Tài liệu Circuits & Electronics P8 pptx

6.002 – Fall 2002: Lecture 8 1 6.002 CIRCUITS AND ELECTRONICS Dependent Sources and Amplifiers 6.002 – Fall 2002: Lecture 8 2  Nonlinear circuits — can use the node method  Small signal trick resulted in linear response Today  Dependent sources Reading: Chapter 7.1, 7.2 Review  Amplifiers 6.002 – Fall 2002: Lecture 8 3 Dependent sources + – v R i R v i = Resistor 2-terminal 1-port devices + – v i Ii = I Independent Current source Seen previously control port output port I i I v O i O v + – + – New type of device: Dependent source 2-port device E.g., Voltage Controlled Current Source Current at output port is a function of voltage at the input port )v(f I 6.002 – Fall 2002: Lecture 8 4 Dependent Sources: Examples independent current source Example 1: Find V 0 II = + – V R RIV 0 = 6.002 – Fall 2002: Lecture 8 5 voltage controled current source Example 2: Find V () V K VfI == + – V R I i I v O i O v + – + – + – V R () I I v K vf = Dependent Sources: Examples 6.002 – Fall 2002: Lecture 8 6 voltage controled current source R V K IRV == KR V = 2 KRV = 33 1010 ⋅= − Volt1= or or Example 2: Find V () V K VfI == + – V R e.g. K = 10 -3 Amp·Volt R = 1kΩ Dependent Sources: Examples 6.002 – Fall 2002: Lecture 8 7 Another dependent source example I v + – ( ) IND vfi = L R + – S V e.g. ( ) IND vfi = () 2 IN 1v 2 K −= for v IN ≥ 1 IN i IN v D i O v + – + – otherwise 0i D = Find v O as a function of v I . 6.002 – Fall 2002: Lecture 8 8 Another dependent source example I v + – ( ) IND vfi = L R S V e.g. ( ) IND vfi = () 2 IN 1v 2 K −= for v IN ≥ 1 IN i IN v D i O v + – + – otherwise 0i D = Find v O as a function of v I . 6.002 – Fall 2002: Lecture 8 9 Another dependent source example Find v O as a function of v I . I v + – I v S V O v L R () 2 IND 1v 2 K i −= for v IN ≥ 1 otherwise 0i D = 6.002 – Fall 2002: Lecture 8 10 Another dependent source example 0 =++− OLDS vRiV KVL LDSO RiVv −= () LISO Rv K Vv 2 1 2 −−= for v I ≥ 1 SO Vv = for v I < 1 I v + – I v S V O v L R () 2 IND 1v 2 K i −= for v IN ≥ 1 otherwise 0i D = Hold that thought . 2002: Lecture 8 1 6.002 CIRCUITS AND ELECTRONICS Dependent Sources and Amplifiers 6.002 – Fall 2002: Lecture 8 2  Nonlinear circuits — can use the node. vRiV KVL LDSO RiVv −= () LISO Rv K Vv 2 1 2 −−= for v I ≥ 1 SO Vv = for v I < 1 I v + – I v S V O v L R () 2 IND 1v 2 K i −= for v IN ≥ 1 otherwise 0i