DC Circuits Lab
ECE 002
Professor Ahmadi
George Washington University
Outline
 Basic Components of a Circuit
 Series Circuit
 Parallel Circuit
 Ohm’s Law
 Lab Overview
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Basic Circuit Components
We represent real electrical components with symbols
1.5V
1.5 V
A Battery…
…can be represented with this symbol
…called a “DC voltage source”
A DC Voltage Source
• Provides Power for our circuit
• Battery or Lab ‘power supply’ is an example
• DC voltage is supplied across the two terminals
• Its voltage is VOLTS (V)
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Basic Circuit Components
We represent real electrical components with symbols
RΩ
A Light Bulb…or any ‘device’…
…can be represented with this symbol
…called a “resistor”
A Resistor
• Represents any device that requires power to operate
• Could be a light bulb, your computer, a toaster, etc.
• Each device has a certain amount of ‘resistance’, R, in
the unit called: OHMS (Ω)
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Basic Circuit Components
We represent real electrical components with symbols
The Earth…
…can be represented with this symbol
…called the “ground” symbol
The Ground
• Represents 0 volts
• We use it as a ‘reference’ voltage…to measure other
voltages against it
• The ‘Earth’ is at 0 volts, so we call this ground
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Basic Circuit Components
We represent real electrical components with symbols
A Tollbooth…or any ‘barrier’
…can be represented with this symbol
…called a “diode”
The Diode
• Controls the flow of current.
• Has two ends called the anode and cathode.
• Charges a ‘toll’ or voltage penalty of ~0.7V for passing
through it.
• If the anode voltage is not at least 0.7V, no current will flow
to the cathode.
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Basic Circuit Components
The diode is like a switch that takes ~0.7V to turn on
Anode
Cathode
The Diode Has Two Modes of Operation
•Negative DC Voltage Source
•When the Anode is at least ~0.7V. Replace the diode by a -0.7V DC Source.
=
0.7V
•Open Circuit
•When the Anode is less than ~0.7V, the diode is an open circuit. This means
no current can flow through it!
=
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Building a Circuit…
 We wish to ‘power’ our
flashlight’s light bulb…
 We need a battery…
1.5 V
 We need to attach the
light bulb to the
battery…
 We use wires to
connect the light bulb
to the battery…
Instead…let's represent the real components with their symbols
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Building a Circuit…creating a schematic
.5 Ω
1.5V
1.5 V
Since this “node” is at GND (OV) this node
must be 1.5Volts higher
 Replace the battery
with a ‘DC Voltage
Source’ symbol
 Replace the light bulb
with a ‘Resistor’
symbol
 Mark the symbol’s
values (V=, R=, etc.)
 Add the Ground
reference
0V
Instead…let's represent the real components with their symbols
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Analyzing the Circuit…using Ohm’s Law
.5K Ω
1.5V
 When we attach the
resistor to the DC
voltage source,
current begins to flow
 How much current will
flow?
 Ohm’s Law (V=IR)
->Describes the relationship
between the voltage (V),
current (I), and resistance
(R) in a circuit
 Using Ohm’s Law, we can determine
how much current is flowing
our circuit
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Washington University
0V
Analyzing the Circuit…using Ohm’s Law
 How much current will
flow?
I = 3 mA
.5K Ω
1.5V
 Use Ohm’s Law:
V
=I
x R
1.5V
=I
x .5K
Ω
Solve for I:
0V
I = 1.5V / .5 KΩ = 3 mA
So, 3 mA will flow through the .5kΩ
when 1.5 Volts are across it
George resistor,
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Resistors in Series
R1 = .5K Ω
1.5V
Req
R2 = .5K
1K ΩΩ
 Resistors connected by only
1 terminal, back-to-back,
are considered to be in
‘series’
 We can replace the two
series resistors with 1
single resistor, we call Req
 The value of Req is the SUM of
R1 & R2:
Req=R1+R2=.5K Ω + .5K Ω = 1KΩ
0V
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Resistors in Series
I = 1.5 mA
 Now we can find the
current through the circuit
using Ohm’s Law
 Use Ohm’s Law:
1.5V
Req = 1K Ω
V
=I
x Req
1.5V
=I
x 1K
Solve for
Ω
I:
I = 1.5V / 1K Ω = 1.5 mA
0V
The bigger the resistance in the
circuit,
the
harder it is for current to flow
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Washington
University
I = 1.5 mA
Resistors in Series
 The current is the SAME
