6.00 CIRCUITS AN ELECTRONICS ependent Sources and Amplifiers
Review Nonlinear circuits can use the node method Small signal trick resulted in linear response Today ependent sources Amplifiers Reading: Chapter 7.1, 7.
ependent sources Seen previously Resistor Independent Current source + v i R + v i I i = v R i = I -terminal 1-port devices New type of device: ependent source control port i I + v I f ( v I ) i O + output port -port device E.g., Voltage Controlled Current Source Current at output port is a function of voltage at the input port
ependent Sources: Examples Example 1: Find V independent current source R + V I = I 0 V = I0R
ependent Sources: Examples Example : Find V voltage controled current source + R V ( V ) I = f = V R + V + v I i I ( v ) f = I v I i O +
ependent Sources: Examples Example : Find V voltage controled current source R + V ( V ) I = f = e.g. = 10-3 Amp Volt R = 1kΩ V V V = R V = IR = or or R V = R = 10 = 1Volt 3 10 3
Another dependent source example R L V S + v + I i IN + v IN i + e.g. i = f ( ) v IN ( ) i = f v IN = ( v ) IN 1 for v IN 1 i = 0 otherwise Find as a function of v I.
Another dependent source example V S R L v + I i IN + v IN i + i = f ( ) v IN e.g. i = f ( ) v IN = ( v ) IN 1 for v IN 1 i = 0 otherwise Find as a function of v I.
Another dependent source example V S R L vi v + I i ( ) = vin 1 for v IN 1 i = 0 otherwise Find as a function of v I.
Another dependent source example V S R L vi v + I i ( ) = vin 1 for v IN 1 i = 0 otherwise VL V v S O + i = V S R L i + v R O L = 0 vo = VS ( vi 1) RL for v I 1 v = for v I < 1 O V S Hold that thought
Next, Amplifiers
Why amplify? Signal amplification key to both analog and digital processing. Analog: AMP IN OUT Input Port Output Port Besides the obvious advantages of being heard farther away, amplification is key to noise tolerance during communcation
Why amplify? Amplification is key to noise tolerance during communcation No amplification useful signal 1 mv noise 10 mv huh?
Try amplification noise AMP not bad!
Why amplify? igital: Valid region 5V 0V V V IH IL IN igital System OUT 5V 0V V OH V OL 5V IN 5V OUT V IH V OH V IL 0V t V OL 0V t
Why amplify? igital: Static discipline requires amplification! Minimum amplification needed: V V IH IL V OH V OL V V OH IH V V OL IL
An amplifier is a 3-ported device, actually Power port Input port i I i O + v I Amplifier + Output port We often don t show the power port. Also, for convenience we commonly observe the common ground discipline. In other words, all ports often share a common reference point called ground. POWER IN OUT How do we build one?
Remember? V S R L vi v + I i ( ) = vin 1 for v IN 1 i = 0 otherwise VL V v S O + i = V S R L i + v R O L = 0 vo = VS ( vi 1) RL for v I 1 v = for v I < 1 O V S Claim: This is an amplifier
So, where s the amplification? Let s look at the versus v I curve. ma e.g. VS = 10V, =, RL = 5kΩ V v = ( 1) O VS RL vi = ( 1) 10 5 v = I ( 1) 3 3 10 10 5 10 v I V S Δ Δv Δv O I 1 > 1 Δv I amplification v I
Plot versus v I ( ) v O = 10 5 vi 1 0.1 change in v I v I 0.0 10.00 1.0 10.00 1.5 8.75.0 5.00.1 4.00..80.3 1.50.4 ~ 0.00 1V change in Gain! emo Measure.
One nit What happens here? 1 v I Mathematically, v V O = S RL I ( v 1) So is mathematically predicted behavior
One nit v O = VS RL I ( v 1) What happens here? 1 However, from i = ( 1) vi for v I 1 V S v I R L VCCS i For >0, VCCS consumes power: i For <0, VCCS must supply power!
If VCCS is a device that can source power, then the mathematically predicted behavior will be observed i.e. v I v O = VS RL I ( v 1) where goes -ve
If VCCS is a passive device, then it cannot source power, so cannot go -ve. So, something must give! Turns out, our model breaks down. Commonly i = I ( v 1) will no longer be valid when 0. e.g. i saturates (stops increasing) and we observe: 1 v I