Electrician,

Compare the details of the Branch method with the Node method, and you will see that Studiot is using the Branch method.

OK, I looked over the material you linked. One way subtracts the counter voltage and the other way Theveninizes the voltage sources. It appears to me that is a distinction without a difference. They both ultimately balance the current going into a node against the current leaving the node.

Ratch

 

Studiot's analysis had I1, I2, I3, I4, I5, I6 and I7 as unknowns. Those are the currents through the individual resistors, branches of the circuit. That's why the method he used is called the Branch method.

The analysis you provided at the end of post #34, and which you called a node analysis has as unknowns the voltages, Vb, Vc, Vd, Ve, and Vf.

If Studiot had done a node analysis, he would have gotten the same node voltages you did when you did a node analysis. He didn't get the node voltages; he got the branch currents.

There is a distinction with a substantial difference.

Electrician,
OK, I looked over the material you linked. One way subtracts the counter voltage and the other way Theveninizes the voltage sources. It appears to me that is a distinction without a difference. They both ultimately balance the current going into a node against the current leaving the node.
Ratch

That currents are balanced doesn't mean that there is no difference in the methods. At a minimum, the choice of voltages or currents as unknowns is a substantial difference, and often one method will result in fewer equations than the other to solve for some particular unknown, such as the voltage at a particular node, for example.

 

Electrician,

OK, I see what you mean. You are right, there is a difference between obtaining currents between nodes and currents in a loop. I am more comfortable doing loop equations. Ratch

 

I have never used the Branch method except when I was forced to in an EE Circuits class. I think it's the least friendly method.

Concerning this thread, the control law for the controlled source is 2*Vx. Imagine if it were ABS(2*Vx); the absolute value of 2*Vx. Could we have a bistable network?

This is the assumption being made implicitly if one assumes that the upward pointing arrow in the diamond symbol for the controlled source means that the current must always be upwardly directed.

 

What facinates me is the reason Favour's college set this question. What point were they trying to demonstrate?

I have run a SPICE simulation of the circuit in 4 modes.

Twice with the dependent current arrow as in the posted diagram and twice with it reversed.

For each direction of the dependent current arrow I ran both polarities of sensor connection.

The results below are interesting and consistent with earlier calculations presented by Ratch and myself.

With nodes G - H earthed at zero (taken as the reference node)

VCIS arrow As Diag - VCIS arrow As Diag - VCIS reversed - VCIS reversed
-ve sense to C.........+ve sense to C.........-ve sense to C.....+ve sense to C

Vb = 86.92..............Vb = -21.36.............Vb = -21.36.........Vb = 86.92
Vc = 74.77..............Vc = -11.04.............Vc = -11.04.........Vc = 74.77
Vd = 119.20............Vd = -34.00.............Vd = -34.00.........Vd = 119.20
Ve = 84.31..............Ve = -13.76.............Ve = -13.76.........Ve = 84.31
Vf = 87.54...............Vf = -4.40...............Vf = -4.40..........Vf = 87.54
Vg = 0....................Vg = 0....................Vg = 0................Vg = 0
Vh = 0....................Vh = 0....................Vh = 0................Vh = 0

Vout = 24.92...........Vout = -3.68............Vout = -3.68........Vout = 24.92

I1 = 8.07.......................I1 = -3.16
I2 = 6.08.......................I2 = -5.16
I3 = 12.46......................I3 = -1.84
I4 = -6.39......................I4 = -3.32
I5 = 17.45......................I5 = -10.12
I6 = 21.08......................I6 = -3.44
I7 = -1.62......................I7 = -4.68

Idependent = 25.52..........Idependent = -13.28



So what is interesting about these figures?

Well if we look at the set of results with the solution forcing current between D and G in the direction of the arrow; D is 119.2 volts more positive than G and the source is forcing 24.92 amps into it.

What sort of current source is this?

OK so what about the negative solution?

