# Very Simple Discrete Current Source

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).