george washington university dc circuits lab ece 002 professor ahmadi
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George Washington University
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
George Washington University
Basic Circuit Components
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)
We represent real electrical components with symbols
…can be represented with this symbol 1.5 V
1.5V A Battery…
…called a “DC voltage source”
George Washington University
Basic Circuit ComponentsWe represent real electrical components with symbols
…can be represented with this symbol
…called a “resistor”
A Light Bulb…or any ‘device’…
R Ω
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)
George Washington University
Basic Circuit ComponentsWe represent real electrical components with symbols
…can be represented with this symbol
…called the “ground” symbol
The Earth…
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 ComponentsWe represent real electrical components with symbols
…can be represented with this symbol
…called a “diode”
A Tollbooth…or any ‘barrier’
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 ComponentsThe diode is like a switch that takes ~0.7V to turn on
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.
•Open Circuit•When the Anode is less than ~0.7V, the diode is an open circuit. This means no current can flow through it!
Anode Cathode
=0.7V
=
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Building a Circuit…
We wish to ‘power’ our flashlight’s light bulb…
1.5 V
We need a battery…
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…
Replace the battery with a ‘DC Voltage Source’ symbol
1.5 V
Replace the light bulb with a ‘Resistor’ symbol
Mark the symbol’s values (V=, R=, etc.)
Add the Ground reference
Instead…let's represent the real components with their symbols
creating a schematic
Since this “node” is at GND (OV) this node
must be 1.5Volts higher
1.5V .5 Ω
0V
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Analyzing the Circuit…using Ohm’s Law
Ohm’s Law (V=IR)->Describes the relationship between the
voltage (V), current (I), and resistance (R) in a circuit
When we attach the resistor to the DC voltage source, current begins to flowHow much current will flow?1.5V .5K Ω
0VUsing Ohm’s Law, we can
determine how much current is flowing through our circuit
George Washington University
Analyzing the Circuit…using Ohm’s Law
Use Ohm’s Law:
V = I x R
1.5V = I x .5K Ω
Solve for I:
I = 1.5V / .5 KΩ = 3 mA
How much current will flow?
1.5V .5K Ω
0V
I = 3 mA
So, 3 mA will flow through the .5kΩ resistor, when 1.5 Volts are across it
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Resistors in SeriesResistors connected by
only 1 terminal, back-to-back, are considered to be in ‘series’
1.5V R2 = .5K Ω
0V
R1 = .5K Ω
We can replace the two series resistors with 1 single resistor, we call ReqThe value of Req is the SUM
of R1 & R2:Req=R1+R2=.5K Ω + .5K Ω =
1KΩ
Req = 1K Ω
George Washington University
Resistors in SeriesNow we can find the
current through the circuit using Ohm’s Law
1.5V
0V
Req = 1K Ω
Use Ohm’s Law:
V = I x Req
1.5V = I x 1K Ω
Solve for I:
I = 1.5V / 1K Ω = 1.5 mA
I = 1.5 mA
The bigger the resistance in the circuit, the harder it is for current to flow
George Washington University
Resistors in SeriesBack to our original series circuit, with R1 and R2
1.5V R2 = .5K Ω
0V
R1 = .5K Ω
The current is the SAME through each resistor
I =
1.5
mA
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
Current flows like water through the circuit, notice how the 1.5 mA ‘stream of current’ flows through both resistors equally
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Resistors in ParallelResistors connected at 2
terminals, sharing the same node on each side, are considered to be in ‘parallel’
1.5V
0V
R1 = .5K Ω
Unlike before, we cannot just add them. We must add their inverses to find Req:R2 = .5K Ω
2
1
1
1
Re
1
RRq
Req = .25K Ω
KKq 5.
1
5.
1
Re
1
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Resistors in ParallelThis is the equivalent
circuit
1.5V
0V
Req = .25K Ω
Use Ohm’s Law, we find the current through Req:
V = I x Req
1.5V = I x .25K Ω
Solve for I:
I = 1.5V / .25KΩ = 6 mA
I = 6 mA
The smaller the resistance in the circuit, the easier it is for current to flow
George Washington University
Resistors in Parallel
1.5V
0V
R1 = .5K Ω R2 = .5KΩ
Back to our original series circuit, with R1 and R2
The Voltage across each resistor is equal when they are in parallel
The current is NOT the SAME through all parts of the circuit
Current flows like water through the circuit, notice how the 6 mA ‘stream of current’ splits to flow into the two resistors
George Washington University
Resistors in Parallel
1.5V
0V
R1 = .5K Ω R2 = .5K Ω
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 circuit It split equally between the
two legs, because the resistors have the same value
The current will split differently if the resistors are not equal…
I = 6 mA I = 3 m
A
I = 3 m
A
George Washington University
Resistors in ParallelThis is the equivalent
circuit
1.5V
0V
Req = .25K Ω
Use Ohm’s Law, we find the current through Req:
V = I x Req
1.5V = I x .25K Ω
Solve for I:
I = 1.5V / .25K Ω = 6 mA
I = 6 mA
The smaller the resistance in the circuit, the easier it is for current to flow
George Washington University
Including a DiodeSteps to Analyze the
Circuit
1.5V
0V
R = .5K Ω
First, is the anode potential at least 0.7V?
Anode = 1.5V
Yes, it is at 1.5V. So, replace the diode with a -0.7V DC Source.
George Washington University
Including a DiodeSteps to Analyze the
Circuit
1.5V
0V
R = .5K Ω
Voltage sources in series can be combined.
0.7V 1.5V + (-0.7)V = 0.8V
Use that 0.8V value as the V in Ohm’s Law!
George Washington University
Including a DiodeSteps to Analyze the
Circuit
0V
R = .5K Ω
Now, how much current will flow through R?
0.8VUse Ohm’s Law:
V = I x R
0.8V = I x .5K Ω
Solve for I:
I = 0.8V / .5 Ω = 1.6 mA
I = 1.6 mA
George Washington University
Including a DiodeCheck Your Answer
1.5V
0V
R = .5K Ω
The Voltage on the Left (From the DC Source) Should equal the Voltage Drops on the Right.
0.7V
Use Ohm’s Law For the Resistor:
VR = I x R0.8V = 1.6mA x .5K Ω
For the Diode:
VD = 0.7VAdd the Voltage Drops:
VR +VD = 0.8V+0.7V= 1.5V
This matches our voltage source…YAY!
George Washington University
Including a DiodeSteps to Analyze the
Circuit
0.5V
0V
R = .5K Ω
First, is the anode potential at least 0.7V?
Anode = 0.5V
No, it is at 0.5V. Therefore, no current can flow through the resistor.
I = 0 A
mp
s
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In Summary…
Ohm’s Law: V=IRDescribes the relationship between the voltage (V),
current (I), and resistance (R) in a circuit
Current is equal through two resistors in seriesVoltage drops across each resistorReq = R1 + R2 + . . .
Voltage is equal across two resistors in parallelCurrent splits through branches of parallel circuits 1/Req = 1/R1 + 1/R2
George Washington University
In Summary…
DiodesThere 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.
George Washington University
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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