Well at node H I3 and 8 amp are both outward so I8 must be inward.
Therefore I8 (=8 + 1.84 = 9.84) must be outward at G
But I6 is negative so also outward at G.

So the current through the dependent source is inward at G (=3.44 + 9.84 = 13.44) and therefore reversed, relaive to the arrow.

Worse D is at -34 volts relative to G.

So this dependent current source is forcing 13.44 amps of conventional current from negative to positive.

 

studiot,

Idependent = 25.52..........Idependent = -13.28

I believe that should be labeled dependent, not independent. Please change if you agree.

Well if we look at the set of results with the solution forcing current between D and G in the direction of the arrow; D is 119.2 volts more positive than G and the source is forcing 24.92 amps into it.

What sort of current source is this?

A voltage controlled current source. The difference between note C and node F is 12.77 volts. Two times that is 24.54 amps. Going around the loop 119.20-2*17.45-4*21.08 = 0 , so the loop DEG checks out.

OK so what about the negative solution?

Well at node H I3 and 8 amp are both outward so I8 must be inward.
Therefore I8 (=8 + 1.84 = 9.84) must be outward at G
But I6 is negative so also outward at G.

So the current through the dependent source is inward at G (=3.44 + 9.84 = 13.44) and therefore reversed, relaive to the arrow.

independent source = 13.28 inward to G
I6 = 3.44 outward from G
I8 = 9.84 outward from G

Total equals zero, so it checks out.

Worse D is at -34 volts relative to G.

That is what the voltage has to be if it is going to force conventional current to go south.

So this dependent current source is forcing 13.44 amps of conventional current from negative to positive.

That is not the way I am reading the tea leaves.

Ratch

 

Idependent = I dependent, the current through the dependent current source between nodes G and D


*********************************************************
D is at -34 volts
G is at zero volts

Conventional current is flowing from D towards G

Is that not negative to positive?

If you prefer, look at the currents summed at D

I1 and I5 are both negative so inward at D.
So the current from D to G must be outward at D.

The sum of the currents entering the branch at D equal the sum of the currents leaving the branch leaving at G

*******************************************

Finally I ask the question again, what sort of sources are these? I look forward to a practical implementation.

 

VCIS arrow As Diag - VCIS arrow As Diag - VCIS reversed - VCIS reversed
-ve sense to C.........+ve sense to C.........-ve sense to C.....+ve sense to C

Vb = 86.92..............Vb = -21.36.............Vb = -21.36.........Vb = 86.92
Vc = 74.77..............Vc = -11.04.............Vc = -11.04.........Vc = 74.77
Vd = 119.20............Vd = -34.00.............Vd = -34.00.........Vd = 119.20

In the first two cases, for example, if the control law was ABS(2*Vx), apparently either of the solutions could exist. Does this mean if we started things up with the state of the system in either mode, it would stay there? How could we make it flip states?

Consider the network with Ro deleted. Then with the sense leads in each polarity, calculate the resistance to ground (consider nodes G-H to be ground) from each node. Could a reactive element, say a capacitor, make this circuit oscillate?

 

Since the only practical circuit elements I can think of that will offer the negative resistance required are gyrators I wouldn't be at all suprised if the whole durn shootin match oscillated like hell.

But that still begs the question what do these controlled generators look like?

The control law was not defined as the absolute value of the coefficient.

Indeed when voltage controlled curent sources were first invented there was only one polarity and law available. That is because valves (you call them tubes) only work one way.
Other real world vcsources I can think of include FETs and ideal op amps. But none would suffice here.

One of the problems generally in the modern world is to rely to heavily on the numbers you get from a machine, when you don't fully appreciate what is going on and have no 'feel' for reality.

I have seen too many mistakes ensue, both fatalities and costing $millions when this happens.

So I wonder if this wasn't the real lesson for this question.

Since I have set up the circuit I would be happy to run variations anyone suggests.

 

With regard to polarity of the sense terminals.

