Introduction to the Multimeter
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Transcript Introduction to the Multimeter
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 Close
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
Georgethrough
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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
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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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