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EE210 Digital Electronics
Class Lecture 9
April 08, 2009
In This Class
Digital CMOS Logic Circuits
We Will Discuss:
10.3 CMOS Logic-Gate Circuits
2
Chapter 10: Digital CMOS Logic
Circuits
We will Start from
10.3 CMOS Logic-Gate Circuits
But First Home Work# 03…
Home Work# 03…
For the circuit in Fig., consider the application of
inputs of 5 V and 0.2 V to X and Y in any combination
,and find the output voltage for each combination .
• Tabulate your results. How many input
combinations are there? (4 Marks)
•What happens when any input is high? (3 Marks)
• What happens when both inputs are low? (3 Marks)
10.3 CMOS Logic-Gate Circuits
• Using Inverter knowledge we consider
CMOS ckts that realize combinationallogic functions
• In combinational ckts output at any time is
function only of the value of input signal at
that time. Thus, these do not have
memory.
• Combinational-logic circuits are used in
large quantities in many applications.
Indeed, every digital system contains large
numbers of Combinational-logic circuits
10.3.1 Basic Structure
• CMOS logic ckt is
extension or
generalization of the
CMOS Inverter
• As we learned, CMOS
inverter consists of
NMOS pull-down
transistor, and a PMOS
pull-up transistor,
operated by input voltage
in complementary fashion
10.3.1 Basic Structure
• CMOS Logic Gate has two
Networks: Pull-Down
Network (PDN) constructed
of NMOS transistors and
Pull-up Network (PUN)
constructed of PMOS
Transistors
• Two Networks are Operated
by Input Variables, in
Complementary fashion
10.3.1 Basic Structure
• When All Three input
combinations are High
PDN will conduct and will
Pull the output node down
to Ground making Output
Low (Y=0) (Voltage Zero)
• Simultaneously, PUN will
be OFF and no path will
Exists between VDD and
Ground
10.3.1 Basic Structure
• When All Three input
combinations are Low
PUN will conduct and will
Pull the output node Up to
VDD making Output High
(Y=1) (Voltage = VDD )
• Simultaneously, PDN will
be OFF and no path will
Exists between VDD and
Ground
10.3.1 Basic Structure
• PDN and PUN each Utilize Devices in Parallel
to form an OR Function
PDN :
QA will conduct when A is
Hi and will Pull the Output
Down to ground (Y=0)
QB will conduct when B is Hi
and will Pull the Output
Down to ground (Y=0)
Thus Y=0, when A OR B is High
Y A B
Y A B
10.3.1 Basic Structure
• PDN and PUN each Utilize Devices in Parallel
to form an OR Function
PUN :
QA will conduct when A is
Lo and will Pull the Output
Up to VDD (Y=1)
QB will conduct when B is Lo
and will Pull the Output
Up to VDD (Y=1). Thus
Y=1 (Hi), when A OR B is Lo
Y A B
10.3.1 Basic Structure
• PDN and PUN each Utilize Devices in Series to
form an AND Function
PDN :
QA and QB will conduct
ONLY when both A and B
are Hi Simultaneously.
Thus Y=0 (low),
when A is High AND B is High
Y AB
Y AB
10.3.1 Basic Structure
• PDN and PUN each Utilize Devices in Series to
form an AND Function
PUN :
QA and QB will conduct
ONLY when both A and B
are Lo Simultaneously.
Thus Y=1 (High),
when A is High AND B is High
Y AB
10.3.1 Basic Structure
PDN :
Y=0 (low),
when A is High OR
when
A AND B are both
High
Y A BC
Y A BC
10.3.1 Basic Structure
PUN :
Y=1 (low),
when A is Lo OR
when
A AND B are both Lo
Y A BC
10.3.1 Basic Structure
• After understanding structure and
operation of PDNs and PUNs we will
consider complete CMOS gates
• BUT Before that we need to introduce
alternative ckt symbols which are almost
universally used for MOS transistors by
digital-ckt designers
10.3.1 Basic Structure
• Circle at Gate Terminal for PMOS indicate that
the Signal at gate has to be low for it to be
activated (conduct)
• These symbol omit indication of source and
drain
Basic Boolean Identities
Fundamental Laws:
OR
AND
A0A
A0 0
A 1 1
A1 A
AA A
AA A
A A 1
AA 0
Associative Law:
(A B) C A (B C)
(AB)C A(BC)
NOT (Inverter)
A A 1
AA 0
A A
Commutative Law
A B B A AB BA
Basic Boolean Identities
Distributive Law:
A(B C) AB AC
DeMorgan’s Law:
AB... A B ...
