Dependent Voltage and Current Sources
SYRACUSE UNIVERSITY
ELE231 Electrical Fundamentals I
Dependent Voltage and Current Sources by Akhan Almagambetov, [email protected] Office Hours: Wed. 1:15-3:15 p.m., Thurs. 12:30-2:30 p.m in CST 3-204 Recitation Session # 5 Week of September 22nd, 2008
Dependent (otherwise known as “controlled”) sources are sources whose current or terminal voltage depends on (or is “controlled by”) a voltage or current defined at some other location in the circuit. There are four types of controlled sources, as outlined on the next two pages. Although there exist four variations of this device, the most common types are: voltage controlled current sources and current-controlled voltage sources. • Voltage-controlled voltage source (VCVS): a voltage source that is controlled by a voltage somewhere in the circuit, as shown below (µ represents voltage gain, or what the v1 value has to be multiplied by. Naturally, it has no units). 2 + v1 -10
4
1
+ --
µv1
Dependent Sources
• Current-controlled voltage source (CCVS): a voltage source that is controlled by a current somewhere in the circuit, as shown below (r represents transresistance, measured in Ohms, Ω).
2
i1 +
4
10
--
ri1
. • Voltage-controlled current source (VCCS): a current source that is controlled by a voltage somewhere in the circuit, as shown below (g represents transconductance, or 1r , and is measured in Siemens, S, or Mhos, f). 2
+ v1 -10
4
gv1
. • Current-controlled current source (CCCS): a current source that is controlled by a current somewhere in the circuit, as shown below (β represents current gain, a dimensionless unit).
2 i1
10
4
βi 1
. What dependent sources do for us, is allow the design of amplifiers and active filters (which will be studied later in the course, beginning with the operational amplifier, or the OpAmp). Dependent sources are used to isolate one part of a circuit from another, prevent excessive circuit loading, and provide behaviors such as negative resistance1 . 1
Tyner, D. J. et al, Electric Circuit Analysis 2e. New York, NY: Wiley & Sons, 1992.
2
10/14/2008 v.4 (JJHS) c Akhan Almagambetov Copyright May be reprinted without permission.
Dependent Sources
Example Problem # 1 So how do we solve a problem with a dependent source? Let’s take the circuit in the figure below as an example. Our goal for this problem is to find the current i. 2 i
-- v1 + +
6
--
3v1
6
We immediately see that it’s a VCVS with the voltage gain, or µ, of +3. We also see that the controlling voltage in this case is v1 . By applying KCL around the loop in red (from the above figure), we get the following equation: −v1 + 3v1 + 6i = 6
(1)
and by using our well-known Ohm’s law equations, we get the equation for voltage v1 : v1 = 2(−i) = −2i
(2)
Using the second equation, we can eliminate v1 from the first equation by substitution, after which, it’s easy to calculate i to be equal to 3 Amperes: − (−2i) + 3 (−2i) + 6i = 6 =⇒ 2(−2i) + 6i = 6 =⇒ i = 3A
For all intents and purposes, superposition is not considered a valid method of analyzing circuits containing any kind of dependent sources. Many textbooks state this fact, although it has been disproven by Dr. W. Marshall Leach, Jr., a professor at the Georgia Institute of Technology. Those of you curious of his method can read the article: http://users.ece.gatech.edu/~mleach/papers/superpos.pdf Abstract: Many introductory circuits texts state or imply that superposition of dependent sources cannot be used in linear circuit analysis. Although the use of superposition of only independent sources leads to the correct solution, it does not make use of the full power of superposition. The use of superposition of dependent sources often leads to a simpler solution than other techniques of circuit analysis. A formal proof is presented that superposition of dependent sources is valid provided the controlling variable is not set to zero when the source is deactivated.
3
Dependent Sources
Example Problem # 2 Let’s analyze the circuit below, which has a voltage-controlled current source (VCCS) in parallel with some components. Our goal is to find the voltage v1 . 4
vA
vB
+ v1 --
2v1
4
8
2
The first step that we take is label all of the nodes in the circuit (marked in the figure above in red). We have a node at zero potential (ground), vA , and vB . First off, we have to define v1 in terms of what we “know”, which is vA and vB . Since the positive side of the voltage measurement is on the vA node, we write the equation with respect to that node: v1 = vA − vB
(3)
Then, we have to write two nodal equations for each of the non-zero nodes above, into which we substitute the equation for v1 . Note that we treat dependent sources just like independent ones when writing nodal equations: @ vA :
2v1 +
vA −0 4
2 (vA − vB ) +
vA −vB = 4 vA vA −vB 4 + 4
+
0
@ vB :
=0
vB −vA − 8 + vB2−0 = 4 vB −vA −32+2vB =0 4
0
8vA − 8vB + vA + (vA − vB ) = 0
vB − vA − 32 + 2vB = 0
10vA − 9vB = 0
−vA + 3vB = 32
Since you have the two equations, you can easily plug them into a 2x2 matrix and solve: "
10
−9
−1
3
#"
vA vB
#
" =
0
# (4)
32
Therefore, vA = 13.7V and vB = 15.2V , which is reasonable given that we’re using passive sign notation. Using the equation for v1 , we calculate it to be equal to v1 = −1.50V . 4
10/14/2008 v.4 (JJHS) c Akhan Almagambetov Copyright May be reprinted without permission.
