1. 5 March 2024 VLSI Design 1
5 March 2024 1
5 March 2024 1
VLSI Design
Dr. T R Lenka
Asst. Professor
Deptt of Electronics & Comm Engg.
National Institute of Technology Silchar
E-mail: trlenka@ece.nits.ac.in
MOS Inverters
2. Assignment-1
1. Explain the operation principle of MOSFET using
energy band diagrams in accumulation,
depletion & inversion regions.
2. Derive the expression of threshold voltage and
flat band voltage.
3. Derive the expression of drain currents in all
regions of operation (Cut-off, Linear &
Saturation).
4. Find the expression of transconductance.
5. Draw the C~V characteristics of MOSFET.
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3. 5 March 2024 VLSI Design 3
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Chapter Coverage
Static Characteristics
Dynamic Characteristics
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Ideal Voltage Transfer
Characteristic (VTC)
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General Circuit Structure of
an nMOS Inverter
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Voltage Transfer
Characteristic (VTC)
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VTC
• Applying Kirchhoff’s Current Law
(KCL), the Load current is always
equal to the nMOS drain current.
ID (Vin, Vout)=IL(VL)
• Two critical voltage points (VIL,VIH)
defined on this VTC curve, where the
slope of the Vout (Vin) characteristic
becomes equal to -1.
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Critical Voltages
• VOH:Maximum output voltage when the
output level is logic “1”.
• VOL:Minimum output voltage when the
output level is logic “0”.
• VIL:Maximum input voltage which can be
interpreted as logic “0”.
• VIH:Minimum input voltage which can be
interpreted as logic “1”.
• VTH:Threshold voltage of inverter, is
defined as the point, where Vin=Vout.
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Noise Immunity and Noise
Margins
• The ability of an inverter to interpret
an input signal within a voltage
range as either a logic “0” or as a
logic “1”, allows digitals circuits to
operate with a certain tolerance to
external signal perturbations.
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Noise Margins
• Noise tolerances for digital circuits,
called, Noise Margin (NM).
• NML=VIL-VOL :Noise Margin Low
• NMH=VOH-VIH :Noise Margin High
• The noise immunity of the circuit
increases with NM
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Resistive Load Inverter
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Resistive Load Inverter
• Calculation of VOH:
Vout= VDD–RL.IR , where (ID=IR)
• When Vin is low, i.e., smaller than the
threshold voltage of the driver MOSFET,
the driver transistor is cut-off.
• So ID=IR =0
• VOH =VDD
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Resistive Load Inverter
• Calculation of VIL:
• IR= IDSaturation
• (VDD-VOUT)/RL= Kn/2[(Vin-VTO)2]
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Resistive Load Inverter
• Calculation of VIH:
• IR= IDLinear
• (VDD-Vout)/RL= Kn/2[(Vin-VTO).Vout-Vout
2]
• Differenting both sides w.r.t Vin and
substituting the slope=-1
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Resistive Load Inverter
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Resistive Load Inverter
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Layout of Resistive Load
Inverter
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Enhancement-nMOS Load
Inverter
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Depletion-nMOS Load
Inverter
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Depletion-nMOS Load Inverter
Calculation of VOH:
• When Vin is low, i.e., smaller than the
threshold voltage of the driver MOSFET,
the driver transistor is cut-off and does
not conduct any drain current.
• ID, Driver, Cutoff=ID, Load, Lin=0 A
• The Load device which operates in the
linear region also has zero drain current.
• So ID, Load=0 A
• Only valid solution in the linear region is
VOH =VDD
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Depletion-nMOS Load Inverter
Calculation of VOL:
• ID, Driver, Lin = ID, Load, Sat
• ID, Driver, Lin =(Kdriver/2)[2(VOH-VTO).VOL- VOL
2]
• ID, Load, Sat =(KLoad/2)[-VT, Load (VOL)]2
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Depletion-nMOS Load Inverter
Calculation of VIL:
• ID, Driver, Sat = ID, Load, Lin
• ID, Driver, Sat =(KDriver/2)[Vin-VTO]2
• ID, Load, Sat =(KLoad/2){2[VT, Load (Vout)](VDD-
Vout)- (VDD-Vout)2}
• Differenting both sides w.r.t Vin and
substituting the slope=-1
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Depletion-nMOS Load Inverter
Calculation of VIH:
• ID, Driver, Lin = ID, Load, Sat
• ID, Driver, Lin =(KDriver/2)[2(Vin-VTO) Vout-
Vout
2]
• ID, Load, Sat =(KLoad/2)[-VT, Load (Vout)]2
• Differenting both sides w.r.t Vin and
substituting the slope=-1
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VTC of a Depletion-Load
Inverter Circuit
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VTC of Depletion-Load
Inverter Circuits
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Layout of Depletion-Load
Inverters
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VTC of CMOS Inverter
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Region of Operation
Region Vin Vout nMOS pMOS
A <VT0,n VOH Cut-Off Linear
B VIL high≈VO
H
Saturation Linear
C Vth Vth Saturation Saturation
D VIH Low≈VOL Linear Saturation
E >(VDD+VT0,p) VOL Linear Cut-Off
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CMOS Inverter
Calculation of VOH:
• When Vin is low, i.e., smaller than the
threshold voltage of the driver MOSFET,
the driver transistor is cut-off and does
not conduct any drain current.
