Transcript Circuit with voltage source set to zero (SHORT CIRCUITED)

ADDITIONAL ANALYSIS TECHNIQUES
LEARNING GOALS
REVIEW LINEARITY
The property has two equivalent definitions.
We show and application of homogeneity
APPLY SUPERPOSITION
We discuss some implications of the superposition property in
linear circuits
DEVELOP THEVENIN’S AND NORTON’S THEOREMS
These are two very powerful analysis tools that allow us to
focus on parts of a circuit and hide away unnecessary complexities
MAXIMUM POWER TRANSFER
This is a very useful application of Thevenin’s and Norton’s theorems
THE METHODS OF NODE AND LOOP ANALYSIS PROVIDE POWERFUL TOOLS TO
DETERMINE THE BEHAVIOR OF EVERY COMPONENT IN A CIRCUIT
The techniques developed in chapter 2; i.e., combination series/parallel,
voltage divider and current divider are special techniques that are more
efficient than the general methods, but have a limited applicability.
It is to our advantage to keep them in our repertoire and use them when
they are more efficient.
In this section we develop additional techniques that simplify
the analysis of some circuits.
In fact these techniques expand on concepts that we have
already introduced: linearity and circuit equivalence
SOME EQUIVALENT CIRCUITS
ALREADY USED
LINEARITY
THE MODELS USED ARE ALL LINEAR.
MATHEMATICALLY THIS IMPLIES THAT THEY
SATISFY THE PRINCIPLE OF SUPERPOSITION
THE MODEL y  Tu IS LINEAR IFF
T (1u1   2 u2 )  1Tu1   2Tu2
for all possible input pairs u1 , u2
and all possible scalars 1 , 2
v IS A VECTOR CONTAINING ALL THE NODE
VOLTAGES AND f IS A VECTOR DEPENDING
ONLY ON THE INDEPENDENT SOURCES.
IN FACT THE MODEL CAN BE MADE MORE
DETAILED AS FOLLOWS
Av  Bs
AN ALTERNATIVE, AND EQUIVALENT,
DEFINITION OF LINEARITY SPLITS THE
SUPERPOSITION PRINCIPLE IN TWO.
THE MODEL y  Tu IS LINEAR IFF
1. T (u1  u2 )  Tu1  Tu2 , u1, u2 additivity
2. T (u)  Tu,  , u
USING NODE ANALYSIS FOR RESISTIVE
CIRCUITS ONE OBTAINS MODELS OF THE
FORM Av  f
HERE, A,B, ARE MATRICES AND s IS A VECTOR
OF ALL INDEPENDENT SOURCES
FOR CIRCUIT ANALYSIS WE CAN USE THE
LINEARITY ASSUMPTION TO DEVELOP
SPECIAL ANALYSIS TECHNIQUES
homogeneit y
NOTICE THAT, TECHNICALLY, LINEARITY CAN
NEVER BE VERIFIED EMPIRICALLY ON A SYSTEM.
BUT IT COULD BE DISPROVED BY A SINGLE
COUNTER EXAMPLE.
IT CAN BE VERIFIED MATHEMATICALLY FOR THE
MODELS USED.
FIRST WE REVIEW THE TECHNIQUES
CURRENTLY AVAILABLE
A CASE STUDY TO REVIEW PAST TECHNIQUES
DETERMINE VO
SOLUTION TECHNIQUES AVAILABLE??
Redrawing the circuit may help us
in recognizing special cases
The procedure can be made entirely algorithmic
USING HOMOGENEITY
1. Give to Vo any arbitrary value (e.g., V’o =1 )

2. Compute the resulting source value and call it V’_s
3. Use linearity.
REQ
V1
4. The given value of the source (V_s)
corresponds to
k

