1. DC Circuits Lab
ECE 002
Professor Ahmadi

2. Outline Basic Components of a Circuit Series Circuit Parallel Circuit
Ohm’s Law
Lab Overview

3. Basic Circuit Components
We represent real electrical components with symbols
1.5 V
1.5V
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)

4. Basic Circuit Components
We represent real electrical components with symbols
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)
A Light Bulb…or any ‘device’…
R Ω
…can be represented with this symbol
…called a “resistor”

5. Basic Circuit Components
We represent real electrical components with symbols
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
The Earth…
…can be represented with this symbol
…called the “ground” symbol

6. Basic Circuit Components
We represent real electrical components with symbols
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.
A Tollbooth…or any ‘barrier’
…can be represented with this symbol
…called a “diode”

7. 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.
Open Circuit
When the Anode is less than ~0.7V, the diode is an open circuit. This means no current can flow through it! =
0.7V =

8. 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

9. Building a Circuit… creating a schematic
Replace the battery with a ‘DC Voltage Source’ symbol
1.5 V
1.5V
Replace the light bulb with a ‘Resistor’ symbol
.5 Ω
Mark the symbol’s values (V=, R=, etc.)
Since this “node” is at GND (OV) this node
must be 1.5Volts higher
Add the Ground reference 0V
Instead…let's represent the real components with their symbols

10. Analyzing the Circuit…using Ohm’s Law
When we attach the resistor to the DC voltage source, current begins to flow
1.5V
How much current will flow?
.5K Ω
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 through our circuit 0V

11. Analyzing the Circuit…using Ohm’s Law
How much current will flow?
I = 3 mA
1.5V
Use Ohm’s Law:
V = I x R
1.5V = I x .5K Ω
Solve for I:
I = 1.5V / .5 KΩ = 3 mA
.5K Ω 0V
So, 3 mA will flow through the .5kΩ resistor, when 1.5 Volts are across it

12. Resistors in Series
Resistors connected by only 1 terminal, back-to-back, are considered to be in ‘series’
R1 = .5K Ω
1.5V
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
Req = 1K Ω
R2 = .5K Ω 0V

13. Resistors in Series
Now we can find the current through the circuit using Ohm’s Law
I = 1.5 mA
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
1.5V
Req = 1K Ω 0V
The bigger the resistance in the circuit, the harder it is for current to flow

14. Resistors in Series
Back to our original series circuit, with R1 and R2
The current is the SAME through each resistor
I = 1.5 mA
R1 = .5K Ω
Current flows like water through the circuit, notice how the 1.5 mA ‘stream of current’ flows through both resistors equally
1.5V
R2 = .5K Ω
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 0V

15. Resistors in Parallel
Resistors connected at 2 terminals, sharing the same node on each side, are considered to be in ‘parallel’
1.5V
Unlike before, we cannot just add them. We must add their inverses to find Req:
R1 = .5K Ω
R2 = .5K Ω
Req = .25K Ω 0V

16. Resistors in Parallel This is the equivalent circuit
I = 6 mA
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
1.5V
Req = .25K Ω 0V
The smaller the resistance in the circuit, the easier it is for current to flow

17. 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
1.5V
Current flows like water 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

18. Resistors in Parallel 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
I = 6 mA
I = 3 mA
I = 3 mA
1.5V
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
R1 = .5K Ω
R2 = .5K Ω 0V
The current will split differently if the resistors are not equal…

19. Resistors in Parallel This is the equivalent circuit
I = 6 mA
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
1.5V
Req = .25K Ω 0V
The smaller the resistance in the circuit, the easier it is for current to flow

20. Including a Diode Steps to Analyze the Circuit
Anode = 1.5V
First, is the anode potential at least 0.7V?
1.5V
Yes, it is at 1.5V. So, replace the diode with a -0.7V DC Source.
R = .5K Ω 0V

21. Including a Diode Steps to Analyze the Circuit
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!
1.5V
R = .5K Ω 0V

22. Including a Diode Steps to Analyze the Circuit
Now, how much current will flow through R?
I = 1.6 mA
0.8V
Use Ohm’s Law:
V = I x R
0.8V = I x .5K Ω
Solve for I:
I = 0.8V / .5 Ω = 1.6 mA
R = .5K Ω 0V

23. Use Ohm’s Law For the Resistor:
Including a Diode
Check Your Answer
The Voltage on the Left (From the DC Source) Should equal the Voltage Drops on the Right.
0.7V
1.5V
Use Ohm’s Law For the Resistor:
VR = I x R
0.8V = 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 source…YAY!
R = .5K Ω 0V

24. 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
No, it is at 0.5V. Therefore, no current can flow through the resistor.
I = 0 Amps 0V

25. 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 + R
Voltage is equal across two resistors in parallel
Current splits through branches of parallel circuits
1/Req = 1/R /R2

26. 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.

27. 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!