Page -16.002 Circuits and SystemsFinal ReviewJason KimCIRCUITS AND SYSTEMSFINAL REVOLUTIONSPage 06.002 Circuits and Systems: Outline of Topics• Resistor Networksconcepts: Node Method, KCL, KVL, Superposition, Thevenin and Norton equivalent circuit models, Power.•1st Order Circuitsconcepts: Consituitive Laws, RC netwroks, RL networks, Homogenous and Particular solution, Time constant , response to an impulse, power, Low/High/Band/Notch pass filters, sinusodial steady-state, transients, Impedance Model, Bode Plots(Magnitude and Phase).•2nd Order Circuitsconcepts: LC networks, LRC networks, damping coefficient , natural frequency , Underdamped/Critical/Overdamped Systems, Resonance, Q factor, Half-power point, Sinusodial steady-state, transients, Power(real/reactive), Impedance Model, ELI ICE, Bode Plots(Magnitude and Phase).• Digital Abstractionconcepts: Boolean Logic(truth table, formula, gates, and transistor level), primitive laws, DeMorgan’s Law(formula and gate equivalent), MSP(minumum sum of products), MPS(minimum product of sums).• MOSFET Transistorsconcepts: Large Signal Model(S, SR, SCS, SVR), Small Signal Model, Saturation/Linear Region, Ron, Loadline, Operating Point, Input/Output Resistence, Current Gain, Power Gain, Noise Margin.• Op-Amp Circuitsconcepts: Single/Multiple input op-amp circuits, differentiator, integrator, adder, subtractor, op-amp with R,L,C, negative feedback, positive feedback, Bode Plots(Magnitude and Phase), Input/Output resistence, Schmitt Trigger, Hysteresis, Cascaded stages.• Diodesconcepts: i-v characteristics, model, diode w/ R, L, C, and op-amp, Peak detection, Clipper circuits, Incremental Analysis.• Energy and Powerconcepts: Energy and Power relationships, nMOS and CMOS logic inverter power dissipation, static and dynamic power, power reduction techiniques.ταωoPage 1Resistor: time domain: freq. domain: Series: Parallel: • Resistor NetworkN-unknowns & N-equations (most primitive approach)Node Analysis (KCL, KVL)*KCL: Sum of all currents entering and leaving a node is zero.*KVL: Sum of all voltages around a closed loop path is zero. (be careful with polarities)1) LABEL all current directions and voltage polarities. (Remember, currents flow from + to -)2) write KCL & KVL Equation.3) substitute and solve.Simplification (by inspection)• Thevenin Equivalent Circuit Model Norton Equivalent Circuit Model * Three variables: Vth=Voc, Rth=RN, and IN=Isc . They are related by Voc=Isc*Rth .* Vth=Voc: Leave the port open(thus no current flow at the port) and solve for Voc. For a resistive network, this gives you a point on the V-axis of the i-v plot.* IN=Isc: Short the port(thus no voltage across the port) and solve for Isc flowing out of + and into -. For a resistive network, this gives you a point on the I-axis of the i-v plot.* Rth: Set all sources to zero, except dependent sources. Solve the resistive network. When setting sources to zero, V source becomes short and I source becomes open. Rth may also be found by attaching Itest and Vtest and setting Rth = (Vtest)/(Itest). If dependent sources are present, set only the independent sources to zero and attach Itest and Vtest. Use KCL and KVL to find the expression (Vtest)/(Itest)=Rth. * For both Thevenin & Norton E.C.M.’s, they have the same power consumption at the port as the original circuit, but not for its individual components. Power dissipated at Rth does not equal power dissipated at the resistors of the original circuit. Same holds for the power delivered by the sources in the Thevenin model and the original circuit. Thevenin and Norton E.C.M.’s are for terminal i-v characteristics only.• Power = Real Power = IV = I2R = V2/R (energy dissipated as heat)•VIR= ZR=R1R2+ R1R2||R1R2R1R2+------------------=+-R1R2VoVi+-R2I2ViR1IiVoR2R1R2+--------------------Vi=I2R1R1R2+--------------------Ii=Voltage DividerCurrent Divider+-RthVth Voc=VthRNINIsc=INsame i-v characteristics at the ports+-+-NItest+-VtestOriginal Circuitnote the direction of Itestand polarity of Vtest.RthVtestItest-----------=Power〈〉1T--- It()Vt()td0T∫=1. Primitive ElementsPage 2• Superposition"In a linear network with a number of independent sources, the response can be found by summing the response to each independent sources acting alone, with all other independent sources set to zero." 1) Leave one source on and turn off all other sources.(Voltage source "off" = short & Current source "off" = open)2) Find the effect from the "on" source.3) Repeat for each sources.4) Sum the effect from each sources to obtain the total effect.For cases where a linear dependent source is present along with multiple independent sources, DO NOT turn off the dependent source. Leave the dependent source on and carry it in your expressions. Tackle the dependent source term last by solving linear equations. Remember that the variable which the dependent source is depended on is affected by the individual independent sources that you are turning on and off.Capacitor: time domain: freq. domain: High Frequency = Short & Low Frequency = OpenSeries: Parallel: Vc(0-)=Vc(0+) except when the input source is an impulse.Vc(t) is continuous while Ic(t) may be discontinuous."I C E" : Current(I) LEADS Voltage(EMF) by 90o (mnemonic: ELI ICE)Energy Conservation: .Energy Stored: . (no energy dissipation)Note: Two identical capacitors, C1 and C2, in series may act like an open circuit after a long time where the total voltage across the capacitors are split between the two cap’s. For this case, Vc does not discharge to zero but VC1=VC2.Inductor: time domain: freq. domain: High Frequency = Open & Low Frequency = ShortSeries: Parallel: IL(0-)=IL(0+) except when the input source is an impulse.IL(t) is continuous while VL(t) may be discontinuous.Linear Networkwith n sourcesR+_VRVRVR1V2 n∼0=VR2V1V,3 n∼0=… VRnV1 n 1–∼0=++=ICtddV=Z1Cs------1jwC----------==C1C2C1C2+------------------- C1C2+Itd∫QCV==E12---CV2=VLtddI=ZLsjwL==L1L2+L1L2L1L2+------------------VR when only V1 is on.C1C2R+_VC2+_VC1VC1(0-) = VC2(0-)1. Primitive ElementsPage 3"E L I" : Voltage(EMF) LEADS Current(I) by 90o (mnemonic: ELI ICE)Energy Conservation: .Energy Stored: . (no energy dissipation)Note: Two identical inductors, L1 and L2, in parallel may act like a short circuit after a long time and have current flow through this loop indefinitely. For this case, iL is not zero but
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