Module 1 Circuit Loading - Purdue University

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Purdue University© ECET 17700 DAQ & Systems Control

Electrical Engineering TechnologyECET 17700 - DAQ & Control Systems

Lecture # 9 – Loading, Thévenin Model & Norton Model

Professors Robert Herrick & J. Michael Jacob

Module 1

Circuit Loading

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Lecture 9 Modules

1. Circuit Loading

2. Thévenin Circuit Model – What’s in the Box

3. Norton Circuit Model

4. Model Conversion

5. What’s in the Box – Practical Measurements

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Circuit Loading

Based on load current.

Iload

Source circuit must deliver more current.

Rload

Increased loading effect on circuit.

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IDEAL Voltage Supply – Fixed Output

E12 V

R12 k

+12 V−

E12 V

R12

+12 V−

1 mA 1 A

(c)(b)

E12 V

open+12 V−

0 mA

(a)

E = 12 V with varying loads (a) I = 0 A(b) I = 1 mA(c) I = 1 A

Ideal source maintains E with varying loads

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Never SHORT a Voltage Supply

I = 12V / 0Ω = ∞AE 0 V

∞ AE

12 V Bad News? Short 0V

Excessive Loading. Many power supplies current limited.

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Never SHORT a Voltage Supply

Watch your P/S voltmeter when hooking up circuit.

Voltage suddenly drops toward 0 V.Current suddenly pegs.

Driving SHORT circuit.

Turn power supply off.

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UNLOADED Real Voltage Supplies

VNL - No Load VoltageVOC - Open Circuit VoltageVsupply = VNL = VOC = Esupply

Rsupply

Externalsupplyterminals

Rsupply

+

Vsupply

−

IsupplyEsupply Esupply

= ETH

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REAL Supply Resistance – Series Circuit

Esupply

Rsupply

+

Vsupply

−

Isupply

Rload

Iload +Vload−

+ VRsupply −20V1Ω

100Ω

Isupply = Iload =

Vsupply = Vload =

VRsupply = LOST

198 mA

19.8 V0.2V

198mΑ × 100 Ω = 19.8V

20V / 101Ω = 198mΑ

20V – 19.8V = 0.2V

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Electrical Engineering TechnologyECET 17700 - DAQ & Control SystemsLecture # 9 – Loading, Thévenin & Norton


Module 2Thévenin Model

or

What’s in the Box

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Lecture 9 Modules

1. Circuit Loading



4. Model Conversion


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Why Model?

Model Complex Devices and

Circuits into Simple Circuit.

Simplify Circuit Analysis

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Thévenin’s TheoremLinear

BilateralCircuit

Supplies, resistors,lamps, switches, etc.

Thévenin Model

ETH

RTH

Not diodes or LEDs !Not bilateral !

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Thévenin’s Theorem

Thévenin Model

ETH

RTH

Look Familiar ?A real voltage source !

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Measuring Thévenin Voltage

Open Circuit or No Load voltage

ETH = VOC = VNL

LinearBilateralCircuit


voltmeter

red lead

black lead

ETH

RTH

voltmeter

red lead

black lead

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Finding Thévenin ResistanceTest Load Resistor: Measure two of the following

Iload, Vload, Rload

Then calculate RTH

LinearBilateralCircuit Rload

Iload+

Vload− ETH

RTH

Rload

Iload+

Vload−

+ VRTH −

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Example – Find Thévenin VoltageOpen Circuit Voltage

Test Circuit

+

10 V

−

ETH = VOC = VNL = 10 V

LinearBilateralCircuit


voltmeter

red lead

black lead

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Example – Find Thévenin Resistance

Now just a simple Series Circuit

Loaded circuit

Vload = 8 V

Rload = 4 kΩ

ETH

10 V

RTH

Rload

4 k

Iload +Vload

8 V−

+ VRTH −

2 mA

2V

Iload = 8V / 4kΩ = 2 mA

VRTH = 10 V – 8 V = 2 V

RTH = 2V / 2mA = 1 kΩ

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Example –Thévenin Model

Now

Model works for any load !

