The Operational Amplifier

Transcription

1 The Operational Amplifier The operational amplifier i a building block of modern electronic intrumentation. Therefore, matery of operational amplifier fundamental i paramount to any practical application of electronic circuit. Specialit in electronic intrument find employment in medical chool, hopital, reearch laboratorie, aircraft indutrie, and thouand of other indutrie where electronic intrument are routinely ued. The operational amplifier i often called the op amp for hort. 1

2 Op Amp The op amp i an electronic deice coniting of a complex arrangement of reitor, tranitor, capacitor, and diode. A full dicuion of what i inide the op amp i the ubject of another coure. In thi coure, it will uffice to treat the op amp a a circuit building block and imply tudy what take place at it terminal. 2

3 Op Amp The op amp i an actie circuit element that behae like a oltage-controlled oltage ource. It can alo be ued in making a oltage- or current-controlled current ource. An op amp i deigned to perform mathematical operation of addition, ubtraction, multiplication, diiion, differentiation, and integration. The ability of the op amp to perform thee mathematical operation i the reaon it i called an operational amplifier. 3

4 Op Amp Pin Configuration Op amp are commercially aailable in integrated circuit package in eeral form. The figure how a typical op amp package. The eight-lead Dual Inline Package (DIP) of an op amp. 4

5 Op Amp Circuit Symbol A typical op amp: pin configuration and circuit ymbol. 5

6 Powering the Op Amp A an actie element, the op amp mut be powered by a oltage upply. Although the power upplie are often ignored in op amp circuit diagram for the ake of implicity, the power upply current mut not be oerlooked. i i i i i p+ n = 0 c c Powering the op amp. 6

7 Practical Limitation Poitie aturation, o = V CC Linear region, V CC o = A d V CC Negatie aturation, o = V CC Op amp output oltage o a a function of the differential input oltage d = p n A i the o-called open-loop gain Note: If A i large, then d = p n mut be ery mall or the Op Amp will immediately go into poitie or negatie aturation. 7

8 Op Amp Equialent Circuit The input reitance i i the Theenin equialent reitance een at the input terminal The output reitance o i the Theenin equialent reitance een at the output Equialent circuit of the non-ideal op amp. 8

9 Open Loop Gain The differential input oltage d i gien by d = p n The op amp ene the difference between the two input, multiplie it by the gain A, and caue the reulting oltage to appear at the output. Thu, the output o i gien by o = A d = A( p n ) A i called the open-loop oltage gain becaue it i the gain of the op amp without any external feedback from output to input. How to tabilize an amplifier with a ery large gain? 9

10 Cloed Loop Gain The concept of feedback i crucial to our undertanding of op amp circuit. A negatie feedback i achieed when the output i fed back to the inerting terminal of the op amp. When there i a feedback path from output to input, the ratio of the output oltage to the input oltage i called the cloed-loop gain. The cloed-loop gain i almot inenitie to the open-loop gain A of the op amp. For thi reaon, op amp are ued in circuit with feedback path. 10

11 Negatie Feedback + 1 F + F + + i A F + o Feedback factor Open loop gain Cloed loop gain K = F F = F = F o o A = i o 1 o Cloed loop gain K o o o i A A = = = = = + + F + 1 FA+ 1 1 F i F i i i o i 1 1 K =» if A F F+ 1 A F Aume a paie feedback network with F 1, then A= 10 F and 1 K = ³ 1 F 5 11

12 Ideal Op Amp An op amp i ideal if it ha the following characteritic: 1. Infinite open-loop gain, A 2. Infinite input reitance, i 3. Zero output reitance, o 0 The current into both input terminal are zero The oltage acro the input terminal i negligibly mall i» = -» 0 d p n i p = 0, i = 0 p n = n Ideal op amp model 12

13 Inerting Amplifier i = 0 i =-i n f But The oltage gain A i - - =- n n o n f = = 0 for an ideal op amp p o o =- o f =- f =- f An inerting amplifier reere the polarity of the input ignal while amplifying it. 13

14 80k Example 5.1 Aume that the op amp i ideal. Calculate the output oltage o for the following alue of : 0.4V, 2.0V, 3.5V, -0.6V, -1.6V, -2.4V. 16k 10V -15V 14

15 Example 1 Find the output oltage of the op amp circuit. Calculate the current through the feedback reitor. i 15

16 Example 2 Determine o. a-0 6V- = 40k 20k - = 12V-2 a 0 0 a a = 3-12V a a = = b 2V 0 = 6V- 12V=-6V 16

17 Example 5.2 a) Deign an inerting amplifier with a oltage gain of -12. Ue 15V power upplie. b) What range of input oltage allow the op amp to operate in it linear region? 15V -15V f a) Since o =-, chooe = 1k = 12k f ( )( ) b) 15V ³ V 12 = 1.25V 17

