# ENGR 100 1 Introduction to Analog Electrical Circuits Richard J. Kozick Electrical Engineering Department.

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ENGR 100 1 Introduction to Analog Electrical Circuits Richard J. Kozick Electrical Engineering Department

ENGR 100 2 Outline for Today’s Lecture Fundamental quantities,concepts & units: –Charge, current, voltage, power Battery and light bulb: –Show actual circuit versus “circuit model” Resistance and Ohm’s Law Kirchhoff’s Laws; series & parallel circuits Voltage divider –Light dimmer, volume control, sensors,...

ENGR 100 3 Technical Subdivisions of EE Computer Systems Electronics Electromagnetics Electric Power Systems Signal Processing and Control Systems Communication Systems

ENGR 100 4 Talk with neighbors and define... What is electric charge? What is electric current? What is electric voltage? [Note these are things we can’t see or feel directly!]

ENGR 100 5 Charge Property of matter Two kinds, + and - Electrical forces: –Opposite charges attract, like charges repel –Force varies as inverse square of distance between charges (like gravitational force) Basis for all electrical phenomena Unit: coulomb (C)

ENGR 100 6 Current Charges can move Current = flow rate of charge Unit: ampere (A) = C/s Example: –A battery is a supply of charges –Larger current drains the battery faster

ENGR 100 7 Voltage Potential energy per unit charge –Arises from force between + and - charges Unit: volt (V) = Joule/coulomb = J/C Analogy with gravitational potential energy: –P.E. = m g h –P.E. per unit mass = g h Need a reference to measure voltage: –Analogous to the floor in auditorium –Common voltage reference is ground (earth)

ENGR 100 8 Power Power = flow rate of energy (W = J/s) Current = flow rate of charge (A = C/s) Voltage = P.E. per unit charge (V = J/C) Say we have a flow of charges (current) that are “giving up” their P.E.: –Power = ??? (W = J/s)

ENGR 100 9 Power Power = flow rate of energy (W = J/s) Current = flow rate of charge (A = C/s) Voltage = P.E. per unit charge (V = J/C) Say we have a flow of charges (current) that are “giving up” their P.E.: –Power = Voltage × Current (W = J/s)

ENGR 100 10 Battery and Light Bulb Operation of actual circuit Circuit model: –Ideal voltage source for battery (9 V always) –“Resistor” to model light bulb (R ohms) –Ideal wires (0 resistance) 9 V IrIr R + - VrVr Ground

ENGR 100 11 Ohm’s Law Resistance: –Characterizes “ease” of charge flow (current) –Depends on material and geometry of wire Ohm’s Law: V r = I r R 9 V IrIr R + - VrVr Ground

ENGR 100 12 Georg Simon Ohm (1826): –First clear definition of voltage and current –Showed voltage and current are related –Then he lost his job and was ridiculed! –Finally, he became a university professor in 1849

ENGR 100 13 More on Battery and Light Bulb V r = _____ Measurement: I r = ______ Power dissipated by bulb: P = _____________ Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

ENGR 100 14 More on Battery and Light Bulb V r = 9 V Measurement: I r = ______ Power dissipated by bulb: P = _____________ Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

ENGR 100 15 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = _____________ Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

ENGR 100 16 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W Ohm’s Law: V r = I r R R = _____________ 9 V IrIr R + - VrVr Ground

ENGR 100 17 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W Ohm’s Law: V r = I r R R = V r / I r = 277 ohms  9 V IrIr R + - VrVr Ground

ENGR 100 18 More on Battery and Light Bulb V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W Ohm’s Law: V r = I r R R = V r / I r = 277 ohms  What if we use an 18 V battery? 9 V IrIr R + - VrVr Ground

ENGR 100 19 Kirchhoff’s Current Law (KCL) “The total current entering a node equals the total current leaving a node.” Why? Because charge is conserved (neither created nor destroyed), and charge is not accumulated at nodes. Find I 1, I 2, I 3 : 9 V I1I1 Ground I2I2 I3I3 4 A 2 A1 A

ENGR 100 20 Kirchhoff’s Current Law (KCL) “The total current entering a node equals the total current leaving a node.” Why? Because charge is conserved (neither created nor destroyed), and charge is not accumulated at nodes. Find I 1, I 2, I 3 : 9 V 2 A Ground 1 A 4 A 2 A1 A

ENGR 100 21 Kirchhoff’s Voltage Law (KVL) “Around any closed loop, the sum of voltage rises equals the sum of voltage drops.” Why? Energy is conserved! Find V a and V b 9 V +-5 V+-1 V + + - - VaVa VbVb

ENGR 100 22 Kirchhoff’s Voltage Law (KVL) “Around any closed loop, the sum of voltage rises equals the sum of voltage drops.” Why? Energy is conserved! Find V a and V b 9 V +-5 V+-1 V + + - - 4 V 3 V

ENGR 100 23 More Light Bulb Circuits Bulbs in series Bulbs in parallel How does power per bulb compare with single bulb? 9 V R R + - Ground R + - R

ENGR 100 24 Single Bulb 9 V IrIr R + - VrVr Ground V r = 9 V Measurement: I r = 32.5 mA Power dissipated by bulb: P = V r I r = 0.29 W R = V r / I r = 277 ohms 

ENGR 100 25 More Light Bulb Circuits Bulbs in series P = (V r / 2) (I r / 2 ) = 1/4 power Bulbs in parallel P = V r I r = same power For parallel, battery provides twice as much power. 9 V I r / 2 R 9 V IrIr R + - VrVr Ground R + - V r / 2 R

ENGR 100 26 Voltage Divider Important building block of analog circuits –Behind most “knob” and “slider” controls! –Light dimmer, volume control, treble/bass, … –Used for “filters” (equalizers, crossovers) –Basis for sensors (temperature, light, …) Easy to derive equations using KCL, KVL, and Ohm’s Law (please try it if interested)

ENGR 100 27 Voltage Divider VsVs +-V1V1 + - V2V2 Source Voltage R1R1 R2R2 Describes the “split” of source voltage across series resistors:

ENGR 100 28 Application: Light Dimmer Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) If R POT = 0.2 ohms: –V POT ~ _______ –V r ~ _______ If R POT = 5 k ohms: –V POT ~ _______ –V r ~ _______ 9 V +-V POT + - VrVr Battery R POT R = 277  Light Bulb

ENGR 100 29 Application: Light Dimmer Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) If R POT = 0.2 ohms: –V POT ~ 0 V –V r ~ 9 V, Bulb is ON If R POT = 5 k ohms: –V POT ~ _______ –V r ~ _______ 9 V +-V POT + - VrVr Battery R POT R Light Bulb

ENGR 100 30 Application: Light Dimmer Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) If R POT = 0.2 ohms: –V POT ~ 0 V –V r ~ 9 V, Bulb is ON If R POT = 5 k ohms: –V POT ~ 9 V –V r ~ 0 V, Bulb is OFF 9 V +-V POT + - VrVr Battery R POT R Light Bulb

ENGR 100 31 Application: Heat and Light Sensors Sensor resistance R sensor varies with physical property –Thermistor (temperature) –Photoresistor (light) R 1 is a fixed resistor Then V sensor changes with temperature or light! Bonus on HW: how to choose R 1 ? V sensor 9 V + - Battery R1R1 R sensor

ENGR 100 32 Concluding Remark Hopefully electric circuits are a little bit less mysterious to you now!

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