Module 1 Circuit Loading - Purdue University

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

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

Circuit Loading

Based on load current.

Source circuit must deliver more current.

Increased loading effect on circuit.

IDEAL Voltage Supply – Fixed Output

+12 V−

1 mA 1 A

(c)(b)

open+12 V−

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

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.

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.

UNLOADED Real Voltage Supplies

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

Rsupply

Externalsupplyterminals

Rsupply

Vsupply

IsupplyEsupply Esupply

REAL Supply Resistance – Series Circuit

Esupply

Rsupply

Vsupply

Isupply

Iload +Vload−

+ VRsupply −20V1Ω

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

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

Module 2Thévenin Model

What’s in the Box

Lecture 9 Modules

1. Circuit Loading

4. Model Conversion

Why Model?

Model Complex Devices and

Circuits into Simple Circuit.

Simplify Circuit Analysis

Thévenin’s TheoremLinear

BilateralCircuit

Supplies, resistors,lamps, switches, etc.

Thévenin Model

Not diodes or LEDs !Not bilateral !

Thévenin’s Theorem

Thévenin Model

Look Familiar ?A real voltage source !

Measuring Thévenin Voltage

Open Circuit or No Load voltage

ETH = VOC = VNL

LinearBilateralCircuit

voltmeter

red lead

black lead

voltmeter

red lead

black lead

Finding Thévenin ResistanceTest Load Resistor: Measure two of the following

Iload, Vload, Rload

Then calculate RTH

LinearBilateralCircuit Rload

Iload+

Vload− ETH

Iload+

Vload−

+ VRTH −

Example – Find Thévenin VoltageOpen Circuit Voltage

Test Circuit

ETH = VOC = VNL = 10 V

LinearBilateralCircuit

voltmeter

red lead

black lead

Example – Find Thévenin Resistance

Now just a simple Series Circuit

Loaded circuit

Vload = 8 V

Rload = 4 kΩ

Iload +Vload

8 V−

+ VRTH −

Iload = 8V / 4kΩ = 2 mA

VRTH = 10 V – 8 V = 2 V

RTH = 2V / 2mA = 1 kΩ

Example –Thévenin Model

Model works for any load !

Attach any load and find Iload & Vload

1 k RTH

+ VRTH −

Example – Use Thévenin Model

Attach 1kΩ load.Find

Vload & Iload

1 k RTH

Iload +Vload

+ VRTH −

5 mA +

Vload = 1kΩ • 5mA = 5 V

Same load values as if solved with original circuit

Find Thévenin Model

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

Find ETH

Find Thévenin Model

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

Find RTH

Find Thévenin ResistanceLoaded circuit

Iload = 4 mA

Rload = 2 kΩ

Iload +Vload

8 V−

+ VRTH −

Vload = 4mA × 2kΩ = 8 V

VRTH = 12 V – 8 V = 4 V

RTH = 4V / 4mA = 1 kΩ

12V 2k8V

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.

Example – Use Thévenin Model

Same load value as if solved with original circuit

Attach 5kΩ load.Find Iload

1 k RTH

Iload +Vload

+ VRTH −

12V 5k

Use Thévenin Model

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

Use Thévenin Model

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

Module 3 – Intro onlyWatch Class Video

Norton Model

Lecture 9 Modules

1. Circuit Loading

4. Model Conversion

Norton Model

Original circuit Norton Model

ElectronicCircuit IN

IN = original circuit’s short circuit current

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

Norton Model

• Valid for any load• Same resulting load

voltages and currents

Original circuit Norton Model

ElectronicCircuit IN

IN Norton Current

ElectronicCircuit

ShortISC Circuit

Current

Careful

IN Norton Current

ElectronicCircuit

ShortISC Circuit

Current

IN = ISC

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

RN Norton Resistance

ElectronicCircuit RN

RN = RTH

Ohm’s Law – find RN

RN = VOC / ISC

ElectronicCircuit

RN = ETH / IN

Norton Model Analysis

ElectronicCircuit

NortonModel

Same IL and VL

Series-Parallel Circuit Analysis - example

6Ω24V+

R// = 3Ω // 6Ω = 2Ω

+VL−

6Ω24V+

2 ΩR// = 3Ω // 6Ω = 2Ω

+VL−

VL = 2 Ω2Ω+6Ω

× 24V = 𝟔𝟔𝟔𝟔

𝟔𝟔𝟔𝟔

𝑉𝑉𝑉𝑉𝑉𝑉

6Ω24V+

IL +VL−

IL = 6V / 6Ω = 𝟏𝟏𝟏𝟏

𝟔𝟔𝟔𝟔𝟏𝟏𝟏𝟏

Could you now find the rest of the circuit values?

