Download We analyze circuits for several reasons • Understand how they work

Introduction to Circuit Analysis
Getting Started
We analyze circuits for several reasons
•
Understand how they work
•
Learn how to design from other people’s work
•
Debug our own designs
•
Troubleshoot circuit or system that may have failed
Observe these are same reasons we analyze a system in any field
Chemistry
Mechanical engineering
Civil engineering
Computing science
Physics
Any other field of science
We learning to solve problems
Concentrating on electronics for the moment
Circuit analysis so far has identified circuit comprise
Number of branches
Number of loops
Have commented that the loops not all independent
Same holds true for branches
Let’s try to develop a formalism
Establishing a Basis
Consider the following generalized simple circuit
Note ....this does not have to be an electronic circuit
Can build identical model in
Fluids
Mechanical engineering
Thermodynamic systems
In following table we show the equivalents for electrical and mechanical systems
Electrical
V=
1
idt
C∫
V =L
i=C
di
dt
V = RI
dV
dt
i=
i
dV
L∫
i=
V
R
Mechanical
dV
dt
V=
1
M
V=
1 df
K dt
f = DV
1
f
D
f = K ∫ Vdt
V =
∫ fdt
f =M
mass
spring
damper
With a simple change of symbol and independent and dependent variables
All of the of the tools we learn for electrical circuits
Apply to all the major engineering disciplines
Nodal Equations
Let’s analyze from point of view of Kirchoff’s Current Law - a KCL analysis
a
2
+ V2
+
V1
1
+
V3
+ V4
3
d
For this circuit
We have
5 branches
4 nodes
c
4
+
V5
5
We can write 4 nodal equations relating them
-i1 - i2 = 0
a.
-i2 - i3 - i4 = 0
b.
i4 - i5 = 0
c.
i1 + i3 + i5 = 0
d.
We observe these are not independent
Any one equation can be found by combining the other 3
a.
b.
c.
d.
-i1 - i2 = 0
-i2 - i3 - i4 = 0
i4 - i5 = 0
i1 + i3 + i5 = 0
⇒ i1 + i2 = 0
In general
The number of independent KCL node equations is
given as N - 1, where N is the number of nodes in the
system
As noted we can eliminate any of the equations
Typically we do not write an equation for the reference node
Ground
Loop Equations
Let’s analyze from point of view of Kirchoff’s Voltage Law - a KVL analysis
For this circuit
We have
3 loops
We can write 3 loop equations relating them
Loop1:
V1 - V2 - V3= 0
Loop2:
V3 - V4 - V5= 0
Loop3:
V1 - V2 - V4 - V5= 0
We observe these are not independent
Any one equation can be found by combining the other 2
Loop3:
Loop2:
V1 - V2 - V4 - V5= 0
-V3 + V4 + V5= 0
⇒
V1 - V2 - V3= 0
Intuitively - following reasoning from branch analysis
We would think we would have N-1 independent loop equations
Not quite as simple
Consider
Circuit with B branches
Has 2 B unknowns
B currents through the branches
B voltages across those branches
From linear algebra we need 2 B independent equations
Where do such equations come from
The answer is
•
Ohm’s law
With B branches we have B equations of following form
V = IR
•
Kirchoff’s node equations give N -1
As we’ve seen
•
Kirchoff’s voltage equations must supply remainder
2 B - B - (N-1) - #KVL equations = 0
#KVL equations = B - N-1
In general
The number of independent KVL loop equations is
given as B - N - 1, where N is the number of nodes in
the system
For the given circuit
B=5
N=4
Therefore the number of independent KVL equations
5-4+1=2
As we determined earlier
In general finding set of independent KVL equations
Requires techniques from linear algebra
Fortunately for most simple circuits
Task is clear by inspection
When dealing with more complex networks
Other tools are available
Mesh Equations
Let’s look at what we have been calling loop analysis in more detail
Planar Networks
Let N be the set of all networks
Define subset P of N as
Set of all networks that can be drawn in two dimensions without lines crossing
Such a network is called planar
Most circuits we will deal with care planar
For a planar network
We can define a special type of loop
We call it a mesh
A mesh is a loop that does not enclose any other loops
The following network contains 3 loops
a
2
+ V1
+
V1
1
c
4
+
V1
+ V1
3
+
V1
5
d
Loop1:
Loop2:
Loop3:
= V1, V2, and V3
= V3, V4, and V5
= V1, V2, V4, and V5
Observe
•
L3 encloses L1 and L2
•
L1 does not enclose any loops
• L2 does not enclose any loops
Thus
The network contains 2 meshes
Equal number of independent KVL equations
This is not a co-incidence
From this observation
If we confine the KVL equations in planar networks to meshes
We will always get a complete set of independent loop (mesh) equations
We can restate our earlier formula as a special case
The number of independent KVL loop equations
equals the number of meshes in the system
In general
A network can be completely solved
For all currents and voltages using
1. Ohm’s Law and
2. Kirchoff’s Current Law for branch variables or
3. Kirchoff’s Voltage Law for loop variables
Example
I1
10
I3
i1
40
R2
I2
R4
i2
Use branch variables: I1, I2, I3, I4
Node a:
Is - I1 = 0
Node b:
I1 - I 2 - I 3 = 0
Node c:
I1 - I 2 - I 3 = 0
We also know
V
1. I 2 = b
R2
2. I 3 =
Vb − Vc
R3
3. I 4 =
Vc
R4
4. I s − I1 = 0
5. I1 −
6.