18 Noninerting Amplifier The oltage gain A Since i = 0 = = p p g n Ue oltage diider to get : = = n o g f + f + æ ö f = = 1+ ç çè ø i o g g o g = 1+ f n An noninerting amplifier proide a poitie gain which i greater or equal unity. The erie reitance for g i often omitted. Why i it ueful in practice? 18

19 Voltage Follower o f æ ö f = 1 ç + çè ø i = 0, o = i The oltage follower. Such a circuit ha a ery high input impedance and i therefore ueful a an intermediate-tage (or buffer) amplifier to iolate one circuit from another. The oltage follower minimize interaction between the two tage and eliminate inter tage loading. A oltage follower ued to iolate two cacaded tage of a circuit. 19

20 Example 3 For the op amp circuit, calculate the output oltage o. Method 1: Uing uperpoition = + 0 0a 0b 0a 0b 10k =- 6V =- 15V 4k æ 10k ö = 1+ ç 4V= 14V çè 4k ø = + =- 15V+14 V =-1V 0 0a 0b Method 2: Applying KCL 6V-a a-0 = 4k 10k 6V-4V 4V-0 = 4 10 =-1V a before 0 20

21 Example 4 Calculate o. By oltage diiion: 1 1 = 8k 3V 2V 8k + 4k = Noninerting amplifier: æ 5k ö = 1+ ç 2k çè ø 0 1 = 3.5 2V = 7 V 21

22 Summing Amplifier i a A ummer can hae more than three input. KCL: = = 0 a b c n a n b n c n n o n a b c f p i + i + i = i = æ ö ç ø f f f o =- a+ b+ c çèa b c An umming amplifier combine eeral input and produce an output that i the weighted um of the input. 22

23 Example 5 Find o and i o. 0 æ ö =- ç 1.5V+ 2 V 1.2 V =-3.8V çè ø i V -3.8V = + = + =-1.425mA 8k 4k 8k 4k 23

24 Digital-to-Analog Conerter Four-bit DAC: (a) block diagram (b) binary weighted ladder type 24

25 Example 6 In the op amp circuit, let f = 10 kω, 1 = 10 kω, 2 = 20 kω, 3 = 40 kω, and 4 = 80 kω. Obtain the analog output for binary input [0000], [0001], [0010],..., [1111]. 25

26 Difference Amplifier - - = a n n o a - = - b a b n a n a ( + ) = + a b n b a a b 0 0 = n b a a + + a b 0 p = d d b n = p ( a + b ) b = - + ( + ) c d o b a a c d a A difference amplifier amplifie the difference between two input but reject any ignal common to the two input. Note: If If = = b- a c 0 b b d a ( ) = and = = - a b c d 0 b a a 26

27 Example 7 Deign an op amp circuit with input 1 and 2 uch that o = ( + ) o = ( 3+ 4) 1 (1 + ) o = ( ) 1 Deign 1 27

28 Example 7 cont d Deign a different op amp circuit with input 1 and 2 uch that o = Deign 2 28

29 Example 8 Deign a difference amplifier with gain of 4. Figure 5.24 (1 + ) o = ( ) 1 if = = - ( ) Since the gain i 4, 2 1 = 4 = But 1 3 = =

30 Intrumentation Amplifier Show that: An intrumentation amplifier i an amplifier of low-leel ignal ued in proce control or meaurement application and i commercially aailable in ingle-package unit. 30

31 Example 9 Obtain i o in the intrumentation amplifier circuit of Fig. A. ecall: Figure A compare 0 i 0 40k =- ( ) V = 0.02 V 20k = V 2μA 10k = 10k = 31

32 Summary 32

33 Summary 33

34 Op Amp Circuit Analyi With PSPICE PSpice doe not hae a model for an ideal op amp. PSpice ha four nonideal, commercially aailable op amp in it eal.lb library. Note that each of them require dc upplie, without which the op amp will not work. Nonideal op amp model aailable in PSpice. 34

35 Ue PSpice to ole for o /. Example 10 o V = = V o 20k Uing the ideal op-amp model: =- = k 35

36 Controlled Source Voltage Controlled Voltage Source (VCVS) Non-inerting amplifier: o i 2 = 1+ independent of 1 L 36

37 Controlled Source Voltage Controlled Current Source (VCCS) edraw the circuit: o = + = - = + = 2 2 o 2 i-o 2 -o KCL: ii + i2 - io = io = 0 i o o o i io = = 1 io = i independent of L 37

38 Controlled Source Current Controlled Voltage Source (CCVS) æ ö = ç 1 +, = i ince i = 0 o ç çè 2 i 1 ø i 3 i 0 æ ö 2 3 ç i çè 1 ø = 1+ i independent of + L 38

39 Controlled Source Current Controlled Current Source (CCCS) edraw the circuit: KVL: i - - i = 0, = 0 11 D 2 2 D i = i i = i i = i 1 2 1, 1 i, 2 2 i i 1 o = i independent of L 2 o 39