Tài liệu Circuits & Electronics P19 pdf
... ) −+ −= vvAv O ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + −= 21 2 RR R vvA OIN IN 21 2 O Av RR AR 1v = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + 21 2 IN O RR AR 1 Av v + + = 6.002 Fall 2000 Lecture 1 19 6.002 CIRCUITS AND ELECTRONICS The Operational Amplifier Abstraction 6.002 Fall 2000 Lecture 7 19 Let us build ... for more complex circuits (of course, need to be careful about input and output). Today Introduce a more powerful ampli...
Ngày tải lên: 22/12/2013, 19:17
... 13 12 t C v V5 V0 VV I 5 = VV O 0 = 5 0 VV I 0 = VV O 5 = 5 0 t C v V5 V0 RC t e − + 55 RC t e − 5 RC = τ Remember demo B Examples 6.002 Fall 2000 Lecture 1 12 6.002 CIRCUITS AND ELECTRONICS Capacitors and First-Order Systems 6.002 Fall 2000 Lecture 8 12 Example… Method ... −= () RC t I0IC eVVVv − −+= 6.002 Fall 2000 Lecture 2 12 5V 0V C A B 5V A B C 5 0 5 0 5 0 Reading: Chapters 9 &...
Ngày tải lên: 12/12/2013, 22:15
... the same KVL, KCL equations BOUTA VVV ++++ """ so, we can cancel them out BOUTA vvv ++++++ """" 1 bouta vvv ++++ """ Leaving 2 Since small signal models ... circuit. Foundations: (Also see section 8.2.1 of A&L) KVL, KCL applied to some circuit C yields: III The Small Signal Circuit View bBoutOUTaA vVvVvV +++++++ """ Replace...
Ngày tải lên: 12/12/2013, 22:15
Tài liệu Circuits & Electronics P7 pdf
... are approximating A with B A B d v d i Graphical interpretation 6.002 Fall 2000 Lecture 1 7 6.002 CIRCUITS AND ELECTRONICS Incremental Analysis 6.002 Fall 2000 Lecture 6 7 Result v d very small D i D v d i D I D V 6.002 ... Taylor’s Expansion to expand f(v D ) near v D =V D : () D Vv D D DD v dv vdf Vfi DD ∆⋅+= = )( "+∆⋅+ = 2 2 2 )( !2 1 D Vv D D v dv vfd DD large DC increment a...
Ngày tải lên: 12/12/2013, 22:15
Tài liệu Circuits & Electronics P6 pdf
... Graphical method X Introduction to incremental analysis 6.002 Fall 2000 Lecture 1 6 6.002 CIRCUITS AND ELECTRONICS Nonlinear Analysis 6.002 Fall 2000 Lecture 8 6 Combine the two constraints 1 4 1 1 1 = = = = b a R V e.g. Ai Vv D D 4.0 5.0 = = D v D i 5.0~ 4.0~ R V V 1 1 a ¼ called ... when v D is negative) 6.002 Fall 2000 Lecture 3 6 Discretize value t Digital abstraction X Subcirc...
Ngày tải lên: 12/12/2013, 22:15
Tài liệu Circuits & Electronics P22 pdf
... T 2 >> R 2 C So, capacitor discharges ~fully in T 2 So, energy dissipated in R 2 during T 2 2 S2 CV 2 1 E = E 1 , E 2 independent of R 2 ! Initially, v C = V S (recall T 1 >> ... Lecture 13 22 We can show (see section 12.2 of A & L) () () 2 ONL 2 L 2 S ONL 2 S RR R fCV RR2 V P + + + = fCV R2 V P 2 S L 2 S += when R L >> R ON What is for gate? P r e m e m b...
Ngày tải lên: 22/12/2013, 19:17
Tài liệu Circuits & Electronics P20 pdf
... resistor i R v I → O O v d t dv RC >> when I O v d t dv RC ≈ dtv RC 1 v t IO ∫ ∞− ≈ or IO O vv d t dv RC =+ R v larger the RC, smaller the v O for good integrator ωRC >> 1 I v + – i + – O v C R v + – R Demo 6.002 ... 6.002 Fall 2000 Lecture 1 20 6.002 CIRCUITS AND ELECTRONICS Operational Amplifier Circuits 6.002 Fall 2000 Lecture 3 20 Consider this circuit: − + ≈ + = v R...
Ngày tải lên: 22/12/2013, 19:17
Tài liệu Circuits & Electronics P16 pptx
... v P is particular response to V i cos ωt . 6.002 Fall 2000 Lecture 16 11 6.002 CIRCUITS AND ELECTRONICS Sinusoidal Steady State 6.002 Fall 2000 Lecture 16 1 Usual approach… 1 ... + – R i C + v I v C – v I ( t ) = V i cos ω t for t ≥ 0 ( V i real) = 0 for t < 0 v C (0) = 0 for t = 0 I v 0 t 6.002 Fall 2000 Lecture 16 4 1 1 1 1 e c t u r...
Ngày tải lên: 12/12/2013, 22:15
Tài liệu Circuits & Electronics P15 docx
... conditions to solve for the remaining constants. Solving 6.002 Fall 2000 Lecture 19 15 RLC Circuits See A&L Section 12.2 add R no R + – C R L + – )( tv )(tv I )( ti )( tv t Damped sinusoids ... [ZSR] I v 0 V 0 t Let’s solve I vv d t vd LC =+ 2 2 For input 6.002 Fall 2000 Lecture 1 15 6.002 CIRCUITS AND ELECTRONICS Second-Order Systems 6.002 Fall 2000 Lecture 11 15 Total soluti...
Ngày tải lên: 12/12/2013, 22:15
Tài liệu Circuits & Electronics P14 pptx
... 1 14 6.002 CIRCUITS AND ELECTRONICS State and Memory 6.002 Fall 2000 Lecture 11 14 Input resistance R IN B Second attempt buffer R IN store buffer d IN d OUT C * 5 ln OH IN V CRT −= LIN RR >> Better, ... + 6 + 5 + 3 + 8 M+ 6.002 Fall 2000 Lecture 10 14 A Stored value leaks away store pulse width >> R ON C Building a memory element … CR t C L ev − ⋅= 5 5 ln OH L V CRT...
Ngày tải lên: 12/12/2013, 22:15