George Washington University DC Circuits Lab ECE 002 Professor Ahmadi

George Washington University

DC Circuits Lab

ECE 002

Professor Ahmadi


Outline

Basic Components of a Circuit

Series Circuit

Parallel Circuit

Ohm’s Law

Lab Overview


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”


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)




…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




…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.


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

=


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


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


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


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


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


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


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


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


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


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


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


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


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.



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!



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


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!



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


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


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.


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!

Documents

George Washington University DC Circuits Lab ECE 002 Professor Ahmadi