# Lecture 11 - Last Week Closed Loop Op-Amp Currents i and i...

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1 ECTE290 Spring 2007 Lecture 11 2 Last Week ± Closed Loop Op-Amp V POS –V NEG r in = V neg < V OUT < V pos Currents i + and i to (or from) input terminals are zero V + = V i + = 0 i = 0 Defines the Linear Region of operation Virtual short Virtual open 1. i in =0 Assumptions for analysis of ideal 2. V + =V - op-amp with negative feedback feedback 3 This Week ± Number Systems ± Binary number ± Hexadecimal number ± Binary arithmetic ± Signed binary numbers ± Logic Gates & Truth Tables ± Boolean Algebra 4 Numbering Systems ± We use the decimal numbering system ± 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ± For example: 12 ± All computers use the binary numbering system ± Only two digits: 0, 1 ± For example: 10, 10001, 10110 ± Similar to decimal, except uses a different base ± Binary (base-2): 0, 1 ± Decimal (base-10): 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ± Octal (base-8): 0, 1, 2, 3, 4, 5, 6, 7 ± Hexadecimal (base-16): 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F (A=10, . .., F=15) ± What do we mean by a base?

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2 5 Decimal Number System ± Numbers consist of a bunch of digits, each with a weight: ± The weights are all powers of the base, which is 10. We can rewrite the weights like this: ± To find the decimal value of a number, multiply each digit by its weight and sum the products. 1 6 2 . 3 7 5 Digits 100 10 1 1/10 1/100 1/1000 Weights ( 1 x 10 2 ) + ( 6 x 10 1 ) + ( 2 x 10 0 ) + ( 3 x 10 -1 ) + ( 7 x 10 -2 ) + ( 5 x 10 -3 ) = 162.375 162. 375 Digits 10 2 1 0 -1 -2 -3 Weights Base 162.375 6 Binary Number System ± We can use the same trick to convert binary, or base 2, numbers to decimal. The only difference is that the weights are powers of 2. ± For example, here is 1101.01 in binary: ± The decimal value is: 1 1 0 1 . 0 1 Binary digits, or bits 2 3 2 2 2 1 2 0 2 -1 2 -2 Weights (in base 2) Powers of 2: 2 0 = 1 2 4 = 16 2 8 = 256 2 1 = 2 2 5 = 32 2 9 = 512 2 2 = 4 2 6 = 64 2 = 1024 2 3 = 8 2 7 = 128 ( 1 x 2 3 ) + ( 1 x 2 2 ) + ( 0 x 2 1 ) + ( 1 x 2 0 ) + ( 0 x 2 -1 ) + ( 1 x 2 -2 ) = 8 + 4 + 0 + 1 + 0 + 0.25 = 13.25 7 ± Convert the following binary numbers to decimal 101 2 = 1×2 2 +0×2 1 +1×2 0 =4+1=5 10 1010.01 2 = 1×2 3 +0×2 2 +1×2 1 +1×2 0 +0×2 -1 +1×2 -2 = 8+2+0.25=10.25 10 Exercises 8 Converting Decimal to Binary ± To convert a decimal integer into binary, keep dividing by 2 until the quotient is 0. Collect the remainders in reverse order. ± To convert a fraction, keep multiplying the fractional part by 2 until it becomes 0. Collect the integer parts in forward order. ± Example: 162.375: ± So, 162.375 10 = 10100010.011 2 162 / 2 = 81 rem 0 81 / 2 = 40 rem 1 40 / 2 = 20 rem 0 20 / 2 = 10 rem 0 10 / 2 = 5 rem 0 5 / 2 = 2 rem 1 2 / 2 = 1 rem 0 1 / 2 = 0 rem 1 0.375 x 2 = 0 .750 0.750 x 2 = 1 .500 0.500 x 2 = 1 .000
3 9 Exercises ± Convert the following binary numbers to decimal a) 111 b) 1101.1011 ± Convert the following decimal into binary c) 39 d) 39.75 10 Answer a) 111 b) 1101.01 =1×2 2 +1×2 1 +1×2 0 =7 =4+2+1 =13.25 10 =1×2 3 +1×2 2 +0×2 1 +1×2 0 +0×2 -1 +1×2 -2 +1×2 -2 = 8+4+1+0.25 11 Answer c) Convert the decimal integer 39 10 to binary Quotient Remainder 39/2 = 19 1 19/2 = 9 1 Read binary equivalent in reverse order: 9/2 = 4 1 Stop when quotient is zero!

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## This note was uploaded on 10/27/2009 for the course ECTE 290 taught by Professor Zhengli during the Three '09 term at University of Wollongong, Australia.

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Lecture 11 - Last Week Closed Loop Op-Amp Currents i and i...

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