Introduction

The Method Of Node Voltages
Applied To A Voltage Divider
Applied To A Ladder Circuit

Solving Node Equations With Back Substitution

The Conductance Matrix

Problems

 
Puzzle  1
Puzzle  2
Puzzle  3
Puzzle  4
Puzzle  5
Puzzle  6
Puzzle  7
Puzzle  8

        In this lesson you will encounter general electrical circuits for the first time.  You need to consider what you expect to learn from this lesson.  Here are our goals.

   Given a simple electrical circuit,

   Be able to write the node equations for the circuit.
   Be able to write the loop equations for the circuit.
   Be able to know which of the two approaches above are appropriate in any given situation.
   Be able to solve the resulting equations to determine voltages and/or currents in the circuit.

        There are a number of tools you can use for very small electrical circuits.  For example, given a small electrical circuit you can do any or all of the following.

• Look for resistor combinations - either series or parallel,
• Look for sourcesthat you can find a Thevinin or Norton equivalent for,
• Look for embedded voltage dividers,
• Hope that all of the above will lead to a very simple circuit that you can analyze.

• You need some general method that you know will get you to a result.

• The method of node voltages,
• The loop equation method.

        In this lesson you will begin to learn some general methods for analysis of electrical circuits.  When you learn those analysis methods you will have increased the number of tools in you analysis toolbox.  We ask you to be thoughtful as you do that.  If your regular toolbox is empty and someone gives you a hammer, then pretty soon everything will start to look like a nail.  Not only do you have to learn the tools for analysis, you also have to learn when you need to use them, and how to choose from the tools in your toolbox. 

        We are going to start with the method of Node Voltages.  The name of the method tells you exactly what you are going to do.

• The method of Node Voltages starts with writing the KCL equations for the nodes in a circuit. Those equations are written in terms of node voltages (even though you are writing the Current Law, the variables are voltages!) thus the name!

• There are two nodes in this circuit.  One is marked with a red dot, the other is marked with an green dot.
• There is a difference between the two nodes.  If we know V_s - the source voltage - then we know the voltage at the node marked with an green dot.
• We don't know the voltage at the node marked with a red dot unless we do some analysis.

• Since there is only one node where we do not know the voltage, when we write the equations for this circuit, we really only need one equation.
• There's one unknown and there's going to be one equation.
• The equation we need to write is KCL at the node with the unknown voltage

• We need to define a symbol for the unknown voltage.  We'll call it V_x.
• Then we write KCL at the node where V_x appears.
• Finally, we solve whatever equation results from writing KCL.

        Now, we will write the equation at the node with the unknown voltage.

• To write KCL we need to define two currents for this case.  They are shown below.

• Then KCL is written as:
• I_a + I_b = 0    since both currents are leaving the node.
• We need to get expressions for I_a and I_b.  The expressions we need are:
• I_a = (V_x - V_s)/R_a
• I_b = (V_x - 0)/R_b

• Note that one end of R_b is connected to ground, so a zero appears in that expression.
• It is important to be sure you understand how each current is calculated, and the exact expression for the current.  The current through a resistor, R, connected to a node is:

• Then, since:
• I_a = (V_x - V_s)/R_a and
• I_b = (V_x - 0)/R_b
• We can write the complete KCL equation.
• KCL gives us: I_a +  I_b = 0.
• So, we have:
• (V_x - V_s)/R_a +  (V_x - 0)/R_b  = 0

• Define a symbol for the unknown voltage.  We've done that.
• Write KCL at the node where Vx appears.  Done that also.
• Solve whatever equation results from writing KCL.  We need to do that.

The result is:

        Actually, this circuit was pretty easy.

• This circuit has one unknown node voltage.
• We wrote one KCL equation.
• We had one equation with one unknown.

        Here is a more complicated circuit.  This ladder circuit is more complex for one simple reason.

• This circuit has two unknown node voltages.
• The nodes with unknown voltages are marked with red dots.
• Once again, we have a node where the voltage is known, and it is marked with an green dot.

        We use the same solution algorithm as before.

• Define symbols for the unknown voltages.  We'll use V_x and V_y.
• Then we write KCL at the nodes where V_x and V_y appear.
• Finally, we solve whatever equations result from writing KCL.

• Define symbols for the unknown voltage - V_x and V_y.  They are shown on the diagram below.

       V_x and V_y are shown at the respective nodes.  Notice again that the voltage and the node with the orange dot is already know to be V_s volts when measured with respect to ground.

        For the second step:

• Write KCL at the nodes where V_xand V_yappear.

 I_a+  I_b + I_c = 0.

• Then express each current in terms of the node voltages.
• I_a = (V_x - V_s)/R_a
• I_b = (V_x - V_s)/R_a
• I_c = (V_x - V_s)/R_a

Now, we need to solve this equation.  This is one equation, but it has two unknowns, V_x and V_y.  We need a second equation, and clearly it is obtained by writing KCL at node y.  Writing KCL at node y we get:

• We use I_c' because it is not the same as I_c.  Actually I_c' = - I_c.
• Still, using these definitions, we get:
• I_c' = (V_y - V_x)/R_c
• I_d =(V_y - 0)/R_d

 And, the equation we had from node x is:

Now, we're going to write out both equations and combine V_x and V_y terms.  Here's the result of rearranging the equations that way.

        These are close to what we want.  We are going to rearrange these equations to show the coefficients of V_x and V_y  better next.

        There are several things to note here, even though we only have a set of equations to solve, and we haven't solved those equations yet..  First, the equations can be easily obtained from the circuit diagram.  We'll work on that in a while, but you may already have noticed that there is a kind of symmetry to the equations, and there are clear relationships between the nodes the resistors are connected to, and how those resistance values enter into the equations.

