n+(2-)×0.75_5V+0.75'=1.92y 2+1 3 V=VDp-V3=5V-1.92V=3.1VV VDD V V V 1.92 3 5 0.75 2 1 (2 1) 0.75 = + = + + −  = V1 =VDD −V2 = 5V −1.92V = 3.1V