Nodal Analysis (AC Analysis)

Nodal analysis is one of the many methods in solving and finding a specific value of a parameter in electronic circuit analysis. The aim of using nodal analysis is to determine the voltage in each node that’s relative to the reference node, which is the ground GND where voltage is equal to 0. This means that all the other nodes present in the circuit are referred to as the non-reference nodes; the ones that has voltage you are trying to solve for. Depends, of course, if you do need the voltage present in them or not.

Let’s do a quick recap about the parts of an electronic circuit; A node is a point of connection between two or more branches. A branch represents a single element such as a voltage source, or a resistor, etc. And a loop is any closed path in a circuit.

Here are the steps on how to determine node voltages:

Determine the nodes of the circuit, and then select a node as the reference node (ground GND). Then assign the non-reference nodes to voltages V1, V2, Vx, or whatever you feel comfortable with.
Apply KCL (Kirchhoff’s Current Law) to each of the non-reference nodes. Use Ohm’s law (V=IR) to express the branch currents in terms of node voltages. I use the shortcut method, though the same principles are still applied.
Solve the resulting simultaneous equations formed from the non-reference nodes to obtain the unknown node voltages.
Always remember that the number of equations formed should be equal to the number of unknowns. So, taking this simple circuit as an example:

This sample circuit has an AC voltage source, a current source, a resistor, an inductor, and a capacitor. We can also observe that there are four nodes present in this circuit. Assuming that we need to find the voltage across the capacitor, what node will we choose as our reference node GND that will make the problem easier? Note that the lesser the unknowns, the easier the problem will be. It’s like the unknowns determine the difficulty of the problem.

So, there are four nodes. If we were to select the top-left node as the GND, V1 as the top-mid node, V2 as the top-right node, and V3 as the bottom node, it would mean that we can get the voltage across the capacitor with V1 – V3 where V3 = -Vs (voltage source, since the negative terminal of the voltage source is connected to node V3). This would be a fine option but it’s kind of — maybe, unethical — to have the GND connected to the positive terminal of the voltage source.

If we were to select the top-mid node as the GND, V1 as the top-right, and the same position for V2 and V3, then a supernode (formed by enclosing a voltage source, either dependent or independent, connected between two non-reference nodes and any elements connected in parallel with it) would be present which adds more difficulty in solving the circuit. Same situation goes if the top-right node is the GND.

But if we were to select the bottom node as the GND, and V1, V2, and V3 as the top-left, top-mid, and top-right nodes, respectively, then we can get the voltage across the capacitor with V2 – GND = V2 – 0 = V2 (since current flows from a higher potential to a lower potential in a resistor), while V1 = Vs (since the positive terminal of the voltage source is connected to node V1). That makes two unknowns (V2 and V3), though we only need to solve for V2. And it’s more ethical compared to having the GND on top of the circuit.

Now to label the nodes and elements, assuming that all the elements’ respective values are given and have been converted to their equal impedances already. If you want to know how to solve for the impedance in a resistor, inductor, and a capacitor,

There we go! So, what we are trying to find is the voltage across the capacitor Z3 and we have discussed earlier that Vz3 = V2, since Vz3 = V2 – GND and that the voltage at the GND is equal to 0. Now to form the equation at node V2 using the shortcut method. Brace yourself for I am about to make up names that i’ll be using to better explain the shortcut method.

@node V2:

V2( ) = 0

To form the equation at node V2, you must locate V2 (imagine V2 as a person or an animal or any object) and look around its surroundings. And I mean the lines or pathways that are connected to it. In this case, there are three pathways connected to V2 (path to V1, to V3, and to GND). And each path is connected to an element (resistor, inductor, and capacitor), or what I will be naming as a bridge. And this is how you start the equation:

V2( 1/bridge1 + 1/bridge2 + 1/bridge3) = 0

..or..

V2( 1/z1 + 1/z2 + 1/z3 ) = 0

It’s like connecting impedances in parallel. Continuing to the path across one of the bridges, you’ll encounter another node (it could either be the GND or not), or what i’ll be naming as your neighbour. This is how neighbours are treated in the equation:

V2( 1/z1 + 1/z2 + 1/z3 ) – neighbour1/bridge1 – neighbour2/bridge2 – neighbour3/bridge3 = 0

..or..

V2( 1/z1 + 1/z2 + 1/z3 ) – V1/z1 – V3/z2 – GND/z3 = 0

..and since V1 = Vs, and GND = 0..

V2( 1/z1 + 1/z2 + 1/z3 ) – Vs/z1 – V3/z2 – 0 = 0

..and since Vs/z1 is a constant, we transpose it to the other side..

V2( 1/z1 + 1/z2 + 1/z3 ) – V3/z2 = Vs/z1

And that’s it for the first equation, with V2 and V3 as unknowns. As I mentioned earlier, the number of unknowns should be equal to the number of equations. So, we are going to need one more equation, and that’d be the equation at node V3.

@node V3:

V3( ) = 0

Following the same procedures with the bridges and neighbours, we get–

V3( 1/z2 ) – V2/z2 = 0

As you can see, node V3 is connected to a current source. A current source in nodal analysis is considered as a constant unless it’s a dependent source. When the current source’s direction is away from the node, you add the current to the equation. If its direction is towards the node, you subtract the current to the equation. And since the current source in the sample circuit is directed towards node V3, it goes like this:

V3( 1/z2 ) – V2/z2 – Is = 0

..and since Is is already a given constant, we transpose it to the other side..

