Download 3. Voltage divider, current divider and superposition

Lecture 3. Single Loop Circuits &
Superposition Method
Single Loop Circuits
* with a current source
* with a voltage source
* with multiple sources
* voltage divider circuits
* Equivalent resistance
Superposition method
* Principle
* Procedures
* How to apply
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Single Loop Circuits
• The same current flows through each element of the circuit—the
elements are in series.
I
V1
V1
I
R1
R1
R2
+
–
V2
R2
V2
Rn
V3
IS
VS
Rn
V3
With an independent voltage source
With an independent current source
2
Single Loop Circuits – with a Current Source
• What is I?
I  Is
V1
I
• In terms of I, what is the
voltage across each resistor?
R1
V1  IR1  I s R1
V2  IR2  I s R2
…
R2
V2
Rn
V3
IS
Vn  IRn  I s Rn
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Single Loop Circuits – with a Voltage Source
• In terms of I, what is the
voltage across each resistor?
V1  IR1 V2  IR2
Vn  IRn
I + I R1 –
…
• To solve for I, apply KVL
around the loop.
IR1 + IR2 + … + IRn – VS = 0
Vs
Vs
I

R1  R2  ...  Rn  Ri
+
R1
R2
+
–
I R2
–
+
VS
Rn
I Rn
–
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With Multiple Voltage Sources
• The current i(t) is:
 VSi sum of voltage sources
i (t ) 

 Rj
sum of resistances
• Resistors in series
Requivalent  R1  R2   RN   R j
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Voltage Division
• Consider two resistors in series with a voltage v(t) across them:
+
R1
v1 (t )  v(t )
R1  R2
+
R1
–
+
v(t)
R2
–
• If n resistors in series:
v1(t)
v2(t)
R2
v2 (t )  v(t )
R1  R2
–
Ri
vRi (t )   vSk (t )
 Rj
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Voltage Divider: A Practical Example
Electrochemical Fabrication of
Quantum Point Contact
or Atomic-scale wire
Molecular Junction
7
Voltage Divider: An Example
E
+
+
+
+
+
+
+
-
+
+
+
+
+
+
+
Anode: Etching delocalized, but
Cathode: Deposition localized at sharpest point,
due to:
• Self-focusing – directional growth
Decreasing Gap!
-
Voltage Divider: An Example
Vgap
Vext
+
+
+
+
+
Rext
+
Vgap 
Rgap
Rgap  Rext
A
V0
• Initially, Rgap >> Rext, Vgap ~ V0
• Finally, Rgap << Rext, Vgap ~ 0
full speed deposition.
deposition terminates.
• The gap resistance is determined by Rext.
V0
Voltage Divider: An Example
3
1
• Growth starts
after applying 1.5 V
• Self-terminates after
forming a tunneling gap
• Two electrodes with
10 mm initial separation
2
Voltage Divider: An Example
G (2e2/h)
Stepwise increase in Conductance
Time (sec.)
Ohmic behavior
Example: Two Resistors in Parallel
I
I1
R1
I2
R2
+
V
–
How do you find I1 and I2?
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Example: Two Resistors in Parallel
• Apply KCL with Ohm’s Law
1
V V
1 
I  I1  I 2  
 V   
R1 R2
 R1 R2 
I1
R1
I
I2
R2
+
V
–
V I
1
1
1

R1 R2
I
R1 R2
R1  R2
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Equivalent Resistance of Parallel Resistors
• Two parallel resistors is often equivalent to a single
resistor with resistance value of:
R1R2
Req 
R1  R2
• n-Resistors in parallel:
1
1 1
1
1
 


Rpar R1 R2
Rn
Ri
What are I1 and I2 ?
R1 R2
I
V
R2
R1  R2
I1 

I
R1
R1
R1  R2
R1 R2
I
V
R1
R1  R2
I2 

I
R2
R2
R1  R2
• This is the current divider formula
• It tells us how to divide the current through parallel resistors
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Circuits with More Than One Source
Is1
Is2
I1
R1
I2
R2
+
V
–
How do we find I1 or I2?
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What if More Than One Source?
• Apply KCL at the Top Node
I s1  I s 2
Is1
 1
V
V
1 

 I1  I 2 

 V  
R1 R2
 R1 R2 
Is2
I1
R1
I2
R2
+
V
–
R1 R2
V  I s1  I s 2 
R1  R2
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Class Examples
• Example: P1-33 (page 43).
• Drill Problem P1-34 (page 43).
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Superposition Method
– A More General Approach to Multiple Sources
“In any linear circuit containing multiple independent
sources, the current or voltage at any point in the circuit
may be calculated as the algebraic sum of the individual
contributions of each source acting alone.”
19
How to Apply Superposition
• To find the contribution due to an individual independent
source, zero out the other independent sources in the
circuit
– Voltage source  short circuit
– Current source  open circuit
• Solve the resulting circuit using your favorite
technique(s)
Superposition of Summing Circuit
1kW
V
+
+
–
Vou
t–
1
1kW
V
1
+
–
1kW
1kW
+
–
2
1kW
1kW
+
V’ou
t–
V
1kW
+
1kW
+
V’’ou
1kW
–
t
+
–
V
2
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Superposition of Summing Circuit (cont’d)
1kW
1kW
1kW
+
V1
+
–
V’ou
t–
1kW
+
1kW
+
V’’out
1kW
+
–
V2
–
V’out = V1/3
V’’out = V2/3
Vout = V’out + V’’out = V1/3 + V2/3
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Superposition Procedure
1. For each independent voltage and current source
(repeat the following):
a) Replace the other independent voltage sources with
a short circuit (i.e., V = 0).
b) Replace the other independent current sources with
an open circuit (i.e., I = 0).
Note: Dependent sources are not changed!
c) Calculate the contribution of this particular voltage or
current source to the desired output parameter.
2. Algebraically sum the individual contributions (current
and/or voltage) from each independent source.
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Class Examples
• Example 2-9 (page 70).
• Drill Problem 2.7.
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