through each resistor
R1 = .5K Ω
1.5V
 Back to our original series
circuit, with R1 and R2
R2 = .5K Ω
0V
 Current flows like water
through the circuit, notice
how the 1.5 mA ‘stream of
current’ flows through both
resistors equally
 Ohm’s Law shows us voltage
across each resistor:
V(R1) = 1.5mA x .5K Ω = .75V
V(R2) = 1.5mA x .5K Ω = .75V
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Resistors in Parallel
1.5V
R1 = .5K Ω
 Resistors connected at 2
terminals, sharing the same
node on each side, are
considered to be in
‘parallel’
 Unlike before, we cannot
just add them. We must
add their inverses to find
1
1
1
R2 = .5K Ω Req:


0V
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Re q R1 R 2
1
1
1


Re q .5K .5K
Req = .25K Ω
Resistors in Parallel
I = 6 mA
 This is the equivalent
circuit
 Use Ohm’s Law, we find
the current through Req:
1.5V
Req = .25K Ω
V
=I
x Req
1.5V
=I
x .25K
Solve for
Ω
I:
I = 1.5V / .25KΩ = 6 mA
0V
The smaller the resistance inGeorge
the circuit,
easier it is for current to flow
Washington the
University
Resistors in Parallel
 Back to our original series
circuit, with R1 and R2
 The current is NOT the
SAME through all parts of
the circuit
 Current flows like water
1.5V
through the circuit, notice
how the 6 mA ‘stream of
current’ splits to flow into
the two resistors
R1 = .5K Ω R2 = .5KΩ
 The Voltage across each
resistor is equal when they
are in parallel
0V
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Resistors in Parallel
1.5V
I = 3 mA
I = 3 mA
I = 6 mA
 The voltage is 1.5 V across
each resistor
 Ohm’s Law tells us the
current through each:
I(R1)=V/R= 1.5V /.5KΩ = 3mA
I(R2)=V/R= 1.5V /.5KΩ = 3mA
 The 6mA of current has split
down the two legs of our
R1 = .5K Ω R2 = .5K Ω
circuit
0V
 It split equally between the
two legs, because the resistors
have the same value
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The current will split differently
if the University
resistors are not equal…
Resistors in Parallel
I = 6 mA
 This is the equivalent
circuit
 Use Ohm’s Law, we find
the current through Req:
1.5V
Req = .25K Ω
V
=I
x Req
1.5V
=I
x .25K
Solve for
Ω
I:
I = 1.5V / .25K Ω = 6 mA
0V
The smaller the resistance inGeorge
the circuit,
easier it is for current to flow
Washington the
University
Including a Diode
 Steps to Analyze the Circuit
Anode = 1.5V
 First, is the anode
potential at least 0.7V?
1.5V
R = .5K Ω
 Yes, it is at 1.5V. So,
replace the diode with a
-0.7V DC Source.
0V
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Including a Diode
 Steps to Analyze the Circuit
 Voltage sources in series
can be combined.
0.7V
1.5V
 1.5V + (-0.7)V = 0.8V
 Use that 0.8V value as the
V in Ohm’s Law!
R = .5K Ω
0V
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Including a Diode
 Steps to Analyze the Circuit
 Now, how much current
will flow through R?
I = 1.6 mA
0.8V
R = .5K Ω
 Use Ohm’s Law:
V
=I
x R
0.8V
=I
x .5K
Ω
Solve for I:
0V
I = 0.8V / .5 Ω = 1.6 mA
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Including a Diode
 Check Your Answer
0.7V
1.5V
 The Voltage on the Left
(From the DC Source)
Should equal the Voltage
Drops on the Right.
Use Ohm’s Law For the Resistor:
R = .5K Ω
VR
0.8V
=I
x R
= 1.6mA x .5K
Ω
For the Diode:
VD = 0.7V
Add the Voltage Drops:
VR +VD = 0.8V+0.7V= 1.5V
This matches our voltage
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source…YAY!
0V
Including a Diode
 Steps to Analyze the Circuit
Anode = 0.5V
 First, is the anode
potential at least 0.7V?
0.5V
R = .5K Ω
I = 0 Amps
 No, it is at 0.5V.
Therefore, no current
can flow through the
resistor.
0V
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In Summary…
 Ohm’s Law: V=IR
 Describes the relationship between the voltage (V),
current (I), and resistance (R) in a circuit
 Current is equal through two resistors in series
 Voltage drops across each resistor
 Req = R1 + R2 + . . .
 Voltage is equal across two resistors in parallel
 Current splits through branches of parallel circuits
 1/Req = 1/R1 + 1/R2
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In Summary…
 Diodes
 There is a voltage cost associated with every diode.
 Current will only flow through the diode if the
voltage at the anode is ≥ to that cost.
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In Lab Today
 You will build series circuits
 Build parallel circuits
 Work with a breadboard
 Verify Ohm’s Law by measuring voltage using a
multimeter
 And yes, there is HW!
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