Topic 11 Super-Nodes (4.4)

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 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

A More Complex Circuit 20 i 50 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! 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

Supernodes? 5iφ 11 11.8 6Ω v1+ 5iφ 2Ω v1 25 22.8 Yes iφ 8Ω 25v 4Ω 5A x24 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

4 Work on this part first xR1R3(R2-6) 5/10/2019 Super Nodes

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  5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

Algebra 1 2 3 3 1 into 4 4 5 Gather terms 6 Solve for vc 5/10/2019 Super Nodes Solve for vc

What happens if we make β very large? Solve for vc 6 (dimensions check) What happens if we make β very large? 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

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 5/10/2019 Super Nodes

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 & 5/10/2019 Super Nodes

…and the problem falls apart v 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 5/10/2019 Super Nodes