LED switching circuit using N-Channel(BSS123) MOSFET
Find R1 by using Ohm's Law.  That says that the resistance of a resistor is the voltage across it divided by the current thru it.  In common units:

&nbsp; &nbsp; &Omega; = V / A

You know you want the current to be 70 mA.  You know that the LED drops 2.3 V, and that overall the string of resistor, LED, and switch will drop 3.3 V.  If the switch were perfect, then there would be (3.3 V)-(2.3 V) = 1.0 V across the resistor when the switch is on.  The ideal resistance value in that case would be (1.0 V)/(70 mA) = 14.3 &Omega;.

The best way to deal with the drop across Q3 is to make sure it is small enough so that you can ignore it.  There are plenty of FETs available for which that would be true.  These MOSFETs don't have a fixed voltage drop, and look more like a resistance when fully on.  That is something you look up in the FET datasheet.  It is usually called RDSON, for "Resistance, drain to source, on-state".

The voltage the FET will drop is its RDSON times the current thru it (again from Ohm's law).  As long as this is small compared to the total of 1 V available after the LED drop, you can ignore it.

For example, let's say the FET has a maximum RDSON of 30 m&Omega;.  (30 m&Omega;)(70 mA) = 2.1 mV, which is well below the level at which you can simply ignore it.

There are many MOSFETs available with such low on resistance.  I often use the IRLML2502 in such roles, but that's just one example of many.