1 ENGR 100

Introduction to Analog Electrical Circuits

Richard J. Kozick

Electrical Engineering Department

2 ENGR 100

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, ...

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Technical Subdivisions of EE

• Computer Systems

• Electronics

• Electromagnetics

• Electric Power Systems

• Signal Processing and Control Systems

• Communication Systems

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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!]

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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)

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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

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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)

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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)

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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)

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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

Ir

R

+

-Vr

Ground

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Ohm’s Law

• Resistance:– Characterizes “ease” of charge flow (current)– Depends on material and geometry of wire

• Ohm’s Law:

Vr = Ir • R 9 V

Ir

R

+

-Vr

Ground

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• 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

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More on Battery and Light Bulb

• Vr = _____

• Measurement: Ir = ______

• Power dissipated by bulb:P = _____________

• Ohm’s Law: Vr = Ir • R

• R = _____________

9 V

Ir

R

+

-Vr

Ground

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More on Battery and Light Bulb

• Vr = 9 V

• Measurement: Ir = ______

• Power dissipated by bulb:P = _____________

• Ohm’s Law: Vr = Ir • R

• R = _____________

9 V

Ir

R

+

-Vr

Ground

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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 = _____________

9 V

Ir

R

+

-Vr

Ground

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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 = _____________

9 V

Ir

R

+

-Vr

Ground

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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

9 V

Ir

R

+

-Vr

Ground

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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

• What if we use an 18 V battery?

9 V

Ir

R

+

-Vr

Ground

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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, I3:9 V

I1

Ground

I2 I3

4 A 2 A 1 A

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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, I3:9 V

2 A

Ground

1 A 1 A

4 A 2 A 1 A

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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 9 V

+ -5 V + -1 V

+ +

- -

Va Vb

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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 9 V

+ -5 V + -1 V

+ +

- -

4 V3 V

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More Light Bulb Circuits

• Bulbs in series • Bulbs in parallel

How does power per bulb compare with single bulb?

9 VR

9 V R

+

-

Ground

R

+

- R

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Single Bulb

9 V

Ir

R

+

-Vr

Ground

Vr = 9 V

Measurement: Ir = 32.5 mA

Power dissipated by bulb:P = Vr • Ir = 0.29 W

R = Vr / Ir = 277 ohms

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More Light Bulb Circuits

• Bulbs in series

• P = (Vr / 2) • (Ir / 2 ) = 1/4 power

• Bulbs in parallel

• P = Vr • Ir = same power

For parallel, battery provides twice as much power.

9 V

Ir / 2R

9 V

Ir

R

+

-Vr

Ground

R

+

-Vr / 2

R

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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)

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Voltage Divider

Vs

+ -V1

+

-

V2

SourceVoltage

R1

R2

Describes the “split” of source voltage across series resistors:

ss VRR

RVV

RR

RV

21

22

21

11

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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 ~ _______

9 V

+ -VPOT

+

-

Vr

BatteryRPOT

R =

277

LightBulb

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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 ~ _______

9 V

+ -VPOT

+

-

Vr

BatteryRPOT

R LightBulb

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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

9 V

+ -VPOT

+

-

Vr

BatteryRPOT

R LightBulb

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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 ?

Vsensor9 V

+

-

Battery

R1

Rsensor

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Concluding Remark

• Hopefully electric circuits are a little bit less mysterious to you now!