ELECTRIC CIRCUITSECSE-2010

Spring 2003 Class 7

ASSIGNMENTS DUE• Today (Tuesday/Wednesday):

• Will do Experiment #2 In Class (EP-2)• Activities 7-1, 7-2, (In Class)• Today is a Very Busy Day!

• Thursday:• Will do Experiment #3 In Class (EP-3)• Activities 8-1, 8-2, (In Class)

• Next Monday:• Homework #3 Due• Activity 9-1, Op-Amp ILM

REVIEW• Node Equations:

• Technique to Solve Any Linear Circuit• Label Unknown Node Voltages, v1, v2, v3, etc.• Write KCL at Each Unknown Node • Sum of Currents Out of Node= 0• Relate Currents to Node Voltages using Ohm’s

Law• Always get same number of equations as

unknowns• Solve linear, algebraic equations for v1, v2, v3,

etc.

REVIEW• Mesh Equations:

• Another Technique to Solve Any Linear Circuit• Define All Mesh Currents• Unknown Mesh Currents (i1, i2, i3, etc.) and Current

from Current Sources (Independent and Controlled)• Write a KVL around Each Unknown Mesh Current• Sum of Voltages around Mesh due to all Mesh

Currents = 0• Relate Voltages to Mesh Currents using Ohm’s Law• Always get same number of equations as unknowns• Solve linear, algebraic equations for i1, i2, i3, etc.

REVIEW• Add Controlled Sources:

• Things get harder!• Must find a Constraint Equation:

• Relate Controlling Voltage or Current to Unknown Node Voltages or Unknown Mesh Currents

• Must Do By Inspection; THINK!• Find Constraint Equation directly from circuit• Try Ohm’s Law, KCL, KVL, etc.• It’s there; you just have to find it!

NODE OR MESH?

• # Unknown Node Voltages = # Nodes - # Voltage Sources - 1 (Ref):

• # Unknown Mesh Currents = # Meshes - # Current Sources:

• Choose the Method with the Smaller # of Unknowns:

• If Same: Choose Method that Gives Desired Output (Node Equations for v or Mesh Equations for i):

MORE CIRCUIT ANALYSIS TECHNIQUES

• Source Conversions:• Will Use Many Times During Course

• Superposition:• Will Practice with Activity 7-1

• Thevenin and Norton Equivalent Circuits:• Will Practice with Activity 7-2 today and

Activity 8-1 Tomorrow• Maximum Power Transfer:

• Tomorrow – Activity 8-2 using PSpice

SOURCE CONVERSIONS

• Any Voltage Source, vs, in Series with a Resistor, Rs, May be Replaced by a Current Source, is = vs / Rs, in Parallel with Same Resistor, Rs: (And Vice Versa)• Sometimes a Useful Technique• Often Simplifies a Problem• Not Always the Best Technique to Use• Be Very Careful with Controlled Sources• Sometimes “lose” controlling v, i, when make

source conversion

SOURCE CONVERSIONS

• Define v, i using Active Convention

Sv

SR

v

i

Si SR

i

v

SS

S

viR

Sometimes Source Conversions Make Circuits Easier to Solve

Not Always!

EXAMPLE

2 A 4

6

x3vxv

xv

6

x3v

4

2 x 4 8 V

Source Conversion

2 Meshes

1 Mesh

xBut Where is v ?

Be Careful!

SOURCE CONVERSIONS• Source Conversion Reduces Ckt from 2

Meshes to 1 Mesh in Example:• Source Conversion Appears to Simplify Ckt

• vx is No Longer the Voltage Across the 4 Ohm Resistor:• vx still exists in this case• Just need to be careful• Sometimes controlling v,i will completely

disappear

SUPERPOSITION• Technique to use when there is more than

1 Independent Source in a circuit:• Not always the best technique to use• Will learn lots of techniques; Experience

helps us learn which technique to choose• Find Output due to each independent

source with all other independent sources set = 0; then Add to find Total Output:• Source of 0 is called a “dead source”• “Dead” voltage source = 0 V = Short Circuit• “Dead” current source = 0 A = Open Circuit

SUPERPOSITION• Total Output = Sum of all Outputs due to

each independent source with all other independent sources “dead”:• Simply Add them • Works only for Linear Circuits; Only kind we

will consider• Leave Controlled Sources Alone!!:

• Find Outputs due to Independent Sources only!!