ELE231 Electrical Fundamentals I
Dependent Voltage and Current Sources by Akhan Almagambetov, [email protected] Office Hours: Wed. 1:15-3:15 p.m., Thurs. 12:30-2:30 p.m in CST 3-204 Recitation Session # 5 Week of September 22nd, 2008
Dependent (otherwise known as “controlled”) sources are sources whose current or terminal voltage depends on (or is “controlled by”) a voltage or current defined at some other location in the circuit. There are four types of controlled sources, as outlined on the next two pages. Although there exist four variations of this device, the most common types are: voltage controlled current sources and current-controlled voltage sources. • Voltage-controlled voltage source (VCVS): a voltage source that is controlled by a voltage somewhere in the circuit, as shown below (µ represents voltage gain, or what the v1 value has to be multiplied by. Naturally, it has no units). 2 + v1 -10
4
1
+ --
µv1
Dependent Sources
• Current-controlled voltage source (CCVS): a voltage source that is controlled by a current somewhere in the circuit, as shown below (r represents transresistance, measured in Ohms, Ω).
2
i1 +
4
10
--
ri1
. • Voltage-controlled current source (VCCS): a current source that is controlled by a voltage somewhere in the circuit, as shown below (g represents transconductance, or 1r , and is measured in Siemens, S, or Mhos, f). 2
+ v1 -10
4
gv1
. • Current-controlled current source (CCCS): a current source that is controlled by a current somewhere in the circuit, as shown below (β represents current gain, a dimensionless unit).
2 i1
10
4
βi 1
. What dependent sources do for us, is allow the design of amplifiers and active filters (which will be studied later in the course, beginning with the operational amplifier, or the OpAmp). Dependent sources are used to isolate one part of a circuit from another, prevent excessive circuit loading, and provide behaviors such as negative resistance1 . 1
Tyner, D. J. et al, Electric Circuit Analysis 2e. New York, NY: Wiley & Sons, 1992.
2
10/14/2008 v.4 (JJHS) c Akhan Almagambetov Copyright May be reprinted without permission.
Dependent Sources
Example Problem # 1 So how do we solve a problem with a dependent source? Let’s take the circuit in the figure below as an example. Our goal for this problem is to find the current i. 2 i
-- v1 + +
6
--
3v1
6
We immediately see that it’s a VCVS with the voltage gain, or µ, of +3. We also see that the controlling voltage in this case is v1 . By applying KCL around the loop in red (from the above figure), we get the following equation: −v1 + 3v1 + 6i = 6
(1)
and by using our well-known Ohm’s law equations, we get the equation for voltage v1 : v1 = 2(−i) = −2i
(2)
Using the second equation, we can eliminate v1 from the first equation by substitution, after which, it’s easy to calculate i to be equal to 3 Amperes: − (−2i) + 3 (−2i) + 6i = 6 =⇒ 2(−2i) + 6i = 6 =⇒ i = 3A
For all intents and purposes, superposition is not considered a valid method of analyzing circuits containing any kind of dependent sources. Many textbooks state this fact, although it has been disproven by Dr. W. Marshall Leach, Jr., a professor at the Georgia Institute of Technology. Those of you curious of his method can read the article: http://users.ece.gatech.edu/~mleach/papers/superpos.pdf Abstract: Many introductory circuits texts state or imply that superposition of dependent sources cannot be used in linear circuit analysis. Although the use of superposition of only independent sources leads to the correct solution, it does not make use of the full power of superposition. The use of superposition of dependent sources often leads to a simpler solution than other techniques of circuit analysis. A formal proof is presented that superposition of dependent sources is valid provided the controlling variable is not set to zero when the source is deactivated.
3
Dependent Sources
Example Problem # 2 Let’s analyze the circuit below, which has a voltage-controlled current source (VCCS) in parallel with some components. Our goal is to find the voltage v1 . 4
vA
vB
+ v1 --
2v1
4
8
2
The first step that we take is label all of the nodes in the circuit (marked in the figure above in red). We have a node at zero potential (ground), vA , and vB . First off, we have to define v1 in terms of what we “know”, which is vA and vB . Since the positive side of the voltage measurement is on the vA node, we write the equation with respect to that node: v1 = vA − vB
(3)
Then, we have to write two nodal equations for each of the non-zero nodes above, into which we substitute the equation for v1 . Note that we treat dependent sources just like independent ones when writing nodal equations: @ vA :
2v1 +
vA −0 4
2 (vA − vB ) +
vA −vB = 4 vA vA −vB 4 + 4
+
0
@ vB :
=0
vB −vA − 8 + vB2−0 = 4 vB −vA −32+2vB =0 4
0
8vA − 8vB + vA + (vA − vB ) = 0
vB − vA − 32 + 2vB = 0
10vA − 9vB = 0
−vA + 3vB = 32
Since you have the two equations, you can easily plug them into a 2x2 matrix and solve: "
10
−9
−1
3
#"
vA vB
#
" =
0
# (4)
32
Therefore, vA = 13.7V and vB = 15.2V , which is reasonable given that we’re using passive sign notation. Using the equation for v1 , we calculate it to be equal to v1 = −1.50V . 4
10/14/2008 v.4 (JJHS) c Akhan Almagambetov Copyright May be reprinted without permission.
