# Lab Report Nodal Analysis DC Circuit

Discussion in 'Homework Help' started by arbrog, Sep 26, 2016.

1. ### arbrog Thread Starter New Member

Sep 26, 2016
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As part of an assignment, I am required to perform nodal analysis on a DC circuit consisting of resistors and a DC voltage source. I am a bit unsure of where to begin as when I usually attempt it I end up doing mesh analysis. Any help would be much appreciated and the circuit diagram is below. The nodes I am focusing on are labeled Va, Vb, and Vout+.

2. ### WBahn Moderator

Mar 31, 2012
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6,980
Okay, so don't use mesh analysis. Instead, use nodal analysis.

Why aren't you focusing on the node labeled Vout-?

Post your best attempt to set up your node equations and that will give us a starting point to help you along.

3. ### arbrog Thread Starter New Member

Sep 26, 2016
6
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I didn't focus on Vout- because the load resistor and R5 could be combined into a single 2kΩ resistor. The point of the exercise was to determine the voltage difference between Vout+ and Vout- which is why it's labeled.
The picture below shows my attempt, apologies for 1) not using Latex and 2) having awful handwriting. I can rewrite any part(particularly the sketch) if it's too hard to read.
http://i.imgur.com/kE85QI9.jpg

4. ### RBR1317 Active Member

Nov 13, 2010
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81
Node equations which are poorly formed are difficult to check for errors. Unless there are current sources in the circuit there should be no reference to current in a node equation. Do not try to guess the direction of current flow because it will just become a source of errors.

The node equation for any node should have the simple form: SUMMATION for all adjacent nodes (Vnode-Vadjacent)/(Resistance to adjacent node) + {current sources} = 0. Every term in the node equation begins with the voltage of that node. If there are current sources attached to the node, current flowing out of the node is positive. That's all there is to writing a node equation.

Consider the following circuit and the demonstration of how to write a set of node equations for it. I find it quite easy to check for errors when the node equations are written in this manner.

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5. ### WBahn Moderator

Mar 31, 2012
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Your reason for not including Vout- is perfectly valid.

In your work the problem (which you alluded to in your first post) is that you are too wedded to mesh analysis and so you are trying to develop your node equations starting with mesh equations. But that largely defeats the purpose -- if you are going to do that, then do mesh analysis.

Remember that nodal analysis is nothing more than a formalized application of KCL in which you systematically apply KCL at each node assuming that the currents are flowing outward from each node. With just a little bit of practice you can write them down by inspection, in a form ready to solve, for most circuits.

Taking just your Node A, this would yield

$
\frac{V_A \; - \; V_1}{R_2} \; + \; \frac{V_A \; - \; V_{out+}}{R_3} \; + \; \frac{V_A \; - \; V_B}{R_6} \; = \; 0
$

But notice how, with a bit of rearranging, this becomes

$
V_A $$\frac{1}{R_2} \; + \; \frac{1}{R_3} \; + \;\frac{1}{R_6}$$ \; - \; V_B $$\frac{1}{R_6}$$ \; - \; V_{out+} $$\frac{1}{R_3}$$ \; = \; V_1 $$\frac{1}{R_2}$$
$

If you apply this pattern to the other two nodes you will end up with a set of linear equations in the unknown node voltages in which the diagonal terms are positive, the off-diagonal terms are negative, and the coefficient matrix of the left-hand side is symmetric.

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6. ### arbrog Thread Starter New Member

Sep 26, 2016
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Thank you both for your help. It makes much more sense now. I needed to forgo the mesh equations and focus on KCL making sure to stick consistently with the convention that current flows out of the nodes.

7. ### arbrog Thread Starter New Member

Sep 26, 2016
6
0
I got my system of equations to look like this.

However when I solve the system of equations I get
VA=7.182V
VB=3.425V
Vout+=8.933V
which don't exactly match up with the voltage values I obtained from my PSPICE simulation:

Last edited: Sep 27, 2016
8. ### arbrog Thread Starter New Member

Sep 26, 2016
6
0
Sry image didn't show

9. ### WBahn Moderator

Mar 31, 2012
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6,980
Remember what I said about the coefficients being symmetric?

Your top-right coefficient (last term of left hand side of equation #1) is (1/1 kΩ), but the corresponding term in the other half, the bottom-left term, is (1/5.6 kΩ). So you know right away that you have made a mistake. No point solving anything until this is corrected.

Then look at your diagonal terms.

I see that Node A is connected to two 1 kΩ resistors and one 5.6 kΩ resistor. Your first equation has it connected to three 1 kΩ resistors.

Then your last equation has Vout+ connected to a 1 kΩ resistor, a 2 kΩ resistor, and a 5.6 kΩ resistor. Does that agree with the circuit?

Get in the habit of breaking your work into two pieces (when possible). The first part is setting up the equations. This is were all of the electrical engineering is done. So once you have them set up, spend time reviewing them and asking as many sanity check questions as you can to ensure that little mistakes haven't gotten by. It is usually pretty easy to validate every equation there in your head very quickly. After this is nothing but math, which can get pretty involved. But if the setup isn't right, then all you will have accomplished is guaranteeing that all of the time spent to slug through the math was completely wasted.

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10. ### arbrog Thread Starter New Member

Sep 26, 2016
6
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That makes a lot of sense. Of course, now that I see the mistakes I can't help but wonder how I missed them. I fixed it and solved again. This time a got values that were almost spot on thank you very much for all the help!

11. ### WBahn Moderator

Mar 31, 2012
23,180
6,980
Good!

It's actually very easy to miss them unless you explicitly look for them -- but that's hard to do until you know the kind of things to look for. As you gain experience you will develop a little mental checklist of things to look for and of different ways to ask if the answer makes sense. If you make the conscious effort to ALWAYS apply them, then after a while it will become such an engrained habit that you will do it automatically and not even realize you are doing it most of the time. After that, you will reach a point where you can often immediately spot issues in many instances just because the form of the set of equations doesn't look right at a superficial level and you won't even know what is catching your attention until you walk through them.

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