have an $\omega$ that is isomorphic to the standard $\omega$. These models are called $\omega$-models.

*The membership relation of the model could be well founded when viewed from the outside. These models are called well founded models.

*The model could use the standard membership relation $\in$ rather than some other membership relation. These models are sometimes called standard models. 

*The model could be a transitive set, and use the standard membership relation $\in$. A transitive set is a set $A$ such that, for all $B \in A$, every member of $B$ is also a member of $A$. These models are called transitive models.
For any given model, (4) implies (3) implies (2) implies (1), and none of the converse implications hold in general.  
The Mostowski collapse lemma shows that every model of ZF satisfying (2) is isomorphic to a model satisfying (4). 