Attach any load and find Iload & Vload

ETH

10 V

1 k RTH

+ VRTH −

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Example – Use Thévenin Model


Attach 1kΩ load.Find

Vload & Iload

ETH

10 V

1 k RTH

Rload

1 k

Iload +Vload

−

+ VRTH −

5 mA +

5 V

-

Vload = 1kΩ • 5mA = 5 V

Same load values as if solved with original circuit

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Find Thévenin Model

Open circuit voltage = 12 VA 2 kΩ load produces 4 mA

Find ETH

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Find Thévenin Model

Open circuit voltage = 12 VA 2 kΩ load produces 4 mA

Find RTH

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Find Thévenin ResistanceLoaded circuit

Iload = 4 mA

Rload = 2 kΩ

ETH

10 V

RTH

Rload

4 k

Iload +Vload

8 V−

+ VRTH −

4 mA

4V

Vload = 4mA × 2kΩ = 8 V

VRTH = 12 V – 8 V = 4 V

RTH = 4V / 4mA = 1 kΩ

12V 2k8V

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Use Thévenin Model

Draw the Thévenin model from previous two quiz questions with ETH = 12V and RTH = 1kΩ

A 5 kΩ load is attached.

Find the circuit load current.

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Example – Use Thévenin Model


Same load value as if solved with original circuit

Attach 5kΩ load.Find Iload

ETH

10 V

1 k RTH

Rload

1 k

Iload +Vload

−

+ VRTH −

12V 5k

2 mA

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Use Thévenin Model

Q9.6

Circuit loading increases asA. Load resistance increasesB. Load current increasesC. None of the aboveD. All of the above

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Use Thévenin Model

Circuit loading increases asA. Load resistance increasesB. Load resistance decreasesC. None of the aboveD. All of the above

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Module 3 – Intro onlyWatch Class Video

Norton Model

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Lecture 9 Modules

1. Circuit Loading



4. Model Conversion


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Norton Model

Original circuit Norton Model

ElectronicCircuit IN

RN

IN = original circuit’s short circuit current

RN = original circuit’s output resistance with sources 0’d

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Norton Model

• Valid for any load• Same resulting load

voltages and currents

Original circuit Norton Model

ElectronicCircuit IN

RN

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IN Norton Current

ElectronicCircuit

ShortISC Circuit

Current

Careful

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IN Norton Current

ElectronicCircuit

ShortISC Circuit

Current

IN = ISC

Note: If ISC is down, current source must be up.IN

RNISC

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RN Norton Resistance

ElectronicCircuit RN

RN = RTH

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Ohm’s Law – find RN

RN = VOC / ISC

ElectronicCircuit

+VOC-

ElectronicCircuit

ISC

RN = ETH / IN

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Norton Model Analysis

ElectronicCircuit

NortonModel

Load+

VL-

IL

Same IL and VL

Load+

VL-

IL

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Series-Parallel Circuit Analysis - example

3Ω+-

6Ω

6Ω24V+

VL-

IL

R// = 3Ω // 6Ω = 2Ω

+VL−

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3Ω+-

6Ω

6Ω24V+

VL-

IL

2 ΩR// = 3Ω // 6Ω = 2Ω

+VL−

VL = 2 Ω2Ω+6Ω

× 24V = 𝟔𝟔𝟔𝟔

𝟔𝟔𝟔𝟔

𝑉𝑉𝑉𝑉𝑉𝑉

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3Ω+-

6Ω

6Ω24V+

VL-

IL +VL−

IL = 6V / 6Ω = 𝟏𝟏𝟏𝟏

𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏

Could you now find the rest of the circuit values?

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Norton Model Analysis - example

Find IL and VL using the Norton model approach !

3Ω+-

6Ω

6Ω24V+

VL-

IL

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IN Norton Model Current - example

IN = ISC

3Ω+-

6Ω

24VShortcircuitcurrent

ISC

= 24V / 6Ω = 4A

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ETH Thévenin Voltage - example

ETH = VOC

3Ω+-

6Ω

24V+VOC-

VDR – Voltage Divider Rule

= 3 Ω3Ω+6Ω

× 24V = 𝟖𝟖𝟔𝟔

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RN Norton Resistance - example

RN = ETH / IN

= 2ΩRN = 8V / 4A

RTH =

Or simply find RN directly since E supply acts like short.