Vb Vb − Vc
−
=0
R2
R3
Vb − Vc Vc
−
=0
R3
R4
I4
R3
R1
5mA
20
30
Some algebra gives
7. I s − I1 = 0
8. R2 R3 I1 − (R2 + R3 )Vb + R2Vc = 0
9. R4Vb − (R3 + R4 )Vc = 0
Substituting variables and solving
1
Vb = v
9
1
v
15
From which we get
I1 =
5mA
I2 =
2.78 mA
I3 =
2.22 mA
I4 =
2.22 mA
Vc =
Equivalent Systems
Circuits
Often times in engineering we use tools to simplify our analysis
Replace a complex circuit or system
Equivalent but simpler
Can make such replacement as long as behaviour is same
In simple arithmetic
We can replace the two integers in the following equation by a single one
X + 3 + 5 = ...
Is equivalent to
X + 8 = ...
We think nothing of such a substitution
In electronics we do the same thing
Let’s look at a couple of simple circuits
i1
+
V1
i1
R1
V1
R1
R2
2R2
2R2
In the first circuit the two resistors are connected in series
These give a total resistance between nodes A and B as
R1 + R2
If we inject a current I into node A
We will have a voltage drop from A to B of
V = I / (R1 + R2)
In the second circuit notice we have two resistors in parallel
In parallel those two resistors will have a resistance of R2
Combining all three resistors
We see that we have a total resistance between nodes C and D of
R1 + R2
If we inject a current I into node C
We will have a voltage drop from C to D of
V = I / (R1 + R2)
Observe that the behaviour of the two circuits - from their ports - is identical
That is they have an identical V-I relationship
Based upon such a relationship we can say the two circuits are equivalent
Let’s generalize
First consider a one port abstraction of a (electrical) system
i1
+
V1
N1
Based upon the contents of the system
For each current we inject into the port
We measure a voltage across the port
Once again note this same analysis holds as well for
Mechanical systems
Fluidics
Thermal systems
We can plot the voltage vs current
We call this the V-I characteristic
When any two networks have identical V-I
characteristics, we say they are equivalent
For the following two networks
i1
i2
+
+
N1
V1
V2
N2
N1 may contain only a few components
N2 may contain thousands
None-the-less if the V-I characteristics are identical
The circuits are considered equivalent
Example
Let’s examine a more complex circuit
I
+
V1
R3
R1
R2
R4
Analyzing
R3 and R4 are in series - replace these by RC
I
+
R1
V1
R2
RC
R2 and RC are in parallel - replace these by RB
I
+
R1
V1
RB
R1 and RB are in series - replace these by RA
I
+
V1
RA
The resulting circuit is equivalent to the original network
Sources
We can apply a similar analysis to voltage and current sources
Voltage
Series
Voltage
For the following circuit
+
+
+
+
V1
V
V
Veq
+
V2
We know from Kirchoff’s voltage law that
V - V1 - V2 = 0
Thus we see
V = V1 - V2
+
V1
Now we have
V
+
For this circuit we have
V + V1 + V2 = 0
Thus we see
V = - ( V1 + V2 )
