LM317 with digital output selection

Hello everybody
I hope this is not an already answered question but I didn't find it.
In the application note for LM317 and similar devices there is the application for digitally control the output voltage using some parallel resistors on the reference path toward ground to alter the output value.
To switch the resistors on and off BJTs are used (I think that MOS could be used too, but this is not relevant now). My question is: the calculation is considered the same apparently as it is just the equivalent resistence of the enabled parallel. But... shouldn't that be corrected to take into account the voltage drop Vce of the saturated BJT used as switch?
From the datasheet (http://www.ti.com/lit/ds/symlink/lm117.pdf)
the output voltage is expressed as:
Vout = Vref(1 + R1/R2) + Iadj*R2
I'm missing somenthing (or just doing wrong considerations)?
Thank you everybody

You are correct. I looked at the data sheet and I did not see a formula for the output voltage for the case you describe. In any event, you must account for any voltage drop across the switch.

Thank you for your answer John, I thought the same, I just find it a bit strange that I wasn't able to find any hint on this topic like noone has even noticed that

The drop across each transistor will be about 0.25V (Vce-sat) and therefore the o/p voltage will be higher by that much. For most applications this should be fine and therefore the formula to calculate the o/p voltage remains as it is for the voltage divider mechanism. If indeed the transistor's saturation voltage is an issue, simply recalculate using the voltage divider method, but taking the output voltage to be 0.25V less than the actual required voltage. That way, the Vce-sat drop will add to the o/p thereby bringing it back to the target value.

Thank you, I was thinking that the "missing part" of the calculation was related to the fact that the effect of the BJTs was neglectable. Thanks for confirming this idea