Elementary embeddings,V,set theory,L,cardinals I would like to make clear some properties of $j,M,L$ and $V$ in ZFC.
let $j:V \to M$ denote a (nonidentity) elementary emedding and $M$ a transitive $\in$-model of ZFC.Is there an example of moderately large cardinal, whose existence in $V$ ensures that for $M=L$,such $j$ exists? At best, which LC is equivalent to this assertion?
Next, what is some inutition (good explanation) behind the fact, that taking $M$ too large prevents existentce of such a $j$? 
One would say, that taking $M$ larger, then V is more easily embedded into it.
Is it true that, if such a $j$ exists for some $M$,then it exists for $M=L$?
 A: You've asked a bunch of questions; let me give an overview of the answers. For more detailed information, you should look at Jech's book on set theory, which contains a ton of excellent information on large cardinals.
So let's start with an elementary embedding $j: V\rightarrow M$. It's easy to see that, if $j$ is nontrivial, then there is some least ordinal $\kappa$ such that $j(\kappa)>\kappa$; this is called the critical point of $j$.  Now, if such a $j$ exists, we can consider the following family of subsets of $\kappa$: $$\mathcal{U}_j=\{X\subseteq\kappa: \kappa\in j(X)\}.$$ You can check that $\mathcal{U}_j$ is a $\kappa$-complete ultrafilter on $j$, so $\kappa$ is measurable. 
In the other direction, if $\kappa$ is measurable, then let $N=\prod_\kappa V/\mathcal{U}$ be the ultrapower of the universe mod a measure $\mathcal{U}$ on $\kappa$. $N$ is well-founded, so we may take its Mostowski collapse; this is a class $M\subset V$ which is an inner model. Composing the ultrapower embedding with the Mostowski collapse gives an elementary embedding $j: V\rightarrow M$; this embedding is nontrivial, with critical point $\kappa$ (a good exercise).
So we have: The existence of a nontrivial elementary embedding of $V$ into $M$, for some inner model $M$, is equivalent to the existence of a measurable cardinal.
Note: the fact that the thing we're embedding is "$V$" itself is crucial here. For much cheaper we can get a nontrivial elementary embedding of $L$ into $L$ - this just takes $0^\#$.

Now moving on to your second question: Is there a  large cardinal which implies that there is a nontrivial elementary embedding from $V$ into $L$, specifically? The answer turns out to be no - the existence of such an embedding is inconsistent! For, if there were one, $V$ would contain a measurable cardinal (by the above), so by elementarity, $L$ would have to contain a measurable cardinal. But that's not possible. This was proved by Scott, and is a good exercise: show that if $j: V\rightarrow M$ is the elementary embedding coming from a measure on $\kappa$, then $(2^\kappa)^M\supsetneq 2^\kappa$, and so $M$ is a strictly smaller inner model - this means $V\not=L$, since $L$ has no proper inner models.

The above shows that making $M$ too small can prevent an embedding from $V$ into $M$. And, this is for very good intuitive reasons: in order to have an embedding of $V$ into $M$, $M$ must at least have all the basic large cardinal structure of $V$, and modest large cardinals can't exist in "too small" models.
Your question indicates that you're aware of the dual fact (the Kunen inconsistency), which states that we cannot have an elementary embedding of $V$ into $V$ - that is, making $M$ too big also poses a problem. So, what's the intuition behind this one?
Well, the Kunen inconsistency is a bit complicated. A good proof can be found on page 82 of http://math.bu.edu/people/aki/d.pdf (I prefer this to the exposition in Kanamori). Basically, the idea is this:


*

*Suppose we have $j: V\rightarrow V$ a nontrivial elementary embedding with critical point $\kappa$.

*Then look at $\lambda=\sup\{j^n(\kappa): n\in\omega\}$. Note that $j(\lambda)=\lambda$, so this is the point where $j$ first "catches its tail".

*We can find a combinatorial gadget $f$ - an $\omega$-Jonsson function - on $\lambda$, which lets us nicely code the countable subsets of $\lambda$: $$f: ^\omega\lambda\rightarrow\lambda\quad\mbox{and for $y\subseteq \lambda, \vert y\vert=\lambda$, we have }f"{}^\omega y=\lambda.$$

*By elementarity, $j(f)$ is again $\omega$-Jonsson on $\lambda$. By nontriviality of $j$, $\kappa\not\in ran(j)$. But let $y=j"\lambda$. Then $\vert y\vert=\lambda$, so $\kappa\in j(f)"{}^\omega y$ via some $x\in {}^\omega y$. Let $\alpha_n$ be such that $x(n)=j(\alpha_n)$; then $$\kappa=j(f)(x)=j(f)(j(\langle \alpha_n\rangle_{n\in\omega}))=j(f(\langle \alpha_n\rangle_{n\in\omega})),$$ so $\kappa\in ran(j)$, contradiction.
The construction of an $\omega$-Jonsson function on an arbitrary infinite cardinal isn't hard; in fact, the proof takes only a couple paragraphs.
A: If $j: V \to L$ were elementary, there would be a measurable cardinal in $V$ hence $L$ would think so too. But Scott showed that this is impossible.  
