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ECEN3714 Network Analysis Lecture #27 21 March 2016 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 Read 15.4 Problems 15.8, 9, & 12 Exam #2 Friday 13 – 15 EXCEPT Convolution Chapter ECEN3714 Network Analysis Lecture #28 23 March 2016 Dr. George Scheets www.okstate.edu/elec-eng/scheets/ecen3714 Problems: Old Exam #2 Exam #2 this coming Friday, 25 March 13 – 15 EXCEPT Convolution Chapter HKN Test Review Thursday, 24 March, CLB 101, 5:30 pm Op Amp Characteristics vp(t) Zin vn(t) +Vcc + vout(t) = Av(vp(t)-vn(t)) Av - -Vcc Zin? In Hopamp(f) f3dB? In M ohms XX or XXX MHz Voltage gain Av? On order of 104 - 106 Op Amps: No Feedback +Vcc + vin(t) vout(t) = Av(vp(t)-vn(t)) Av - -Vcc Output likely to hit rails Unless tiny voltage Use: Comparator Compares two voltages Yields binary output Op Amps: Positive Feedback & No Negative Feedback 0v 0v5v + Av vin(t) - vout(t) 0v Output likely to hit rails May get stuck there Use: None Suppose |Vcc| = 15 v Op Amps: Positive Feedback 10 v if feedback & input resistors = 5v + Av vin(t) - 15 v Output likely to hit rails May vout(t) get stuck there Suppose |Vcc| = 15 v Op Amps: Positive Feedback 7.5 v if feedback & input resistors = 0v + Av vin(t) - 15 v Output likely to hit rails May vout(t) get stuck there Suppose |Vcc| = 15 v Op Amps: Positive Feedback ≈ 0 v if feedback & input resistors = -15 v + Av vin(t) - 15 v Output likely to hit rails May vout(t) get stuck there Suppose |Vcc| = 15 v Op Amps: Negative Feedback - vin(t) 0v Av vout(t) + Safe to assume vp(t) = vn(t) Safe to assume no current enters Op Amp If low Z outside paths exist Op Amps: Negative Feedback - vin(t) 0v vout(t) + OpAmp feedback makes these two = Analyze This Op Amps: Negative Feedback 0v 0v5v vin(t) - Av + vout(t) = Av[v+(t) – v-(t)] 0v Stable System LTI so long as don't hit Power Supply Limits & get Clipped Output Suppose |Vcc| = 15 v Op Amps: Negative Feedback (Actual) 0.0004999 v 5v vin(t) - Av + vout(t) = 10,000[v+(t) – v-(t)] -4.999 v Suppose… OpAmp Voltage Gain = 10,000 OpAmp Input Impedance = ∞ External elements are Resistors of R Ω Op Amps: Negative Feedback (Ideal) 0.0+ v 5v vin(t) - Av + vout(t) -5.0 v Suppose… OpAmp Voltage Gain = ∞External elements are Resistors of R Ω V+ = V- Op Amps: Output Load - vin(t) Av + vout(t) Zload Ideally, load does not effect characteristics Practically, load may effect characteristics If Op Amp output can't source or sink enough current OpAmps: External Resistors 1.020a -.1020v 1.020a Note: +&feedback is OK 1Ω |Vcc| = 25v Av = 10,000 Zin = 1 MΩ - 5v 5Ω + -11/49*5v = -1.122v -.1020v 1Ω .1020a 10 Ω .1020a Generally not too small: X or XX Ohms. Current flows too large. OpAmps: External Resistors < 10 μa 10 μa 1 MΩ |Vcc| = 25v Av = 10,000 Zin = 1 MΩ 5 MΩ Generally not too large: X or XX M Ohms. + 1 MΩ 10 MΩ Current flows in/out inputs cannot be ignored. OpAmps? External Resistors 1.020 ma -.1020v 1.020 ma 10K Ω |Vcc| = 25v Av = 10,000 Zin = 1 MΩ - 5v 5K Ω + -11/49*5v = -1.122v -.1020v 1K Ω 102.0 μa 10K Ω 102.0 μa Just Right if in X or XX K Ohm range OpAmps (Analysis Assumptions) Is there negative feedback? Generally, must be "Yes" to be LTI Yes? Then V+ = V|Vcc| = 25v Av = 10,000 vin 5Ω 1Ω 1Ω + vout 10 Ω OpAmps (Analysis Assumptions) If LTI, analyze circuit with OpAmp removed. No current enters OpAmp inputs Current can enter/exit OpAmp outputs |Vcc| = 25v Av = 10,000 vin 5Ω 1Ω 1Ω + vin 1Ω - 5Ω vout 10 Ω OpAmp negative feedback makes V- = V+ 1Ω + vout 10 Ω Analyze this circuit. Ideal OpAmps Default Assumption on Quiz or Test … unless specifically stated otherwise Voltage Gain = ∞ Output can source or sink any current Input impedance = ∞ No current enter or exits inputs OpAmp BW = ∞ V+ = V- if negative feedback Output doesn't hit Power Supply rails 2nd Order Low Pass Filter (Two back-to-back 1st order active filters) |H(ω)| 3dB break point changes. 1 0.707 1st order 2nd order 1 ω