How To Calculate Node Voltages In A Circiut
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1 Matrices & Their Applications: Nodal Analysis Introduction Nodal analysis is a method applied to electrical circuits to determine the nodal voltages. In electrical circuits nodes are points where two or more electrical elements such as resistors meet. For two nodes to be different their voltages should be different. In Figures 1 and 2, the nodes are indicated with numbers. Figure 1, 2, nodes in electrical circuits This nodal analysis method uses the Germen scientist s, Kirchhoff s rules, to determine the voltage of each node in a circiut. Kirchhoff s Current Law states that the amount of current entering a node must equal to the amount of current leaving the node. And that the algebraic sum of all currents entering and exiting a node must sum to zero. To understand this, picture a bottle with a hole in each end. Now imagine we are pouring water through these holes. To determine the amount of water you have within the bottle, it is reasonable to conclude that it would be the total amount of water flowing into each hole. In electrical circuits currents are flowing in and out of nodes. In Figure 3, I1, I2, I3 represent the currents flowing in and out of a node.
2 Form Kirchhoff s Law of Current, the sum of the currents entering a node, equal the sum of the currents leaving that node. So, I enter = I leaving Node equation: I1 + I2 = I3, As you can see I1 and I2 are entering the node and I3 is exiting the node. If we move I3 to the left side of the Node equation, then the node equation becomes, Node equation: I1 + I2+ (- I3) Figure 3, currents flowing into and out of node Now that we have an idea of what a node is and how to construct its Node equation, we can use this knowledge to find the voltages at each node using matrices. Procedure Deriving Nodal Analysis Equations: The number of nodes you have directly relates to the number of equations you will have to solve. Looking at the schematic in Figure 1, there are three nodes so after applying nodal analysis I should receive three nodal equations, which in this application is a system of three related equations. Since the equations are all related (the node voltage values depend on one another) you can solve them simultaneously using the matrix operations learned in your math class but there s one method included at the end of the lab. Rule 1: If a node is directly connected to the end of a voltage source (i.e. 8.7V or 1.2V battery) the voltage at that node is the voltage of the battery, you will see this in the following nodal analysis example. This lab will utilize 9V sources (9V battery) and 1.2V sources (AA batteries). NOTE: The voltage the batteries will output will be slightly less than what it reads on the label.
3 Nodal Analysis Example 1. Label all the nodes in the circuit (e.g. 1, 2, 3...). A node is where one end of the resistor shares the same point with at least one end of another resistor (you need at least 2 resistors connected at one point for it to be considered a node. (For examples of what node looks like please refer to figure 1&2). Figure 3: Example of a node on a protoboard Figure 4: labeled nodes and voltages
4 2. Assign a variable to represent the voltage of each previously labeled node (e.g. V 1, V 2, V 3...). So in the previous step, the node you marked as node 1, the voltage at this node is V 1. The node you labeled as node 2, the voltage at this node is V 2. Do this until all the nodes in the circuit are labeled with a variable and corresponding voltage variable. After applying the nodal analysis steps, V 1, V 2 and V 3 will yield the nodal equations mentioned above. Refer to back to Figure 4, it shows the labeled nodes and their corresponding voltages. Figure 5: Example Circuit 1(schematic and protoboard) To derive the nodal equations for the circuit in Figure 5, we first look at Node 1 and derive node equation #1.
5 NODE EQUATION #1: Since Node 1 is directly connected to a voltage source, we can safely assume and or derive using Kirchhoff s current law, that Node 1 has the same voltage as the voltage source V1, in this case Node1 (V1) = 8.7 Volts. Yielding node equation #1, NODE EQUATION #2: Node equation #1: V1 = 8.7Volts For node 2, visualize that all the currents are leaving from this node to other parts of the circuit; it may make it simpler to draw them. Now applying Kirchhoff s Current Law (KCL), the sum of the currents equal zero. Since current is, I = Voltage Resistance, you can substitute this expression in for the currents leaving node 2. Writing these equations in terms of the node voltage variables instead of I1, I2 and I3 allows us to write node equation #2, where V 1 is the voltage corresponding to node 1, V 2 corresponds to node 2 and V 3 to node 3. Node equation #2: VV2 VV1 RR1 + VV2 0 RR2 + VV2 VV3 RR3 As you can see the equation is set up beginning with node 2 voltage V 2 and visualizing all the currents leaving this node. The currents are leaving node 2 but are going towards another node in the circuit, the voltage across the resistors in-between the nodes is V = (V 2 V 1 ) for the current that is leaving node 2 and going towards node 1. This is idea is what is used to derive the other voltage expressions in Node equation #2, (V 2-0) and (V 2 V 3 ). NODE EQUATION #3: Next we set up the node equation for Node 3, looking at the circuit from the perspective of node 3. Again, visualize that all the currents are leaving from this node. Since there is only two directions the currents can leave from node 3, only two currents exist. Applying KCL,, Node equation #3: VV3 VV2 RR3 + VV3 0 RR4 -Solving Node Equations: Now that we have all 3 equations: Node equation #1: V1 = 8.7Volts Node equation #2: VV2 VV1 + VV2 0 RR1 RR2 Node equation #3: VV3 VV2 RR3 + VV3 0 RR4 + VV2 VV3 RR3
