Circuits & Electronics P4

6.002 Fall 2000 Lecture 1 4 6.002 CIRCUITS AND ELECTRONICS The Digital Abstraction 6.002 Fall 2000 Lecture 2 4 Review z Discretize matter by agreeing to observe the lumped matter discipline zAnalysis tool kit: KVL/KCL, node method, superposition, Thévenin, Norton (remember superposition, Thévenin, Norton apply only for linear circuits) Lumped Circuit Abstraction 6.002 Fall 2000 Lecture 3 4 Discretize value Digital abstraction Interestingly, we will see shortly that the tools learned in the previous three lectures are sufficient to analyze simple digital circuits Reading: Chapter 5 of Agarwal & Lang Today 6.002 Fall 2000 Lecture 4 4 Analog signal processing But first, why digital? In the past … By superposition, The above is an “adder” circuit. 2 21 1 1 21 2 0 V RR R V RR R V + + + = If , 21 RR = 2 21 0 VV V + = 1 V 1 R 2 R + – 2 V + – 0 V and might represent the outputs of two sensors, for example. 1 V 2 V 6.002 Fall 2000 Lecture 5 4 Noise Problem … noise hampers our ability to distinguish between small differences in value — e.g. between 3.1V and 3.2V. Receiver: huh? add noise on this wire t 6.002 Fall 2000 Lecture 6 4 Value Discretization Why is this discretization useful? Restrict values to be one of two HIGH 5V TRUE 1 LOW 0V FALSE 0 … like two digits 0 and 1 (Remember, numbers larger than 1 can be represented using multiple binary digits and coding, much like using multiple decimal digits to represent numbers greater than 9. E.g., the binary number 101 has decimal value 5.) 6.002 Fall 2000 Lecture 7 4 Digital System sender receiver S V R V noise S V “0” “0” “1” 0V 2.5V 5V HIGH LOW t R V “0” “0” “1” 0V 2.5V 5V t VV N 0= N V S V “0” “0” “1” 2.5V t With noise VV N 2.0= S V “0” “0” “1” 0V 2.5V 5V t 0.2V t 6.002 Fall 2000 Lecture 8 4 Digital System Better noise immunity Lots of “noise margin” For “1”: noise margin 5V to 2.5V = 2.5V For “0”: noise margin 0V to 2.5V = 2.5V 6.002 Fall 2000 Lecture 9 4 Voltage Thresholds and Logic Values 1 0 1 0 sender receiver 1 0 0V 2.5V 5V 6.002 Fall 2000 Lecture 10 4 forbidden region V H V L 3V 2V But, but, but … What about 2.5V? Hmmm… create “no man’s land” or forbidden region For example, sender receiver 0V 5V 1 1 0 0 “1” V 5V “0” 0V V H L [...]... Boolean equation OR C OR gate More gates B B Inverter X Y Z NAND Z = X • Y 6.002 Fall 2000 Lecture 4 19 Boolean Identities X X X X • 1 = X • 0 = X + 1 = 1 +0 = X 1 = 0 0 = 1 AB + AC = A • (B + C) Digital Circuits Implement: B C output = A + B • C B•C output A 6.002 Fall 2000 Lecture 4 20 . 6.002 Fall 2000 Lecture 1 4 6.002 CIRCUITS AND ELECTRONICS The Digital Abstraction 6.002 Fall 2000 Lecture 2 4 Review. three lectures are sufficient to analyze simple digital circuits Reading: Chapter 5 of Agarwal & Lang Today 6.002 Fall 2000 Lecture 4 4 Analog signal