Download Current Sources

Lesson 7: Current Sources /
Source Conversion
1
Learning Objectives
• Analyze a circuit consisting of a current source, voltage
source and resistors.
• Convert a current source and a resister into an equivalent
circuit consisting of a voltage source and a resistor.
• Evaluate a circuit that contains several current sources in
parallel.
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Ideal Sources
• An ideal source is an active element that provides a
specified voltage or current that is completely
independent of other circuit elements.
DC Voltage
Source
DC current
source
3
Current Sources
• In general, a current source determines the direction
and magnitude of the current in the branch where it
is located.
• Furthermore, the magnitude and the polarity of the
voltage across a current source are each a function
of the network to which the voltage is applied.
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Constant Current Sources
• A constant current source will maintain the same
current in branch of circuit.
− It doesn’t matter how components are connected
external to the source.
• The direction of the current source indicates direction
of the current flow in branch.
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Constant Current Sources
• Series circuit
− Current must be same everywhere in circuit.
• Current source in a series circuit
− Value of the current for that circuit.
• For the circuit shown
Is = 2 mA
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Constant Current Sources
• If we wanted to find Vs, V1 and V2; we can simply
apply Ohm’s Law and KVL:
V1  2mA *1000  2V
V2  2mA * 2000  4V
KVL=> Vs  2V  10V  4V  0
Vs  2V  10V  4V  4V
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Constant Current Sources
The voltage across the current source (VS) is dependent on
how other components are connected to it.
Changing R2 from 2kΩ to 6kΩ causes the voltage across
the current source to change polarity to maintain KVL.
Current source voltage polarity does NOT have to follow
the current source’s arrow!
KVL=> E  Vs  V1  V2  0
V1  2mA *1000  2V
Vs  V1  V2  E
V2  2mA *6000  12V
V2  2mA * 2000  4V
KVL=> Vs  2V  10V  4V  4V
V1  2mA *1000  2V
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KVL=> Vs  2V  12V  10V  4V
Practical Voltage Sources
• A real or practical source supplies its rated voltage
when its terminals are not connected to a load (opencircuited) but its voltage drops off as the current it
supplies increases.
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Practical Voltage Source
• A real (or practical) voltage source supplies its rated
voltage when its terminals are open-circuited (NOT
connected to a load), but its voltage drops off as the
current supplied increases.
• We can model a practical voltage source using an
ideal source vs in series with an internal resistance Rs.
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Practical Current Source
• A practical current source supplies its rated current
when its terminals are short-circuited but its current
drops off as the load resistance increases.
• We can model a practical current source using an
ideal current source is in parallel with an internal
resistance Rs.
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Source Conversions
Source conversion.
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Source Transformation
Voltage Source to Current Source
• Replace a voltage source vs in series with a resistor R
by a current source is in parallel with the SAME
resistor R.
• How do you calculate the new current source
value?
vs 100V
is  
 2A
− Just apply Ohm’s Law:
R 50
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Source Transformation
Current Source to Voltage Source
• Replace a current source is in parallel with a resistor
R by a voltage source vs in series with the SAME
resistor R.
• How do you calculate the new voltage source
value?
− Just apply Ohm’s Law: vs  I s * R  2 A *50  100V
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Example Problem 1
Use source transformation to determine vo.
V 12V
Is 

 3A
Rs 4
1
1 1 1
 
R eq
 2 
Io  I s
 3A* 4 8 8  3A* 
  750mA
Ro
8
8



Vo  I o * Ro  750mA *8  6V
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Example Problem 2
Use source transformation to determine vo.
Vs  I s * Rs  3 A * 4  12V
RT  4  2  8  14
IT 
Vs  12V 

  857mA
RT  14 
Vo  I s * Ro  857mA *8  6.856V
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Current Sources – A Rule
• We found that voltage sources of different terminal
voltages cannot be placed in parallel because of a
violation of Kirchhoff’s voltage law.
• Similarly, current sources of different values cannot
be placed in series due to a violation of Kirchhoff’s
current law.
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Current Sources – A Parallel Rule
• However, current sources can be placed in parallel just as
voltage sources can be placed in series.
− In general, two or more current sources in parallel can be replaced by
a single current source having a magnitude determined by the
difference of the sum of the currents in one direction and the sum in
the opposite direction. The new parallel internal resistance is the total
resistance of the resulting parallel resistive elements.
Take care to note the
polarity of the
sources!
IT  I1  I 2  I 3  I 4
 2  (7)  5  3
IT  3 A
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Example Problem 3
Use source transformation and parallel current source rule to
simplify the circuit and determine vo.
- Convert the voltage source into a current source:
- Calculate the Is needed:
Is 
Vs
12V
*
 4A
Rs
3
- Redraw the circuit with the new current source
- Redraw again to show the added current sources to
get IT = 7A.
- Use CDR to calculate for Io:
4A
1 1 1
   
Req
4 8 3
I o  IT
 7A* 
Ro
8
1
 1.24 A
- Use Ohm’s Law to calculate for Vo:
7A
4Ω
Vo  I o * RR 8  1.24 A *8  9.88V
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Source Transformation
Use source transformation to determine V0
• Unfortunately, ONLY one source transformation is possible!
• The voltage source is in parallel with the 6Ω resistor (thus not a source resistance)
and the 3A current source is in series with the 3Ω resistor (again not a source
resistor).
• Changing only the 9A source into a voltage source doesn’t help you solve the
problem….
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Example Problem 5
Determine I2 and V3.
- Redraw the circuit with the
combined current source (6A2A=4A).
- Now calculate RT.
RT 
1
 10
1
1
1
 
60 20 30
- Now calculate I2 using CDR.
4A
I2  Is
R eq
R2
 4A*
10
 2A
20
- Finally calculate V2 using
Ohm’s Law.
V2  I 2 * R2  2 A * 20  40V
V2  V3  40V  V1
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QUESTIONS?
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