Now we can determine the value for the W/L ratio of the NMOS switching device from the MOS drain current expression using the circuit conditions in Fig. 6.12(b). In this case, the input is set equal to V = 2.5 V, and the output of the inverter should then be at V. The expression for the drain current in the triode region of the device is used because vGS - VTN = 2.5 V-0.6 V = 1.9 V, and vps = V = 0.20 V, yielding vps < vGs – VyN- VDD = 2.5 V IpD 28.8 k2 o vo = VL 80 μΑ Ms + Vos = 0.20 V 2.22 Uj = Vµ=2.5 V EXERCISE: Redesign the logic gate in Fig. 6.12 to operate at a power of 0.4 mW while main- taining V = 0.20 V. %3D ANSWER: (W/L)s= 4.44/1; R= 14.4 k2

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Now we can determine the value for the W/L ratio of the NMOS switching device from the
MOS drain current expression using the circuit conditions in Fig. 6.12(b). In this case, the input is
set equal to V = 2.5 V, and the output of the inverter should then be at V. The expression for the
drain current in the triode region of the device is used because vGS - VTN = 2.5 V-0.6 V = 1.9 V,
and vps = V = 0.20 V, yielding vps < vGs – VyN-
VDD = 2.5 V
IpD
R
28.8 k2
80 μΑ
A = 0a o
Ms +
Vos = 0.20 V
2.22
Uj = Vµ = 2.5 V
EXERCISE: Redesign the logic gate in Fig. 6.12 to operate at a power of 0.4 mW while main-
taining V = 0.20 V.
ANSWER: (W/L)s= 4.44/1; R= 14.4 k2
Transcribed Image Text:Now we can determine the value for the W/L ratio of the NMOS switching device from the MOS drain current expression using the circuit conditions in Fig. 6.12(b). In this case, the input is set equal to V = 2.5 V, and the output of the inverter should then be at V. The expression for the drain current in the triode region of the device is used because vGS - VTN = 2.5 V-0.6 V = 1.9 V, and vps = V = 0.20 V, yielding vps < vGs – VyN- VDD = 2.5 V IpD R 28.8 k2 80 μΑ A = 0a o Ms + Vos = 0.20 V 2.22 Uj = Vµ = 2.5 V EXERCISE: Redesign the logic gate in Fig. 6.12 to operate at a power of 0.4 mW while main- taining V = 0.20 V. ANSWER: (W/L)s= 4.44/1; R= 14.4 k2
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