From Gates To Circuits Ii Sequential Circuits

From-Gates-to-Circuits-II-Sequential-Circuits

Sequential Circuits

Sequential circuits’ output depends not only on its current inputs, but also its previous inputs (current state)

$Sequential$ $Circuits$ = $Combinational$ $Circuits$ $+$ $Memory$ tylAV6

One bit memory

  • It should be able to hold a single bit, 0 or 1.
  • You should be able to read the bit that was saved.
  • You should be able to change the bit. There are only two choices:
    • Set the bit to 1
    • Reset, or clear, the bit to 0.

SR NOR Latch (or Flip-flop)

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  • Input:
    • $S$: set
    • $R$: rese
    • $Q,\overline Q$
  • Output:
    • $Q,\overline Q$
  • Set Function: When $S=1,R=0$ $\Rightarrow {\overline {Q}=0, Q=1}$ Then change $S=0,R=0$ lock the state $\Rightarrow {\overline {Q} = 0, Q=1}$

  • Reset Function: When $S=0,R=1$ $\Rightarrow \ {\overline{Q}=1,Q=0}$ Then change $S=0,R=0$ lock the state $\Rightarrow {\overline {Q} = 1, Q=0}$

When $S=1,R=1$ $Illegal$ $State$ $\Rightarrow \ {\overline{Q}=1,Q=0}$ $if$ $\overline {Q}$ reach first $\Rightarrow \ {\overline{Q}=0,Q=1}$ $if$ ${Q}$ reach first This results in a J-K flip-flop PRXb7m

  • For inputs SR = 00, the next value of Q could be either or 1, depending on the current value of Q.
  • So the same inputs can yield different outputs, depending on whether the latch was previously set or reset.
  • This is very different from the combinational circuits that we’ve seen so far, where the same inputs always yield the same outputs.

  • 4-bit latch

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Gated SR Latch
  • Enable 实现锁存功能
    • 当Enable为0时: S,R皆为0电路锁存

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SR-NAND Latch

  • SR—NOR Latch 类似

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Glitch

  • There is a finite time delay between a change in the inputs of a gate and any change in the output. This time is called gate delay.(栅极延迟)
  • In order to avoid glitches, we want to design storage elements that only accept input when ordered to so We use a clock to be the control input that gives orders to the circuit about when to change states

Clock

A clock is a special circuit that produces electrical pulses Clock speed is generally measured in megahertz (Mhz), or millions of pulses per second A clock is used by a sequential circuit to decide when to update the state of the circuit 决定何时更新电路状态 inputs to the circuit can only affect the storage element at given, discrete instances of time

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Edge-Triggered vs Level-Triggered 边沿触发与电平触发
  • Edge-triggered: allowed to change their states on either the rising or falling edge of the clock signal
  • Level-triggered: allowed to change state whenever the clock signal is either high or low
  • Technically, a latch is level triggered, whereas a flip-flop is edge triggered 锁存器是电平触发,触发器是边缘触发
Clocked S-R Latch
  • When C is 1, the circuit acts just like the NOR gate S-R latch.
  • When C is 0, the Set and Reset inputs are disabled
  • The latch can change only when C is true HphHwT

J-K flip-flop

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When {J=1,K=0,C=1} work like a SR latch When {J=0,K=1,C=1} work like a SR latch When {J=1,K=1} Q 状态反转

D flip-flop

  • It stores one bit of information
  • The output changes only when the value of D changes
  • an output value of 1 means the circuit is currently “storing” a value of 1
  • A D flip-flop is a true representation of physical computer memory jNaCls

  • D 实现设置/重置 0/1
  • EN 实现 锁存 当EN为0 电路锁存 D-SR-Latch

4-bit Register

  • 4 input lines, 4 output lines and a WE (Write Enable) line (also called clock)
  • WE实现锁定输出状态

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Binary Counter

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B0 (t+1) = JK (1, 1) = NOT B0(t) B1 (t+1) = JK (B0(t), B0(t)) B2 (t+1)= JK (B0(t)B1(t), B0(t)B1(t)) B3 (t+1)= JK (B0(t)B1(t)B2(t), B0(t)B1(t)B2(t))

4 x 3 Memory

可以说是三个4-bit register组合 Decoder 选择 ${word_0,word_1,word_2,word_3}$其中一条线路 输出则为${In_0,In_1,In_2}$ 的 ${word_0,word_1,word_2,word_3}$中的一个 以选择${word_0}$为例,输出${In_0,In_1,In_2}$ 的 ${word_0}$

  • The inputs lines: In0, In1, and In2
  • Address lines: S0 and S1

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  • Write a word to the memory
    1. An address is asserted on S0 and S1.
    1. WE (write enable) is set to high.
    1. The decoder using S0 and S1 enables only one AND gate, selecting a given word in memory.
    1. The line selected in Step 3 combined with the clock and WE select only one word.
    1. The write gate enabled in Step 4 drives the clock for the selected word.
    1. When the clock pulses, the word on the input lines is loaded into the D flip-flops.

Memory Addressing

  • Let’s assume a very simple microprocessor with 10 address lines (1KB memory)
  • Let’s assume we wish to implement all its memory space and we use 128 x 8 memory chips
  • SOLUTION
    • We will need 8 memory chips (8x128=1024)
    • Chip selection: We will need 3 address lines to select each one of the 8 chips
    • Memory location selection inside each chip: Each chip will need 7 address lines to address its internal memory cells
    • Address: 000 0000000 ($2^3$)($2^7$)

ivrwmk 具有10根地址线的简单微处理器(可寻址1KB内存)使用128 x 8存储芯片的方法。

具体来说:

  1. 10根地址线的微处理器(1KB内存):
    • 这个微处理器有10根地址线,这意味着它可以寻址 (2^{10}) 个不同的内存位置,可访问1KB(1千字节)的内存。
  2. 存储芯片:
    • 为了覆盖整个内存空间,选择了128 x 8存储芯片。每个芯片的容量为128个存储位置(128个地址),每个存储位置可存储8位(1字节)。
  3. 所需存储芯片数量:
    • 由于总内存空间为1KB,每个芯片的容量为128个存储位置,因此需要总共8个芯片((1 \text{KB} / 128 \text{每芯片的位置}))来容纳整个内存空间((8 \text{芯片} \times 128 = 1024 \text{位置}))。
  4. 芯片选择:
    • 为了选择8个存储芯片中的每一个,使用3根地址线。使用3根地址线可以选择8个芯片中的任意一个(因为 (2^3 = 8))。
  5. 每个芯片内存位置的选择:
    • 在每个芯片内部,需要7根地址线来寻址内部存储单元。这7根线帮助选择每个芯片中128个存储位置中的特定一个。
  6. 地址:000 0000000:
    • 这是一个示例地址。前三位数(000)用于选择8个存储芯片中的一个。剩下的7位数(0000000)用于在选定的芯片内选择特定的存储位置,其中每个芯片有128个可用的存储位置。

Summary

  • Sequential circuits can remember their previous inputs
  • Sequential circuits require clocks to control their changes of states
  • The basic sequential circuit unit is the flip-flop: SR, JK and D flip-flop
  • Examples: registers, binary counter and memory

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