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EECS 465: Digital Systems Design Lecture Notes Logic Design Using Compound Components: Multiplexers SHANTANU DUTT Department of Electrical and Computer Science University of Illinois, Chicago Phone: (312) 355-1314; e-mail: dutt@eecs.uic.edu URL: http://www.eecs.uic.edu/~dutt. PLD.

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slide1

EECS 465: Digital Systems Design

Lecture Notes

Logic Design Using Compound Components: Multiplexers

SHANTANU DUTT

Department of Electrical and Computer Science

University of Illinois, Chicago

Phone: (312) 355-1314; e-mail: dutt@eecs.uic.edu

URL: http://www.eecs.uic.edu/~dutt

the world of integrated circuits lsi vlsi

PLD

PLA/PAL

CPLDs

The World of Integrated Circuits (LSI/VLSI)

Full-Custom

ASICs

Semi-Custom

ASICs

User

Programmable

Simple

gates (nand/

nor/xor/xnor..)

PLA/PAL

(layout aspect; not

programmable)

Complex

gates/cells

Muxes

FPGA

slide3

Logic Design Using Multiplexers

A Transmission Gate ( T-Gate )

Out

Steering gate.

In=B

A

When A=1, ‘In’ is “Steered” to ‘Out’.

[I.e., the T-gate conducts]

Thus Out = B when A=1

Out = AB

In

Out

Symbolic for T-gate:

A

A is the control input (CI)

Normal CI connected to A

Bubbled CI connected to

T-gate conducts

when A=1.

slide4

Reversing the connection of A: A

bubble CI

A

normal CI

In

Out

T-gate conducts when A=0.

Multiplexer (MUX) Design: A 2:1 MUX.

S

I0

Out

I0

Z

Z

2:1

MUX

I1

Out

I1

S

S

Z= I0 when S=0

Z= I1 when S=1

slide5

Generalization of the TT:

S Z

S Z

0 1

1 1

0 1

1 0

S Z

1 1

0 0

= S

S Z

0 I0

1 I1

is the function

implemented by the above 2:1 MUX. Such a TT in general provides a decomposition of the final function Z into constituent functions I0 & I1

Thus, the 2:1 MUX can also be implemented by logic gates:

I0

z

2:1 MUX :

S

I1

slide6

I0

A 2:1 MUX selects input Ii if S0 = I

[If S0 = 0, Z = I0

S0 = 1, Z = I1]

Z

2:1

MUX

I1

S0

I0

The same can be said about

a 4:1 MUX:

Input Ii is selected (Z=Ii)

if S1S0 combination represents

the number i in binary.

Z

I1

4:1

MUX

I2

I3

S1

S0

slide7

In general, # of data inputs (Iis) is 2n

# of control I/Ps = n

[If S1S0 = 00 (#0), Z = Io

S1S0 = 01 (#1), Z = I1

S1S0 = 10 (#2), Z = I2

S1S0 = 11 (#3), Z = I3]

A generalized or symbolic TT

S1 S0 Z

0 0 I0

0 1 I1

1 0 I2

1 1 I3

slide8

I0

I1

Z

2n : 1

Sn-1

S1

S0

In general for a 2n:1 MUX with control signals/inputs

Sn-1 ··· S1S0 & “data” inputs I0, ···, ,

Z = Ii when Sn-1 ··· S1S0 combination represents #i in binary.

slide9

I0

2:1

MUX

I1

Z

2:1

MUX

S0

I2

2:1

MUX

S1

I3

MSB

S0

Design of MUXes using Divide-&-Conquer

  • • Saw the design of a 2:1 MUX using T-gates, as well as logic gates
  • Messy and expensive to design larger MUXes using a flat TT based approach
  • • A 4:1 MUX can be hierarchically constructed using 2:1 MUXes
  • Idea: Divide the selection problem by bits of the select/control variables

These inputs should

have different lsb or S0 values, since their sel. is based on S0. All other bits should be equal.

Inputs selected are those w/ the same lsb or S0 values. So further selection needs to be based on the non-lsb bits.