3Ω+-

6Ω

24V

RN

RN = 6 Ω // 3 Ω = 2 Ω

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Norton Model – our example values

NortonModelIN

RTH

4A2Ω RN

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Norton Model - with load attached

RT = 2Ω // 6Ω = 1.5 Ω

VL = 4A × 1.5Ω = 6V

6Ω4A 2Ω+

VL-

IL

model load

1 A

6V

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Norton Model - with load attached - CDR

IL =

6Ω4A 2Ω+

VL-

IL

model load

1 A

2Ω2Ω+6Ω

× 4Α = 1Α

2 BranchCDR – Current Divider Rule

6V

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ISC = 40 V / 20 Ω = 2A

3Ω+-

6Ω

6Ω24V

ISC20

540V 4Q9.7 Find ISC (A)

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VOC = V5Ω =

3Ω+-

6Ω

6Ω24V

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540V 4Q9.8 Find VOC

+VOC-

( 5 Ω / 25 Ω ) 40 V = 8 V

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RN = RTH = ΕΤΗ / ΙΝ = 8 V / 2 A = 4 Ωor simply

RN = 20 Ω // 5 Ω = 4 Ω

3Ω+-

6Ω

6Ω24V

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540V 4Q9.9 Find RN

RN Recall ISC = 2AVOC = 8V

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3Ω+-

6Ω

6Ω24V

Norton Model Analysis - exampleIL20

540V 4Q9.10 Use Nortonmodel, attach load, and find IL

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Norton Model - with load attached

IL = 4Ω / 8Ω × 2A = 1A 2-branch CDR

VL = 1A × 4Ω = 4V

6Ω2A 4Ω+

VL-

IL

model load

4Ω

1A

4 V

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Module 4 – Watch Class Video

Source Conversion

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Lecture 9 Modules

1. Circuit Loading



4. Model Conversion


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Equivalent Models - supply conversion

+-ETH

RTH

INRN

Thévenin model Norton model

Same loadresults !

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RTH & RN - supply conversion

Zero the sources

+-ETH

RTH

INRN

Thevenin model Norton model

openRTH RN

RTH = RN

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IN - supply conversion

Short Circuit Current

+-ETH

RTH

INRN


IN = ETH / RTH

ISC

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ETH = IN × RN

+-ETH

RTH

INRN


+ETH

−

Open Circuit Voltage

ETH - supply conversion

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Thevenin Resistance

RTH = RN = VOC / ISC

ElectronicCircuit

+VOC-

ElectronicCircuit

ISC

Warning - SHORT can cause smoke or loading !

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Equivalent Circuit - source conversion

RTH = RN = 6 kΩETH = 3mA × 6 kΩ = 18V

R16 kΩ3 mA

R2 3 k

9 V

Convert to Thevenin

+

−

18 V

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Example model - source conversion

ENET = 18V − 9V = 9VRT = 6 kΩ + 3 kΩ = 9 kΩIR2 = 9V / 9kΩ = 1 mA

6 kΩ

18 V

R2 3 k

9 V

1 mΑ +3V-

+12V−

Thevenin Model

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Switch Back to Original Circuit

IR1 = 3 mA - 1mA = 2 mA

R16 kΩ3 mA

R2 3 k

9 V

1 mΑ +3V-

+12V-

2 mΑ

VR1 = VIsupply = 2mA × 6kΩ = 12V

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Module 5

What’s in the BoxPractical Measurements

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Lecture 9 Modules

1. Circuit Loading



4. Model Conversion


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Thévenin Resistance Measurements

Four basic techniques

1. Measure ETH & IN, then RTH = ETH / IN

2. Live circuit, matched load3. Live circuit, partial loading4. Dead circuit, Ohmmeter with supplies zeroed

Not all methods work for all circuits

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ElectronicCircuit

Removeload

Load+

VL-

IL

ElectronicCircuit

+ETH

-voltmeter

1. RTH Measurements - Using VOC and ISC

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careful

IN

ElectronicCircuit ammeter

ammeter SHORT


ISC not always possible !

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RTH = VOC / ISC

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2. RTH Measurements - matched load

ElectronicCircuit

+ETH-

voltmeterElectronicCircuit

+½ ETH

-

Attach & adjust pot. VPOT = ½ ETH

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2. RTH Measurements - matched load

+-ETH

RTH

+½ ETH-

+ ½ ETH -

Measure Potentiomenter with Ohmmeter RPOT = RTH

RPOT

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3. RTH Measurements - unmatched load

voltmeterElectronic

Circuit

+VL

-

IL = IRth = VL / RLVRth = ETH – VLRTH = VRth / IL

RL

+-ETH

RTH

RL

+VL−

IL

+ VRth −

Series Circuit Analysis

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4. RTH Measurements – circuit not powered

ohmmeterElectronicCircuit

Replace sources with equivalent resistances

• Current source Replace with Open

• Voltage source Replace with Short

• Measure at output terminals with ohmmeter

Module 1 Circuit Loading - Purdue University

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