Tài liệu Circuits & Electronics P15 docx

... 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 13 6.002 CIRCUITS AND ELECTRONICS Digital Circuit 6.002 Fall 2000 Lecture 10 13 Falling Delay t f Equivalent circuit

Ngày tải lên: 12/12/2013, 22:15

... feedback + – 2 R o v 1 R i v 21 1 RR Rv v o + = + 21 1 RR RV v S + = + 21 1 RR RV v S + − = − So Vv += 15 So Vv −= 15− 15 e. g. 21 = = S V RR 5.7v v)vv( i > >= − +− 5.7−< < − + − v vv i v 6.002 Fall 2000 Lecture 11 21 Why is hysteresis useful? e.g., ... disturbance to v o (noise). Now, let’s build some useful circuits with positive feedback. + > − γγif stableeKv positiveisT...

Ngày tải lên: 22/12/2013, 19:17

... 6.002 Fall 2000 Lecture 1 17 6.002 CIRCUITS AND ELECTRONICS The Impedance Model 6.002 Fall 2000 Lecture 10 17 The Big Picture… tV i ω cos [] pp VtV ∠+ ω cos set up DE usual circuit model nightmare trig. 6.002

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 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

... 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

... 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 CIRCUITS AND ELECTRONICS Amplifiers Small Signal Model 6.002 Fall 2000 Lecture 10 1 Graphically ... ) 2 V O = V S − R L K ( V I −V T ) 2 2 2  substituting v I = V I + v i v i << V I v O = V S − R L K ( [ V I + v i ] − v T ) 2 2 = V S − R L K ( [ V I −V T ... corresponding v I −V T interesting v I −V T region for v O v I −V T S...

Ngày tải lên: 12/12/2013, 22:15

... MODEL DS v DS i Demo TGS Vv < TGS Vv < 2GS v 3GS v + – DS v + – DS i GS v TGSDS Vvv −= Cutoff region 6.002 Fall 2000 Lecture 7 9 Graphically DS v DS i TGS Vv ≥ TGS Vv < TGS Vv ≥ 1GS v Saturation region T r i o d e ... resistor model would have you believe. D S G D S TGS Vv < G TGS Vv ≥ ? 6.002 Fall 2000 Lecture 6 9 Graphically DS v DS i TGS Vv ≥ TGS Vv < TGS Vv ≥ 1GS v Sat...

Ngày tải lên: 12/12/2013, 22:15