Actie Circuits: Life gets interesting Actie cct elements operational amplifiers (OP AMPS) and transistors Deices which can inject power into the cct External power supply normally comes from connection to the oltage supply rails Capable of linear operation amplifiers and nonlinear operation typically switches Triodes, pentodes, transistors 67
Actie Cct Elements Amplifiers linear & actie Signal processors Stymied until 97 and Harold Black Negatie Feedback Amplifier Linearity Control rescues communications Telephone relay stations manageable against manufacturing ariability Output signal is proportional to the input signal Note distinction between signals and systems which transform them Yes! Just like your stereo amplifier Idea controlled current and oltage sources 68
A Brief Aside Transistors Bipolar Junction Transistors Semiconductors doped silicon ndoping: mobile electrons Si doped with Sb, P or As pdoping: mobile holes Si doped with B, Ga, In Two types npn and pnp Heaily doped Collector and Emitter Lightly doped Base and ery thin Collector and Emitter thick and dopey Need to bias the two junctions properly Then the base current modulates a strong C E current E E p n p B n p n B C C B B C E C E Amplification i C βi B 69
Transistors V CC C Common Emitter Amplifier Stage C B C C B Biasing resistors and Keep transistor junctions biased in amplifying range Blocking capacitors C B and C B Keep dc currents out in B E E C E L out Feedback capacitor C E Grounds emitter at high frequencies 70
Linear Dependent Sources Actie deice models in linear mode Transistor takes an input oltage i and produces an output current i 0 g i where g is the gain This is a linear oltagecontrolled current source VCCS i ri i βi CCVS r transresistance CCCS β current gain µ VCVS µ oltage gain g VCCS g transconductance 7
Linear dependent source (contd) Linear dependent sources are parts of actie cct models they are not separate components But they allow us to extend our cct analysis techniques to really useful applications This will become more critical as we get into dynamic ccts Dependent elements change properties according to the alues of other cct ariables i i S ri ri S 0 Source on i S 0 i ri 0 0 Source off 7
Cct Analysis with Dependent Sources Golden rule do not lose track of control ariables Find i O, O and P O for the 500Ω load 5Ω i x A i S 48i x 500Ω O i y 300Ω io 50Ω Current diider on LHS i x i 3 S Current diider on HS i 3 O 8 ("48)i x "8i x "i S Ohm s law Power O i O 500! 6000i S p i 7,000i O O O S 73
Analysis with dependent sources 5Ω i x A i S 48i x 500Ω O i y 300Ω i O 50Ω Power proided by ICS 50 p (50 5) S i S i 3 S Power deliered to load 7000i S Power gain G p O ps 7000i S 50 3 i S 430 Where did the energy come from? 74
A Nodal Analysis with Dependent Source S _ S B _ C B P i B D βi B E O KCL at node C KCL at node D CCCS element description G ( C! S ) G ( C! S ) G B C G P ( C! D ) 0 G P ( D! C ) G E D!" i B 0 i B G P ( C! D ) Substitute and sole ( G G GB GP ) C! GPD G S GS! (" ) G P C [( " ) G G ] 0 P E D 75
_ S T&, 5th ed, Example 43 p 48 x 3 µ x i O 4 O Find O in terms of S What happens as µ? S 3 A B Node A: i O ( G O G G3 ) A! G3 B G S x 4 µ x Node B: B! µ x! µ A Solution: & ' µ G # O B ' µ A $! G G G S % ( µ ) 3 " For large gains µ: (µ)g 3 >>G G (! µ G % O " 3 S "! ( µ ) G S 3 &' This is a model of an inerting opamp #$ 76
Mesh Current Analysis with Dependent Sources Dual of Nodal Analysis with dependent sources Treat the dependent sources as independent and sort out during the solution x S _ 3 g x 4 O i O in ( 3 g3 ) i A! 3i B S! ( 3 g3 ) i A ( 3 4 ) i B 0 i O x 3 S _ i B 4 O i A g 3 x in 77
B i B B T&, 5th ed, Example 45 BJTransistor C C _ Needs a supermesh Current source in two loops without in parallel Supermesh entire outer loop i i Supermesh equation βi B V i V 0 V γ i E E i C E V CC i E! " B CC Current source constraint i i! " i B Solution i B! i B V CC! V # (" ) E 78
