module 1 circuit loading - purdue university
TRANSCRIPT
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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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Circuit Loading
Based on load current.
Iload
Source circuit must deliver more current.
Rload
Increased loading effect on circuit.
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Electrical Engineering TechnologyECET 17700 - DAQ & Control SystemsLecture # 9 – Loading, Thévenin & Norton
Professors Robert Herrick & J. Michael Jacob
Module 2Thévenin Model
or
What’s in the Box
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Why Model?
Model Complex Devices and
Circuits into Simple Circuit.
Simplify Circuit Analysis
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Purdue University© ECET 17700 DAQ & Systems Control
Thévenin’s TheoremLinear
BilateralCircuit
Supplies, resistors,lamps, switches, etc.
Thévenin Model
ETH
RTH
Not diodes or LEDs !Not bilateral !
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Purdue University© ECET 17700 DAQ & Systems Control
Thévenin’s Theorem
Thévenin Model
ETH
RTH
Look Familiar ?A real voltage source !
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Purdue University© ECET 17700 DAQ & Systems Control
Measuring Thévenin Voltage
Open Circuit or No Load voltage
ETH = VOC = VNL
LinearBilateralCircuit
Supplies, resistors,lamps, switches, etc.
voltmeter
red lead
black lead
ETH
RTH
voltmeter
red lead
black lead
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Example – Find Thévenin VoltageOpen Circuit Voltage
Test Circuit
+
10 V
−
ETH = VOC = VNL = 10 V
LinearBilateralCircuit
Supplies, resistors,lamps, switches, etc.
voltmeter
red lead
black lead
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Example – Use Thévenin Model
Iload = 10V / 2kΩ = 5 mA
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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Purdue University© ECET 17700 DAQ & Systems Control
Find Thévenin Model
Open circuit voltage = 12 VA 2 kΩ load produces 4 mA
Find ETH
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Purdue University© ECET 17700 DAQ & Systems Control
Find Thévenin Model
Open circuit voltage = 12 VA 2 kΩ load produces 4 mA
Find RTH
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Example – Use Thévenin Model
Iload = 12V / 6kΩ = 2 mA
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Electrical Engineering TechnologyECET 17700 - DAQ & Control SystemsLecture # 9 – Loading, Thévenin & Norton
Professors Robert Herrick & J. Michael Jacob
Module 3 – Intro onlyWatch Class Video
Norton Model
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model
• Valid for any load• Same resulting load
voltages and currents
Original circuit Norton Model
ElectronicCircuit IN
RN
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Purdue University© ECET 17700 DAQ & Systems Control
IN Norton Current
ElectronicCircuit
ShortISC Circuit
Current
Careful
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
RN Norton Resistance
ElectronicCircuit RN
RN = RTH
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Purdue University© ECET 17700 DAQ & Systems Control
Ohm’s Law – find RN
RN = VOC / ISC
ElectronicCircuit
+VOC-
ElectronicCircuit
ISC
RN = ETH / IN
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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model Analysis
ElectronicCircuit
NortonModel
Load+
VL-
IL
Same IL and VL
Load+
VL-
IL
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Purdue University© ECET 17700 DAQ & Systems Control
Series-Parallel Circuit Analysis - example
3Ω+-
6Ω
6Ω24V+
VL-
IL
R// = 3Ω // 6Ω = 2Ω
+VL−
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Purdue University© ECET 17700 DAQ & Systems Control
Series-Parallel Circuit Analysis - example
3Ω+-
6Ω
6Ω24V+
VL-
IL
2 ΩR// = 3Ω // 6Ω = 2Ω
+VL−
VL = 2 Ω2Ω+6Ω
× 24V = 𝟔𝟔𝟔𝟔
𝟔𝟔𝟔𝟔
𝑉𝑉𝑉𝑉𝑉𝑉
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Purdue University© ECET 17700 DAQ & Systems Control
Series-Parallel Circuit Analysis - example
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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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model Analysis - example
Find IL and VL using the Norton model approach !
