Tài liệu Circuits & Electronics P21 docx

6.002 Fall 2000 Lecture 1 21 6.002 CIRCUITS AND ELECTRONICS Op Amps Positive Feedback 6.002 Fall 2000 Lecture 2 21 Consider this circuit — negative feedback + – + – 1 R 1 R v IN IN v + – INOUT v R R v 1 2 −= 2 R What’s the difference? Consider what happens when there is a pertubation… Positive feedback drives op amp into saturation: SOUT Vv ±→ and this — positive feedback + – + – 1 R IN v + – 2 R INOUT v R R v 1 2 −= “” s e e a n a l y s is o n n e xt p a g e Negative vs Positive Feedback 6.002 Fall 2000 Lecture 3 21 + – + – 1 R IN v 2 R OUT v ( ) −+ −= vvAv OUT       +⋅ + − = IN1 21 INOUT vR RR vv A IN 21 IN1 OUT 21 1 Av RR vAR v RR AR + + − + = + = Av IN 1 2 IN 21 1 21 1 OUT v R R Av RR AR RR R 1 v −=⋅             + − + − =       + −=       + − 21 1 IN 21 1 OUT RR R 1Av RR AR 1v + – 1 R IN v 2 R OUT v + v − v ( ) −+ − vvA + – Static Analysis of Positive Feedback Ckt 6.002 Fall 2000 Lecture 4 21 Representing dynamics of op amp… + v − v o v * Av + – + – * v )( −+ − vv R C + – 6.002 Fall 2000 Lecture 5 21 Representing dynamics of op amp… Consider this circuit and let’s analyze its dynamics to build insight. + – 1 R 2 R o v 3 R 4 R Let’s develop equation representing time behavior of v o . Circuit model 1 R 2 R 3 R 4 R + – * v )( −+ − vv R C + – + – + – o v + v − v A v o 6.002 Fall 2000 Lecture 6 21 A v vAvv o o == ** or )γγ(A RC Twhere0 T v dt dv oo + − − ==+ or o o v RR Rv v + = + = + γ 21 1 o o v RR Rv v − = + = − γ 43 3 0)γγ( = + − − + o o v RC A dt dv 1 time − 0)γγ( 1 =       + − − ++ o o v RC A RCdt dv or neglect _* * vvv d t dv RC −=+ + o v)γγ( − − + = Dynamics of op amp… _ vv A v d t dv A RC oo −=+ + 0)0(v o = 6.002 Fall 2000 Lecture 7 21 Consider a small disturbance to v o (noise). Now, let’s build some useful circuits with positive feedback. + > − γγif stableeKv positiveisT T t o − = − > + γγif unstableeKv negativeis T T t o = − = + γγif neutralKv largeveryis T o = o v t neutral stable K disturbance unstable 6.002 Fall 2000 Lecture 8 21 One use for instability: Build on the basic op amp as a comparator + – + v o v S V+ S V− − v −+ − vv o v S V+ S V− 0 t 0v → − + v o 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−< < − + − v vv i v 6.002 Fall 2000 Lecture 10 21 Now, use positive 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 [...]... t − v+ Assume Lecture vo = VS vC = 0 21 at t = 0 13 Clocks in Digital Systems We built an oscillator using an op amp t can use as a clock Why do we use a clock in a digital system? (See page 735 of A & L) 1 1 0 sender receiver clock a 1,1,0? b When is the signal valid? common timebase when to “look” at a signal (e.g whenever the clock is high) Discretization of time one bit of information associated . 5.7v −= − 15v o = 15v o −= 15 e. g. 21 = = S V RR 5.7v 5.7)vv( i > >= − − 5.7−< < − + − v vv i v 6.002 Fall 2000 Lecture 10 21 Now, use positive. − 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