Is this correct?
One finds resistance required to get a circuit down to 20mA by V/P=R

Inputing supply voltage minus v-drop across the component then, devide the desired mA to get required R?

say I want to kick my 30mA rated component down to 12ma;

I have a spec'd vdrop of 2.27v.

Supply is spec'd 2.5v

Supply measured 2.545 at full charge and 2.47 at low charge.. Going on the safe side and using full charge as my value for supply voltage.


(VS - Vdrop)x.012 = P
2.545-2.27=.275 "measured .31"
.31x.012=.00372
P=37.2 uW
R power rating minimum I could find was 1/4 watt 5%

then
V/P=R

.31 / .012 A = 25.8 Ohms