+
V2
+
V1
V
For this circuit we have
V - V1 + V2 = 0
+
V2
Thus we see
V = ( V1 - V2 )
V1
For this circuit we have
V
+
+
V2
V + V1 - V2 = 0
Thus we see
V = V2 - V1
Current
For the following circuit
I
I
I1
I2
Ieq
We know from Kirchoff’s current law that at
Node a
I - I1 = 0
Node b
I1 - I 2 = 0
Therefore
I =I1
and
I1 = I2
Thus we have violated KCL if
I1 ≠ I2
Voltage
Parallel
Voltage
For the following circuit
+
+
+
V
V1
+
V2
V
+
Veq
We know from Kirchoff’s voltage law that
V - V1 = 0
V1 - V2 = 0
Thus we see
V = V1 and
V1 = V2
Thus we have violated KVL if
V1 ≠ V2
Current
For the following circuit
I
I
+
I1
Ieq
I2
We know from Kirchoff’s current law that at
I = I1 + I2
Current sources in parallel add as voltage sources did in series
Source Transformation
Consider the following two networks
I1
+
I1
R1
+
1
V
Vs
+
Is
R1
2
N1
1
V
2
N2
Analyzing
N1
N2
VS - I1 R1 - V = 0
V = VS - I1 R1
IS - V/R1 - I1 = 0
V = IS R1 - I1 R1
For these two networks to be equivalent
The following must hold
VS = IS R1
That is
IS = VS / R1
Thus we see:
I1
+
Vs
R1
+
1
V
+
⇒
R1
1
V
Vs / R1
2
2
Similarly
I1
1
+
Is
+
⇒
V
Is R1
V
R1
1
+
R1
2
2
Example
Let’s examine the following circuit
I1
1K
+
R1
+
1
+
V
10V
10mA
R1
V
1K
2
2
15 K
I1
+
5 mA
1
+
R1
+
1
V
75 V
V
15 K
R1
2
2
Example
Let’s use source transformation to solve the following circuit
+
5V
Vin
1
10
30
R1
R3
20
R2
Vo
R4
40
Find Vout
1. Convert Vin and R1 into a current source
30
Vo
R3
R1
10
Ia
R2
20
R4
40
(Vin/R1)
2. R1 and R2 ⇒ R1 || R2 ⇒ Ra = 20/3
30
Vo
R3
Ra
Ia
R4
40
3. Convert Ia, Ra to voltage source Vb = (20/3)(1/2) = 10/3
Vo
R3
Ra
+
R4
Vb
4. Convert Ra and R3 ⇒ Rb = Ra + R3 = 110/3
Vo
+
Ra
R4
Vb
5. Convert Vb, Rb to current source Ic = Vb / Rb = (10/3)(3/110) = 1/11
Vo
Rb
Ic
R4
6. Convert Rb and R4 ⇒ Rc = Rb || R4 ⇒ Rc = 440/23
Vo
Rc
Ic
7. Vout = Ic Rc = (1/11) (440/23) = 1.91V
Voltage and Current Dividers
Voltage Dividers
Let’s look at the following circuit
VR1
+
VR2
R1
R2
Vs
Using KVL
Vs − I R1 − I R2 = 0
I=
Vs
R1 + R2
From Ohm’s law
VR1 = I R1
V R 2 = I R2
Substituting
VR1 =
R1 Vs
VR 2 =
R1 + R2
R2Vs
R1 + R2
Thus we see
The supply voltage divides between the two resistors
According to the formulas gives
Example
In the following circuit
Vo = 5V
R1
+
R2
10 V
If we want Vout = 5V for Vin = 10 V
We know
VO =
∴
R2Vin
R1 + R2
R2
R1 + R2
=
1
2
2 R1 = R1 + R2
R1 = R2
Select 1 K Ω
Current Dividers
Let’s look at the following circuit
Using KCL
Vo
Is
R1
I1
R2
I2
Is = I1 + I2
R1 R2
V = IS
I1 =
R1 + R2
V
R1
I2 =
V
R2
Substituting
∴ I1 =
I S R2
R1 + R2
I2 =
I S R1
R1 + R2
Thus we see
The supply current divides between the two resistors
According to the formulas gives
Example
In the following circuit
We want to find R1 and R2 such that I2 = 4 mA
Is
R1
I1
R2
I2
10 mA
I2 =
I S R1
R1 + R2
4mA =
4(R1 + R2) = 10 R1
4 R2 = 6 R1
let
R1 = 20
R2 = 30
10mA R1
R1 + R2