T&, 5th ed, Example 46 Field Effect Transistor x y g x r ds r ds g y S O _ 3 4 _ S Since cct is linear O Sole ia superposition K S S First S and S 0 then S 0 and S This gies K and K K 79
Operational Amplifiers OpAmps Basic building block of linear analog circuits Package of transistors, capacitors, resistors, diodes in a chip Fie terminals Positie power supply V CC Negatie power supply V CC Noninerting input p Inerting input n Linear region of operation A(! n ) O Ideal behaior 5 p 0 < A <0 8 n p V CC 3 4 Slope A O 8 7 6 5 V CC V O V CC p n Saturation at V CC /V CC limits range V CC 80
eal OpAmp (u74) 8
Equialent linear circuit 0 0 0 " V Dependent source model 6 < < 0! Ω 5 < < < 00! A < 0 8 Need to stay in linear range CC V " A CC Ideal conditions p n i O!! p O p i " n! V n 0 CC V! A CC Ideal OpAmp 0Ω p n i p i n Slope A V CC O O A( p n ) V CC p n i O 8 O
Noninerting OpAmp Feedback What happens now? Voltage diider feedback n O Operating condition p S Linear noninerting amplifier Gain K O S S _ p n O 83
T&, 5th ed, Example 43 Analyze this i 0 p K p S S OpAmp has zero output resistance _ S p n 3 4 O L L does not affect O K AMP O p K Total 3 4 K 4 S K AMP O S 3 4 4 84
Voltage Follower Buffer Feedback path n O i p i n p n O i O Infinite input resistance _ S L i 0, p p S Ideal OpAmp p n Loop gain is O S Power is supplied from the Vcc/Vcc rails i O O L 85
OpAmp Ccts inerting amplifier Input and feedback applied at same terminal of OpAmp is the feedback resistor N! i So how does it work? N S KCL at node A! N O in 0 0, N p O S 0 S _ S _ i p n i N A N p i Inerting amp O O O K S hence the name Noninerting amp 86
Inerting Amplifier (contd) Current flows in the inerting amp i S, in S _ i A i N N p i i L O L i! O S! i O i L "!! L L S 87
OpAmp Analysis T&, 5th ed, Example 44 Compute the inputoutput relationship of this cct Conert the cct left of the node A to its Théenin equialent T OC S T in 3 Note that this is not the inerting amp gain times the oltage diider gain There is interaction between the two parts of the cct ( 3 ) This is a feature of the inerting amplifier configuration S 3 3 O _ T! 4 T T ' B A 4 _ T 3 A 4 4 (! ) %& 3 3 "# %&! 4 S 3 3 $ ' $ "# O L S 88 O L
Summing Amplifier Adder So what happens? Node A is effectiely grounded n p 0 _ i _ i i N A N p F if O Also i N 0 because of in i i So if O F 0 0 This is an inerting summing amplifier O K F K F Eer wondered about audio mixers? How do they work? 89
Mixing desk Linear ccts _ i i m _ O _ i m m O A N p F if O Virtual ground at n Currents add Summing junction O & F # $ '! % " K K & F # $ '! % " L K Permits adding signals to create a composite Stringsbrasswoodwindpercussion Guitarsbassdrumsocalkeyboards m K m & $ ' % F m #! " m 90
T&, 5th ed, Design Example 45 Design an inerting summer to realize ( ) O 5 3 9
T&, 5th ed, Design Example 45 Design an inerting summer to realize Inerting summer with 3KΩ 65KΩ 5KΩ F 5, O F 3 KΩ 4.3KΩ ( ) O 5 3 56KΩ O Nominal alues Standard alues If 400mV and V CC ±5V what is max of for linear op n? Need to keep O >5V " 5 < "(5 5 > 5 3 3 5 " 5! 0.4 < 3 ) V 9
OpAmp Circuits Differential Amplifier i i _ 3 i n n i p O p _ 4 93
OpAmp Circuits Differential Amplifier _ i 3 _ i n n i p p 4 i O Use superposition to analyze 0: inerting amplifier O! 0: noninerting amplifier plus oltage diider O O O K K 3 4 4 & #& # O! % " " 4 $ 3! $ 4 % K inerting gain K noninerting gain 94
T&, 5th ed, Exercise 43 What is O? This is a differential amp is 0V, is 0V KΩ KΩ500Ω 3 4 KΩ 0V _ KΩ KΩ KΩ KΩ O O K K 0 3 0 5V 3 4 4 KΩ 95
Lego Circuits O K O K Noninerting amplifier O K O K Inerting amplifier 96
Lego Circuits (contd) F K O K O Inerting summer K K F F O 3 4 K K O K K 3 4 4 Differential amplifier 97
T&, 5th ed, Example 46: OpAmp Lego 9.7V 0KΩ F 0KΩ 3.3KΩ 0KΩ 0KΩ 9KΩ 5KΩ >> >> C V CC ±5V So what does this circuit do? 98
Example 46: OpAmp Lego 9.7V 0KΩ F 0KΩ 3.3KΩ 0KΩ 0KΩ 9KΩ 5KΩ >> >> C V CC ±5V So what does this circuit do? F 9.7V 3.V 0.33 3. F!5 9 C It conerts tens of ºF to tens of ºC Max current drawn by each stage is.5ma 99
