In The Art of Electronics by Horowitz & Hill, there’s a section about current mirrors with multiple outputs. We're then given the following circuits:
It's then described how the circuit shown in Figure 2.49 doesn't compensate for current hogging, since if load 1 was removed, the corresponding transistor will saturate and pull more current than the other transistors connected to the common reference base.
The solution to this is then given via Figure 2.50, where compensation for current hogging is provided via an extra transistor connected at the collector of the "programming transistor" on the left. This extra transistor effectively pulls the voltage of the programming transistor's collector down to 2 diode drops below \$V_{CC}\$, moving it further into the forward active region of operation.
What confuses me is this: if you remove load 1 from Figure 2.50, the corresponding transistor will still be saturated there, just as in Figure 2.49. With that said, how does the extra transistor resolve this?
