Thevenin Equivalent Circuits (EC 4.10) Thevenin equivalent • Current delivered to any load resistance by a circuit is equal to: • Voltage source equal to open circuit voltage V th at load • In series with a simple resistor R th (the source impedance).
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Thevenin Equivalent Circuits (EC 4.10) Thevenin equivalent • Current delivered to any load resistance by a circuit is equal to: • Voltage source equal to open circuit voltage Vth at load • In series with a simple resistor Rth (the source impedance).
Procedure for Finding Thevenin Equivalent (1) Remove all elements not included in the circuit, • Remove all loads at the output. (2) Find the open circuit voltage = Vth (3) Do one of the following to get Rth: (3a) Find the output resistance Rth: • Turn off all sources • Voltage sources become shorts • Current sources become open • Then calculate resistance of circuit as seen from output. (3b) Find short circuit current: • Calculate the load resistance by
short
thth I
VR =
Example Thevenin Equivalents • Consider the simple voltage divider circuit • Finding Vth by open circuit voltage • This acts a simple voltage divider
V710140006000
14000VRR
RVV21
2openth =
+=
+==
• Setting sources off (short voltage source) • Then input resistance is parallel resistors
mhos1038.26000
114000
1R1
R1 4
tinth
−×=+==
Ω= K2.4Rth
Example Thevenin Equivalents • Using alternate method for Rth • Short the output • Hence R2 removed • Short circuit current is
mA667.1600010
RVI
1short ===
• Then
Ω=== K2.400166.0
7IVRshort
thth
• Note often easier to use short output method.
Norton Equivalent Circuits Norton Equivalent • Current delivered to any load resistance by a circuit is equal to: • Constant current source = to short circuit current IN at the load • Shunt resistor RN = resistance circuit when all sources are stopped.
Procedure for Norton Equivalent Circuits (1) Remove all elements not included in the circuit • Remove all loads at the output. (2) Find the short circuit current = IN (3) Do one of the following: (3a) Find the output resistance: turning off all constant sources • Voltage sources become shorts • Current sources become open • Then calculate resistance of circuit as seen from output. (3b) Find the open circuit voltage Vopen • Calculate the Norton resistance RN by
N
openN I
VR =
Example Norton Equivalent Circuit • Consider a current source with R1 in parallel & R2 on output • Find the short circuit current = IN • When shorted becomes a simple current divider
⎥⎦
⎤⎢⎣
⎡+
⎥⎦
⎤⎢⎣
⎡
==
21
2
11
1
RR
RIII shortN
mA
RR
RIIN 4
00075.00005.0006.0
20001
40001
20001
006.011
1
21
2 ==
⎥⎦⎤
⎢⎣⎡ +
⎥⎦⎤
⎢⎣⎡
=
⎥⎦
⎤⎢⎣
⎡+
⎥⎦
⎤⎢⎣
⎡
=
• Then find the output resistance • Setting the current source off (open) • Now R1 and R2 in series
Ω=+=+== KRRRR outN 64000200021
Example Norton Equivalent Circuit Con’d • Alternate way for RN • Finding the open circuit voltage
VIRVopen 244000006.02 =×==
• Thus the output resistance is
Ω=== KI
VR
N
openN 6
004.024
• Depending on circuit this may be faster then shutting off source
Relationship Between Thevenin and Norton Circuits • Can easily change Thevenin into Norton or vice versa • By definition
Nth RR = • Thus
thNNNth RIRIV ==
N
th
th
thN R
VRVI ==
When to Use Thevenin and Norton Circuits
Thevenin equivalents most useful for: • Where the information wanted is a single number • Such as the current out of a given line. • Circuits similar to non-ideal voltage sources • Eg Battery or simple power supply (50 ohm input) • There Rth provides current limit & internal resistance
Norton equivalents are most useful for: • Circuits that approximate a non-ideal current source. • Eg. current limited power supply • There RN creates the voltage limit & internal resistance
Superposition of Elements • Another way of solving circuits with linear circuit elements • With linear circuits the effect of the combined system is linear • Obtained by adding together effect of each source on circuit • To use superposition analysis: (1) Set all power sources except one to zero • Short circuit all voltage sources • Open circuit all current sources • These can be thought of as simplified circuits (2) Calculate the voltages and currents from that one source (3) Repeat 1-2 for each source individually (4) Final result, at each point in the circuit: • Sum the voltages and currents of all the simplified circuits • Result is the combined voltage/current of the circuit.
Example of Superposition • Solve the circuit below with a current and voltage source
• First turn off (open circuit) the current source • Then really only have two resistors in series:
mA140002000
6RR
VI21
1 =+
=+
=
• Using a voltage divider for the R2 line
V440002000
40006RR
RVV21
21R =
+=
+=
Example of Superposition Continued • Now turn on the current source • Turn off the voltage source (short circuit it) • Then using the current divider formula
⎥⎦
⎤⎢⎣
⎡+
⎥⎦
⎤⎢⎣
⎡
=
21
22
R1
R1
R1
II
mA200075.000025.0006.0
20001
40001
40001
006.0
R1
R1
R1
II
21
22 ==
⎥⎦⎤
⎢⎣⎡ +
⎥⎦⎤
⎢⎣⎡
=
⎥⎦
⎤⎢⎣
⎡+
⎥⎦
⎤⎢⎣
⎡
=
• Similarly for the R1 branch
mA400075.00005.0006.0
20001
40001
20001
006.0
R1
R1
R1
II
21
21 ==
⎥⎦⎤
⎢⎣⎡ +
⎥⎦⎤
⎢⎣⎡
=
⎥⎦
⎤⎢⎣
⎡+
⎥⎦
⎤⎢⎣
⎡
=
• The voltage across the R2 is
V84000002.0RIV 222R =×==
Example of Superposition Continued • Now add the voltages and currents • For the voltage across R2
V1248VVV onV2RonI2R2R =+=+= −−
• For the current in R2
mA312III onV2RonI2R2R =+=+= −−
• For the current in R1 • Note: the V and I source only currents are in opposite directions
mA314III onV1RonI1R1R =−=+= −−
• Voltage drop across R1
V62000003.0RIV 11R1R =×==
• With this we have a full analysis of the circuit