Review of Circuit Analysis Common Symbols Independent Sources Battery • Constant voltage • DC voltage source Voltage Source + • - Can be constant or time varying • +/- indicates polarity Current Source • Can be constant or time varying • Arrow denotes direction of current flow Dependent Sources • Depend upon the voltage/current measured at some location in the circuit to provide its value • Often a linear proportionality term, but can be non-linear + - Dependent voltage source Dependent current source Reference Voltages Ground • Reference potential • 0V V cc / V dd • Supply voltage • Typically highest potential • Often called a supply “rail” 1
Ohm’s Law V = IR R V 1 V 2 I ( ) − = V V IR 1 2 − V V = 1 2 I R More generally, as applied to complex impedances, Z Z V 1 V 2 I ( ) − = V V IZ 1 2 − V V = 1 2 I Z Resistors in Series R 1 R 2 R ser = R 1 + R 2 = 100 Ω 500 Ω R ser = = Ex. 2
Resistors in Parallel R 1 R par = R 1 //R 2 = R 2 1 1 1 = + R par R R 1 2 R R = = 1 2 R par R R + 1 2 R R 1 2 Ex. 300 Ω R par = = 150 Ω The parallel combination will always be less than the smaller of the two resistors Useful cases • Parallel combination of 2 identical resistors • Case where one resistor is much larger than the other → Equivalent resistance looks mostly like the smaller of the two resistors 3
Kirchoff’s Voltage Law (KVL) • “Loop Equations” • Sum of all voltage drops around any loop equals zero ∑ = 0 V x ∑ ∑ = V V up down Ex. Find the current flowing through R 2 and the voltage across R 2 . R 1 + - V in R 2 4
Kirchoff’s Current Law (KCL) • “Node Equations” • Sum of all currents entering [exiting] a node equals zero ∑ = 0 I x ∑ ∑ = I I in out Ex. Find the current flowing through R 2 . I in R 1 R 2 5
Superposition • Only valid for linear circuits • For the case where there are multiple voltage/current sources • The voltage [current] at any node [branch] of a circuit can be found by adding the sum of the contributions of each source, while all others are turned off • Method o Keep only one source on. Turn all others off. Solve for the desired voltage/current. OFF + + V x - - 0V = Short Circuit OFF I x 0A = Open Circuit o Turn first source off. Turn next source on. Solve for the desired voltage/current. o … o Repeat for all N sources o Desired voltage/current is the sum of each case Ex. R 1 =1k Ω R 2 =10k Ω + + + V 1 =1V - V x =? - V 2 =5V - 6
Thevenin / Norton Equivalent Circuits • “Black box” representation of a circuit • Exactly the same when viewed from the terminals • Internal circuitry may be different Thevenin Norton How to determine values – V th , I N , R th =R N R th = R N Turn off all sources (V → short circuit; I → open circuit) Solve for the equivalent resistance V th Disconnect (i.e. open circuit) the circuit at the location of interest Solve for the “open-circuit voltage” at this location I N Short circuit the location of interest Calculate the current that flows through this short (i.e. “short- circuit current”) Ex. Find the Thevenin and Norton equivalent circuits to the left of the amplifier. R 1 R 3 + V in - R 2 Amplifier 7
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Source Transformation • Thevenin and Norton versions are equivalent • You can switch between the two representations without issue = V I R x x x where V = x I x R x 9
Measure Input Impedance • Let the output voltage float V test • Apply a test current source at the input, I test • Measure the change in the input voltage, Let V out float V test I test V = • test R in I test Ex. Find the input resistance of the following circuit. R 1 + + + R 2 0.01V x R 3 V in V x V out - - - 10
Measure Output Impedance • Turn off the input voltage • Apply a test current source at the output, I test • Measure the change in the output voltage, V test V = • test R out I test Ex. Find the output resistance of the following circuit. R 1 + + + R 2 0.01V x R 3 V in V x V out - - - 11
Capacitors • Store energy in the form of an electric field • Typically parallel plates separated by a non-conductive material dV I = C dt 1 ( ) ∫ t = + V I x dx V init C 0 Capacitors in parallel → add like resistors in series Capacitors in series → like resistors in parallel C 1 C 1 C 2 C ser = C 1 // C 2 C par = C 1 + C 2 = = C 2 Impedance of capacitors is frequency dependent (sinusoidal signals) ω → As 0 The capacitor acts like an open circuit ω → ∞ As The capacitor acts like a short circuit Complex Impedance 1 = = Z C ω j C Easier method of using complex impedances → Use Laplace transforms… = dV Laplace I C dt = I sCV Solve Ohm’s Law 1 = Z CL sC = σ + ω Where is the complex frequency s j If you replace the capacitance with its complex impedance, you can treat it like a resistor 12
Inductors • Store energy in the form of an magnetic field • Typically coiled wire/conductive material dI V = L dt = ∫ 1 ( ) t + I V x dx I init L 0 Combine like resistors (series/parallel) Impedance of inductors is frequency dependent ω → As 0 The inductor acts like a short circuit ω → ∞ As The inductor acts like an open circuit Complex Impedance = ω = Z L j L Using Laplace Notation dI = Laplace V L dt = V sLI ∴ L = Z sL = σ + ω Where s j is the complex frequency Treat complex impedance like a resistor and follow normal circuit analysis 13
Time-Domain Response of Circuits with Energy-Storage Elements Ex. Find V out (t) for the following circuit for t > 0 . The switch closes at time t = 0 ( ) = − = Let the capacitor have no initial charge stored on it → V out t 0 0 V Let the input signal be a constant voltage t = 0 R + + - V in C V out - 14
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