I0

Z

I1

4:1

MUX

I2

I3

S0

S1

These inputs should

have different lsb or S0 values, since their sel. is based on S0. All other bits should be equal.

LSB of control variables

slide10

When S0=0, I0, I2 get selected at the 1st level, i.e., Input w/ 0

in LSB.

When S0=1, I1, I3 (LSB=1) get selected at the 1st level.

If S0 = 0, I0, I2, become the 0th & 1st inputs to the next level.

At the next level, the I/P order # is determined by the rest of the

bits of their index after stripping off the LSB.

Thus I0 I0

I2 I1

At level 2

0 0 (# 0)

1 0 (# 2)

strip away for the 2nd level inputs.

slide11

Z= I0 of level 2

(I0 of level 1), S1=0.

= I1 of level 2

(I2 of level1) if S1=1.

I0

I2

I0

I1

Level 2

From level 1

S0=0

strip

0 1

Z

I0

I1

2:1

MUX

I1

I3

Z= I0 of Level 2

(I1 of Level1) if S1=0

= I1 of level 2

(I3 of level 1) if S1=1

Level 1

1 1

S1

strip

S0=1

Thus the design works as a 4:1 MUX.

slide12

An 8:1 MUX is designed similarly.

Selected when S0 = 0

Z

Z

I0

I0

2:1

MUX

I1

S0

I1

I2

I2

2:1

MUX

I3

4:1

MUX

I0

I1

I2

I3

I4

I5

I6

I7

I3

I5

8:1

MUX

S0

I4

I4

2:1

MUX

I5

S2

S1

S0

I6

I6

S2 S1 S0

2:1

MUX

I7

I7

These inputs should

have different lsb or S0 values, since their sel. is based on S0 (all other remaining, i.e., unselected bit values should be the same). Similarly for other i/p pairs at 2:1 Muxes at this level.

S0

Selected when S0 = 1

slide13

Selected when S0 = 0

I0

I1

I2

I3

I4

I5

I6

I7

Z

8:1

MUX

2:1

MUX

2:1

MUX

2:1

MUX

2:1

MUX

S2 S1 S0

S0

S2

S1

Opening up the 8:1 MUX’s hierarchical design

I0

I0

2:1

MUX

Selected when

S0 = 0, S1 = 1, S2=1

I1

S0

I2

I2

I2

2:1

MUX

I6

I3

S1

Z

I6

S0

I4

I4

I5

S0

Selected when S0 = 0, S1 = 1. These i/ps should differ in S2

I6

I6

2:1

MUX

These inputs should

have different lsb or S0 values, since their sel. is based on S0 (all other remaining, i.e., unselected bit values should be the same). Similarly for other i/p pairs at 2:1 Muxes at this level.

These inputs should

have different S1 values, since their sel. is based on S1 (all other remaining, i.e., unselected bit values should be the same). Similarly for other i/p pairs at 2:1 Muxes at this level.

I7

slide14

Selected when S2 = 1

I0

I1

I2

I3

I4

I5

I6

I7

Z

8:1

MUX

2:1

MUX

2:1

MUX

2:1

MUX

2:1

MUX

S2 S1 S0

S0

S2

S1

8:1 MUX’s: Input groupings for a different control variable order (S2, S0, S1)

Seed input

I4

I0

2:1

MUX

+/- bit 2

Selected when

S0 = 0, S1 = 1, S2=1

I4

S2

+/- bit 0

I1

I5

I4

2:1

MUX

I6

+/- bit 1

+/- bit 2

I5

S0

Z

S2

I6

I6

I2

+/- bit 2

I6

+/- bit 1

S2

Selected when S2 = 1, S0 = 0. These i/ps should differ in S1

+/- bit 0

I3

I7

2:1

MUX

These inputs should

have different lsb or S2 values, since their sel. is based on S2 (all other remaining, i.e., unselected bit values should be the same). Similarly for other i/p pairs at 2:1 Muxes at this level.

These inputs should

have different S0 values, since their sel. is based on S0 (all other remaining, i.e., unselected bit values should be the same). Similarly for other i/p pairs at 2:1 Muxes at this level.