Since I have employed SPICE, I have had to employ and connect sense terminals.
Hence I have had to get these the 'right way round'. So I have displayed both ways round.

This requirement in inherent in the SPICE implementation of a controlled source, not in the source itself.

For instance you can only connect a (working) triode one way round.

SPICE also allows more complicated and even non linear control laws, but I don't think Favour has reached that stage yet.

 

studiot,

D is at -34 volts
G is at zero volts

Conventional current is flowing from D towards G

Is that not negative to positive?

For conventional current sources, the end that the charge leaves is considered the positive end. Therefore the voltage will show more negative at D than at G.

I1 and I5 are both negative so inward at D.
So the current from D to G must be outward at D.

And so it is. No question about that.

Finally I ask the question again, what sort of sources are these? I look forward to a practical implementation.

Before I give the same answer again, tell me what you find unable to understand about the previous answer.

Ratch

 

The Electrician,

In the first two cases, for example, if the control law was ABS(2*Vx), apparently either of the solutions could exist. Does this mean if we started things up with the state of the system in either mode, it would stay there? How could we make it flip states?

First of all, circuit would be nonlinear across Vx=0. You would use one set of solutions for Vx>0 and the other set for Vx<0. One way to implement this would be to use a DPDT switch that changes the sensing when the Vx voltage is positive or negative.

Consider the network with Ro deleted. Then with the sense leads in each polarity, calculate the resistance to ground (consider nodes G-H to be ground) from each node. Could a reactive element, say a capacitor, make this circuit oscillate?

Where are we going with this? What does knowing the resistance to ground all nodes with one loop missing buy anyone?

Ratch

 

For conventional current sources, the end that the charge leaves is considered the positive end. Therefore the voltage will show more negative at D than at G.

Consider branch FH. H is at zero, F is at -4.4 volts. Hence H is 4.4 volts more positive than F. 8 amps flows via the independent source from H to F; from positive to negative.

This is as it should be. The whole circuit analysis is done on the basis of conventional current flowing from positive to negative.

Your suggestion about the source in DG reverses this convention. However the rest of the sources follow it.

Before I give the same answer again, tell me what you find unable to understand about the previous answer.

I didn't ask for a glib answer on this I asked for a real circuit comprised of real components that could actually achieve this feat, as I have enever seen such.

I thought we were co-operating at last.

 

But that still begs the question what do these controlled generators look like?

They look like this:
http://en.wikipedia.org/wiki/Operational_transconductance_amplifier

The control law was not defined as the absolute value of the coefficient.

I didn't say it was; I said "...IF the control law was ABS(2*Vx)...", with the emphasis on IF.

For instance you can only connect a (working) triode one way round.

SPICE also allows more complicated and even non linear control laws

Spice allows voltage controlled sources that have 2 terminals to sense the controlling voltage, and an output which may be a current or a voltage, to be applied elsewhere in the circuit.

 

Where are we going with this? What does knowing the resistance to ground all nodes with one loop missing buy anyone?

Calculate them and then use your imagination.

 

studiot,

Consider branch FH. H is at zero, F is at -4.4 volts. Hence H is 4.4 volts more positive than F. 8 amps flows via the independent source from H to F; from positive to negative.

This is as it should be. The whole circuit analysis is done on the basis of conventional current flowing from positive to negative.

Your suggestion about the source in DG reverses this convention. However the rest of the sources follow it.

It appears possible for a current source to pump 8 amps into a negative voltage node. The voltage within the current source is indeterminate and it has a theoretical resistance of infinity, but its current is defined. We both assumed that the voltage at node F would be positive, but evidently it was not. The voltages around GEF and GFC are all zero.

I didn't ask for a glib answer on this I asked for a real circuit comprised of real components that could actually achieve this feat, as I have enever seen such.

I asked a honest question because I did not know what you wanted. Are you asking how to implement the dependent source? I don't know what you mean. Please elucidate.

Ratch