1 module 9 - thévenin and norton equivalent circuits in this module, we’ll learn about an...
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Module 9 - Thévenin and Norton Equivalent Circuits
In this module, we’ll learn about an important propertyof resistive circuits called Thévenin Equivalence.
M. Leon Thévenin (1857-1926), published his famous theorem in 1883.
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Thévenin’s Theorem applies to circuits containingresistors, voltage sources, and/or current sources
Thêvenin Equivalent Circuit
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Thévenin Equivalent Circuit
VTh
RTh
Thévenin’s Theorem: A resistive circuit can be represented by one voltage source and one resistor:
Resistive Circuit
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Definition of a “Port”
Resistive Circuit
Port: Set of any two terminals
PORT
PORT
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Simple resistive circuit
Vo
R1
R2vX
+
_
iX
Illustrate concept with a simple resistive circuit:
• Any two terminals can be designated as a port.
Define portvariables vX and iX
• Our objective: Find the equivalent circuit seen looking into the port
ix flows to some load
(not shown)
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Vo
R1
R2 vX
+
_
iX
Find an equation that relates vx to ix
i1
i2
KVL: i1R1 + i2R2 = Vo
(Each resistor voltage expressed using Ohm’s Law)
Also note: vX = i2R2
KCL: i1 = i2 + iX
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Solve these equations for vX versus iX :
i1R1 + i2R2 = Vo i1 = iX + i2 vX = i2R2
(iX + i2) R1 + i2R2 = Vo
(iX + vX/R2) R1 + vX = Vo
iX R1 + vX (R1 /R2 + 1) = Vo
Rearrange the variables…
or Vo – iX R1
vX = ––––––––– 1 + R1 /R2
R2 R1 R2
vX = Vo ––––––– – iX ––––––– R1 + R2 R1 + R2
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R1 R2 RTh = –––––––
R1 + R2
R2 R1 R2
vX = Vo ––––––– – iX ––––––– R1 + R2 R1 + R2
Vo
R1
R2 vX
+
_
iX
Examine this last equation:
It has the form vX = VTh – iX RTh
R2
VTh = Vo ––––––– R1/ + R2
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Constructing the Thévenin Equivalent Circuit
VTh
RTh
vX
+
_
iX
Write down KVL for this circuit:
vX = VTh – iXRTh
+ –iXRTh
“Output voltage = voltage source – voltage drop across RTh”
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vX = VTh – iXRTh
VTh
RTh
vX
+
_
iX
Vo
R1
R2 vX
+
_
iX
R2 R1 R2
vX = Vo ––––––– – iX ––––––– R1 + R2 R1 + R2
Choose model parameters VTh and RTh:
Actual Circuit: Model:
R2
VTh = Vo ––––––– R1 + R2
R1 R2 RTh = ––––––= R1 || R2 R1 + R2
and
• From the point of view of vX and iX, the Thévenin circuit models the actual circuit in every way.
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Vo
R1
R2
Actual Circuit:
R1 || R2
R2
Vo ––––––– R1 + R2
PORT
PORT
vX
vX
iX
iX
+
_
+
_
Thévenin Equivalent:
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Significance of RTh
Vo
R1
R2
• Set all independent sources in the actual circuit to zero.
• For a voltage source, that means substituting a short circuit.
Equivalent resistance
• Equivalent resistance RTh= R1||R2
RTh is the equivalent resistance seen looking into the port with all independent sources set to zero.
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Setting a Voltage Source to Zero
Voltage betweennodes fixed at Vo
Current determined bywhat’s connected…
Vo
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Setting a Voltage Source to Zero
Voltage betweennodes fixed at 0 Vby short circuit
LOAD
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Setting a Current Source to Zero
Voltage betweennodes determined bywhat’s connected
Current through branch set to Io
Ioopen circuit
x
x
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Significance of VTh
Vo
R1
R2
Open Circuit Voltage
• Connect nothing to the port
iX = 0
Open Circuit Voltage
• VTh represents the open circuit voltage of the actual circuit
iX = 0
• iX automatically set to zero.
• Port voltage is called the open circuit voltage.
+
_
VTh
RTh
+
_
+ 0 V –
KVL
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Example: Resistor Network
Balanced audio microphone system
R3 = 10 kR1=100 k
R2 = 30 k
R4 =10 k
Vmic
10 mV
What voltage is developed across a 50 k resistive load?
50 k = Input resistance of typical audio amplifier.
Microphone networkLoad
50 k
+vLOAD
–
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Solution Method: Find the Thévenin Equivalent of the Microphone Network
• Find Thevenin Equivalent remaining circuit.
• Reconnect the load.
• Find vLOAD from simplified circuit.
• Disconnect the load.
R1=100 k
R2 = 30 k
R3 = 10 k
R4 =10 k
Vmic
10 mV
Load Load
Find VTh
and RTh
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Step 1: Find the Equivalent Resistance
• Set the voltage source to zero. (Substitute a short circuit.)
