Tài liệu Circuits & Electronics P13 docx

... 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

... 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...

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... 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 9 21 Now, use positive feedback + – 2 R o v 1 R i v 21 1 RR Rv v o + = + 5.7v = + 5.7v −= − 15v o = 15v o −= 15 e. g. 21 = = S V RR 5.7v 5.7)vv( i > >= − − 5.7−<...

Ngày tải lên: 22/12/2013, 19:17

... Picture… tV i ω cos [] pp VtV ∠+ ω cos set up DE usual circuit model nightmare trig. 6.002 Fall 2000 Lecture 1 17 6.002 CIRCUITS AND ELECTRONICS The Impedance Model 6.002 Fall 2000 Lecture 13 17 Back to LCRCj1 RCj V V 2 i r ωω ω −+ = (

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...

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... 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

... 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 &...

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... 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

... some point (V I , V O ) … looks quite linear ! 6.002 Fall 2000 Lecture 10 6 6.002 CIRCUITS AND ELECTRONICS Amplifiers Small Signal Model 6.002 Fall 2000 Lecture 10 1 I 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

Ngày tải lên: 12/12/2013, 22:15

... each model in 6.002? DS v DS i TGS Vv ≥ TGS Vv < TGS Vv ≥ 1GS v Saturation region T r i o d e re g i o n DS v DS i DS v DS i TGS Vv < TGS Vv < 2GS v 3GS v TGSDS Vvv −= 6.002 Fall 2000 ... 6 9 Graphically DS v DS i TGS Vv ≥ TGS Vv < TGS Vv ≥ 1GS v Saturation region T r i o d e re g i o n S MODEL DS v DS i SR MODEL DS v DS i Demo TGS Vv < TGS Vv < 2GS v 3GS v + – DS v +...

Ngày tải lên: 12/12/2013, 22:15