A B ... AB...
Auxiliary Identities:
A AB A
A AB A B
( A B)( A C ) A BC
PDN :
Y=0 (Low), when A OR B is High
Y A B
Y A B
PUN :
Y=1 (High),
when A is Low AND B is Low
Y AB
10.3.2 Two Input NOR Gate
Combining both PDN and PUN
realizes Complete CMOS NOR Gate
with NOR Function
Y A B AB
Input
A
0
0
1
1
B
0
1
0
1
Output
Y
1
0
0
0
PDN :
Y=0 (low), when
A is High AND B is High
Y AB
Y AB
PUN :
Y=1 (Hi), when A OR B is Lo
Y A B
10.3.3 Two Input NAND Gate
Combining both PDN and PUN
realizes Complete CMOS NAND
Gate with NAND Function
Y AB A B
Input
A
0
0
1
1
B
0
1
0
1
Output
Y
1
1
1
0
10.3.4 A Complex Gate
Consider More Complex Logic Function
Y A( B CD)
Y A( B CD)
Y should be Low for A High AND
Simultaneously either B High
OR C AND D both High.
The PDN for this is.
To get PUN we need to
Express Y in terms of
Complemented variables
10.3.4 A Complex Gate
So we use DeMorgan’s Law
Y A( B CD)
A B CD
A BCD
A B(C D)
Thus, Y is High for A OR B Low AND either
C OR D Low. Thus PUN for this is.
10.3.4 A Complex Gate
Combining both PDN
and PUN realizes
Complete CMOS
Complex Gate
Function
Y A( B CD)
10.3.5 Obtaining PUN from PDN
• So far, we have seen that PDN and PUN
are dual networks: A series branch exist
in one and Parallel branch exist in other.
• Thus, we can obtain one from the other –
a simple process than using Boolean
expressions.
• For Complex Gate we found PDN
relatively easy Y (bar) in terms of uncomplemented inputs. We could obtain
PUN using this duality method instead of
Boolean Expression.
10.3.5 Obtaining PUN from PDN
10.3.5 Obtaining PUN from PDN
Complex Gate using
duality of both PDN
and PUN
Y A( B CD)
10.3.6 Exclusive OR Function (XOR)
• An important Function that is often used in
logic design is the Exclusive-OR (XOR)
function:
Y AB AB
• Y instead of Y(bar) is given so we can
synthesize PUN easily.
• Unfortunately Y is not function of
complemented variables only, thus we will
need additional inverters.
10.3.6 Exclusive OR Function (XOR)
PUN Obtained directly
from:
Y AB AB
Note that we have used
two inverters to generate
A(bar) and B(bar)
PDN can be synthesized
from PUN using duality
or developing the Y(bar)
expression.
10.3.6 Exclusive OR Function (XOR)
First: PDN from PUN
using duality
Second: develop the
Y(bar) expression using
DeMorgan Law on
Y AB AB
Gives Y AB A B
10.3.6 Exclusive OR Function (XOR)
So the complete XOR
using PUN and PDN
Y AB AB
Note that we have used
two inverters to generate
A(bar) and B(bar) which
are not shown.
XOR requires 12
transistors
10.3.7 Synthesis Method Summary
• To synthesize PDN we need Y(bar) expression
in terms of uncomplemented variables. If
complemented variables appear in expression
we need inverters.
• To synthesize PUN we need Y expression in
terms of complemented variables and then apply
uncomplemented variables to the gates of
PMOS transistors. If uncomplemented variables
appear in expression we need inverters.
• PDN can be obtained from PUN (and vice versa)
using duality
In Next Class
We Will Continue to Discuss:
CMOS Logic Gates