R1=100 k
R2 = 30 k
R3 = 10 k
R4 =10 k
RThVmic
• Find the equivalent resistance RTh
• RTh = R3 + R1||R2 + R4 = 10 k + 23 k + 10 k = 43 k
43 k
RTh
Note: R1||R2 = (100 k)||(30 k) = 23 k
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Step 2: Find the Open Circuit Voltage
• Analyze the circuit under no-load conditions.
R1=100 k
R2 = 30 k
R3 = 10 k
R4 =10 k
Vmic
10 mV
• Voltage across port terminals will be VTh
• From KVL around the inner loop*: v2 = VmicR2/(R1 + R2) = 2.3 mV
• Note that no current flows through R3 and R4. Voltage across these resistors is zero.
VTh = 2.3 mV
+
–
*basically, voltage division
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Step 3: Reconnect the Load to the Thévenin Equivalent Model
RTh = 43 k
VTh
2.3 mV
Thévenin equivalent ofmicrophone network
From simple voltage division:
vLOAD = VTh (RLOAD/(RLOAD + RTh)
= 2.3 mV (50 k)/(93 k) = 0.54 mV Answer
50 k
+vLOAD
–RLOAD
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More Examples:The Norton Equivalent Circuit
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Short Circuit Current
Another important parameter of a circuit is its short circuit current
The short circuit current of a port is defined as the current that will flow if:
The load is disconnected
A short circuit is connected instead
RTh
VTh Isc = VTh /RTh
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Circuit Containing a Current Source
Consider the following simple circuit:
I1 R1
What is the Thévenin equivalent circuit seen looking intothe port?
Port
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Step 1: Find the open circuit voltage:
I1 R1
+VTh
–
• From Ohm’s Law:
VTh = I1R1
(That part is simple…)
+
–
Current is zero
• Open circuit conditions All of I1 flows through R1
• Voltage develops across R1 with polarity shown.
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Step 2: Find the equivalent resistance
I1R1
• Trivially, by inspection: RTh = R1
• Set the current source to zero.
• Find the resistance looking into the port.
RTh
• Set the current source to zero open circuit
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The Thévenin Equivalent Circuit:
R1
I1R1
Thévenin Equivalent
VTh = I1R1
RTh = R1
Actual Circuit:
Done!
I1 R1
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Norton Equivalent Circuit
• The Norton and Thévenin equivalents of a circuit are interchangeable.
• The equivalent resistance is the same: RN = RTh
• The open circuit voltage is the same: VTh = INRN
Norton Circuit.
RN
INRNIN RN
Thévenin Circuit.
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INRN Isc = IN
+VN = 0
–
What about the short-circuit current from a Norton Circuit?
• Apply a short circuit:
• The voltage across the Norton resistance becomes zero.
• No current flows through the Norton resistance (I = V/R).
• All the current flows through the short circuit.
• The short circuit current is the source current IN.
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Norton Equivalent Circuit
• The short circuit current is the same in each circuit: IN = VTh/RTh
Norton Circuit
VTh = INRN
IN =VTh/RTh
RTh = RN
Thévenin Circuit
RN = RTh
IN IN
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Example: Resistive Network
R1=100 k
R2 = 30 k
R3 = 10 k
R4 =10 k
Vmic
10 mV
Find the Norton Equivalent of the following circuit using theshort-circuit current method
Step 1: Find RTh (same as RN) by setting the source to zero.
RTh or RN
By inspection, RTh = R3 + R1||R2 + R4 = 10 k + 23 k + 10 k = 43 k
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R2 = 30 k
R3 = 10 k
R4 =10 k
R1=100 k
Vmic
10 mV
Step 2: Apply a short circuit to the port and compute the short-circuit current.
R1=100 k
Vmic
10 mV
RP = R2 || (R3 + R4) = 12 k
IP = Vmic/(R1 + RP) = 0.9 A
From current division:
R2
[R2 + (R3 + R4)]ISC = IP = 0.9 A
30 k50 k = 0.54 A = ISC
= 0.54 AISC
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Find the Norton Equivalent of the Circuit
ISC = 0.54 A
RN = 43 k
RN = 43 kIN = 0.54 A = 23 mV
+vOC
–
“Open Circuit Voltage”
vOC = IN RN = (0.54 A)(43 k) = 23 mV
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Construct the Thévenin Equivalent of the Circuit
ISC = 0.54 A
RTh = 43 k
VTh = ISC RTh = (0.54 A )(43 k) = 23 mV
VTh = 23 mV
RTh = 43 k
This result is the same one obtained in the previous example!
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A circuit that can be represented by a Thévenin Equivalentcan also be represented by its corresponding Norton circuit
VTh= INRN
RTh = RN
INRN
Norton Equivalent Thévenin Equivalent
RTh
VTh
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End of This Module
Do the Homework Exercises