        You may also have noted that the equations that result are simultaneous linear equations.  Basically, you have two unknown node voltages, and the equations you write result in two equations in those two unknown node voltages.  You know everything else in the equations, including resistance and source voltage values.

        Finally, you may already know several algorithms for solving simultaneous linear equations.

 Given these two equations:

        One possible approach to solving these equations is back substitution.  The method of back substitution is fairly simple in concept although you will need to be careful with details in the execution of the algorithm.  Here is the process.

• Solve an equation for one of the unknowns.
• You can pick any equation you want, but pick one.
• You will get an expression for the chosen unknown that involves the rest of the unknowns and the  circuit parameters.
• Then, plug your expression for the chosen unknown into all of the other equations wherever the chosen unknown appears.
• V_x(1/R_a+ 1/R_b + 1/R_c) -V_y/R_c = V_s/R_a
• -V_x/R_c+ V_y(1/R_c+ 1/R_d)  = 0
• In this set of equations, we are going to solve the second (lower) equation for V_x.  Doing that, we get:
• V_x  =  V_yR_c(1/R_c+ 1/R_d)
• Now, we can take this expression for V_x and use it in the first (top) equation.  Doing that we get:
• V_yR_c(1/R_c+ 1/R_d)*(1/R_a+ 1/R_b + 1/R_c) - V_y/R_c = V_s/R_a
• This equation can be solved easily for V_y.  Collect all the terms on the left hand side and solve.
• V_y = V_s/[R_a/R_c(1/R_c+ 1/R_d)*(1/R_a+ 1/R_b + 1/R_c)-1/R_c]

        Given a set of equations arising from applying the node voltage method, there are several solution options.  So far we can see the following.

• With one unknown node voltage, one equation results and it can be solved easily.
• With two unknown node voltages, two equations result, and they can be solved with the method of back substitution.
• With three unknown node voltages, three equations would result.  We could use back substitution to eliminate one of the variables, getting two equations, then use back substitution again to get to one variable and solve.  It looks and sounds messy, but we could do it.

• As long as we have linear resistors  - and no diodes, transistors, etc. - we will always get a set of simultaneous linear equations when we write the node equations.
• Back substitution is a well-known way of solving sets of linear equations, but other methods based on determinants and matrices are widely used.
• Many mathematical analysis packages - like Mathcad and Matlab, for example - have built-in routines for solving sets of simultaneous linear equations.   If we want to use methods based on determinants and matrices, and especiallly if we want to use the packages available, we will need to be able to represent our node equations in a way that is different.
• We need to be able to represent our equations in a vector-matrix format.

• V_x(1/R_a+ 1/R_b + 1/R_c) -V_y/R_c = V_s/R_a
• -V_x/R_c+ V_y(1/R_c+ 1/R_d)  = 0

The vector-matrix formulation has several important characteristics.

• The information about the circuit connections are contained in a matrix that has numerous reciprocal resistance values.  Actually, they are conductances, and this matrix is called the conductance matrix.
• The unknown node voltages are represented as components in a voltage vector.
• The voltage source gets embedded in a source vector on the right hand side of the vector-matrix equation for the circuit.

• The matrix, G, is the conductance matrix.  It has units of ohms^-1.
• The vector, v, is the node voltage vector.  It has units of volts.
• The vector, I, is the source vector.  It has units of current.

Getting it is another story.  How many ways do you know to take the inverse of a matrix?  How do you invert a matrix?  Let us count the ways.

• The inverse of a matrix, G, is given by:
• Adj(G)T/|G| so, you can compute an inverse by hand.
• Mathcad and Matlab have matrix inversion functions.  Use them.

        We wanted to show you this first because we wanted to build a case that it may not be a good idea to compute this inverse symbolically by hand.  This is not a large circuit, and the analytic inverse looks fairly horrendous.

• Even a circuit as simple as this points to the fact that numerical solutions to circuits are very important.  In turn, that implies the you need to be familiar with numerical ways to solve the kinds of equations that come out of linear circuits.  That means you need to understand the algorithms and you need to be able to use those algorithms in various incarnations especially as they appear in analysis applications like Mathcad and Matlab.

• R_a, R_b  and R_c are all attached to the first node, where V_x appears.
• In the conductance matrix, the reciprocals of all those resistances are added together to give element (1,1).
• Similarly, R_c and R_d are attached to the second node, and their reciprocals are added to give element (2,2) in the conductance matrix.
• Finally, note that R_c is shared between the two nodes, and the negative reciprocal of R_c appears in elements (1,2) and (2,1) in the conductance matrix.

• If you have a resistance between two nodes, node m and node n,
• Then that resistance contributes a term that is the nevative reciprocal of the resistance to the (n,m) entry in the conductance matrix, and to the (m,n) entry in the conductance matrix.
• All such contributions are added together.
• If you have a resistance attached to node n,
• Then that resistance makes a positive contrbution to the (n,n) entry.

E1   Here is a small ladder network.

And, here is the conductance matrix equation for the circuit above.  Notice how the conductance matrix is built up from the conductances attached to each node using the method we outlined above.

        When we have a voltage source with one end connected to ground, then the right hand side of the circuit equation set takes on a particularly simple form.

• The voltage source appears in the equation for any node with which it shares a resistor.  In the example circuit, V_s shares R_a with the first node - V_x.  The voltage source appears on the right hand side of the equation for this node, being multipled by the reciprocal of the shared resistance.

       Now, here is something that you can putter with to see how the equations are built up.  If you click on an element - a resistor or the voltage source - you will see how the conductance matrix and source vector are built up symbolically.  You can add elements to the conductance matrix and the source vector (on the right hand side of the vector-matrix equation) in any order.  When you are finished, the result is independent of the order in which you account for the elements in the circuit.