Number and Code

Numerical Systems

• radix decimal(d/10), binary(b/2), octal(o/8), hexadecimal(h/16)
• conversion (Integral parts) repeated division-by-2/8/16, (Fractions part) repeated multiplication by 2/8/16
• signed number

sign-magnitude	1’s complement(反码)	2’s complement(补码)
$\pm 0$	$-2^{n-1}+1\sim 2^{n-1}-1, \pm 0$	$-2^{n-1}\sim 2^{n-1}-1$
0 101	0101	0101
1 101	1010	1011

• arithmetic operations addition&subtraction (overflow!), multiplication (direct addition, partial products)
• floating-point numbers single-precision = sign bit (1S) + exponential (8E) + mantissa (23F) = $\pm$(1+.F)$\times2^{E-127}$

Coding

• BCD, 8421, excess-3 cycle, gray code
• parity error detection code (P=0 when even/odd 1s)
• hamming code $2^P\ge P+D+1$ e.g. $P_1P_2D_1P_3D_2D_3D_4$, $P_1=(1,3,5,7), P_2=(2,3,6,7), P_3=(4,5,6,7)$

  Combinational Logic

  Basics

• logic gates NOT, AND, OR, NAND/negative-OR($\overline{AB}=\bar{A}+\bar{B}$), NOR/negative-AND($\overline{A+B}=\bar{A}\bar{B}$), XOR($A\bar{B}+\bar{A}B$), XNOR($AB+\bar{A}\bar{B}$)
• simplication, standard forms

Applications

• adder
  • half-adder $C_\text{out}=AB, \Sigma=A\oplus B$, no input carry bit
  • full-adder $C_\text{out}=AB+(A\oplus B)C_\text{in}, \Sigma=(A\oplus B)\oplus C_\text{in}$
  • parallel binary adder
  • ripple carry adder cascading
  • look-ahead carry adder an output carry bit can be either generated by inputs $C_g=AB$, or propagated by input carry $C_p=A+B$, $C_\text{out}=C_g+C_pC_\text{in}$
• comparator
  • 1-bit \(\begin{aligned}&Y_{1}(A<B)=\overline{A}B\\&Y_{2}(A=B)=\overline{A\oplus B}\\&Y_{3}(A>B)=A\overline{B}\end{aligned}\)
  • higher bit? cascading LSB->MSB->result
• decoder&encoder expansion of bits by cascading chips and enabling one at the same time
• multiplexer&demultiplexer

Race and Hazard

Race: time delay between parallel signals Hazard: race causing incorrect transient output Hazard in $C+\bar{C}, C\bar{C}$ happens when switching from a variable to its complement (time delay). $Y=A\bar{C}+\bar{A}B+\bar{A}C$, both $A$ and $C$ have race; e.g. when $B=1,C=0$, $A$ arises hazard. Tangent circles in the K-map arises hazard.

How to avoid static hazard?

1. modify logic design, add redundant terms $Y=B\bar{C}+AC\rightarrow Y=B\bar{C}+AC+AB$ (eliminate tangent circles)
2. enable signal
3. filtering

Sequential Logic

Latches and Flip-flops

• all logics for S-R latch are reversed when using NAND (negative-OR)
• pay special attention to negative inputs: gated S-R latch inverts inputs so that they’re still active-high
• inverting bubbles can move along a line and cancel each other
• 对于negative input,可以$\overline{\text{input}}=\text{input_new}$,直接化为对应的negative output
• 可以通过工作时段判断类型: posedge-triggered FF对应slave在反时钟的low/posedge工作, 即为NOR型, 反之则为NAND型; master相反

What is sequential logic?

实际上是将时序逻辑按状态机考虑,分离为迭代过程($Q^n\rightarrow Q^{n+1}$)与组合逻辑(f(状态,输入/控制变量), 表达式, 列表与k-map是不同的表示方式). 复杂问题建议用K-map, 输入为自变量, 输出为所求因变量.

