Title: Circuit Theorems
1Circuit Theorems
- Dr. Mustafa Kemal Uyguroglu
2Circuit Theorems Overview
- Introduction
- Linearity
- Superpositions
- Source Transformation
- Thévenin and Norton Equivalents
- Maximum Power Transfer
3INTRODUCTION
A large complex circuits
Simplify circuit analysis
Circuit Theorems
?Thevenins theorem ? Norton theorem ?Circuit
linearity ? Superposition ?source
transformation ? max. power transfer
4Linearity Property
- A linear element or circuit satisfies the
properties of - Additivity requires that the response to a sum
of inputs is the sum of the responses to each
input applied separately. - If v1 i1R and v2 i2R
- then applying (i1 i2)
-
- v (i1 i2) R i1R i2R v1 v2
5Linearity Property
- Â
- Homogeneity  Â
- If you multiply the input (i.e. current) by some
constant K, then the output response (voltage) is
scaled by the same constant. - If v1 i1R then K v1 K i1R
6Linearity Property
- A linear circuit is one whose output is linearly
related (or directly proportional) to its input.
Suppose vs 10 V gives i 2 A. According to the
linearity principle, vs 5 V will give i 1 A.
7Linearity Property - Example
Solve for v0 and i0 as a function of Vs
8Linearity Property Example (continued)
9Linearity Property - Example
Ladder Circuit
3 A
5 A
1 A
3 V -
6 V -
2 A
2 A
5 V -
8V -
14 V -
This shows that assuming I0 1 A gives Is 5 A
the actual source current of 15 A will give I0
3 A as the actual value.
10Superposition
- The superposition principle states that the
voltage across (or current through) an element in
a linear circuit is the algebraic sum of the
voltages across (or currents through) that
element due to each independent source acting
alone.
11Steps to apply superposition principle
- Turn off all independent sources except one
source. Find the output (voltage or current) due
to that active source using nodal or mesh
analysis. - Turn off voltages sources short voltage
sources make it equal to zero voltage - Turn off current sources open current sources
make it equal to zero current - Repeat step 1 for each of the other independent
sources. - Find the total contribution by adding
algebraically all the contributions due to the
independent sources. - Dependent sources are left intact.
12Superposition - Problem
132mA Source Contribution
I0 -4/3 mA
144mA Source Contribution
I0 0
1512V Source Contribution
I0 -4 mA
16Final Result
I0 -4/3 mA I0 0 I0 -4 mA I0 I0
I0 I0 -16/3 mA
17Example
- find v using superposition
18one independent source at a time, dependent
source remains
KCL i i1 i2 Ohm's law i v1 / 1
v1 KVL 5 i (1 1) i2(2) KVL 5 i(1 1)
i1(2) 2v1 10 i(4) (i1i2)(2) 2v1 10
v1(4) v1(2) 2v1 v1 10/8 V
19Consider the other independent source
KCL i i1 i2 KVL i(1 1) i2(2) 5
0i2(2) 5 i1(2) 2v2Ohm's law i(1)
v2v2(2) i2(2) 5 0 gt i2 -(52v2)/2i2(2)
5 i1(2) 2v2-2v2 (i - i2)(2) 2v2-2v2
v2 (52v2)/2(2) 2v2-4v2 2v2 5
2v2-8v2 5 gt v2 - 5/8 V from
superposition v -5/8 10/8 v 5/8 V
20Source Transformation
- A source transformation is the process of
replacing a voltage source vs in series with a
resistor R by a current source is in parallel
with a resistor R, or vice versa
21Source Transformation
22Source Transformation
23Source Transformation
- Equivalent sources can be used to simplify the
analysis of some circuits. - A voltage source in series with a resistor is
transformed into a current source in parallel
with a resistor. - A current source in parallel with a resistor is
transformed into a voltage source in series with
a resistor.
24Example 4.6
- Use source transformation to find vo in the
circuit in Fig 4.17.
25Example 4.6
Fig 4.18
26Example 4.6
- we use current division in Fig.4.18(c) to get
-
- and
27Example 4.7
- Find vx in Fig.4.20 using source transformation
28Example 4.7
- Applying KVL around the loop in Fig 4.21(b) gives
-
(4.7.1) - Appling KVL to the loop containing only the 3V
voltage source, the resistor, and vx yields
-
(4.7.2)
29Example 4.7
- Substituting this into Eq.(4.7.1), we obtain
- Alternatively
-
- thus
30Thevenins Theorem
- Any circuit with sources (dependent and/or
independent) and resistors can be replaced by an
equivalent circuit containing a single voltage
source and a single resistor. - Thevenins theorem implies that we can replace
arbitrarily complicated networks with simple
networks for purposes of analysis.
31Implications
- We use Thevenins theorem to justify the concept
of input and output resistance for amplifier
circuits. - We model transducers as equivalent sources and
resistances. - We model stereo speakers as an equivalent
resistance.
32Independent Sources (Thevenin)
33No Independent Sources
34Introduction
- Any Thevenin equivalent circuit is in turn
equivalent to a current source in parallel with a
resistor source transformation. - A current source in parallel with a resistor is
called a Norton equivalent circuit. - Finding a Norton equivalent circuit requires
essentially the same process as finding a
Thevenin equivalent circuit.
35Computing Thevenin Equivalent
- Basic steps to determining Thevenin equivalent
are - Find voc
- Find RTh
36Thevenin/Norton Analysis
- 1. Pick a good breaking point in the circuit
(cannot split a dependent source and its control
variable). - 2. Thevenin Compute the open circuit voltage,
VOC. - Norton Compute the short circuit current,
ISC. - For case 3(b) both VOC0 and ISC0 so skip step
2 -
37Thevenin/Norton Analysis
- 3. Compute the Thevenin equivalent resistance,
RTh - (a) If there are only independent sources, then
short circuit all the voltage sources and open
circuit the current sources (just like
superposition). - (b) If there are only dependent sources,
then must use a test voltage or current source in
order to calculate - RTh VTest/Itest
- (c) If there are both independent and
dependent sources, then compute RTh from VOC/ISC.
38Thevenin/Norton Analysis
- 4. Thevenin Replace circuit with VOC in series
with RTh - Norton Replace circuit with ISC in parallel
with RTh - Note for 3(b) the equivalent network is merely
RTh , that is, no voltage (or current) source. - Only steps 2 4 differ from Thevenin Norton!
-