3Ω+-
6Ω
6Ω24V+
VL-
IL
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Purdue University© ECET 17700 DAQ & Systems Control
IN Norton Model Current - example
IN = ISC
3Ω+-
6Ω
24VShortcircuitcurrent
ISC
= 24V / 6Ω = 4A
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Purdue University© ECET 17700 DAQ & Systems Control
ETH Thévenin Voltage - example
ETH = VOC
3Ω+-
6Ω
24V+VOC-
VDR – Voltage Divider Rule
= 3 Ω3Ω+6Ω
× 24V = 𝟖𝟖𝟔𝟔
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model – our example values
NortonModelIN
RTH
4A2Ω RN
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model Analysis - example
ISC = 40 V / 20 Ω = 2A
3Ω+-
6Ω
6Ω24V
ISC20
540V 4Q9.7 Find ISC (A)
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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model Analysis - example
VOC = V5Ω =
3Ω+-
6Ω
6Ω24V
20
540V 4Q9.8 Find VOC
+VOC-
( 5 Ω / 25 Ω ) 40 V = 8 V
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Purdue University© ECET 17700 DAQ & Systems Control
Norton Model Analysis - example
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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Purdue University© ECET 17700 DAQ & Systems Control
3Ω+-
6Ω
6Ω24V
Norton Model Analysis - exampleIL20
540V 4Q9.10 Use Nortonmodel, attach load, and find IL
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Electrical Engineering TechnologyECET 17700 - DAQ & Control SystemsLecture # 9 – Loading, Thévenin & Norton
Professors Robert Herrick & J. Michael Jacob
Module 4 – Watch Class Video
Source Conversion
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Equivalent Models - supply conversion
+-ETH
RTH
INRN
Thévenin model Norton model
Same loadresults !
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Purdue University© ECET 17700 DAQ & Systems Control
RTH & RN - supply conversion
Zero the sources
+-ETH
RTH
INRN
Thevenin model Norton model
openRTH RN
RTH = RN
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Purdue University© ECET 17700 DAQ & Systems Control
IN - supply conversion
Short Circuit Current
+-ETH
RTH
INRN
Thevenin model Norton model
IN = ETH / RTH
ISC
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Purdue University© ECET 17700 DAQ & Systems Control
ETH = IN × RN
+-ETH
RTH
INRN
Thevenin model Norton model
+ETH
−
Open Circuit Voltage
ETH - supply conversion
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Purdue University© ECET 17700 DAQ & Systems Control
Thevenin Resistance
RTH = RN = VOC / ISC
ElectronicCircuit
+VOC-
ElectronicCircuit
ISC
Warning - SHORT can cause smoke or loading !
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
Electrical Engineering TechnologyECET 17700 - DAQ & Control SystemsLecture # 9 – Loading, Thévenin & Norton
Professors Robert Herrick & J. Michael Jacob
Module 5
What’s in the BoxPractical Measurements
62
Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
ElectronicCircuit
Removeload
Load+
VL-
IL
ElectronicCircuit
+ETH
-voltmeter
1. RTH Measurements - Using VOC and ISC
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Purdue University© ECET 17700 DAQ & Systems Control
careful
IN
ElectronicCircuit ammeter
ammeter SHORT
1. RTH Measurements - Using VOC and ISC
ISC not always possible !
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Purdue University© ECET 17700 DAQ & Systems Control
1. RTH Measurements - Using VOC and ISC
RTH = VOC / ISC
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Purdue University© ECET 17700 DAQ & Systems Control
2. RTH Measurements - matched load
ElectronicCircuit
+ETH-
voltmeterElectronicCircuit
+½ ETH
-
Attach & adjust pot. VPOT = ½ ETH
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Purdue University© ECET 17700 DAQ & Systems Control
2. RTH Measurements - matched load
+-ETH
RTH
+½ ETH-
+ ½ ETH -
Measure Potentiomenter with Ohmmeter RPOT = RTH
RPOT
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Purdue University© ECET 17700 DAQ & Systems Control
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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Purdue University© ECET 17700 DAQ & Systems Control
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