Download University of Waterloo 1 Nodal Analysis 2 Loop/Mesh Analysis

1
University of Waterloo
Department of Electrical & Computer Engineering
ME 123: Electrical Engineering for Mechanical
Engineers
DC CIRCUITS AND OPERATIONAL
AMPLIFIERSS
Dr. Claudio Ca~
nizares
June 2001
1 Nodal Analysis
The necessary equations to solve a resistive circuit may be written by inspection as follows:
1. Transform all voltage sources in the circuit to current sources. This
step is optional.
2. Dene one of the (nontrivial) nodes as a reference node. Thus, the
voltages at the other nodes (nodal voltages V1, V2, . . . , Vn ) can be
dened with respect to this reference node.
3. Apply KCL to all node but one in the circuit, considering that the
current in a resistor R connecting two nodes i and j is dened as
iR = Vi ;R Vj
2 Loop/Mesh Analysis
The necessary equations to solve a resistive circuit may also be obtained by
loop/mesh analysis as follows:
1. Transform all current sources in the circuit to voltage sources. This
step is optional.
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2. Dene the mesh currents I as the currents circulating T in an internal
loop (mesh).
3. Apply KVL to all loops in the circuit, considering that the voltage in
a resistor R common to two loops i and j is equal to
vR = R (Ii ; Ij )
3 Source Equivalents
Independent and dependent voltage sources may be transformed into equivalent current sources, and vice versa, as follows:
iL
iL
RS
v
vL
S
L
O
A
D
v
S
RS
RS
iL
vL
L
O
A
D
iL
RS
v = αx
S
vL
L
O
A
D
vs = Rs is
α x
RS
()
RS
vL
L
O
A
D
is = Rvs
s
4 Superposition
Any voltage and current in a linear circuit, i.e., a circuit made up of voltage
sources, current sources, R's, L's and C's may be computed as the sum of the
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corresponding voltages and currents due to each independent source. This
does not apply to powers (nonlinear variable). Thus,
X
vx = vx due to each independent source
X
ix = ix due to each independent source
X
px 6= px due to each independent source
To turn o independent sources in the circuit, the voltage sources are shortcircuited (vs = 0) and the current sources are open-circuited (is = 0). Dependent (controlled) sources cannot be eliminated.
5 Thevenin Equivalent
Any resistive circuit can be reduced at its terminals to an equivalent voltage
source:
iL
iL
R TH
vL
LINEAR CIRCUIT
L
O
A
D
v
TH
vL
L
O
A
D
THEVENIN EQUIVALENT
where:
vTH = vT with load open-circuited (iT = 0)
RTH = viTH
N
iN = iT with load short-circuited (vT = 0)
For circuits with only independent sources, the Thevenin resistance may be
computed as
RTH = Equivalent resistance with all sources and load o
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6 Norton Equivalent
Any resistive circuit can be reduced at its terminals to an equivalent current
source:
iL
iL
vL
LINEAR CIRCUIT
L
O
A
D
i
N
R TH
vL
L
O
A
D
NORTON EQUIVALENT
where:
iN = iT with load short-circuited (vT = 0)
RTH = viTH
N
vTH = vT with load open-circuited (iT = 0)
Observe that the Norton current source is the equivalent of the Thevenin
voltage source.
7 Maximum Power Transfer
A resistive circuit delivers maximum power to a load when the resistance of
the load is the same as the Thevenin resistance of the circuit, i.e.,
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pT
iL
R TH
v
RL
vL
TH
THEVENIN EQUIVALENT
LOAD
2
RL = RTH ) pTmax = 4vRTH
TH
8 OpAmps
An Operations Amplier (OpAmp) is an integrated linear electronic device,
made up of several semiconductor transistors, diodes, resistors and capacitors. The OpAmp structure and equations, when the device is in its linear
region (not saturated), are as follows:
i+
+
+
+
Ro
vi
Av
-
v+
+
i-
io
Ri
d
+
-
+
vo
v-
-
-
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vi
i+
vi
vo
=
=
=
=
v+ ; v;
;i;
Ri i+
A vi ; Ro io
Typical OpAmps have the following constructive characteristics:
Ri ! 1
Ro ! 0
A ! 1
which make the input voltage and currents
vi 0
i+ 0
i; 0
Thus, usually the OpAmp is studied using the ideal assumptions
vi = i+ = i; = 0
That is, the input behaves like a short-circuit and open-circuit at the same
time. Due to these particular characteristics, this device is typically used
with \feedback" connections, i.e., with an element connected between the
output and the input.
OpAmps are usually used as ampliers; hence, one is typically interested
in determining the gain or \transfer function"
Av = vvo
s
This ratio is easily determined for each circuit assuming ideal OpAmps and
applying KCL and KVL to an input and output loops.