Tài liệu Circuits & Electronics P11 pdf

... 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

... 6.002 Fall 2000 Lecture 1 12 6.002 CIRCUITS AND ELECTRONICS Capacitors and First-Order Systems 6.002 Fall 2000 Lecture 10 12 2 Homogeneous ... 55 RC t e − 5 RC = τ Remember demo B Examples 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 & 10 Demo 5V Expected Observed Expect this, right? But observe this! Delay! Motivation 6.002

Ngày tải lên: 12/12/2013, 22:15

... 6.002 Fall 2000 Lecture 1 7 6.002 CIRCUITS AND ELECTRONICS Incremental Analysis 6.002 Fall 2000 Lecture 10 7 () D Vv D D DD v vd vfd Vfi DD ∆⋅+≈ = )( equating

Ngày tải lên: 12/12/2013, 22:15

... 6.002 Fall 2000 Lecture 1 6 6.002 CIRCUITS AND ELECTRONICS Nonlinear Analysis 6.002 Fall 2000 Lecture 10 6 Problem: The LED is nonlinear ... Superposition m5 X Thévenin, Norton any circuit linear circuits Review 6.002 Fall 2000 Lecture 3 6  Discretize value t Digital abstraction X Subcircuits for given “switch” setting are linear! So, ... Introduction to incremental analysis 6.002...

Ngày tải lên: 12/12/2013, 22:15

... 1 22 6.002 CIRCUITS AND ELECTRONICS Energy and Power 6.002 Fall 2000 Lecture 10 22 Putting the two together: Energy dissipated in each cycle 2 S 2 S CV 2 1 CV 2 1 += 21 EEE += C gdischargin & ... 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...

Ngày tải lên: 22/12/2013, 19:17

... 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 ... converters Filters Clock generators Amplifiers Adders Integrators & Differentiators Reading: Chapter 15.5 & 15.6 of A & L. + – Review  ∞ input resistanc...

Ngày tải lên: 22/12/2013, 19:17

... 6.002 Fall 2000 Lecture 1 19 6.002 CIRCUITS AND ELECTRONICS The Operational Amplifier Abstraction 6.002 Fall 2000 Lecture 10 19 e.g. v IN = ... for more complex circuits (of course, need to be careful about input and output).  Today Introduce a more powerful amplifier abstraction and use it to build more complex circuits. Reading: ... powerful amplifier abstraction and use it t...

Ngày tải lên: 22/12/2013, 19:17

... 6.002 CIRCUITS AND ELECTRONICS Sinusoidal Steady State 6.002 Fall 2000 Lecture 16 1 2 Fourth try to find ... + – 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 Example:

Ngày tải lên: 12/12/2013, 22:15

... 6.002 Fall 2000 Lecture 1 15 6.002 CIRCUITS AND ELECTRONICS Second-Order Systems 6.002 Fall 2000 Lecture 10 15 0 2 2 =+ H H v d t vd LC Solution ... back and forth between the Capacitor and the inductor 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 ... Method 1 2 3 Write DE for circuit by a...

Ngày tải lên: 12/12/2013, 22:15

... 6.002 Fall 2000 Lecture 1 14 6.002 CIRCUITS AND ELECTRONICS State and Memory 6.002 Fall 2000 Lecture 10 14 A Stored value leaks away store pulse width >> R ON C Building a memory element ... 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, but still not perfect. Demo Building a memory element … 6.002 Fall 20...

Ngày tải lên: 12/12/2013, 22:15