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VTC Logic Symbol
Load resistor
switch
Put a transistor here, what kind of that?
Load resistor
switch
vI below the reference voltage VREF
abruptly
Large change at in Small change at output
Wave form?
VDD
What is the behavior of this circuit?
VDS Output characteristics
Load line
ID
VDS
= VGS
VDS =
Deeper look
?
?
0.82V
1.42V
VI
Problem
Problem
1.42V
Triode region
Need to know ID to find W/L and how?
R=?
SUMARY
t
simple discrete component logic gate
used to introduce the concepts associated with static logic gate design
IC realizations do not use resistive loads
NMOS inverter with resistive load:
resistor would take up far too much area.
Static power dissipation
Resistor with a thickness of 1 m in a silicon region with a resistivity of 0.001 .cm. A 95-k load resistor would require the ratio of L/W to be
Transistor Alternatives to the Load Resistor
Not used
Saturated load inverter
Linear load inverter.
VGS2 = VDS2 = VDD VDS1
VO,max
Keep in mind: VO,max = VG2 VTN2
VG2
VI = VDD
VG2
?
Check assumption
VIH = 1.12V; VOL = 0.38 VGS1 VTN1 > VDS1
Need to update iteratively VTNL and VOL
VDS2 = VDD VDS1 VDS2 = 0
Pinchoff, VDS2 = -VTN2
VDS2 = VDD VDS1 VDS2 = 0
Pinchoff, VDS2 = -VTN2
VI = VGS1
VO= VDS1
switching device is saturated load device is in its linear region
For VI near VIL, VDS of Ms will be large and that of ML will be small
VTNL = -0.5V VIL= 0.9V
Check operating region assumptions:
VGS1 VTN1 = 0.9V - 0.6V = 0.3V < VDS1 = 2.37 M1: Saturation
VGS2 VTN2 = 0V (- 0.5V) = 0.5V > VDS2 = 2.5V - 2.37V M2: Triode
vSB = VDD
vSB = VDD
(1)
(2)
VTNL = -1V; VIL = 0.965V; VO = 2,41V
VTNL = -0.52V; VIL = 0.79V; VO = 2,456V
VTNL = -0.51V; VIL = 0.787V; VO = 2,456V
1. Choose a starting value for VTNL. 2. Calculate the corresponding value of VIL using Eq. (1). 3. Use the values VTNL and VIL to calculate a new estimate of vO using Eq. (2). 4. Use vO to calculate an updated value for VTNL. 5. Repeat steps 2 and 3 until the convergence is achieved.
Initiate with VTNL = -1V
MS in the linear region ML in the saturation region
+
Iterative Update of VIH and vO 1. Choose an initial value of VTNL. 2. Calculate the corresponding values of VO and VIH from Eq. (1) 3. Use the new value of VO and Eq. (2) to improve the estimate of VTNL. 4. Repeat steps 2 and 3 until the process converges.
(1)
(2)
+
VTNL = -1V; VO = 0.52V; VIH = 1,64V
VTNL = -0.858V; VO = 0.447V; VIH = 1,493V
VTNL = -0.875V; VO = 0.456V; VIH = 1,511V
VTNL = -0.873V; VO = 0.455V; VIH = 1,51V
Initiate with VTNL = -1V
MS in the linear region ML in the saturation region
Check operating region assumptions:
MS : VGS VTNS = 1.51V 0.6V = 0.91V > VDSS = VO = 0.46V Linear
ML : VGS VTNL = 0 (-0.873V) < VDSL = VDD - VO = 2.5V -0.46V Saturation
0.33
0.93
depletion-mode NMOS logic was the dominant technology for many years in the design of microprocessors.
large static power dissipation inherent in NMOS logic eventually limited further increases in IC chip density
rapid shift took place to the more complex CMOS technology
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