Tài liệu Circuits & Electronics P9 doc

... 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

... 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

... 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

... 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 18 6.002 CIRCUITS AND ELECTRONICS Filters 6.002 Fall 2000 Lecture 10 18 Another example: + – + – L R i V C o V i o V V o ω ω BPF C ... Q 6.002 Fall 2000 Lecture 2 18 Review + – C v I v + – R C Reading: Section 14.5, 14.6, 15.3 from A & L. + – c V i V + – R Z C Z i RC C c V ZZ Z V ⋅ + = RCj1 1 R Cj 1 Cj 1 V V i c ω ω ω + = + = 6.002

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

... mechanical condition. 184 2. Adjust the fuel system in accordance with the manufacturer's instructions. 3. Operate the engine within the temperature limits specified in the instructions. ... vertical planes and may in turn be the cause of vibration of surrounding structures, such as the ship's hull in marine installations. This type of vibration does not depend on the...

Ngày tải lên: 12/12/2013, 06:15

... 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