Direct Current (DC): In a DC circuit the current and voltage are constant as a function of time. Power (P): Rate of doing work P = dw/dt units = Watts

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1 Lecture 1: Introduction Some Definitions: Current (I): Amount of electric charge (Q) moving past a point per unit time I dq/dt Coulombs/sec units Amps (1 Coulomb 6x10 18 electrons) oltage (): Work needed to move charge from point a to b Work Q olt Work/Charge Joules/Coulomb oltage is always measured with respect to something "ground" is defined as zero olts Direct Current (DC): In a DC circuit the current and voltage are constant as a function of time Power (P): Rate of doing work P dw/dt units Watts 1

2 Ohms Law: Linear relationship between voltage and current I R R Resistance (Ω) units Ohms nonlinear (diode) (olts) linear (resistor) slope d/di resistance Joules Law: When current flows through a resistor energy is dissipated W Q P dw/dt dq/dt + Qd/dt d/dt 0 for DC circuit and averages to 0 for AC Power dq/dt I Using Ohms law I (Amps) P I 2 R 2 / R 100 Watts 10 and 10 Amps or 10 through 1 Ω 2

3 Simple Circuits Symbols: Battery Resistor Ground + Solder paste Dimension of surface mount components (e.g. 1206): 4700 Lab resistors Stick the leads into bread board to make connections length: 12 x 250 µm mm width: 6 x 250 µm 1.5 mm Use in computers/cell phones Place with pick & place machine Surface tension automatically aligns the component on their pads! smallest available: (0.4 mm 0.2 mm, power rating 0.0 W) slightly less insane version: 0201 (0.6 mm 0. mm, power rating 0.05 W)

4 Simple(st) Circuit: I + R Convention: Current flow is in the direction of positive charge flow When we go across a battery in direction of current ( è +) + oltage drop across a resistor in direction of current (+ è ) IR Conservation of Energy: sum of potential drops around the circuit should be zero IR 0 or IR!! 4

5 Next simple(st) circuit: two resistors in series I R 1 A R 2 I Conservation of charge: I 1 I 2 I at point A I(R 1 +R 2 ) IR R R 1 + R 2 Resistors in Series Add: R R 1 + R 2 + R... What's voltage across R 2? 2 I 2 R 2 R 2 /(R 1 + R 2 ) "oltage Divider Equation" Two resistors in parallel I A I 1 R 1 I 2 R 2 I I 1 + I 2 /R 1 + /R 2 /R 1/R 1/R 1 + 1/R 2 R R 1 R 2 R 1 + R 2 Parallel Resistors add like: 1/R 1/R 1 + 1/R 2 + 1/R + 5

6 In a circuit with resistors (series and parallel), what's I 2 2 /R 2? I 1 + R R 2 I 2 R I reduce to a simpler circuit: I + R1 I + R R 2 I /R /(R 1 + R 2 ) R 2 R 2 R R 2 R R 2 + R 2 IR 2 R 1 + R 2 R R 2 + R R 2 R R 2 + R R 2 R R 1 R 2 + R 1 R + R 2 R I 2 2 R 2 R If R then I 2 I /(R 1 + R 2 ) as expected! K.K. Gan R 1 R 2 + R 1 R + R 2 R L1: Introduction 6

7 Kirchoff's Laws We can formalize and generalize the previous examples using Kirchoff's Laws: 1. ΣI 0 at a node: conservation of charge 2. Σ 0 around a closed loop: conservation of energy example node B: I 1 I 2 + I è I 1 I 2 I 0 loop ABEF: I 1 R 1 I 2 R 2 0 loop ACDF: I 1 R 1 I R 0 linear equations with unknowns: I 1, I 2, I always wind up with as many linear equations as unknowns! use matrix methods to solve these equations: RI " " R 1 R 2 0 " $ ' $ ' $ $ ' $ R 1 0 R ' $ # $ 0 &' # $ 1 1 1& '# $ I 1 I 2 I ' ' &' K.K. Gan L1: Introduction 7

8 # R 1 0 & ( det R 1 R ( $ 1 0 1' ( R I 2 # R 1 R 2 0 & R 1 R 2 + R 1 R + R 2 R ( det R 1 0 R ( $ 1 1 1' ( the same solution as in page 5! Measuring Things oltmeter: Always put in parallel with what you want to measure If no voltmeter we would have: " R AB L $ ' # R S + R L & If the voltmeter has a finite resistance R m then circuit looks like: K.K. Gan L1: Introduction 8

9 From previous pages we have: " * R AB m R L $ ' # R S + R m R L & R m R L R S R L + R m R L + R S R m R L R L + R S + R S R L R m AB if R L << R m good voltmeter has high resistance (> 10 6 Ω) Ammeter: measures current Always put in series with what you want to measure Without meter: I /(R S + R L ) With meter: I* /(R S + R L + R m ) good ammeter has R m << (R S + R L ), i.e. low resistance (0.11 Ω) 9

10 Thevenin's Equivalent Circuit Theorem Any network of resistors and batteries having 2 output terminals may be replaced by a series combination of resistor and battery Useful when solving complicated (!?) networks Solve problems by finding eq and R eq for circuit without load, then add load to circuit. Use basic voltage divider equation: L eq R L R L + R eq Two rules for using Thevenin's Thereom: 1. Take the load out of the circuit to find eq : eq R R 1 + R 10

11 2. Short circuit all power supplies (batteries) to find R eq : Can now solve for I L as in previous examples: I L eq R eq + R L " R $ ' # R 1 + R & 1 R 1 R R 1 + R + R L R eq R 1 R R 1 + R R R 1 R L + R 1 R + R L R Same answer as previous examples! 11