BJT Current Mirror Problem
Your question is about the so-called &quot;compliance voltage&quot; - a very important property of current sources (despite the respectful name &quot;current mirror&quot;, here Q2 is just a simple transistor current source... or, more correctly, a sink).
When you begin increasing RL, the voltage drop VRL across it will begin increasing as well. To compensate this increase, Q2 will begin decreasing its collector voltage with the same rate. Figuratively speaking, you can think of the Q2 collector-emitter  part as of a variable (dynamic) &quot;resistor&quot; that complements the RL resistance to a constant value; hence the constant output current.
Finally, the Q2's Vc will reach VCEsat = 0.6 V. This will happen when VRL = +Vin - 0.6 V and, as they say, Q2 becomes saturated.
After that, the &quot;magic&quot; of Q2 constant-current source ceases. It cannot decrease its &quot;resistance&quot; anymore... and the current will begin decreasing (according to Ohm's law) when you continue increasing RL.