How is the equation for Compliance voltage?
Please commit this to memory forever. You will need to thoroughly understand it, literally over and over and over again. 

$$\begin{align*}
\textrm{Shockley equation applied} \\ \textrm{to simplified BJT model}\\ \textrm{ignoring the Early Effect:} \\
I_C&amp;=I_S\cdot\left(e^{\cfrac{V_{BE}}{V_T}}-1\right) \\ \textrm{move }I_S\textrm{ over to the other side,}\\ \cfrac{I_C}{I_S}&amp;=\left(e^{\cfrac{V_{BE}}{V_T}}-1\right) \\ \textrm{Drop the insignificant -1 term,} \\ \cfrac{I_C}{I_S}&amp;\approx e^{\cfrac{V_{BE}}{V_T}} \\\\  \textrm{Take the logarithm of both sides,} \\ \\ \operatorname{ln}\left(\cfrac{I_C}{I_S}\right)&amp;\approx\cfrac{V_{BE}}{V_T} \\ \\ \textrm{Solve for }V_{BE}, \\ \\ V_{BE}&amp;\approx V_T\cdot \operatorname{ln}\left(\cfrac{I_C}{I_S}\right)
\end{align*}$$

You should be able to do the above in your sleep. You should understand the meaning of any of the above in your sleep.

And you should also know that the thermal voltage \$V_T=\frac{n k T}{q}\$ and is approximately \$26\:\textrm{mV}\$ at room temps, that for small signal BJTs \$n\approx 1\$ (but for diodes it is usually larger), and that \$I_S\$ is, itself, highly temperature dependent (roughly proportional to the 3rd power of T) and that it dominates the temperature behavior of \$I_C\$ because its effect is not only larger but of opposite sign to the temperature effects caused by \$V_T\$. (The temperature-dependent \$I_S\$ equation is more complicated and not provided here.)



Getting back to the point of the current mirror discussion, this just means that you don't want to saturate the BJTs in a mirror -- that's a bad thing as they stop being much of a mirror then. So \$\vert V_{CE}\vert\ge\vert V_{BE}\vert\$, as computed above. Just make sure that the load you attach, together with the mirrored current, doesn't violate that.