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 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)

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”

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

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”

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 =

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

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

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

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

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

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

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

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

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

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

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…

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

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

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

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

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

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

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

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.

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!