+/- bit 2

I7

slide15

General D&C/Hierarchical Design of a 2n :1 MUX

2:1

I0

S0

2n:1

MUX

2n-1 :1

MUX

2:1

2n-1

2:1

MUXes

S0

Design Strategy:

Sn-1 S0

2:1

  • First select inputs based on S0, using 2n-1 2:1 Muxes; 2n-1 inputs get selected on 2n-1 lines
  • The problem now reduces to that of a 2n-1:1 Mux
  • Continue recursively (to a 2n-2:1, 2n-3:1, …., 4:1, 2:1 Mux design problems) until the final output is obtained.
  • Cost = 2n -1 2:1 Muxes = 6*(2n -1) gate inputs (6 is the gate-i/p cost of a 2:1 Mux)
  • Compare to flat design: (n+1)*2n + 2n = (n+2)*2n gate inputs (will actually require more for n > 4, as large gates required for a 2-level impl. are not desirable (e.g., driving resistance become too large). More expensive for n > 4 (for 2-level design, if at all that is possible), and even more expensive for a multi-level design.
  • Note that both costs are exp. in n, but linear in the # of inputs (2n), which is the important parameter.
  • More realistic hardware cost (complexity in terms of gate i/ps) for flat deign when up to only 4-i/p gates are avail? For simplicity, assume n is a power of 4.

Sn-1

S1

S0

slide16

Using MUXes to Realize Logic Circuits

Example:

-- Use A,B,C as control inputs to an 8:1 MUX.

-- Treat the data inputs I0, ···, I7 as possible minterms of f:

Input Ii is a minterm if #i appears in the  m notation of f.

If Ii is a minterm of f, it should be connected to a ‘1’, otherwise

it should be connected to a ‘0’.

1

0

1

0

0

0

1

1

I0

I1

I2

I3

I4

I5

I6

I7

8:1

MUX

f

S2 S1 S0

A

B

C

slide17

A 3-variable function can always be implemented by only an 8:1 MUX.

  • Interestingly, a 3-var. function can also always be implemented by only a 4:1 MUX (assuming vars & their complements are avail.).
  • Note: Can represent a function in terms of all combinations of amy subset of variables. E.g., any f (A,B,C) = A’B’(I0) + A’B(I1) + AB’(I2) + AB (I3), where the Ij’s can be 0, 1, C or C’
    • generalization of Shannon’s expansion theorem: f(X) = xi*f(X,xi=1) + xi’*f(X,xi=0)
  • Example f(A,B,C) =  m(0,2,6,7)
  • Can determine Ij’s using a K-Map: Select any 2 variables, say, A,B, for the 2 control inputs
  • In a 3-var. K-map, group together squares into 2-squares in which the other variable C varies, but A,B remains constant.
  • Form implicants only within 2-squares and write out the function as sum of a min. of these implicants (choosing essential implicants first, etc.)

AB

2-squares

00 01 11 10

C

1

1

1

0

1

1

AB

slide18

• The function now is in terms of combinations of A, B

ANDed with either C, , 1, or 0

Note: 0 is ANDed with an A,B combination (e.g., above) when the

2-square corresponding to that combination does not have any 1’s,

i.e., when none of the product terms obtained from the K-map in the

above manner has that combination of A, B in them.

• Connect either C, , 1, 0 to the appropriate data input

corresponding to the A,B combination they are ANDed with in the

expression for f.

• Thus

• Note that a 4:1 MUX implements the function

Thus for the above f,

and

slide19

Corresponding generalized TT

A B f

0

f

4:1

MUX

1

0 0

0 1

1 0 0

1 1 1

0

1

2

S1 S0

3

A B

-- A general n-variable function f(An-1, An-2, ···, A0)

Implementation possible using a 2n: 1 MUX (w/o requiring any

extra gates)? Yes.

Implementation possible using a 2n-1: 1 MUX (w/o requiring

any extra gates) ? Yes (assuming complemented variables are

available).

slide20

E.g.: A 4-var. function f(A,B,C,D)

Choose A, B, C as the control inputs

AB

00 01 11 10

CD

1

1

00

01

0

1

2

3

4

5

6

7

1

0

f

1

1

0

0

0

1

11

10

1

1

A B C

slide21

-- In general, a 2i : 1 MUX, where i < n-1, can be used to implement

an n-var. function f(An-1, ••• ,A0) by choosing any i variables, say,

Ai-1, ••• , A0 ( This is just an example of i variables; you can choose

any i of the n variables) as the control inputs of the MUX. However,

for i < n-1, extra logic gates may be required.

