I'm trying to work out why someone would choose the right circuit over the left one. The only thing I can think of is, that one on the right can be made an adjustable current source by making \$R1\$ or \$R2\$ a pot. If using the circuit on the left, how does one know what to set the voltage at the base of the BJT to? I would've thought one would set it as high as possible so that it is \$V_B \gg V_\mathit{BE}\$.


simulate this circuit – Schematic created using CircuitLab


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).


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