Maximum depth of a B+ tree Given $K$...# key values, $n$...# pointers in a node.

I read somewhere, that the maximum depth is defined as $\lceil\log_{\lceil\frac{n}{2}\rceil}(K)\rceil$. However, it is not correct, as I can come up with a counterexample. When the tree is minimum filled, it won't work.
E.g.:

 

This is a valid $B^+$-tree, the root has at least two childs, each inner node has at least $\lceil n/2 \rceil$ childs and each leaf has at least $\lceil \frac{n-1}{2}\rceil$ record. So, $n=3$ and $K=4$, then $\log_2(4) = 2$. Now, when you fill up the leafs: [1,1,2,2,3,3,4,4], then it is again a valid tree and $K=8$, hence $\log_2(8) = 3$, but same depth.

Notice: I am looking for a formula or explanation but for a $B^+$-tree not a $B$-tree. A source would be nice.
Since you're not sure where you read it, is it possible you are misremembering slightly what the result was?

In a B+ tree, we require that every node has between $n/2$ and $n$ children.  In other words, every node is required to have at least $\lceil n/2 \rceil$ children.  This means that any tree of depth $d$ where all leaves are at depth $d$ must have at least $\lceil n/2 \rceil^d$ leaf nodes.  Thus, any such tree must have depth at most $\lceil \log_{\lceil n/2 \rceil} L \rceil$, where $L$ is the number of leaf nodes.  Perhaps this is what you read.

Note that the number of key value is related  to the number of leaf nodes: $\lceil b/2 \rceil -1 \le L/K \le b-1$.  This would let you get a similar result in terms of $K$, for a full tree with all leaves at the same level.

Detail: Here I have ignored that the root is usually allowed to have as few as 2 children.  This has only a small effect on the answer.