Vs( 1/z2 ) – V2/z2 = Is

And there’s your second equation. Now you can solve for Vz3 by fusing the two equations using matrices, substitution, elimination, or whatever method you know. Though our professor requires us to use the matrix method.

Phasor (AC Analysis)

Phasor   is a complex number that represents the amplitude and phase of a sinusoid. Phasor in our understanding is the simplified form of the sinusoidal function. Which means it can be used provided that the angular frequencies if faced with two or more sinusoidal input functions, are all equal. Also it is simpler to input in our calculators, there is a saying that I made up, although it might exist already, that simpler is better. Hoho.

PHASOR:

It has three forms:

 

Although we rarely to never use the exponential form, learning its existence can also help in some type with different specification for different approach and derivation in order to provide the correct solution of circuit models. 

On the other hand, polar form is what we often use in solving because it shows the angle of the sinusoid. Which means we can actually track down if our answers lags or leads the other values. 

There are rules in relating two phasors with each other: 

Time domain and phasor domain: 

Time domain is the general expression of the sinusoid, while phasor domain is a more simplified one. 

Rules for transforming Time domain into Phasor domain:

1. Time domain must be in cosine function
2. Vm or the magnitude of the function must be positive.

- amplitude and phase difference fare two principal concerns in the study of voltage and current sinusoids.
- phasor will be defined form the cosine-function in all our proceeding study. If a voltage or expression is in the form of a sine, it will be changed to cosine (function) by substracting from the phase.

Phasor is also called a complex number, which we are about to take in Advance Mathematics. We are grateful because we can double the learnings we get from both subject and apply each others' learnings with one another, if that make sense.

SINUSOIDS (AC Analysis)

SINUSOIDS 

Here we are, the basic of all learnings we are about to pass through in this program. Since we are dealing with alternating power (AC analysis), we have to first have the foundation of the structures and understand its components. Because unlike direct current (DC), alternating current (AC) is oscillating and periodic, which mean it is in a form of a sinusoid, a signal that has the form of the sine or cosine function. Sinusoidal functions are the basis for study of all periodic functions. This periodic pattern contains several sinusoidal functions.

Amplitude - highest and lowest point of the wave.

Period (T) - the amount of time the wave has to consume in order to get back to a specified point of origin.

As mentioned, the period (T) of the periodic function is the time of one
complete cycle or the number of seconds per cycle. The reciprocal of
this quantity is the number of cycles per second, known as the cyclic
frequency f of the sinusoid. Thus, 

However, sinusoids in AC analysis will have an angle. Thus:

Showing a figure that will clearly illustrate and show the lags and leads of the two given sinusoids.

A sinusoid can be expressed in either sine or cosine form. When
comparing two sinusoids, it is expedient to express both as either sine
or cosine with positive amplitudes. This is achieved by using the following trigonometric identities:

And there we have it! The start of it all, understanding the alternating electricity (AC) is just the beginning. We know, however far we will get, there is always something more to learn :)

Start of the new chapter of Electrical Circuits: Impedances (AC Analysis)

IMPEDANCES:

Let’s start the lessons from here — one of the major (or maybe minor) changes when it comes to analyzing an AC circuit from a DC circuit: the impedance.

Now, the impedance of an electronic circuit is the ratio of the phasor voltage V to the phasor current I. It is declared with the variable Z and is measured in Ohms (Ω). In some ways, you could say that the impedance acts the same as the resistance in a circuit, but not completely.

Resistors, capacitors, and inductors are the three elements that cause impedance. Though you have to convert the capacitance C and the inductance L to get their equivalent impedances Z, by applying these formulas:

RESISTORS: 

 Z=R

INDUCTORS: 

Z=j•ω•L

Z=s•L , s=j•ω

CAPACITORS: 

Z=1/(j•ω•C)

Z=1/s•C , s=j•ω

From the stated formulas above, you can observe that the resistance of a resistor can already be considered as its impedance. While for inductors, you have to multiply its inductance by j (imaginary number*) and ω (angular frequency). And lastly for capacitors, you have to divide 1 by the capacitance, j, and ω.

*equal to the square root of -1.

Here are some important notes to remember for inductors and capacitors when it comes to analyzing circuits at DC and high frequencies:

Inductors are considered as short circuit at DC equivalent circuits, while as open circuit at high frequency. The opposite goes for the capacitors, since they are considered as open circuit at DC, while as short circuit at high frequency.

Don’t forget that impedances Z are also measured in Ohms (Ω) like resistances R. So, when you solve for impedances in series and in parallel, all you have to do is to follow the same formulas used when solving for resistors in series and in parallel.

SERIES IMPEDANCES: 

Zeq=Z1+Z2 + ... +Zn

PARALLEL IMPEDANCES: 

1/Zeq = 1/Z1 + 1/Z2 + … + 1/Zn

That’s how impedances look like in a schematic circuit diagram. Just rectangles. Pretty creative, huh?

In solving for the voltage V and current I, you will still apply the same formulas but replacing the resistance R with impedance Z. So, they would now go like:

V = I · Z

I = V / Z

Though for solving power P in AC circuit analysis, it’s a different story compared to solving it in DC. We’ll deal with that sooner or later. Or not.

And that’s about everything there is in the basics of analysing the impedance in an electronic circuit! Though its only the start for another chapter of our life learning the field we chose to walk.