• ID, Driver, Cutoff=ID, Load, Lin=0 A
• The Load device which operates in the
linear region also has zero drain current.
• So ID, Load=0 A
• Only valid solution in the linear region is
VOH =VDD
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CMOS Inverter
Calculation of VOL:
• ID, Driver, Lin = ID, Load, Cut-off
• ID, Driver, Lin =(Kdriver/2)[2(VDD-VTO).VOL- VOL
2]
• ID, Load, Cut-off =0A
• VOL =0 V
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CMOS Inverter
Calculation of VIL:
• ID, nMOS, Sat = ID, pMOS, Lin
• ID, nMOS, Sat=(Kn/2)[VGS,n-VTO,n]2
• ID, pMOS, Lin=(KP/2)[2(VGS,p-VTO,p)VDS,p-VDS,p)2]
• Differenting both sides w.r.t Vin and
substituting the slope=-1
• VIL=(2Vout + VTO,p- VDD+ KRVTO,n)/(1+KR)
• Transconductance Ratio(KR)
• KR = Kn/KP
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CMOS Inverter
Calculation of VIH:
• ID, Driver, Lin= ID, Load, Sat
• ID, nMOS, Lin= (Kn/2)[2(VGS,n-VTO,n)VDS,n-VDS,n)2]
• ID, pMOS, Sat=(Kp/2)[VGS,p-VTO,p]2
• Differenting both sides w.r.t Vin and
substituting the slope=-1
• VIH=(VDD+VTO,p+KR (2Vout+VTO,n))/(1+KR)
• Transconductance Ratio(KR)
• KR = Kn/KP
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Design of CMOS Inverter
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Design of CMOS Inverter
• CMOS inverter doesn’t draw any
significant current from power supply,
except for small leakage and sub-
threshold currents.
• These currents exist when input voltage is
either smaller than VTO,n or larger than
(VDD+VTO,p) repectively.
• The nMOS and pMOS transistors conduct a
non-zero current, during low-to-high and
high-to-low transitions, i.e. in Regions B,
C, D.
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Design of CMOS Inverter
• The current drawn from the power supply
during transition reaches its peak value
when Vin=Vth.
• In other words, the maximum current is
drawn when both transistors are operating
in saturation mode.
• VTC of a CMOS inverter and the power
supply current, as a function of the input
voltage, is shown.
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VTC & Power Supply Current
of a CMOS Inverter
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• The threshold voltage Vth is identified as
one of he most important parameter that
characterize the steady-state input-output
behavior of the CMOS inverter circuit.
• The CMOS inverter, provides a full output
voltage swing between 0 and VDD and
therefore the noise margins (NM) are
relatively wider.
• So, the problem of designing a CMOS
inverter can be reduced to setting the
inverter threshold (Vth) to a desired
voltage value.
Design of CMOS Inverters
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• Switching threshold voltage of an
ideal inverter is, Vth, ideal= VDD/2.
• The Inverter threshold voltage Vth
shifts to lower values with increasing
KR ratio.
• For a symmetric CMOS inverter with
VT0,n =VT0,p and KR=1.
• VIL= 1/8(3VDD+2 VT0,n)
• VIH= 1/8(5VDD-2 VT0,n)
Design of CMOS Inverters
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• For a symmetric CMOS inverter with
VIL+ VIH= VDD
• The noise margins NML and NMH for
this symmetric CMOS inverter are
now calculated as:
• NML=VIL-VOL=VIL
• NMH=VOH-VIH=VDD-VIH
• NML=NMH=VIL
Design of CMOS Inverters
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Design of CMOS Inverters
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Supply Voltage Scaling in
CMOS Inverters
VDD
min=VT0,n + VT0,p
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VTC of CMOS Inverter at
Lower VDD
min
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Layout of CMOS Inverter
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Design of D-nMOS Load
Inverter
75. Calculation of Interconnect
Delay
• RC Delay Models
– An interconnect line can be modeled as
a lumped RC network if the time of
flight across the interconnection line is
significantly shorter than the signal
rise/fall times.