6 We need to simplify them in order to put them into matrix form to solve for unknown node voltages (V 1, V 2, V 3 ). 1. Substitute all known values into Node equation #2, for this example, plug in (V1 =8.7V): VV ,000 + VV2 0 10,000 + VV2 VV3 1, Simplify equation #2 above further by equalizing the denominator to have the same values (i.e multiplying each fraction by 10,000). This yields, 10V V2 + 10V2 10V3 You always want to rearrange your final equation in the order V 1, V 2, V 3 etc. After rearranging and simplifying, node equation 2 becomes, Node equation #2 (simplified): 21V V3 = Look back on Node equation #3, substitute in the known resistance values: Node equation #3 = VV3 VV2 1,000 + VV3 0 10,000 For simplicity we again equalize the denominators as we did above (multiplying each fraction by the largest denominator): Rearranging yields, 10V3 10V2 + V3 Node equation #3 (simplified): -10V2 + 11V3 4. Put the simplified Node equations 2&3 into a matrix to solve for the 2 unknown node voltages (V2, V3), we already know that V1 = 8.7V. As you can see from the matrix below that placing the node equations into a matrix just means place the constants in front of V2 and V3 in matrix form. Node equation #2 (simplified): 21V V3 = 87 Node equation #3 (simplified): -10V2 + 11V * VV2 VV3 = 87 0
7 Matrix equation: A * X = B Where A is a matrix of the constants in from of the voltage variables on the left-side of the node equation, X is a column matrix of the node voltage variables and B is a column matrix of the right-side of the node equations. Where A = , B = 87, X = VV VV3 We can rearrange the Matrix equation to obtain the node voltage variables, X = A -1 * B We solved the matrix using a calculator, obtaining V2 = Volts V3 = 6.64 Volts You will be using methods learned in your class to solve these matrices. You should obtain the same values for V2 & V3. Problem 1 Below we demonstrate nodal analysis on another circuit. See if you can follow the node equations. If you forget how an equation was derived refer to the example above. Figure 6: Circuit 2 (schematic & protoboard)
8 NODE EQUATION #1: Node equation #1: VV1 VV2 RR8 + VV1 0 RR6 As you can see the equation is set up beginning with V1 and seeing all the nodes connected to node 1 through a resistor. The two values are always divided by the resistor connected the two of them. NODE EQUATION #2: Node equation #2: VV2 VV1 RR8 + VV2 1.2VV RR7 + VV2 VV3 RR9 As you can see above the equation for V2 has a part where V2 is connected to the 1.2V battery through a 1k resistor labeled R7. Since the node it is connected to a battery, we can actually use the value of the battery directly instead of creating a new node. NODE EQUATION #3: Node Equation #3: VV3 VV2 RR9 + VV3 0 RR10 -Solving Node Equations: 1. first input all the known values into equation #1: VV1 VV2 + VV1 8.7VV 10,000 1, Simplify equation #2 above further by equalizing the denominator to have the same values. i.e to have 10,000 as the denominator to get:
9 V1-V2 + 10V1-87 OR Node equation #1 (simplified): 11V1 V2 = Look back on Node equation #2, substitute the resistance values: Now we simplify as above: Reorganized: VV2 VV1 + VV2 1.2VV 10,000 1,000 + VV2 VV3 10,000 V2 V1 + 10V V2 - V3 Node equation #2 (simplified): -V1 + 12V2 -V3 = Now we look back on equation #3, we input the values in there: Now we simplify as above: Reorganized: VV3 VV2 + VV3 0 10,000 1,000 V3 V2 + 10V3 Node equation #3 (simplified): -V2 + 11V3 5. Put the 3 simplified equations into matrix equation to solve for the 3 unknown voltages ( V1, V2, V3) VV * VV2 = VV3 0 Solve the matrix above. What did you obtain for V1, V2 & V3? V1 = V2 =
10 V3 = Your answers should be close to these values: V1 = Volts V2 = Volts V3.153 Volts or 153 millivolts Problem 2 Figure 7: Circuit 3 (schematic & protoboard)
11 Since Node 1 is directly connected to a voltage source, we can safely assume that Node 1 has the same voltage as the source, in this case V1 = 1.2 Volts. Node Equation #1: V1 = 1.2Volts Next we set up the node equation for Node 2: Node Equation #2: VV2 1.2VV RR11 + VV2 VV3 RR13 + VV2 VV4 RR12 Next we set up the node equation for Node 3: Node Equation #3: VV3 VV2 RR13 Next we set up the node equation for Node 4: Node Equation #4: VV4 VV2 RR12 + VV3 8.7VV RR14 + VV4 0 RR15 SOLVING NODE EQUATIONS 1. Substitute all the known values into equation #2: VV2 1.2VV 1,000 + VV2 VV3 1,000 + VV2 VV4 10,000 From this, we can simplify equation #2 above further by equalizing the denominator to have the same values. i.e to have 10,000 as the denominator to get: OR 10V V2 10V3 + V2 V4 Node equation #2 (simplified): 21V2 10V3 V4= Look back on equation #3, we input the values in there: Now we simplify as above: VV3 VV2 + VV3 8.7VV 1,000 1,000 Node equation #3 (simplified): -V2 + 2V3 = 8.7
12 3. Look back on equation #4, we input the values in there: Now we simplify as above: VV4 VV2 + VV4 0 10,000 10,000 Node equation #4 (simplified): -V2 + 2V4 4. Put simplified equations 2, 3 & 4 into matrix equation to solve for 3 unknowns (V2, V3, and V4) VV * VV3 = VV4 0 Solve the matrix above. What did you obtain for V1, V2 & V3? V2 = V3 = V4 = Your answers should be close to these values: V1 = 1.2V V2 = 3.51V V3 =6.14V V4 = 1.79V Conclusion Now that you have gone through the lab, you should have an idea on how to apply nodal analysis- applying Kirchhoff s rules (Current and Voltage Law) to derive node equations and utilizing matrices to effectively solve for the nodal analysis on a circuit to determine the voltage at each node in a circuit.
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