Lecture #40: Review of circuit concepts
• This week we will be reviewing the
material learned during the course
• Today: review
– passive devices
– circuit concepts
– Load lines
– RC transients
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EE 42 fall 2004 lecture 40
1
BRANCHES AND NODES
Circuit with several branches connected at a node:
branch (circuit element)
i2
i1
i3
i4
KIRCHOFF’s CURRENT LAW
(Sum of currents entering node)  (Sum of currents leaving node) = 0
q = charge stored at node is zero. If charge is stored, for
example in a capacitor, then the capacitor is a branch and the
charge is stored there NOT at the node.
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EE 42 fall 2004 lecture 40
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GENERALIZATION OF KCL TO SURFACES
Sum of currents entering and leaving any “black box” is zero
Could be a big chunk of
circuit in here, e.g.,
could be a “Black Box”
In other words there
can be lots of nodes
and branches inside
the box.
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KIRCHHOFF’S CURRENT LAW USING SURFACES
Example
surface
5 A  2 A  i
entering
5 A
2 A
i=?
leaving
Another example
50 mA
i=7A
i?
i must be 50 mA
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BRANCH AND NODE VOLTAGES
The voltage across a circuit element is defined as the difference
between the node voltages at its terminals
a
v1

+
v2  vd  va
v2

+
c
b
e
d
v e  0 (since it’s the
select as ref. 
reference)
“ground”
Specifying node voltages: Use one node as the implicit reference
(the “common” node … attach special symbol to label it)
Now single subscripts can label voltages:
e.g., vb means vb  ve, va means va  ve, etc.
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KIRCHHOFF’S VOLTAGE LAW (KVL)
The algebraic sum of the “voltage drops” around any “closed loop” is zero.
Why? We must return to the same potential (conservation of energy).
Voltage drop  defined as the branch voltage if the + sign is encountered first;
it is (-) the branch voltage if the  sign is encountered first … important
bookkeeping
Path
Path
+
V1
“drop”
V2
+
-
“rise” or “step up”
(negative drop)
Closed loop: Path beginning and ending on the same node
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FORMAL CIRCUIT ANALYSIS USING KCL:
NODAL ANALYSIS
1 Choose a Reference Node
2 Define unknown node voltages (those not fixed by
voltage sources)
3 Write KCL at each unknown node, expressing current
in terms of the node voltages (using the constitutive
relationships of branch elements)
4 Solve the set of equations (N equations for N unknown
node voltages)
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NODAL ANALYSIS USING KCL
–Example: The Voltage Divider –
i1
1 Choose reference node
R1
V2
+
2 Define unknown node voltages
VSS
+

R2
V2
i2

3 Write KCL at unknown nodes
VSS  V2 V2  0

R1
R2
4 Solve:
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R2
V2  VSS 
R1  R 2
This is of course the voltage
divider formula and is by
itself very useful.
EE 42 fall 2004 lecture 40
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GENERALIZED VOLTAGE DIVIDER
(solved without Nodal Analysis)
Circuit with several resistors in series
I
R1
VSS
+

+
 V1 ?
R2
R3
R4
+
 V3?
• We know
I  VSS/(R1  R 2  R 3  R 4 )
• Thus,
V1 
R1
 VSS
R1  R 2  R 3  R 4
and
V3 
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R3
 VSS
R1  R 2  R 3  R 4
etc.. etc..
EE 42 fall 2004 lecture 40
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WHEN IS VOLTAGE DIVIDER FORMULA CORRECT?
I
I
R1
R1
+
R2
V SS
+


V2
R3
+
VSS
V
 Z
R2
+

X
R3
R4
i
3
R5
R4
V
2
R
2
R R R R
1
2
3
4
Correct if nothing else
connected to nodes
V
SS
What is VZ?
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VZ 
R2
 VSS
R1  R 2  R 3  R 4
because R5 removes condition of
resistors in series – i.e. i 3  I
Answer:
R2
 VSS
R1  R 2  R 5 ( R 3  R 4 )
EE 42 fall 2004 lecture 40
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RESISTORS IN PARALLEL
VX
1 Select Reference Node
2 Define unknown node voltages
Note: Iss = I1 + I2, i.e.,
V
V
ISS  X  X
R1 R 2
I2
I1
ISS
R1
1
 VX  ISS 
1
1

R1 R 2
 ISS 
R1R 2
R1  R 2
RESULT 1 EQUIVALENT RESISTANCE: R ||  R1 || R 2 
RESULT 2 CURRENT DIVIDER:
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I1 
R2
R1R 2
R1  R 2
VX
R2
 ISS 
R1
R1  R 2
VX
R1
I2 
 ISS 
R2
R 1  R 2 11
EE 42 fall 2004 lecture 40
IDENTIFYING SERIES AND PARALLEL COMBINATIONS
Use series/parallel equivalents to simplify a circuit before starting KVL/KCL
I
R1
R2
R3
R eq
R1
V
R1  R 2
R3
R2
+