OpAmp Nodal Analysis OpAmp Cct Analysis Use dependent oltage source model Identify node oltages Formulate input node equations Sole using ideal characteristic p n est of circuit est of circuit p i p i O n i n A( p n ) O est of circuit 00
OpAmp Analysis T&, 5th ed, Example 48 A _ C D < < O B E _ 3 F 4 0
OpAmp Analysis T&, 5th ed, Example 48 A _ B _ C D E 3 F 4 < < Seemingly six nonreference nodes: AE O O Nodes A, B: connect to reference oltages and Node C, E: connected to OpAmp outputs (forget for the moment) Node D: ( G G ) D G C GE Node F: ( G 3 G4) F G3 E 0 OpAmp constraints A D B, F C G C G G G G 3 E E G ( G ( G 3 G G 4 G G ) ) 3 4 G G G3 0 0
OpAmp Analysis T&, 5th ed, Exercise 44 C 4 A B 3 S _ O D 03
OpAmp Analysis T&, 5th ed, Exercise 44 C 4 Node A: A S Node B: (G G ) B G A G C 0 Node C: (G G 3 G 4 ) C G B G 4 D 0 Constraints B p n 0 Sole G C! G O D S O S A _ ( ) G B ( ) G G 3 3 4 G 4 3 G G 4 3 4 S S 3 D O 04
Comparators A Nonlinear OpAmp Circuit We hae used the ideal OpAmp conditions for the analysis of OpAmps in the linear regime, i i 0 if A "! V n p What about if we operate with p n? That is, we operate outside the linear regime. We saturate!! n O O Without feedback, OpAmp acts as a comparator There is one of these in eery FM radio! p V V CC CC if if p p p > < n n n CC 05
Analogtodigital conerter comparators 3 8V O3 O S _ V CC 5V V CC 0V O 06
Analogtodigital conerter comparators 8V 3 O3 O Current laws still work i p i n 0 Parallel comparison Flash conerter 3bit output Not really how it is done Voltage diider switched S _ V CC 5V V CC 0V O Input O3 > S 0 0 0 3> S > 5> S >3 S >5 O O 5 0 0 5 5 0 5 5 5 07
OpAmp Circuit Design the whole point Gien an inputoutput relationship design a cct to implement it Build a cct to implement O 5 0 0 3 08
OpAmp Circuit Design the whole point Gien an inputoutput relationship design a cct to implement it Build a cct to implement O 5 0 0 3 Inerting summer followed by an inerter 0KΩ 0KΩ 5KΩ 00KΩ 00KΩ 00KΩ > O 3 _ Summer Inerter 09
3 _ 0KΩ 0KΩ 5KΩ.94KΩ T&, 5th ed, Example 4 O 00KΩ How about this one? 0
3 O m _ O K 0KΩ 0KΩ 5KΩ.94KΩ m 00KΩ Example 4 O O (K) How about this one? Noninerting amp p O 00 0 3.94 3 O K 0 p p 35 3 p.94 0 KCL at pnode with i p 0 p 3 p p 0 4 0 4 0.5 0 4 3.5 p 0.5 3 Noninerting summer Fewer elements than inerting summer eq eq eq L m eq L 3 m m 0
Digitaltoanalog conerter MSB /8 /4 3 4 LSB / F O 3 8 4! F O 4 Parallel Digital Inputs MSB LSB DAC O Single Analog Voltage Conersion of digital data to analog oltage alue Bit inputs 0 or 5V Analog output aries between min and max in 6 steps
Signal Conditioning Your most likely brush with OpAmps in practice Signal typically a oltage representing a physical ariable Temperature, strain, speed, pressure Digital analysis done on a computer after Antialiasing filtering data interpretation Adding/subtracting an offset zeroing Normally zero of ADC is 0V Scaling for full scale ariation quantization Normally full scale of ADC is 5V Analogtodigital conersion ADC Maybe after a few more tricks like track and hold Offset correction: use a summing OpAmp Scaling: use an OpAmp amplifier Antialiasing filter: use a dynamic OpAmp cct 3
Théenin and Norton for dependent sources Cannot turn off the ICSs and IVSs to do the analysis This would turn off the DCSs and DVSs Connect an independent CS or VS to the terminal and compute the resulting oltage or current and its dependence on the source T _ T S i S Compute S in response to i S : i S T S T 4
Where hae we been? Where to now? Nodal and mesh analysis Théenin and Norton equialence Dependent sources and actie cct models OpAmps and resistie linear actie cct design Where to now? Laplace Transforms and their use for ODEs and ccts (Ch.9) Capacitors, inductors and dynamic OpAmp ccts (Ch.6) sdomain cct design and analysis (Ch.0) Frequency response (Ch.) and filter design (Ch.4) We will depart from the book more during this phase 5