BASIC ELECTRICAL CIRCUITS AND ANALYSIS. Ref: Horowitz, P, & W. Hill, The Art of Electronics, 2nd. ed., Cambridge (1989).

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1 BASC ELECTCAL CCUTS AND ANALYSS ef: Horowitz, P, & W. Hill, The Art of Electronics, nd. ed., Cambridge (989). CHAGE - symbol C, unit is coulomb; coulomb electrons CUENT - symbol or i, unit is amp(ere) (A); flow of electrons per unit time A coul/sec VOLTAGE - symbol E or V, unit is volt; potential difference, joule of work is needed to move coulomb of charge through a potential difference of 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. ESSTANCE - symbol, unit is ohm OHM S LAW: a potential difference of volt will cause a current flow of ampere through a resistance of ohm. E ESSTOS N SEES - T n ESSTOS N PAALLEL T 3 n TWO ESSTOS N PAALLEL: POWE: P E E T + VOLTAGE DVDE: V o V B + V B V o

2 KCHHOFF ULES 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 n 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. V 3 V V V V n 0

3 SUPEPOSTON: 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 SOUCES: (for ideal voltage source, s 0) s V DEAL 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. V th BLACK BOX V th To determine the Thevenin equivalents, you can either calculate them or make two measurements.. Open-circuit voltage V oc. This is the Thevenin equivalent voltage.. Short-circuit current sc. The Thevenin resistance is then V oc / sc. CUENT SOUCES: (for ideal current source, sh DEAL sh V

4 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 BOX n n 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 constantcurrent circuits. Maximum power transfer theorem: For a given voltage source V B and series (source) resistance S, the value of load resistance L that maximizes the power delivered to the load is equal to the source resistance. PL L V B S L VB + S L VBL PL ( S + L) Set dp L /d L 0 to find L S

5 CAPACTANCE Capacitance of two parallel plates: C C 0.4KA d K dielectric constant A area d spacing Capacitors in series: C C C C C T 3 n The total charge is distributed among the series capacitors, and the total capacitance is decreased For two series capacitors: C T CC C + C Capacitors in parallel CT C + C + C Cn All circuits have "stray" capacitance. EX: MOST Coaxial cables have capacitance ~00 pf/meter.

6 CAPACTVE EACTANCE X C jωc 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. gnoring the phase term, X C ωc XC at ω 0 (DC) XC 0 when ω

7 CHAGNG A CAPACTO mmediately upon closing the switch, Applying Kirchhoff's loop law dq dt Q dq Q VB V + VC + + C dt C This differential equation has as a solution C is time constant τ ( t/ C ) Q CV e V B ( t/ C e )

8 CAPACTO CHAGE vs. TME at t 0, as t, Q 0 VC 0 Q CVo VC Vb V Vb 0 V o at t C, capacitor is 63% charged

9 CAPACTO DSCHAGE Place resistance across capacitor C. Close switch at t0.

10 LOW-PASS FLTE Every system has/is one. t is a voltage divider with a frequency-dependent component. C Simple one characterized by a time constant τ C TME DOMAN ESPONSE (to a step function) SE DECAY V() t V ( e τ ) 0 t Vt () 0 t Ve τ 0.8 ESPONSE TME / TME CONSTANT

11 LOW-PASS FLTE FEQUENCY DOMAN TANSFET FUNCTON T(ω) ( ) T ω + ω τ τ C ω c τ Cut-off frequency ωc where ω τ. T(ω) is down 3dB Falloff at high frequencies is 6dB/octave 0dB per decade. 0.8 ESPONSE ELATVE FEQUENCY ESPONSE ELATVE FEQUENCY

12 HGH-PASS FLTE Many systems have a high-pass filter to block response at DC and low frequencies. Simple HPF also characterized by τ C, ω c τ Cut-on frequency ωc where ω τ. T(ω) is down 3dB ise at low frequencies is 6dB/octave 0dB per decade. C FEQUENCY DOMAN ESPONSE T ωτ + ω τ ωτ ω + ω c ESPONSE ELATVE FEQUENCY ESPONSE ELATVE FEQUENCY

13 BANDPASS FLTE Combine LPF and HPF C + C f they do not interact, T ωτ + ω τ + ω τ 0.8 LP HP ESPONSE ELATVE FEQUENCY ESPONSE ELATVE FEQUENCY

14 BANDPASS FLTE f CLP CHP AND LP HP O τhp τlp τ /ω, T ωτ + ωτ ωτ ω + ω c ESPONSE ELATVE FEQUENCY ESPONSE ELATVE FEQUENCY

15 SGNAL BANDWDTH Signal bandwidth of a low-pass filter Frequency where filter transmission (voltage or current) falls to -3dB ( 0.707) of transmission at DC (typically ) Also referred to as half-power point, where power transmission has dropped to 0.5 (-6dB) Power transmission [voltage transmission] Signal bandwidth of a band-pass filter Frequency where filter transmission (voltage or current) falls to -3dB ( 0.707) of transmission at the maximum transmission Without a high-frequency cutoff, the signal bandwidth of a highpass filter is undefined.

16 TANSFOME Some inevitable losses (heating) mpedance ratio goes as square of turns ratio NS/NP.

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