Find $S,R$ or $D$ or $J,K$ for the logic relationship to organize it in form of the drive equation

Conversion between FFs: $\text{drive variable}=f(Q(\text{combination logic}))=g(\text{drive variable’})$ 可以出现反馈回路形式的输入, 只要组合逻辑关系符合即可

	JK	T	D
JK	$J=Q^{n+1},K=\bar{Q^{n+1}}$	$J=K=T$	$J=D,K=\bar{D}$
T	$T=J\bar{Q^n}+KQ^n$	$T=Q^n\oplus Q^{n+1}$	$T=D\oplus Q^n$
D	$D=J\bar{Q^n}+\bar{K}Q^n$	$D=T\oplus Q^n$	$D=Q^{n+1}$

Propagation Delay

Applications

• data storage (parallel)
• frequency division 偶数次分频, 广义的counter
• register (serial)
  • DFF in series, synchronous triggered $(Q_0Q_1Q_2Q_3)^{n+1}=(S_{in}Q_0Q_1Q_2)^n$
• counter $M$ modulus = number of effective states
  • frequency dividing asynchronous counter ($2^{n-1}<M\le 2^n$, TFF) $CP_{i+1}=Q_i, T=1$ and $R=S=0$ / transition state $R=\text{modulus}, S=\text{modulus}-1$

    看起来反手反脚的方法有望避免glitch

  • frequency dividing synchronous counter $CP_{i+1}=CLK, T_i=f(Q_0,\dots,Q_{i-1})$ 更低位全为0/1时翻转

  • up/down counter (see 10-45)
  • register shift counter (DFF) 只变首尾,逐级传输
    • Johnson Counter (M=2N) $D_0=\bar{Q_{n-1}}$ 各级FF占空比恒定
    • Ring Counter (M=N) $D_0=Q_{n-1}$ FF输出个数恒定
  • common counter + decoder generate pulse as register shift ones (m序列)

    Integrated Chips

• adder
• comparator
• register
  • 74LS194

  $S_1S_0$	operation
  00	keep
  01	right shift, $D_{SR}$ in
  10	left shift, $D_{SL}$ in
  11	parallel data $ABCD$ in

• counter 把transition state 的$\sum{Q_i}$连到clear(R)或load(S); 引脚特性决定syn./asyn. 通过因子分解,以反馈设置对应R/S触发条件,级联多个芯片,可以实现更丰富的modulus

  • asynchronous decade counter 74LS290 (JK FF)

    pin	meaning	operation
    $\bar{A},Q_A$	clk and output for $M=2$
    $\bar{B}, \left{Q_DQ_CQ_B\right}$	clk and outputs for $M=5$

    可以用来实现编码: 1. BCD(8421) 5*2; 2. 5421 2*5 低位信号有时候不一定需要复位信号 注意异步需要多一个!

  • synchronous counter 74LS163 (syn. clr)

    pin	meaning	operation
    $DCBA$
    $\overline{CLR}, \overline{LD}$	clear, load	priorty
    $P$	parallel load enable	$PT=11$ enable
    $T$	toggle enable	$PT=01/x0$ keep
    $RCO$	ripple carry output (end mark)

    先级联后反馈: 1–P1–P2, 1–T1, RCO1–T2, LD–1/RCO2 先反馈后级联: 1–P1–T1, P2–T2–CLR1/LD1

  • synchronous counter 74LS161 (asyn. clr)

Sequential Logic Analysis

analysis steps:

1. combinational: Write down the drive equation of each FF and the output
2. sequential: Substitute the drive equation into each state equation and find out the state transition relationship $Q^{n+1}=f(X,Q^n)$

design steps:

1. define, simplify and assign states
   • simplify: eliminate equivalant states with same output and equivalent next state (隐含表法列举等价条件逐步排除); 注意mealy是X/Z, 而moore是Q/Z
   • assign and encode: state = combinations of outputs (from flip-flop output) $A_{2^{FF}}^k$, $2^{FF}\ge \text{output bits}$; adjacent assignment method (same next state>last state>output)
   • now we have a state transition diagram $Q^{n+1},Z=f(X,Q^n)$
2. get expressions of FF drive variables $g(X,Q^n)$ from $Q^{n+1,n}$ using
   • directly substituting drive equation
   • mechanically listing truth table
   • k-map, just flip x and y; 或者从状态图的k-map中直接挑出每个状态对应的形式,不起效的部分填上x
   • 自启动,就是保证所有x取的次态都接到主环上
3. 对于寄存器式counter状态迁移必须满足移位条件(只变首尾,单向传输),把所有状态连到一个环上

Integrated Circuits

Diode

MOSFET

• 互补逻辑门(对偶网络) for NMOS, series=AND, parallel=OR
• output impedance varies ($0.5r_o, r_o, 2r_o$)$\rightarrow$ 输入端越多$V_{OL},V_{OH}$整体偏高/低; 缓冲级(inverter)解决上述问题
• OD门,线与(并联的CS?)
• 三态门: EN=0, Y=Z; EN=1, 反相

  BJT

• 输入级(并联->AND)+倒相级+输出级(推挽式)
• OC门,线与(并联的CE?)