Tài liệu Circuits & Electronics P23 pptx

... switching capacitor. independent of f. MOSFET ON half the time. STATIC P DYNAMIC P constant time "RC" 2 T RR ONL >> >> Square wave input f T 1 = Demo Review In standby mode, half the gates ... MOSFET (NFET) D S G on when v GS ≥ V TN off when v GS < V TN e.g. V TN = 1V • P-channel MOSFET (PFET) on when v GS ≤ V TP off when v GS > V TP e.g. V TP = -1V S D G ON...

Ngày tải lên: 22/12/2013, 19:17

... 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

... 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

... Fall 2002: Lecture 8 22 One nit … D i S V O v L R VCCS 1 I v O v For v O >0 , VCCS consumes power: v O i D For v O <0 , VCCS must supply power! () 2 1 2 −−= ILSO vR K Vv () 2 1 2 −= ID v K i for ... −−= IO vv () 2 33 110510 2 2 10 −⋅⋅⋅−= − I v 1 I v S V O v 6.002 – Fall 2002: Lecture 8 1 6.002 CIRCUITS AND ELECTRONICS Dependent Sources and Amplifiers 6.002 – Fall 2002: Lecture...

Ngày tải lên: 12/12/2013, 22:15

... 6.002 Fall 2000 Lecture 1 25 6.002 CIRCUITS AND ELECTRONICS Violating the Abstraction Barrier 6.002 Fall 2000 Lecture 5 25 R i V instantaneous ... wires 3. avoid big current swings V S V S 6.002 Fall 2000 Lecture 6 25 Question: So why did our circuits work? More in 6.014 t V5 V5.2 3. Termination P a r a l l e l t e r m i n a t i o n D E M O a d d

Ngày tải lên: 22/12/2013, 19:17

... Conversion Circuits (PCC) Power efficiency of converter important, so use lots of devices: MOSFET switches, clock circuits, inductors, capacitors, op amps, diodes Reading: Chapter 16 and 4.4 of A & ... segment”: R + – I v + – + – V6.0 6.0v I < R O v + – + – I v () R/6.0vi ID −= 6.0vv IO −= “Short segment”: R + – + – + – V6.0 6.0v I ≥ I v Example V6.0 + – Equivalent circuit 6.002 Fa...

Ngày tải lên: 22/12/2013, 19:17

... 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

... Lecture 15 13 Why? Consider … 1Case 1 R 0 R pin1 ok Demo 6.002 Fall 2000 Lecture 1 13 6.002 CIRCUITS AND ELECTRONICS Digital Circuit 6.002 Fall 2000 Lecture 8 13 GSL CR t SSOH eVVv − −= Or Find

Ngày tải lên: 12/12/2013, 22:15

... 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