– This is usually the case in most on-chip
interconnects.
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76. RC Delay Models
• Assuming that the capacitance is
discharged initially, and assuming that the
input signal is a rising step pulse at time t
= 0, the output voltage waveform of this
simple RC circuit is found as
• The rising output voltage reaches the
50%-point at t = TPLH, thus, we have
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77. RC Delay Models
• The propagation delay for the simple
lumped RC network is found as
• The accuracy of the simple lumped RC
model can be significantly improved by
dividing the total line resistance into two
equal parts (the T-model)
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78. RC Delay Models
• The transient behavior of an interconnect
line can be more accurately represented
using the RC ladder network, as shown.
• Here, each RC-segment consists of a
series resistance (R/N), and a capacitance
(C/N) connected between the node and
the ground.
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79. RC Delay Models
• It can be expected that the accuracy of
this model increases with increasing N,
where the transient behavior approaches
that of a distributed RC line for very large
values of N.
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80. The Elmore Delay
• Consider a general RC tree network.
– There are no resistor loops in this circuit
– All of the capacitors in an RC tree are
connected between a node and the
ground.
– There is one input node in the circuit.
– There is an unique path resistive path
from the input node to any other node
in the circuit.
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81. The Elmore Delay
• Assuming that the input signal is a step
pulse at time t = 0 and the Elmore delay
at node i of this RC tree is given by the
following expression.
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82. The Elmore Delay
• the Elmore delay at node 7 can be
found as
• Similarly, the Elmore delay at node 5
can be calculated as
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83. The Elmore Delay
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• As a special case of the general RC tree
network, consider now the simple RC
ladder network as shown.
• Here, the entire network consists of one
single branch, and the Elmore delay from
the input to the output (node N) is found
as:
84. The Elmore Delay
• If we further assume an uniform RC ladder
network, consisting of identical elements
of (R/N) and (C/N) then the Elmore delay
from the input to the output node
becomes:
• For very large N (distributed RC line
behavior), this delay expression reduces
to:
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85. The Elmore Delay
• Thus, the propagation delay of a
distributed RC line is considerably smaller
than that of a lumped RC network.
• If the length of the interconnection line is
sufficiently large and the rise/fall times of
the signal waveforms are comparable to
the time of flight across the line, then the
interconnect line must be modeled as a
transmission line.
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86. Switching Power Dissipation of
CMOS Inverters
• As discussed the static power
dissipation of the CMOS inverter is
quite negligible.
• The dynamic power consumption of
the CMOS inverter is derived.
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87. • During switching events the output
load capacitance is alternatively
charged up and charged down.
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Dynamic Power Dissipation Analysis
88. • The output load capacitance Cload is
being charged up through the pMOS
transistor; therefore, the capacitor
current equals the instantaneous
drain current of the pMOS transistor.
• The average power dissipated by the
inverter over one period can be
found as:
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Dynamic Power Dissipation Analysis
89. Dynamic Power Dissipation Analysis
• Typical input and output voltage waveforms
and the capacitor current waveform during
switching of the CMOS inverter.
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90. • Since during switching, the nMOS
transistor and the pMOS transistor in a
CMOS inverter conduct current for one-
half period each.
• The average power dissipation of the
CMOS inverter can be calculated as the
power required to charge up and charge
down the output load capacitance.
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Dynamic Power Dissipation Analysis
91. • Evaluating the integrals, we obtain
• The average power dissipation of the
CMOS inverter is proportional to the
switching frequency “f”.
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Dynamic Power Dissipation Analysis
92. Summary
• Therefore, the low-power advantage of CMOS
circuits becomes less prominent in high-speed
operation, where the switching frequency is
high.
• Also note that the average power dissipation
is independent of all transistor characteristics
and transistor sizes.
• Consequently, the switching delay times have
no relevance to the amount of power
consumption during the switching events.
• switching power is solely dissipated for
charging and discharging the output
capacitance from VOL to VOH, and vice versa.
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93. Summary
• The switching power expression derived
for the CMOS inverter also applies to all
general CMOS circuits.
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94. Power-Delay Product
• The power-delay product (PDP) is a
fundamental parameter which is often
used for measuring the quality and the
performance of a CMOS process and gate
design.
