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BASIC ELECTRICAL CIRCUITS AND ANALYSIS Ref: Horowitz, P, & W. Hill, “The Art of Electronics,” 2nd. ed., Cambridge (1989). CHARGE - symbol C, unit is coulomb; 1 coulomb = 6.2415×1018 electrons CURRENT - symbol I or i, unit is amp(ere) (A); flow of electrons per unit time 1 A = 1 coul/sec VOLTAGE - symbol E or V, unit is volt; potential difference, 1 joule of work is needed to move 1 coulomb of charge through a potential difference of 1 volt (V) By convention (thanks to Ben Franklin), current flows from more positive (less negative) to more negative (less positive). Electrons are the real carriers; they go in the opposite direction. RESISTANCE - symbol R, unit is ohm OHM’S LAW: a potential difference of 1 volt will cause a current flow of 1 ampere through a resistance of 1 ohm. E=IR RESISTORS IN SERIES - RT = R1 + R2 + R3 + .. + Rn 1 1 1 1 1 RESISTORS IN PARALLEL - = + + + + RT R1 R2 R3 Rn R1R2 TWO RESISTORS IN PARALLEL: RT = R1 + R2 E2 POWER: P = EI = I R = 2 R VOLTAGE DIVIDER: R1 R2 Vo = VB VB R1 + R2 R2 Vo KIRCHHOFF RULES The law of conservation of energy and conservation of charge lead to two fundamental rules for circuit analysis. These rules are called Kirchhoff's rules, and they are used to determine the value of the electric current in each branch of a multi-loop circuit. The first rule, the node equation, states that the sum of the currents into a specific node (junction) in the circuit equals the sum of the currents out of the same node. [A junction is a place in a circuit where more than two wires join.] Electric charge is conserved: it does not suddenly appear or disappear; it does not pile up at one point and thin out at another. I1 I3 I2 I1 + I2 - I3 = 0 ∑I n =0 The second rule, the loop equation, states that around each loop in an electric circuit the sum of the emf 's (electromotive forces, or voltages, of energy sources such as batteries and generators) is equal to the sum of the potential drops, or voltages across each of the resistances, in the same loop. All the energy imparted by the energy sources to the electrons carrying the current is equal to that lost by the electrons in useful work and/or heat dissipation around each loop of the circuit. R1 R2 V2 V1 R3 V1 - IR1 - IR2 + V2 - IR3 = 0 ∑V n =0 SUPERPOSITION: The total current in any part of a circuit is equal to the algebraic sum of the currents produced by each source separately. To calculate, short all other voltage sources and open all other current sources. VOLTAGE SOURCES: (for ideal voltage source, Rs = 0) I R s IDEAL V V Thevenin's theorem: Any network of linear sources and resistors with two terminals can be replaced by an equivalent circuit comprising an ideal voltage source and a series resistance. Rth BLACK Vth BOX To determine the Thevenin equivalents, you can either calculate them or make two measurements. 1. Open-circuit voltage Voc. This is the Thevenin equivalent voltage. 2. Short-circuit current Isc. The Thevenin resistance is then Voc/Isc. CURRENT SOURCES: (for ideal current source, Rsh=∞ I IDEAL I Rsh V Norton's theorem: Any network of linear sources and resistors with two terminals can be replaced by an equivalent circuit comprising an ideal current source and a shunt resistance. This is analogous to Thevenin’s theorem. BLACK In Rn BOX The Norton resistance is the same as the Thevenin resistance, and the Norton current is the ratio of the Thevenin voltage to the Thevenin resistance. Norton and Thevenin are equivalent. Thevenin is preferred for high-impedance or constant-voltage circuits and Norton is preferred for low-impedance or constant- current circuits. Maximum power transfer theorem: For a given voltage source VB and series (source) resistance