Very Simple Discrete Current Source
On the right hand circuit, \$R_{L}\$ does not have a lot of latitude, because \$Q_{2}\$'s emitter will be at -0.7V, and if \$V_{CE(sat)}\$ is for example 0.3V, the voltage across \$R_{L}\$ can only go from 0 to 0.4V.  

Meanwhile, the circuit on the left hand side gives a lot more latitude, depending on where the \$R2/R1\$ voltage divider puts the base (and hence the emitter voltage, 0.7V below the base).