• Do NOT set Controlled Sources = 0!!• Will Practice with Activity 7-1

ACTIVITY 7-16

6 5 xv

1i

xv A2

16 V

10 A1Find i using SuperpositionTwo Independent Sources

ACTIVITY 7-1

• 2 Independent Sources:• Output defined as i1

• Need to find i1 due to each Independent Source with other Independent Source set = 0

• Add i1-1 + i1-2 = i1

ACTIVITY 7-1

• Let’s First Make the Current Source Active and the Voltage Source “Dead”:• Set 16 V Source = 0 => Short Circuit• Find i1-1

ACTIVITY 7-16

6 5 x-1v

x-1v A2

1-1i

10 ACurrent Source ActiveVoltage Source "Dead"

6 6 3

x-1v 10 A x 3 30 V

x-11-1

vKCL at a: 10 i2

a10 A

ACTIVITY 7-16

6 5 x-1v

x-1v A2

1-1i

10 ACurrent Source ActiveVoltage Source "Dead"

1-1i 10 15 5 A

ACTIVITY 7-1

• Current Source Active:• Set 16 V source = 0 => Short Circuit• 6 // 6 = 3 ohms• vx-1 = 3 ohms x -10 Amps = - 30 Volts• KCL at a: 10 + (-15) = i1-1 => i1-1 = - 5 Amps

ACTIVITY 7-1

• Now Make the Voltage Source Active and the Current Source “Dead”:• Set 10 A source = 0 => Open Circuit• Find i1-2

ACTIVITY 7-16

6 5 16 Vx-2v A2

1-2i

x-2v

1-2 x-2i v 4 A

i 0

Voltage Source ActiveCurrent Source "Dead"

x-26v (16) 8 V

6 6

ACTIVITY 7-1• Voltage Source Active:

• Set 10 A source = 0 => Open Circuit• i = 0; => i1-2 = vx / 2• Voltage Divider Rule• vx-2 = [6/(6+6)] 16 = 8 V• i1-2 = vx-2 / 2 = 4 Amps

• Total Output:• i1-1 + i1-2 = i1 = - 5 + 4 = - 1 Amp

THEVENIN EQUIVALENT CIRCUITS• SOURCE Network:

• At Least 1 Independent Source• R’s, Controlled Sources• Define v, i at Terminals Using Active Convention

• Thevenin’s Theorem:• Any Source Network can be replaced by a Voltage Source,

voc, in Series with a Resistor, RT

• voc is called the “Open Circuit Voltage”• Sometimes also called the “Thevenin Voltage”, vT

• voc = vT = v when i = 0• RT is called the “Thevenin Resistance”• Thevenin Equivalent Circuit

THEVENIN EQUIVALENT CIRCUIT

• Define v, i using Active ConventionSource Network(At least 1 Independent Source)

v v

i

i

ocv

TR

ocv Open Circuit Voltage

TR Thevenin Resistance

NORTON EQUIVALENT CIRCUIT• Norton’s Theorem:

• Any Source Network can be replaced by a Current Source, isc, in Parallel with a Resistor, RT

• isc is called the “Short Circuit Current”• isc = i when v = 0• RT is the same Thevenin Resistance• Norton Equivalent Circuit

• RT = voc / isc:• RT = Req of “Dead Source Network”:

• Source Network with Independent Sources set = 0• Dead Source Network is a Load Network

NORTON EQUIVALENT CIRCUIT

• Define v, i using Active Convention

Source Network

i

v

sciTR v

i

sci Short Circuit Current

TR Thevenin Resistanceoc

scT

viR

THEVENIN/NORTON CIRCUITS

T eqThevenin Resistance R R of Dead Network

Dead Source Network

Source Network withIndependent Sources Set 0

eqR

v

i

vi

THEVENIN/NORTON CIRCUITS

• Find voc, isc, RT using Ckt Analysis:• Need only find 2 out of 3• RT = voc / isc

• Let’s Practice with Activity 7-2

ACTIVITY 7-2

oc sc T

Find Thevenin Parametersv , i , R

4k

12 k

xv

x6v V

v

i

Replace everything to theleft of the terminals with itsThevenin Equivalent Circuit

2 mA

ACTIVITY 7-2

• Find voc:• Open Circuit the Output: i = 0• Defines voc

ACTIVITY 7-2

2 mA 4k

12 k

xv

x6v V

i 0

ocv

oc v 40 V

oc x x x v v 6v 5v

x v 2 4 8 V

Let's Use Units of mA, k and V

Open Circuit the Output

ACTIVITY 7-2

• Find voc:• Open Circuit the Output: i = 0• Defines voc

• KVL: voc = vx - 6 vx = - 5 vx

• vx = 2 mA x 4 kohms = 8 V• voc = - 5 x 8 = - 40 V

ACTIVITY 7-2

• Find isc:• Short Circuit the Output: v = 0• Defines isc

ACTIVITY 7-2

2 mA4 k

xv

x6v

12 k

sciv 0

x scKCL: i 2 i

sci 5 mA

xi

x scv 4(2 i ) x x scKVL: v 6v 12i 0

Short Circuit the Outputsci

ACTIVITY 7-2• Find isc:

