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Introduction to Analog Electrical Circuits

Introduction to Analog Electrical Circuits. Richard J. Kozick Electrical Engineering Department. 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

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Introduction to Analog Electrical Circuits

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

  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, ...

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

  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!]

  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)

  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

  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)

  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) Volunteers ...

  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)

  10. Ir + 9 V Vr R - Ground 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)

  11. Ir + 9 V Vr R - Ground Ohm’s Law • Resistance: • Characterizes “ease” of charge flow (current) • Depends on material and geometry of wire • Ohm’s Law: Vr = Ir • R

  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

  13. Ir + 9 V Vr R - Ground More on Battery and Light Bulb • Vr = _____ • Measurement: Ir = ______ • Power dissipated by bulb:P = _____________ • Ohm’s Law: Vr = Ir • R • R = _____________

  14. Ir + 9 V Vr R - Ground More on Battery and Light Bulb • Vr = 9 V • Measurement: Ir = ______ • Power dissipated by bulb:P = _____________ • Ohm’s Law: Vr = Ir • R • R = _____________

  15. Ir + 9 V Vr R - Ground More on Battery and Light Bulb • Vr = 9 V • Measurement: Ir = 32.5 mA • Power dissipated by bulb:P = _____________ • Ohm’s Law: Vr = Ir • R • R = _____________

  16. Ir + 9 V Vr R - Ground More on Battery and Light Bulb • Vr = 9 V • Measurement: Ir = 32.5 mA • Power dissipated by bulb:P = Vr • Ir = 0.29 W • Ohm’s Law: Vr = Ir • R • R = _____________

  17. Ir + 9 V Vr R - Ground More on Battery and Light Bulb • Vr = 9 V • Measurement: Ir = 32.5 mA • Power dissipated by bulb:P = Vr • Ir = 0.29 W • Ohm’s Law: Vr = Ir • R • R = Vr / Ir = 277 ohms (W)

  18. Ir + 9 V Vr R - Ground More on Battery and Light Bulb • Vr = 9 V • Measurement: Ir = 32.5 mA • Power dissipated by bulb:P = Vr • Ir = 0.29 W • Ohm’s Law: Vr = Ir • R • R = Vr / Ir = 277 ohms (W) • What if we use an 18 V battery?

  19. 4 A 2 A 1 A 9 V I1 I2 I3 Ground 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 I1, I2, I3:

  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 I1, I2, I3: 4 A 2 A 1 A 9 V 2 A 1 A 1 A Ground

  21. + 5 V - + 1 V - + + Va Vb 9 V - - 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 Va and Vb

  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 Va and Vb + 5 V - + 1 V - + + 9 V 4 V 3 V - -

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

  24. Ir + 9 V Vr R - Ground Single Bulb Vr = 9 V Measurement: Ir = 32.5 mA Power dissipated by bulb:P = Vr • Ir = 0.29 W R = Vr / Ir = 277 ohms (W)

  25. Bulbs in series P = (Vr / 2) • (Ir / 2 ) = 1/4 power Bulbs in parallel P = Vr • Ir = same power + Vr / 2 - More Light Bulb Circuits Ir / 2 Ir + R 9 V 9 V Vr R R - R Ground For parallel, battery provides twice as much power.

  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)

  27. + V1 - R1 + Source Voltage R2 V2 Vs - Voltage Divider Describes the “split” of source voltage across series resistors:

  28. Application: Light Dimmer • Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) • If RPOT = 0.2 ohms: • VPOT ~ _______ • Vr ~ _______ • If RPOT = 5 k ohms: • VPOT ~ _______ • Vr ~ _______ + VPOT - RPOT Battery + LightBulb 9 V R = 277 W Vr -

  29. Application: Light Dimmer • Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) • If RPOT = 0.2 ohms: • VPOT ~ 0 V • Vr ~ 9 V , Bulb is ON • If RPOT = 5 k ohms: • VPOT ~ _______ • Vr ~ _______ + VPOT - RPOT Battery + LightBulb 9 V R Vr -

  30. Application: Light Dimmer • Potentiometer (POT) = variable resistor as turn knob (from 0.2 ohms to 5,000 ohms) • If RPOT = 0.2 ohms: • VPOT ~ 0 V • Vr ~ 9 V, Bulb is ON • If RPOT = 5 k ohms: • VPOT ~ 9 V • Vr ~ 0 V , Bulb is OFF + VPOT - RPOT Battery + LightBulb 9 V R Vr -

  31. R1 Battery + 9 V Vsensor Rsensor - Application: Heat and Light Sensors • Sensor resistance Rsensor varies with physical property • Thermistor (temperature) • Photoresistor (light) • R1 is a fixed resistor • Then Vsensor changes withtemperature or light! • Bonus on HW: how to choose R1 ?

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

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