1. Topic 11
Super-Nodes
(4.4)

2. Special Cases R1 R3 R2 Is vi
There are 3 essential nodes in this circuit, implying two unique node equations v1 v2
Choosing a ground node
Would normally leave us with two nodes at which to write KCL
But v1 is already specified by the voltage source
This leaves us with only one equation in one unknown So
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Super Nodes

3. A More Complex Circuit 40Ω 50v 5Ω 50Ω 100Ω 4A 20 v1 v2 50
Now we’ve got 2 voltage sources & 4 essential nodes
We cannot tie one end of both sources to ground
So tie bottom of 50v source to ground
How do we deal with the current through the voltage source?
That gives us one specified and two unknown nodes
Normally, we would write KCL at the two unknown nodes in order to get the two equations we require
But we get the individual current terms for the KCL equations…
…either from node voltages and resistors
…or from current sources
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4. A More Complex Circuit 20 i50 v1 v2
40Ω
50v 5Ω
50Ω
100Ω 4A 20 i
We could define a current i through the source
So i is just introducing an extra unknown
But i is an unknown and we are trying to get all currents in terms of node voltages or current sources.
We also have a strong contraint between v1 and v2
…so if we know one node voltage we know them both!
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5. A More Complex Circuit 50 v1 v2
40Ω
50v 5Ω
50Ω
100Ω 4A 20
Let’s just rewrite i as it affects each of the two nodes i
Then write KCL at both nodes in the normal way
What this says is
if we ignore i (which is the only current between the two nodes)
To eliminate i, just add the equations up
The sum of all the rest of the currents out of the pair of nodes combined is still 0
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Super Nodes

6. 20
The Super Node 60 5Ω 80
Whenever we have two non-ground nodes connected by a voltage source… 50 v1 v2
40Ω
100Ω
50v
50Ω 4A
…we combine them into a super node…
…we eliminate one node voltage by writing it in terms of the other one and the source…
…and then we write KCL at the two nodes combined (ignoring the common current between them)…
x100
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7. Which of These Circuits Have Supernodes?
40Ω
50v 5Ω
50Ω
100Ω
20v 50 v1 20 No
We only have one unknown node
So we only require one equation
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8. Supernodes? 5iφ 11 11.8 6Ω v1+ 5iφ 2Ω v1 25 22.8 Yes iφ 8Ω 25v 4Ω 5A
x24
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9. Symbolic Version of Assesment 4.8
Choose the ground node R1 R2 v1
vs+ 6iφ vs
Label the other nodes iφ R4
Since we know vs we know vs+6iφ and so have no need for a super node vs R3 1
KCL at node 1:
where
& So 2
Putting
into 2 1 4
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10. 4
Work on this part first
xR1R3(R2-6)
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11. Dimension Check  6iφ A useful check on our algebra R2 R1 vs+ 6iφ v1
The 6’s comes from…
The voltage produced by the dependent source is given by
Dimensionless scale factor
Terms which are added or subtracted must have the same dimensions
So the factor 6 has dimensions of volts/amps=ohms 
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12. Node-Voltage Analysis of an Amplifier Circuit
vcc a
Node-Voltage Analysis of an Amplifier Circuit R1 RC
1. Identify the essential nodes
2. Choose the reference ground node
3. The voltage at a becomes vcc
βiB
vCC Vo
4. Label the voltages at b and c vb b vc c
5. Note that b and c form a supernode iB R2
6. Write KCL at the supernode RE d
We have 2 unknowns
We need a handle on ib — the current through the supernode
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13. Algebra 1 2 3 3 1 into 4 4 5 Gather terms 6 Solve for vc 5/10/2019
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Solve for vc

14. What happens if we make β very large?
Solve for vc 6
(dimensions check)
What happens if we make β very large?
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15. R in Series with a V Source
In this case we still have two essential nodes with unknowns 10
v1+10
10k 5k 2k
20k 4k
16k 12 12 v1 v2
So supernode does not eliminate an essential node equation in this case but it does let us deal with the current through the 10 volt source
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16. 10k 5k 2k
20k 4k
16k 12 10 v1 v2
v1+10
10.96
.957
0.91
7.3
9.13
0.45
1.1
0.19
0.45
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17. Problem 3.26 2k
10k
300 1k
15 mA
200
The key to the problem is that the vo terminals are open circuit vo + - i1 i2
It is widely used by experienced engineers in all kinds of engineering
This means the current flowing through the 2k and 10k resistors is 0
This means the two 300Ω resistors are effectively in series
It requires an understanding of the underlying technology that goes beyond memorizing formulae
Which means the current flowing through the two 300Ω resistors is the same
As are the 200Ω and 1k resistors
The technique we are using is known as physical reasoning
As is the current through the 200Ω and 1k resistors
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18. Problem 3.26 2k
10k
300 1k
15 mA
200 v 1
From a node voltage point of view we have this vo + - i2 i1 v 2 v 3 v 4 v 5 v 2 v 3
Because the current through the horizontal 2K & 10k resistors is 0
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19. From the current source’s point-of-view we can reduce this
How does it help us? 2k
10k
300 1k
15 mA
200 v 1
From the current source’s point-of-view we can reduce this vo + - i2 i1 v 2 v 3 v 2 v 3 v 1
to this
1200
15 mA
600 i1 i2
…from whence
&
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20. …and the problem falls apartv 1
How does it help us? 2k
10k
300 1k
15 mA
200
Back annotate
10mA vo + -
5mA v 2 v 3
…and the problem falls apart v 2 v 3
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