Mirror Circuit

Q1: What's a mirror circuit?
A1:

$Q_{ref}$ and $Q_1$ are identical NPN transistors.
This is why current splits evenly to each of their bases.

Q2: What state is $Q_{ref}$ in?
A2: Since $i_{C,ref}>0$, we know that $Q_1$ must be on. Hence, $V_{BE,ref}=V_{D0}$
Note that the base and collector share the same node in $Q_{ref}$.
Hence, $V_{BE,ref}=V_{CE,ref}=VD_{0}$
Therefore, $Q_{ref}$ is always active.

Q3: How is $I_1$ related to $I_{ref}$?
A3: First, let's find an expression for $I_{ref}$:
By KCL: $I_{ref}=2i_B+i_{c,ref}$
Since $Q_{ref}$ active: $I_{ref}=2\frac{i_{C,ref}}{\beta}+i_{C,ref}$
$I_{ref}=i_{C,ref}(\frac{2}{\beta}+1)$

Next, let's examine $I_1$:
$I_1$=$i_C$
$I_1=\beta i_B$
$I_1=\beta \frac{i_{C,ref}}{\beta}$
$I_1=i_{C,ref}$
$I_1=\frac {I_{ref}}{\frac{2}{\beta}+1}$

For $\beta \gg 1$:
$I_1\approx I_{ref}$

By KVL through $Q_{ref}$ (either the BE or CE branch):
$I_1\approx I_{ref}=\frac{V_{CC}-V{D0}+V{EE}}{R}$

Q2: Why use a mirror circuit?
A2: You can use a mirror circuit to behave as a current source in order to bias a BJT.

Q3: Why not use a mirror circuit?
A3: The mirror circuit produces integer multiples of the reference current. For a more flexible option, you might want to look into current-steering circuits.