Transcript Brief Overview of Analog Circuits

Introduction to Analog Electrical
Circuits
Richard J. Kozick
Electrical Engineering Department
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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
Ir
(0 resistance)
+
9V
Vr
-
R
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
9V
+
Vr
-
Ir
R
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:
•
•
•
13
9V
Ir = ______
Power dissipated by bulb:
P = _____________
Ohm’s Law: Vr = Ir • R
R = _____________
ENGR 100
+
Vr
-
Ir
R
Ground
More on Battery and Light Bulb
• Vr = 9 V
• Measurement:
•
•
•
14
9V
Ir = ______
Power dissipated by bulb:
P = _____________
Ohm’s Law: Vr = Ir • R
R = _____________
ENGR 100
+
Vr
-
Ir
R
Ground
More on Battery and Light Bulb
• Vr = 9 V
• Measurement:
•
•
•
15
9V
Ir = 32.5 mA
Power dissipated by bulb:
P = _____________
Ohm’s Law: Vr = Ir • R
R = _____________
ENGR 100
+
Vr
-
Ir
R
Ground
More on Battery and Light Bulb
• Vr = 9 V
• Measurement:
•
•
•
16
9V
Ir = 32.5 mA
Power dissipated by bulb:
P = Vr • Ir = 0.29 W
Ohm’s Law: Vr = Ir • R
R = _____________
ENGR 100
+
Vr
-
Ir
R
Ground
More on Battery and Light Bulb
• Vr = 9 V
• Measurement:
•
•
•
17
9V
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)
ENGR 100
+
Vr
-
Ir
R
Ground
More on Battery and Light Bulb
• Vr = 9 V
• Measurement:
•
•
•
•
18
9V
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?
ENGR 100
+
Vr
-
Ir
R
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:
4A
9V
2A
I1
I2
Ground
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1A
I3
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:
4A
9V
2A
2A
1A
Ground
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1A
1A
Kirchhoff’s Voltage Law (KVL)
• “Around any closed loop, the sum of
•
voltage rises equals the sum of voltage
drops.”
Why? Energy is conserved!
+ 5V -
• Find Va and Vb
+
9V
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Va
-
+ 1V +
Vb
-
Kirchhoff’s Voltage Law (KVL)
• “Around any closed loop, the sum of
•
voltage rises equals the sum of voltage
drops.”
Why? Energy is conserved!
+ 5V -
• Find Va and Vb
9V
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+
4V
-
+ 1V +
3V
-
More Light Bulb Circuits
•
•
Bulbs in series
+
9V
R
-
Bulbs in parallel
+
9V
R
R
-
R
Ground
How does power per bulb compare with single bulb?
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Single Bulb
Vr = 9 V
Measurement:
Ir = 32.5 mA
9V
Power dissipated by bulb:
P = Vr • Ir = 0.29 W
R = Vr / Ir = 277 ohms (W)
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+
Vr
-
Ir
R
Ground
More Light Bulb Circuits
•
9V
•
Bulbs in series
+
Vr / 2
-
Ir / 2
R
9V
R
Bulbs in parallel
+
Vr
-
Ir
R
R
Ground
•
P = (Vr / 2) • (Ir / 2 )
• P = Vr • Ir
= 1/4 power
= same power
For parallel, battery provides twice as much power.
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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
Describes the “split” of source voltage across series
resistors:
+ V 1
R1
Source
V
Voltage s
R2
R1
V1 
 Vs
R1  R2
27
+
V2
-
R2
V2 
 Vs
R1  R2
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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
~ _______
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+ VPOT Battery
9V
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RPOT
R=
277 W
+
Vr
-
Light
Bulb
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
~ _______
29
Battery
9V
ENGR 100
+
VPOTRPOT
R
+
Vr
-
Light
Bulb
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:
Battery
9V
– VPOT ~ 9 V
– Vr
~ 0 V , Bulb is OFF
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ENGR 100
+
VPOTRPOT
R
+
Vr
-
Light
Bulb
Application: Heat and Light Sensors
• Sensor resistance Rsensor varies with
physical property
R1
– Thermistor (temperature)
– Photoresistor (light)
• R1 is a fixed resistor
• Then Vsensor changes with
Battery
9V
•
31
Rsensor
temperature or light!
Bonus on HW: how to choose R1 ?
ENGR 100
+
Vsensor
-
Concluding Remark
• Hopefully electric circuits are a little bit less
mysterious to you now!
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