• Short Circuit the Output: v = 0• Defines isc

• KCL: ix = 2 - isc

• Also: vx = 4 ix = 8 - 4 isc

• KVL: vx - 6 vx - 12 isc = 0• - 5 vx - 12 isc = 0• - 40 + 20 isc = 12 isc => 8 isc = 40• isc = 5 mA

ACTIVITY 7-2• Find RT:

• Create Dead Source Network• Independent Sources = 0• Leave Controlled Sources Alone• Now have Load Network• Find Req = RT

• Connect Test Voltage vt

• Define it using Active Convention• Req = vt / it

ACTIVITY 7-2

4 kxv

x6v

12 k

tv

ti

t oct

t sc

v vR 8 k

i i x tv 4i

x x t tv 6v 12i v

Dead Source Network

ACTIVITY 7-2

• vx = 4 it

• KVL: vx - 6 vx + 12 it - vt = 0• - 5 vx + 12 it = vt

• - 20 it + 12 it = vt = - 8 it

• vt / it = Req = RT = - 8 kohms• Check: RT = voc / isc

• RT = - 40 V / 5mA = - 8 k OK

BRIDGE CIRCUITS

• Bridge Circuits are circuits used to accurately measure Circuit Elements:• A Wheatstone Bridge is used to Measure

Resistance• Will Use a Maxwell Bridge later to Measure

Inductance

WHEATSTONE BRIDGE

3Adjust R such that VM reads 0

"Balancing the Bridge"Connect Voltmeter across "Bridge"

uAccurate Measurement of R

v

1R 2R

3R uR

VM

WHEATSTONE BRIDGE• See Circuit:

• 3 Accurately Known Resistors; 1 Variable• 1 Unknown Resistor; To be Measured• Connect Voltmeter Across “Bridge”

• Adjust R3 to Balance Bridge:• im = 0; vm = 0• (R1 + R3) i1 = (Ru + R2) i2; • R3 i1 = Ru i2 and R1 i1 = R2 i2

WHEATSTONE BRIDGE

When Balanced:

v

1R 2R

3R uR

VM1i 2i

mi 0

1 1 3 2 2 u

1 3 2 u 1 1 2 2

i (R R ) i (R R )i R i R ; i R i R

m v 0

WHEATSTONE BRIDGE

• Solve for Ru:• Ru = (i1 / i2) R3

• => Ru = (R2 / R1) R3 = R2R3 / R1

• Very Accurate Method of Measuring an Unknown Resistance:• Wheatstone Bridges can be very expensive• Will build an “inexpensive” Wheatstone

Bridge in Exp. 2b today

2 3u

1

R RRR

EXPERIMENT 2a

uR

i

v

u

1SlopeR

Best Slope when have Random Errors Method of Least Squares

i

v

EXPERIMENT 2a

• Determine the Value of an Unknown Resistor:• Ru is in Plastic Box in Drawer

• Statistical Treatment of Data:• No measurements are “perfect”• Need to know how to handle “errors”• See Notes in Supplement

EXPERIMENT 2a• Take Several Measurements of v

Across Ru and i Thru Ru:• Will Measure voltage v across Ru directly • Will Measure i thru Ru indirectly• Will Measure voltage vPot across Pot and the

resistance RPot of the Pot using MM• Calculate i = vPot/RPot

• Plot i vs v; (or v vs i) and Use Statistics to calculate “best” value for Ru

• Use 5-6 Significant Digits in all your Measurements!

EXPERIMENT 2a• Know that i vs v for a Resistor should be

Linear and go through i = v = 0:• Slope = 1 / Ru: (Slope = Ru if Plot v vs i)• “Best” Slope When Have Random Errors

=> Method of Least Squares:• See Notes on Statistical Treatment of Data

• Take Data Today - Do Analysis Later

EXPERIMENT 2aPot

Pot

viR

20 V

uR

PotR

v

Potv

i

Digital Pot, 0 10 k

Unknown Resistor

Pot Pot

Pot

Pot

Measure v, v , RvCalculate iR

Plot i vs. v (or v vs. i)

EXPERIMENT 2b

• Wheatstone Bridge:• “Cheap” Version; Not Very Accurate • Use 100k Pot for R2; 10k Pot for R3

• Use same Ru as Exp 2a

EXPERIMENT 2b

3Adjust R to "Balance the Bridge" 2 3u

1

R RDetermine R

R

v

1R 1k 2R 100k Pot

3R 10k Pot uR

VM

Multimeter10k, 20k, 1k, .1k

EXPERIMENT 2b

• Try 4 Different Values for R2:• 10k, 20k, 1k, 0.1k • R2 Affects “Range” of Measurement• Need to be able to make R2 R3 / R1 = Ru

• If R2 is too Small; Cannot Balance Bridge