R4
R5
parallel
R1  R 2  10 K
R 3 R  20 K
3
R6
R4  5  5 K
R 6  10 K
I
( R1  R 2 ) || R 3  R eq
+

RX ?
R X  (R1  R 2 ) || R 3  (R 4  R 5 ) || R 6
 15 K
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IDENTIFYING SERIES AND PARALLEL COMBINATIONS
(cont.)
Some circuits must be analyzed (not amenable to simple inspection)
R1 and R2 are not in ||
R2
R1
R1
V
+
R3

I
R4
R5
V
+
-
R2
R3
R4
Special cases:
R3 = 0 OR R3 = 
R5
Example: R3 = 0  R1 || R2; R4 || R5 in series;
OR IF R3 =   (R1 +EER425)fall
|| 2004
(R2 +
R)
lecture4 40
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R1 and R5 are not
in series
Req = R1 || R2 + R4 || R5
13
TWO-TERMINAL LINEAR RESISTIVE NETWORKS
(“One Port” Circuit)
Model of two-terminal linear resistive elements with only two
“accessible” terminals
Replace a complicated circuit with a simple model
+

a
b
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BASIS OF THÉVENIN THEOREM
• All linear one-ports have linear I-V graph
• A voltage source in series with a resistor can produce any linear I-V
graph by suitably adjusting V and I
THUS
We define the voltage-source/resistor combination that replicates
the I-V graph of a linear circuit to be the Thévenin equivalent of
the circuit. The voltage source VT is called the Thévenin
equivalent voltage and the resistance RT is called the Thévenin
equivalent resistance.
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EE 42 fall 2004 lecture 40
15
I-V CHARACTERISTICS OF LINEAR TWO-TERMINAL NETWORKS
i
5K
What is the easy way to
+
+
find the I-V graph?
5V
v

Associated

Apply v, measure i,
or vice versa
i(mA)
1
First find open-circuit V
.5
v=5V if i = 0
v(V)
Now find Short-circuit I
i = -1mA if v = 0
Associated
(i defined in)
5
-.5
-1
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EE 42 fall 2004 lecture 40
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I-V CHARACTERISTICS OF LINEAR TWO-TERMINAL NETWORKS
i
+
5K
Consider how the graph
changes with differences in V
and R.
+
5V
v


Apply v, measure i,
or vice versa
i(mA)
1
If V = 2.5V
.5
First consider change in
V, eg V= 2.5V, not 5V
Now consider change in R
(with V back at 5V)
5
v(V)
If R = 2.5K
-.5
-1
Clearly by varying V and R we can produce an arbitrary linear graph
… in other words this circuit can produce any linear graph
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Thévenin Equivalent Circuit
RT
+
VT

i
i
+

V
Any
linear
circuit
+

V
This circuit is equivalent to any circuit, that is by suitably
choosing VT and RT it will have the same I-V graph
So how do we choose VT and RT ?
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EE 42 fall 2004 lecture 40
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NORTON EQUIVALENT CIRCUIT
Corollary to Thévenin: I N  ISC (short - circuit current) (associated)
RN is found the same way as for Thévenin equivalent
i
IN
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RN
EE 42 fall 2004 lecture 40
+
V
-
19
EXAMPLE 1, Continued
equivalent to these?
In what sense is this circuit
A
1K
2 K
2 K
+ 2V