-- Express f in terms of all combinations of Ai-1, •••, A0 (using Shannon’s expansion recursively i times for each of these vars in turn) as:

Where is a function of An-1,•••,Ai that ANDs with the

kth product term of variables Ai-1, •••, A0 that represents the

binary # k,

slide22

— Thus we get the implementation:

I0

g0(An-1,···, Ai+1)

Where each

gk may need

extra logic

gates for its

implementation.

2i:1

MUX

Ai-1 A0

— The trick is to choose the “right” i variables so that the total # of logic gates needed for the gk’s is minimum (thus us a hard problem but interesting and beneficial to choose).

slide23

Heuristic Technique for gate-minimized design of a n-var. function using a 2i:1 MUX:

  • “PIs” can only be formed within each groups of squares in which the control vars. are constant (constant areas), which are “4-squares” in the example below.
  • To min. # of gates, choose the icontrol vars such that the largest-size implicants & the smallest # of them can be formed in its set const. areas.
  • Main idea of how to achieve this: Form all PIs in the full K-map, and choose the constant areas that break as few of the EPIs as possible, and then as few of the “obvious” choices of other PIs as possible
  • Then, in each constant-area, form the SOP sub-expression for all the MTs in that area, just like in a full K-Map (i.e., for all “PIs” in each area, determine and select all “EPIs” in that area, cover remaining MTs in that area by the least-cost set of the remaining “PIs”).

— Example : 4-var. function f(A,B,C,D) ; n=4. Let i=2

— For i=2 (2 control variables from A,B,C,D for a 4:1 MUX), the K-map needs to be partitioned into groups of 4-squares, such that within each 4-square the 2 control variables are constant.

AB

AB

AB

AB

00 01 11 10

00 01 11 10

00 01 11 10

00 01 11 10

CD

CD

CD

CD

1

1

1

1

1

1

1

1

00

01

00

01

00

01

00

01

c

d

d

c

d

1

1

1

1

a

b

b

1

1

1

1

11

10

11

10

11

10

11

10

a

b

EPIs

1

1

1

1

1

1

1

1

c

d

d

d

c

AC const. (green)

CD const. (orange)

BD const. (blue)

AD const. (maroon)

slide24

0

f

4:1

MUX

C

1

2

S1 S0

3

0

A B

AB

00 01 11 10

CD

1

1

00

01

a

4-squares where A,B = constant  A,B are the mux control vars.

1

c

c

1

11

10

d

d

b

1

1

AB const. (lt. green)

BC const. (yellow)

  • A study of the 1’s in the above K-map tells us that this is achieved by the set of 4-squares that are the columns of the K-map, i.e., for control variables A, B.
  • Then, in each constant-area, form the SOP sub-expression for all the MTs in that area, just like in a full K-Map (i.e., for all “PIs” in each area, determine and select all “EPIs” in that area, cover remaining MTs in that area by the least-cost set of the remaining “PIs”).
  • We this obtain:

AB

00 01 11 10

CD

1

1

00

01

1

Implementation:

No logic gates needed!

(NOTE: This will not

always be the case)

1

11

10

1

1

slide25

— If we choose A,C as the control variables (which also cut 0 EPIs),

then the 4-squares w/ constant A, C are:

AB

00 01 11 10

CD

1

1

00

01

1

1

11

10

1

1

— Grouping 1’s only within the 4-squares, we get 4 terms

and : all are essential (within their

constant areas)

slide26

-- We thus obtain f in terms of all combinations of A, C as

I2

I3

I0

I1

0

f

4:1

MUX

1

B

2

S1 S0

0

3

A C

-- Thus choosing A,C as control variables, results in 2 extra 2 i/p

gates (gate i/p cost = 4) compared to choosing A, B, for which the cost = 0.