RS, the value of load resistance RL that maximizes the power delivered to the load is equal to the source resistance. PL = I 2 RL RS VB I= RS + RL I VB2 RL VB PL = ( RS + RL ) 2 RL Set dPL/dRL = 0 to find RL = RS CAPACITANCE Capacitance of two parallel plates: C 0.224KA C= d K = dielectric constant A = area d = spacing Capacitors in series: 1 1 1 1 1 = + + + + CT C1 C2 C3 Cn The total charge is distributed among the series capacitors, and the total capacitance is decreased For two series capacitors: C1C2 CT = C1 + C2 Capacitors in parallel CT = C1 + C2 + C3 + ….. + Cn All circuits have "stray" capacitance. EX: MOST Coaxial cables have capacitance ~100 pF/meter. CAPACITIVE REACTANCE 1 XC = jω C j merely indicates that the phase difference between the current and the voltage φ is 90°. The voltage and current are out of phase, and the power dissipation is zero. Ignoring the phase term, 1 XC = ωC XC → ∞ at ω = 0 (DC) XC → 0 when ω → ∞ CHARGING A CAPACITOR Immediately upon closing the switch, dQ I= dt Applying Kirchhoff's loop law Q dQ Q VB = VR + VC = IR + = R + C dt C This differential equation has as a solution Q = CV (1 − e − t / RC ) VB I = (1 − e − t / RC ) R RC is time constant τ CAPACITOR CHARGE vs. TIME at t = 0, as t → ∞, Q=0 Q → CVo VC = 0 VC → Vb VR = Vb I→0 Vo I= R at t = RC, capacitor is 63% charged CAPACITOR DISCHARGE Place resistance R across capacitor C. Close switch at t=0. LOW-PASS FILTER I Every system has/is one. It is a voltage divider with a frequency-dependent component. R C Simple one characterized by a time constant τ = RC TIME DOMAIN RESPONSE (to a step function) RISE DECAY −t −t V (t ) = V0 (1 − e τ ) V (t ) = V0e τ 1 0.8 RESPONSE 0.6 0.4 0.2 0 1 2 3 TIME / TIME CONSTANT LOW-PASS FILTER II FREQUENCY DOMAIN TRANSFERT FUNCTION T(ω) 1 τ = RC T (ω ) = ω c = 1τ 1+ω τ 2 2 Cut-off frequency ωc where ω2τ2 = 1. T(ω) is down 3dB = 0.707. Falloff at high frequencies is –6dB/octave = –20dB per decade. 1 0.8 RESPONSE 0.6 0.4 0.2 0 1 2 3 4 5 RELATIVE FREQUENCY 1 RESPONSE 0.1 0.01 0.1 1 10 100 RELATIVE FREQUENCY HIGH-PASS FILTER Many systems have a high-pass filter to block response at DC and low frequencies. Simple HPF also characterized by τ = RC, ω c = 1τ Cut-on frequency ωc where ω2τ2 = 1. T(ω) is down 3dB = 0.707. Rise at low frequencies is 6dB/octave = 20dB per decade. C R FREQUENCY DOMAIN RESPONSE ωτ ωτ T = = 1+ω τ 2 2 ω2 1+ 2 ωc 1 1 0.8 RESPONSE RESPONSE 0.6 0.1 0.4 0.2 0 0.01 0 1 2 3 0.01 0.1 1 10 100 RELATIVE FREQUENCY RELATIVE FREQUENCY BANDPASS FILTER I Combine LPF and HPF R C C + R If they do not interact, 1 ωτ T= × 1+ω τ 2 2 LP 1 + ω 2τ HP 2 1 0.8 RESPONSE 0.6 0.4 0.2 0 5 10 15 20 RELATIVE FREQUENCY 1 RESPONSE 0.1 0.01 0.01 0.1 1 10 100 RELATIVE FREQUENCY BANDPASS FILTER II If CLP = CHP AND RLP = RHP OR τHP = τLP = τ = 1/ω, ωτ ωτ T= = 1+ω τ 2 2 ω2 1+ 2 ωc 0.6 0.5 RESPONSE 0.4 0.3 0.2 0.1 0 1 2 3 RELATIVE FREQUENCY 1 RESPONSE 0.1 0.01 0.01 0.1 1 10 100 RELATIVE FREQUENCY SIGNAL BANDWIDTH Signal bandwidth of a low-pass filter • Frequency where filter transmission (voltage or current) falls to -3dB (1√2 = 0.707) of transmission at DC (typically 1) • Also referred to as half-power point, where power transmission has dropped to 0.5 (-6dB) • Power transmission = [voltage transmission]2 Signal bandwidth of a band-pass filter • Frequency where filter transmission (voltage or current) falls to -3dB (1√2 = 0.707) of transmission at the maximum transmission Without a high-frequency cutoff, the signal bandwidth of a high- pass filter is undefined. TRANSFORMER Some inevitable losses (heating) Impedance ratio goes as square of turns ratio NS/NP.

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posted: | 5/30/2010 |

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