Shift Register

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Shift Register. Section 6.1-6.2. Register. A register is a group of flip-flops, each one of which is capable of storing one bit of information. Issues of the circuit to the right. You do not have an option hold the output when you don’t want to outputs updated. - PowerPoint PPT Presentation

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Shift Register

Section 6.1-6.2

Register• A register is a group of flip-flops,

each one of which is capable of storing one bit of information.

• Issues of the circuit to the right.– You do not have an option hold

the output when you don’t want to outputs updated.

4 D flip-flops=4 bits of storage=4-bit register

4-bit Register with Parallel Load Control

Load=“1”→Update“1”

“0”

“0”

“1”

“I0”

“I0”

I0 is fed to DFF when Load is a 1.

Load=“0”→Hold!“0”

“1”

“A0”

“0”

“0”

“A0”

A0 is fed to DFF when Load is a 0. So the outputis holding!

We will revisit this ideawhen we study the universalshift register.

Four-Bit Serial Shift Register

1 2 3 4

Q of DFF1 gets SI after the first rising edge of the CLKQ of DFF2 gets SI after the second rising edge of the CLKQ of DFF3 gets SI after the third rising edge of the CLKQ of DFF4 gets SI after the fourth rising edge of the CLK

Linear Feedback Shift Register

1 1 0 1

Exclusive OR

Content of Four-Bit Shift Register1 1 0 1

1 1 1 0

1 1 1 1

0 1 1 1

0 0 1 1

0 0 0 1

1 0 0 0

0 1 0 0

0 0 1 0

1 0 0 1

1 1 0 0

0 1 1 0

1 0 1 1

0 1 0 1

1 0 1 0

Block Diagram of a Universal Shift Register

This is called the universal shift register because it has both shifts and parallel load capabilities.

Functionality of the Universal Shift Register

Clear: to clear the register to 0.CLK: to synchronize the operations.{S1,S0} for mode control.A_par: register outputI_par: register inputMSB_in and LSB_in: serial inputs

Detail Implementation

Four-to-one-line Mux

Four-to-one-line Mux

1

0

11

I2 I2

0

0

00

0

0I2

Mode Control

S0=0, S1=0 [No Change Mode]

S0=0, S1=0

S0=1, S1=0 [Shift Right Mode]

S1=0 , S0=1

S0=0, S1=1 [Shift Left Mode]

S1=1 , S0=0

S0=1, S1=1 [Parallel Load Mode]

S1=1 , S0=1

Breadboard Implementation

Universal shift regsiter

Random Number Generator

Waveform

CLK

Random

A3

A2

A1

A0

4-Bit Universal Shift Register

Behavioral Vs. Structural Description

• Behavioral Description– Behavior model of a shift register• Describe the operation of the register

without a preconceived structure.

– Random number generator• Binary values of msb_in, lsb_in, i_par

• Structural Description–Models the circuits in terms of a

collection of components such as gates, flip-flops…

Behavioral Model of Shift Regsiter

a_par[3] a_par[2] a_par[1] a_par[0]

Test Bench

1. Generate random numberWith matlab2. Read random numberat the neg edge of the clock

Test all input combinations by flipping {S1, S0}

Read numbers to i_par[3:0],msb_in, lsb_in at the negedge of t_clock

[s1,s0=[1,1], Load

i_par=0111a_par=0111

[s1,s0]=[0,0], No Change

i_par=0111a_par=0011

[s1,s0]=[1,0], Shift Left

1101 1011

LSB_in

[s1,s0]=[0,1], Shift Right

Synthesized Schematic

Structural Modeling of a 4-Bit Universal Shift Register

clr

clk

select

Q

i0i1

i2i3

Waveform

Load No Change Shift Right Shift Left

4-bit Universal Shift Register

Verilog Code of Each Stage

In-Class Exercise

Load=“1”→Update“1”

“0”

“0”

“1”

“I0”

“I0”

I0 is fed to DFF when Load is a 1.

Load=“0”→Hold!“0”

“1”

“A0”

“0”

“0”

“A0”

A0 is fed to DFF when Load is a 0. So the outputis holding!

We will revisit this ideawhen we study the universalshift register.

S0=0, S1=0 [No Change Mode]

S0=0, S1=0

S0=1, S1=0 [Shift Right Mode]

S1=0 , S0=1

S0=0, S1=1 [Shift Left Mode]

S1=1 , S0=0

S0=1, S1=1 [Parallel Load Mode]

S1=1 , S0=1

If time permits

Serial Transfer Using Shift Register

Information in A is made to circulate by connecting SO to SI.

Parallel Transfer Versus Serial Transfer

Parallel TransferTransfer all the bit in one clock cycle.Require combinatorial circuits.

(Serial Transfer)

Take multiple clock cyclesto transfer data.

Assume n=4, each shift Register has 4 DFF.

Augend, Addend & Sum

1011+1001______10100

AugendAddend

Sum

Serial Adder

1

1

0

1

Feed “1” to zat the next rising edge of the CLK

Assuming a shift-right register, the left most position becomes availablefor storage after the second rising edge of the clock.

Note thatThe sum can bestored in a thirdregister.

But if you want to save shift register, you can store it in A since more and moreslots in SRA become available.

(Augend)

(Addend)

Serial Adder At the end of T4

A3A2A1A0

B3B2B1B0

________________

CoS3S2S1S0

S2S1S0A3

S3

Co

D2D1D0B3

Allowing the Serial Adder to Accumulate

Co

S3 S2 S1S0

D3D2D1D0

________________

Ro T3 T2 T1 T0

T2T1T0S3

T3

Ro

X2X1X0D3

Accumulate with a Shift Register

• A, B and D, each represents a 4 bit sequence.• We want to perform A+B+D• Store A in shift register A.• Store B in shift register B.• Allow the CLK to go on for a couple of cycles.• Store the sum bits of A+B in Shift A and allow

D to enter shift register B.• Allow more cycles of CLK. • Add D to A+B, and allow A+B+C to enter shift

register A.

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