CS 251 Fall 2019Principles of Programming LanguagesBen Woodλ CS 240 Spring 2020Foundations of Computer SystemsBen Wood

https://cs.wellesley.edu/~cs240/s20/

Sequential Logicand State

Output depends on inputs and stored values.(vs. combinational logic: output depends only on inputs)

Elements to store values: latches, flip-flops, registers, memory

Latch: CC-BY Rberteig@flickr

Sequential Logic 1

ALU

Processor: Data Path Components

Registers MemoryInstruction Fetch and Decode

12 3

Sequential Logic 2

Bistable latches

Q Q

Suppose we somehow get a 1 (or a 0?) on here.

Q Q00

=

Sequential Logic 3

SR latch

Q QRS

Set Reset

S R Q Q' Q (stable) Q' (stable)0 0 0 1 0 10 0 1 0 1 01 0 ? ? 1 00 1 ? ? 0 1

Sequential Logic 4

SR latch

Q QRS

Q

QR

S

Q

QR

S R

S Q

Q

R

S Q

Q

Sequential Logic 5

D latch

if C = 0, then SR latch stores current value of Q.if C = 1, then D flows to Q:

if D = 0, then R = 1 and S = 0, Q = 0if D = 1, then R = 0 and S = 1, Q = 1

Sequential Logic 6

D

C

R

S

Q

Q

Clock

Data bit

Time matters!

D

C

Q

Assume Q has an initial state of 0

ex

Sequential Logic 7

ClocksClock: free-running signalwith fixed cycle time = clock period = T. Clock frequency = 1 / clock period

A clock controls when to updatea sequential logic element's state.

Sequential Logic 8

Clock period

Falling edge

Rising edge

Synchronous systemsInputs to state elements must be valid on active clock edge.

Sequential Logic 9

Stateelement

1

Stateelement

2Combinational logic

D flip-flop with falling-edge trigger

Sequential Logic 10

D

C

QEQLDL

CL

D latch

QL

QFDF

CF

D latch

QF Q

leader follower

Clockleader stores D as E

follower stores E as Q

Can still read Qnow Qnext becomes Qnow

Time

Time matters!

D

C

E

Q

Assume Q and E have an initial state of 0

ex

Sequential Logic 11

Reading and writing in the same cycle

Assume Q is initially 0.QD

CD Flip-Flop

QClock

Sequential Logic 12

D flip-flop = one bit of storage

Sequential Logic 13

QD

CD Flip-Flop

Q

1

A 1-nybble* register(a 4-bit hardware storage cell)

Sequential Logic 14

Write

Clock

0

1

0

1

QD

CD Flip-Flop

QQD

CD Flip-Flop

QQD

CD Flip-Flop

QQD

CD Flip-Flop

Q

*Half a byte!

Register file

Read portsWhy 2?

Read registerselector 1Read registerselector 2

Write register selectorWrite data

Write?

Read data 1

Read data 2

r

r

r

w

w

w

r = log2 number of registersw = bits in word

Array of registers, with register selectors, write/read control,input port for writing data, output ports for reading data.

Write port0 = read1 = write

Sequential Logic 15

Read ports(data out)

Sequential Logic 16

C.8 Memory Elements: Flip-Flops, Latches, and Registers C-55

FIGURE C.8.7 A register fi le with two read ports and one write port has fi ve inputs and two outputs. The control input Write is shown in color.

FIGURE C.8.8 The implementation of two read ports for a register fi le with n registers can be done with a pair of n-to-1 multiplexors, each 32 bits wide. The register read number sig nal is used as the multiplexor selector signal. Figure C.8.9 shows how the write port is implemented.

Read registernumber 1 Read

data 1Read registernumber 2

Read data 2

Writeregister

WriteWritedata

Register file

Read registernumber 1

Register 0

Register 1

. . .

Register n – 2

Register n – 1

M

u

x

Read registernumber 2

M

u

x

Read data 1

Read data 2

AppendixC-9780123747501.indd 55AppendixC-9780123747501.indd 55 26/07/11 6:29 PM26/07/11 6:29 PM

FIGURE C.8.9 The write port for a register fi le is implemented with a decoder that is used with the write signal to generate the C input to the registers. All three inputs (the regis ter number, the data, and the write signal) will have setup and hold-time constraints that ensure that the correct data is written into the register fi le.

Write

01

n-to-2n

decoder

n – 2

n – 1

Register 0

C

D

Register 1

C

D

Register n – 2

C

D

Register n – 1

C

D

...

Register number...

Register data

valid during the time it is read, as we saw earlier in Figure C.7.2. The value returned will be the value written in an earlier clock cycle. If we want a read to return the value currently being written, additional logic in the register fi le or out side of it is needed. Chapter 4 makes extensive use of such logic.

Specifying Sequential Logic in VerilogTo specify sequential logic in Verilog, we must understand how to generate a clock, how to describe when a value is written into a register, and how to specify sequential control. Let us start by specifying a clock. A clock is not a predefi ned object in Verilog; instead, we generate a clock by using the Verilog notation #n before a statement; this causes a delay of n simulation time steps before the execu tion of the statement. In most Verilog simulators, it is also possible to generate a clock as an external input, allowing the user to specify at simulation time the number of clock cycles during which to run a simulation.

The code in Figure C.8.10 implements a simple clock that is high or low for one simulation unit and then switches state. We use the delay capability and blocking assignment to implement the clock.

C-56 Appendix C The Basics of Logic Design

AppendixC-9780123747501.indd 56AppendixC-9780123747501.indd 56 26/07/11 6:29 PM26/07/11 6:29 PM

Write port (data in)

Sequential Logic 17

incoming data

register number

write controlclock

RAM (Random Access Memory)

Similar to register file, except…

A B

Sequential Logic 18

16 x 4 RAM

19

4 to 16decoder

dataout

1101

4-bitaddress

Sequential Logic