Assume that the answer is known. Can we
Compute the input in a very easy way ?!!
R1  R2
V0
R2
… And Vs using a second voltage divider
VS 
R4  REQ
REQ
R4  REQ R1  R2
V1 
V0
REQ
R2
Solve now for the variable Vo
VS
VS'
Hence the desired output value is
If Vo is given then V1 can be computed
using an inverse voltage divider.
V1 
VS'  V0'  kVS'  kV0' , k
V0  kV0' 
VS '
V
' 0
VS
This is a nice little tool
for special problems.
Normally when there is
only one source and
when in our judgement
solving the problem
backwards is actually
easier
SOLVE USING HOMOGENEITY
ASSUME Vout  V2  1[V ]
I1
VO
NOW USE HOMOGENEITY
VO  6[V ]  Vout  1[V ]
VO  12[V ]  Vout  2[V ]
LEARNING EXTENSION
COMPUTE IO USING HOMOGENEITY. USE I  6mA
VS  1.5[mA]  2k  V1  6[V ]
VS
0.5[mA ]
2mA
1.5[mA]
V1  3[V ]
0.5[mA ]
USE HOMOGENEITY
I  2mA  I O  1mA
I  6mA  I O  ____
ASSUME IO  1mA
Source Superposition
This technique is a direct application of
linearity.
It is normally useful when the circuit has only
a few sources.
VS
FOR CLARITY WE SHOW A CIRCUIT
WITH ONLY TWO SOURCES
+ -
IL
Due to Linearity
circuit
V
1
V
2
IS
L Can be computed by setting the current
source to zero and solving the circuit
L
VL
_
VL  a1VS  a2 I S
CONTRIBUTION BY VS
CONTRIBUTION BY I S
1
VL
V L2
+
Can be computed by setting the voltage
source to zero and solving the circuit
Circuit with voltage source
set to zero (SHORT CIRCUITED)
SOURCE SUPERPOSITION
I
I L2
1
L
=
V
1
L
+
Circuit with current
source set to zero(OPEN)
Due to the linearity of the models we must have
I L  I L1  I L2
VL  VL1  VL2
Principle of Source Superposition
The approach will be useful if solving the two circuits is simpler, or more convenient, than
solving a circuit with two sources
We can have any combination of sources. And we can partition any way we find convenient
VL2
LEARNING EXAMPLE
WE WISH TO COMPUTE THE CURRENT
=
i
1
+
Req  3  3 || 6 [k ] R  6  (3 || 3) [k ]
eq
i2"
Loop equations
v2

Req
Contribution of v1
Once we know the “partial circuits”
we need to be able to solve them in
an efficient manner
Contribution of v2
LEARNING EXAMPLE
Compute V0 using source superposition
We set to zero the voltage source
Current division
Ohm’s law
Now we set to zero the current source
Voltage Divider
 2[V ]
6k
3V
V0"
V0  V0'  V0"  6[V ]
+
-
3k
LEARNING EXAMPLE
Compute V0 using source superposition
We must be able to solve each circuit in a very
efficient manner!!!
If V1 is known then V’o is obtained using a voltage divider
V1 can be obtained by series parallel reduction and divider
Set to zero current source

+
-
V1
V1
_

Set to zero voltage source
6k
4k||8k
V1 
2k
V1
_
8/3
(6)
2  8/3
V'0
_
2k
VO' 
6k
18
V1  [V ]
6k  2k
7
The current I2 can be obtained using a current divider
and V”o using Ohm’s law
I2
2k||4k
2mA
+
I2
6k
V"0
I2 
2k  (2k || 4k )
(2)mA
2k  6k  (2k || 4k )
VO"  6kI 2
VO  VO'  VO"
2k
_
WHEN IN DOUBT… REDRAW!
+
+
+
Sample Problem COMPUTE I0 USING SOURCE SUPERPOSITION
1. Consider only the voltage source
I 01  1.5mA
2. Consider only the 3mA
source
3. Consider only the 4mA source
Current divider
I 02  1.5mA
I 03  0
Using source superposition
I 0  I 01  I 02  I 03  3mA
SUPERPOSITION APPLIED TO OP-AMP CIRCUITS
TWO SOURCES. WE ANALYZE ONE AT A TIME
CONTRIBUTION OF V1.
Basic inverter circuit
R2
VO 1   V1
R1
Principle of superposition
VO  VO 1  VO 2  

R2
R 
V1   1  2 V2
R1
R1 

CONTRIBUTION OF V2
Basic non-inverting amplifier V
O2
Notice redrawing for added
clarity

R2 
  1  V2
R1 

I1
I O1  
I1
2
2
2
1
3
1
3
Linearity