Transcript Nodal analysis

Lecture #6 methods of solving circuits
• Nodal analysis
• Graphical analysis
•In the next lecture we will look at circuits which
include capacitors
•In the following lecture, we will start exploring
semiconductor materials (chapter 2).
Reading:
Malvino chapter 2 (semiconductors)
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Nodal analysis
• So far, we have been applying KVL and KCL “as
needed” to find voltages and currents in a circuit.
• Good for developing intuition, finding things
quickly…
• …but what if the circuit is complicated? What if you
get stuck?
• Systematic way to find all voltages in a circuit by
repeatedly applying KCL: node voltage method
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Branches and Nodes
Branch: elements connected end-to-end,
nothing coming off in between (in series)
Node: place where elements are joined—includes
entire wire
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Node Voltages
The voltage drop from node X to a reference node
(ground) is called the node voltage Vx.
Example:
a
+
Va
+
+
_
Vb
_
_
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b
ground
EE 42 fall 2004 lecture 6
Nodal Analysis Method
1. Choose a reference node (aka ground, node 0)
(look for the one with the most connections,
or at the bottom of the circuit diagram)
2. Define unknown node voltages (those not connected to ground by voltage
sources).
3. Write KCL equation at each unknown node.
– How? Each current involved in the KCL equation will either come from
a current source (giving you the current value) or through a device like
a resistor.
– If the current comes through a device, relate the current to the node
voltages using I-V relationship (like Ohm’s law).
4. Solve the set of equations (N linear KCL equations for N unknown node
voltages).
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* with “floating voltages” we will use a modified Step 3
Example
node voltage set
R1

+
-
V1
Va
R
3
R2
What if we
used different
ref node?
Vb
R4
IS
 reference node
•
•
•
•
Choose a reference node.
Define the node voltages (except reference node and the one set
by the voltage source).
Apply KCL at the nodes with unknown voltage.
Vb  Va Vb
Va  V1 Va Va  Vb

 IS


0
R3
R4
R1
R2
R3
Solve for Va and Vb in terms of circuit parameters.
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Example
R1
Va
R3
V
1
R2
I1
R4
R5
V2
Va   V1 Va Va  V2


 I1
R1
R4
R5
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Load line method
• I-V graph of circuits or sub-circuits
• I-V graph of non-linear elements
• Using I-V graphs to solve for circuit voltage, current
(The load-line method)
• Reminder about power calculations with nonlinear
elements.
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Example of I-V Graphs
Resistors in Series
If two resistors are in series the current is the same; clearly the total
voltage will be the sum of the two IR values i.e. I(R1+R2).
Thus the equivalent resistance is R1+R2 and the I-V graph of the series
pair is the same as that of the equivalent resistance.
Of course we can also find the I-V graph of the combination by adding
the voltages directly on the I-V axes. Lets do an example for 1K + 4K
resistors
I (ma)
+
1K V1 +
I
4K
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- V
+
V2
- -
4
2
1K
Combined
1K + 4K
4K
5
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V (Volt)
Example of I-V Graphs
Simple Circuit, e.g. voltage source + resistor.
If two circuit elements are in series the current is the same; clearly the
total voltage will be the sum of the voltages i.e. VS + IR.
We can graph this on the I-V plane. We find the I-V graph of the
combination by adding the voltages VS and I R at each current I.
Lets do an example for =2V, R=2K
I (ma)
+
I
V
2V
S
+
-
V
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2V
4
2K
2K
I
-
Combined
2K + 2V
2
EE 42 fall 20045
lecture 6
V (Volt)
Example of I-V Graphs
Nonlinear element, e.g. lightbulb
If we think of the light bulb as a sort of resistor, then the resistance
changes with current because the filament heats up. The current is
reduced (sort of like the resistance increasing). I say “sort of” because a
resistor has, by definition a linear I-V graph and R is always the same.
But for a light bulb the graph kind of “rolls over”, becoming almost flat.
Consider a 100 Watt bulb, which means at at the nominal line voltage of
117 V the current is 0.85A. (117 V times 0.85 A = 100W).
I (A)
I=0.85A at 117V
+
0.8
V
I
-
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0.4
100
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V (Volt)
I-V Graphs as a method to solve circuits
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) 2.5V
I
+
2.5V
1K
+
4
V
-
4K
-
Solution: I = 0.5mA,
V = 2.5V
Combined
1K + 4K
2
5
V (Volt)
This
is called the LOAD LINE method;
it works
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lecture 6for harder (non-linear) problems.
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!! (The sing of the current is reversed from page 3)
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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V (Volt)
Power of Load-Line Method
We have a circuit containing a two-terminal non-linear element
“NLE”, and some linear components.
The entire linear part of the circuit can be replaced by an equivalent,
for example the Thevenin equivalent. This equivalent circuit consists
of a voltage source in series with a resistor. (Just like the example
we just worked!).
So if we replace the entire linear part of the circuit by its Thevenin
equivalent our new circuit consists of (1) a non-linear element, and (2)
a simple resistor and voltage source in series.
If we are happy with the accuracy of graphical solutions, then we just
graph the I vs V of the NLE and the I vs V of the resistor plus voltage
source on the same axes. The intersection of the two graphs is the
solution. (Just like the problem on page 6)
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The Load-Line Method
We have a circuit containing a two-terminal non-linear element
“NLE”, and some linear components.
First replace the entire linear part of the circuit by its Thevenin
equivalent (which is a resistor in series with a voltage source). We will
learn how to do this in Lecture 11.
Then define I and V at the NLE terminals (typically associated signs)
D
Nonlinear
9mA
element
S
D
250K 1M
+
-
1V
N
L
E
ID
+
VDS + 200K
2V
S
NOTE: In lecture 11 we will show that the circuit shown on
the left is equivalent to a 200K resistor in series with 2V.
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Example of Load-Line method (con’t)
Given the graphical properties of two terminal non-linear circuit
(i.e. the graph of a two terminal device)
And have this connected to a linear
(Thévenin) circuit
D
N
L
E
Whose I-V can also be graphed
on the same axes (“load line”)
S
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200K
+
-
2V
The intersection gives circuit solution
ID (mA)
D ID
VDS+
+
VDS
S
ID
10
The solution !
200K
+
-
2V
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1
2
VDS (V)
Load-Line method
The method is graphical, and therefore approximate
But if we use equations instead of graphs, it could be accurate
It can also be use to find solutions to circuits with three terminal
nonlinear devices (like transistors), which we will do in a later
lecture
ID (mA)
D ID
10
The solution !
200K
S
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+
-
2V
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1
2
VDS (V)
Power Calculation Review
Power is calculated the same way for linear and non-linear
elements and circuits.
For any circuit or element the dc power is I X V and, if associated signs
are used, represents heating for positive power or extraction of energy
for negative signs.
For example in the last example the NLE has a power of +1V X 5mA or
5mW . It is absorbing power. The rest of the circuit has a power of - 1V
X 5mA or - 5mW. It is delivering the 5mW to the NLE.
So what it the power absorbed by the 200K resistor?
Answer: I X V is + 5mA X (5mA X 200K) = 5mW. Then the voltage
source must be supplying a total of 10mW. Can you show this?
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