Norton Model Analysis - example

Find IL and VL using the Norton model approach !

6Ω24V+

IN Norton Model Current - example

IN = ISC

24VShortcircuitcurrent

= 24V / 6Ω = 4A

ETH Thévenin Voltage - example

ETH = VOC

24V+VOC-

VDR – Voltage Divider Rule

= 3 Ω3Ω+6Ω

× 24V = 𝟖𝟖𝟔𝟔

RN Norton Resistance - example

RN = ETH / IN

= 2ΩRN = 8V / 4A

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

RN = 6 Ω // 3 Ω = 2 Ω

Norton Model – our example values

NortonModelIN

4A2Ω RN

Norton Model - with load attached

RT = 2Ω // 6Ω = 1.5 Ω

VL = 4A × 1.5Ω = 6V

6Ω4A 2Ω+

model load

Norton Model - with load attached - CDR

6Ω4A 2Ω+

model load

2Ω2Ω+6Ω

× 4Α = 1Α

2 BranchCDR – Current Divider Rule

ISC = 40 V / 20 Ω = 2A

6Ω24V

540V 4Q9.7 Find ISC (A)

VOC = V5Ω =

6Ω24V

540V 4Q9.8 Find VOC

( 5 Ω / 25 Ω ) 40 V = 8 V

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

RN = 20 Ω // 5 Ω = 4 Ω

6Ω24V

540V 4Q9.9 Find RN

RN Recall ISC = 2AVOC = 8V

6Ω24V

Norton Model Analysis - exampleIL20

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

Norton Model - with load attached

IL = 4Ω / 8Ω × 2A = 1A 2-branch CDR

VL = 1A × 4Ω = 4V

6Ω2A 4Ω+

model load

Module 4 – Watch Class Video

Source Conversion

Lecture 9 Modules

1. Circuit Loading

4. Model Conversion

Equivalent Models - supply conversion

Thévenin model Norton model

Same loadresults !

RTH & RN - supply conversion

Zero the sources

Thevenin model Norton model

openRTH RN

RTH = RN

IN - supply conversion

Short Circuit Current

IN = ETH / RTH

ETH = IN × RN

Open Circuit Voltage

ETH - supply conversion

Thevenin Resistance

RTH = RN = VOC / ISC

ElectronicCircuit

Warning - SHORT can cause smoke or loading !

Equivalent Circuit - source conversion

RTH = RN = 6 kΩETH = 3mA × 6 kΩ = 18V

R16 kΩ3 mA

R2 3 k

Convert to Thevenin

Example model - source conversion

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

R2 3 k

1 mΑ +3V-

+12V−

Thevenin Model

Switch Back to Original Circuit

IR1 = 3 mA - 1mA = 2 mA

R16 kΩ3 mA

R2 3 k

1 mΑ +3V-

VR1 = VIsupply = 2mA × 6kΩ = 12V

Module 5

What’s in the BoxPractical Measurements

Lecture 9 Modules

1. Circuit Loading

4. Model Conversion

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

ElectronicCircuit

Removeload

ElectronicCircuit

-voltmeter

1. RTH Measurements - Using VOC and ISC

careful

ElectronicCircuit ammeter

ammeter SHORT

ISC not always possible !

RTH = VOC / ISC

2. RTH Measurements - matched load

ElectronicCircuit

voltmeterElectronicCircuit

+½ ETH

Attach & adjust pot. VPOT = ½ ETH

2. RTH Measurements - matched load

+½ ETH-

+ ½ ETH -

Measure Potentiomenter with Ohmmeter RPOT = RTH

3. RTH Measurements - unmatched load

voltmeterElectronic

Circuit

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

+VL−

+ VRth −

Series Circuit Analysis

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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