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EE456 HW3 SOLUTIONS
Figure 1 shows an NMOS inverter with resistive load.
Use 0.25um TSMC 2.5V CMOS model
parameters, i.e., for NMOS, VT0=0.456V, UO=0.0339m
2
/Vs, and TOX=5.8e9m.
Figure 1.
Resistiveload inverter
a.
)
Qualitatively discuss why this circuit behaves as an inverter.
SOLUTIONS
For Vin < V
T
, M1 is in cutoff regime, thus I=0 and Vout=2.5V. For Vin > V
T
, M1 is conducting and Vout=2.5V
 (I*R). This in turn gives a low Vout and the input signal is inverted.
b. Calculate V
OH
, V
OL
, V
IH
, V
IL
, V
M
, NM
H
, and NM
L
Assuming negligible leakage, when \Vin < V
T
, transistor M1 is off and
V
OH
=2.5V.
For Vin=2.5V, assume M1 is
in the linear region, and because V
DS
is negligible in the linear region, channellength modulation can be ignored.
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View Full DocumentTherefore
V
OL
=26.89mV.
Checking the assumption: V
GS
V
T
(=2.044V), and V
DS
=26.89mV, thus, M1 was
correctly assumed to be in the linear region.
To find V
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 Spring '09
 Mohammadi
 Integrated Circuit

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