1V
A
A
1 mA
+
-
B
1K
B
B
They have identical I-V characteristics and therefore have
The same open circuit voltage
The same short circuit current
For any voltage, they will produce the same current
And visa versa
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Load line method
We can find the currents and voltages in a simple circuit graphically. For
example if we apply a voltage of 2.5V to the two resistors of our earlier
example:
We draw the I-V of the voltage and the I-V graph of the two resistors on
the same axes. Can you guess where the solution is?
At the point where the voltages of the two graphs AND the currents are
equal. (Because, after all, the currents are equal, as are the voltages.)
I (ma)
I
+
2.5V
-
1K
+
4
V
4K
-
2
2.5V
Solution: I = 0.5mA,
V = 2.5V
Combined
1K + 4K
5
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EE 42 fall 2004 lecture 40
V (Volt)
21
Another Example of the Load-Line Method
Lets hook our 2K resistor + 2V source circuit up to an LED (light-emitting
diode), which is a very nonlinear element with the IV graph shown below.
Again we draw the I-V graph of the 2V/2K circuit on the same axes as
the graph of the LED. Note that we have to get the sign of the voltage
and current correct!!
At the point where the two graphs intersect, the voltages and the currents
are equal, in other words we have the solution.
I (ma)
I
+
2V
-
LED
2K
+
4
LED
Solution: I = 0.7mA,
V = 1.4V
V
-
2
5
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EE 42 fall 2004 lecture 40
V (Volt)
22
Simplification for time behavior of RC Circuits
We call the time period
during which the output
changes the transient
We can predict a lot about the
transient behavior from the pre- and
post-transient dc solutions
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input
time
voltage
Long after the input change
occurs things “settle down” ….
Nothing is changing …. So
again we have a dc circuit
problem.
voltage
Before any input change occurs we have a dc circuit
problem (that is we can use dc circuit analysis to relate the
output to the input).
EE 42 fall 2004 lecture 40
output
time
23
RC RESPONSE
Case 1 – Rising voltage. Capacitor uncharged: Apply + voltage step
V1
Vin
Input node
Vout
0
R
+
Vin
Output node
Vout
C
-
ground
time
• Vin “jumps” at t=0, but Vout cannot “jump” like Vin. Why not?
0
 Because an instantaneous change in a capacitor voltage would
require instantaneous increase in energy stored (1/2CV²), that is,
infinite power. (Mathematically, V must be differentiable: I=CdV/dt)
V does not “jump” at t=0 , i.e. V(t=0+) = V(t=0-)
The dc solution before the transient tells us the capacitor voltage at the
beginning of the transient.
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RC RESPONSE
Case 1 Continued – Capacitor uncharged: Apply voltage step
V1
Vin
Input node
+
Vout
Vin
-
0
0
time
R
Output node
Vout
C
ground
• Vout approaches its final value asymptotically (It never actually
gets exactly to V1, but it gets arbitrarily close). Why?
After the transient is over (nothing changing anymore) it means d(V)/dt
= 0 ; that is all currents must be zero. From Ohm’s law, the voltage
across R must be zero, i.e. Vin = Vout.
 That is, Vout  V1 as t  . (Asymptotic behavior)
Again the dc solution (after the transient) tells us (the asymptotic limit of)
the capacitor voltage during the transient.
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RC RESPONSE
Example – Capacitor uncharged: Apply voltage step of 5 V
5
Vin
Input node
Output node
+
Vout
Vin
-
0
0
R
time
Vout
C
ground
• Clearly Vout starts out at 0V ( at t = 0+) and approaches 5V.
•
We know this because of the pre-transient dc solution (V=0) and
post-transient dc solution (V=5V).
So we know a lot about Vout during the transient - namely its initial value,
its final value , and we know the general shape .
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Review of simple exponentials.
Rising Exponential from Zero
Falling Exponential to Zero
Vout = V1(1-e-t/t)
Vout = V1e-t/t
Vout = 0 , and
at t = 0,
Vout = V1 , and
at t  ,
8
Vout  V1 also
at t  ,
at t = t,
Vout  0, also
Vout = 0.63 V1
at t = t,
Vout = 0.37 V1
8
at t = 0,
Vout
Vout
V1
V1
.63V1
.37V1
0
0
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t
time
0
EE 42 fall 2004 lecture 40
0
t
time
27
Further Review of simple exponentials.
Rising Exponential from Zero
Falling Exponential to Zero
Vout = V1(1-e-t/t)
Vout = V1e-t/t
We can add a constant (positive or negative)
Vout = V1(1-e-t/t) + V2
Vout = V1e-t/t + V2
Vout
Vout
V1 + V2
V1 + V2
.63V1+ V2
.37V1 + V2
V2
0
V2
0
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t
time
0
EE 42 fall 2004 lecture 40
0
t
time
28
Further Review of simple exponentials.
Rising Exponential
Falling Exponential
Vout = V1
(1-e-t/t)
+ V2
Vout = V1e-t/t + V2
Both equations can be written in one simple form: Vout = A + Be-t/t
Initial value (t=0) : Vout = A + B.
Final value (t>>t): Vout = A
Thus: if B < 0, rising exponential; if B > 0, falling exponential
Vout
Vout
A
A+B
Here B > 0
Here B < 0
A+B
0
A
0
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time
0
EE 42 fall 2004 lecture 40
0
time
29
LOGIC GATE DELAY tD
Time delay tD occurs between input and output: “computation” is not
instantaneous
Value of input at t = 0+ determines value of output at later time t = tD
F
A
Logic State
B
Capacitance to Ground
Input (A and B tied together)
1
0
t
0
1
Output (Ideal delayed step-function)
F
0
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tD
Actual exponential voltage versus time.
0
EE 42 fall 2004 lecture 40
t
30
SIGNAL DELAY: TIMING DIAGRAMS
Show transitions of variables vs time
Oscilloscope Probe
A
B
C
D
Logic state
A
Note B changes one gate delay
after A switches
Note that C changes two gate delays
after A switches.
1
0
B
t
t
t 2t
t
t 2t 3t
t
C
Note that D changes three gate delays
D
after A switches.
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t
0
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