• As a physical quantity, the power-delay
product can be interpreted as the
average energy required for a gate to
switch its output voltage from low to high
and from high to low.
• The amount of energy required to switch
the output.
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95. Power-Delay Product
• From a design point-of-view, it is desirable
to minimize the power-delay product.
• Since the PDP is a function of the output
load capacitance and the power supply
voltage, the designer should try to keep
both Cload and VDD as small as possible
when designing a CMOS logic gate.
• where is the average switching power
dissipation at maximum operating
frequency and is the average
propagation delay.
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p
*
avg
p
96. Power-Delay Product
• The factor of 2 accounts for two
transitions of the output, from low to high
and from high to low.
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VLSI Design
Combinational Logic
Circuits
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Combinational Logic Circuit
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NOR Gate using Depletion
type Load
100. Critical Voltages
• Calculation of VOH
– The solution of this equation gives
VOH = VDD.
• Calculation of VOL
– Three cases must be considered
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101. Critical Voltages
• In case (i), where the driver transistor A is
on, the ratio is
• In case (ii), where the driver transistor B
is on, the ratio is
• The output low voltage level VOL in both
cases is found as:
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102. Critical Voltages
• if the (W/L) ratios of both drivers are
identical, i.e., (W/L)A = (W/L)B the output
low voltage (VOL) values calculated for
case (i) and case (ii) will be identical.
• In case (iii), where both driver transistors
are turned on, the saturated load current
is the sum of the two linear-mode driver
currents.
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103. Critical Voltages
• Since the gate voltages of both driver transistors
are equal (VA = VB = VOH), we can devise an
equivalent driver-to-load ratio for the NOR
structure:
• The NOR gate with both of its inputs tied to a
logic-high voltage is replaced with an nMOS
depletion-load inverter circuit with the driver-to-
load ratio (KR) having output voltage level:
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104. 5 March 2024 VLSI Design 104
Generalized NOR Structure with
Multiple Inputs
105. Generalized NOR Structure
• The combined pull-down current can then be
expressed as
• Assuming that the input voltages of all driver
transistors are identical and VGSk = VGS for
k=1,2,...,n. Then the pull-down current
expression can be rewritten as
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106. Generalized NOR Structure
• Thus, the multiple-input NOR gate can
also be reduced to an equivalent inverter
for static analysis. The (W/L) ratio of the
driver transistor here is
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107. 5 March 2024 VLSI Design 107
Transient Analysis of NOR Gate
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Two-Input NAND Gate using
Depletion type Load
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NAND Gate using Depletion Load
110. Two-Input NAND Gate
• Consider the NAND2 gate with both of its
inputs equal to VOH.
• The drain currents of all transistors in the
circuit are equal to each other.
• Neglecting the substrate-bias effect for
driver transistor A for simplicity, we get
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111. Two-Input NAND Gate
• The drain-to-source voltages of both
driver transistors can be
• Let the two driver transistors be identical,
i.e., kdriverA = kdriverB = kdriver
• The output voltage VOL is equal to the sum
of the drain-to-source voltages of both
drivers.
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112. Two-Input NAND Gate
• The following analysis gives a better and
more accurate view of the operation of
two series-connected driver transistors.
• Now consider the two identical
enhancement-type nMOS transistors with
their-gate terminals connected and VTA =
VTB = VT0.
• Since IDA = IDB, this current can also be
expressed as
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113. Two-Input NAND Gate
• Using VGSA = VGSB - VDSB
• Now let VGS = VGSB and VDS = VDSA + VDSB.
The drain-current expression can then be
written as follows.
• Thus, two nMOS transistors connected in
series and with the same gate voltage
behave like one nMOS transistor with keq
= 0.5 kdriver.
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114. Generalized NAND Structure with
Multiple Inputs
• n-series-connected driver transistors,
assuming that the threshold voltages of all
transistors are equal to VT0.
• Hence, the (W/L) ratio of the equivalent
driver transistor is
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115. Two-Input NAND Gate
• If the series-connected transistors are
identical, i.e., (W/L)1 = (W/L)2 =.. .=
(W/L), the width-to-length ratio of the
equivalent transistor becomes
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116. 5 March 2024 VLSI Design 116
n-input NAND Gate using
Depletion Load
117. 5 March 2024 VLSI Design 117
Transient Analysis of NAND Gate
118. Transient Analysis of NAND Gate
• Assume, for example, that the input VA is equal
to VOH and the other input VB is switching from
VOH to VOL. In this case, both the output voltage
Vout, and the internal node voltage Vx will rise,
resulting in
• In its reverse case
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