29-Mar-2016: Day 11

Exercise 1:

Operational amplifiers (op amps)

Op amps are small intricate circuits that essentially do math operations (i.e. sum, multiply, differentiate...etc). 

Generally they are "dumb down" into the small triangular circuit you see below. There is a voltage drop between two terminals (inverting and non-inverting terminals) on the left side which is the voltage input side, and a seemingly disconnected voltage output on the right side. While drawn disconnected, the voltage output is actually dependent on a gain value multiplied by the voltage input. In reality the internal structure is too horrifying to even imagine at this level of electrical analysis. 

We were asked to try and describe the meaning of open and closed loops.

Above, you see our failed attempt at open and closed loop, which actually describe whether the output has a way sending feedback to the input (i.e. open = no feedback, closed = yes feedback). Feedback is great because it allows the user to operate in a linear range.

Exercise 2:

Analyzing circuits with op amps.

Given a simple circuit with an op amp, we were asked to redraw it to its equivalent circuit, as shown below.

Using the redrawn circuit, we are now asked to determine the gain. As it turns out, gain of an op amp is the ratio of the output over the input voltage, as seen below.

Continuing with the same circuit we go a step further in determining the gain value, given what we have.

Exercise 3:

More practice...

Attempting to solve for the gain again, we quickly discovered that mesh grid method is much more of a mess to analyze than node analysis.., With that said, NODE ANALYSIS is the way to go when analyzing these circuits with operational amplifiers.

Exercise 4:

Inverting Voltage Amplifier Lab

It turns out that gain can more simply be expressed as the ratio between the resistor that closes the loop and the resistor that connects the voltage input to the op amp. By selecting a set of resistors that would give us a ratio of 1:2. Seeing as how this is an inverting op amp what we will see is in terms of  voltages is a negative gain.

V_out = -(R2/R1)*V_in

R1 = 1.8k Ohm
R2 = 3.6k Ohm

Above is a horrifying mess of equations that were done just for the sake of doing. Instead we may use MATLAB to as usual compare theoretical and experimental plots of data. Below is the theoretical calculations.

Below is the collected data.

Graphed together, below, it is easy to see that the theoretical and experimental line up almost perfectly... at least for a moment. What happened on the outer ends of the experimental plot? Well, as it turns out, the op amps have saturation points where essentially a no further change in voltage occurs.

Saturation Voltage range
1.5V < V_out < -2V

Gain = 2

22-Mar-2016: Day 9

Exercise 1:

After redrawing a complex circuit into its Thevenin equivalent, then we may easily calculate the power of the load resister. As seen below, by simply converting current into its voltage-resistance equivalent, a value may be calculated.

The question is then, what would the graph of power as a function of the variable load resistor look like? Below is a sketch which turns out to be the product of two more well known functions.

The sketch of the graph is not exactly what is depicted on the right of the image. What was meant to happen was a graph that increases rapidly, slows as down until reaching a maximum power value, then gradually descends until it almost reaches zero... looking more like the following image.

What this tells us is that there exists a value for the load resistor that will provide a maximum value of power. 

It  turns out that when the load resister is equal to that of the Thevenin equivalent resistance, maximum power is achieved.

A simplified equation for power may then be written as below in terms of only the Thevenin voltage and resistance.

Exercise 2:

Given the circuit below we are to:

1. Obtain the Thevenin equivalent at terminals a and b.
2. Calculate the current through the load resistor at 4, 8, 12, and 16 Ohms.
3. Find the load resistance value for maximum power.
4. Determine the maximum power value.

Using some source transformations the circuit may be rewritten. Current throughout is equivalent since we are looking at a circuit that is in series, therefore, it is simply the Thevenin voltage over the total resistance. Previously it was mentioned that maximum power occurs when the load resistor is equivalent to that of the Thevenin resistor. Power is now a simply expressed in terms of Thevenin voltage and resistance.

Exercise 3:

Below, on the upper-right is a diagram with three motors connected in parallel with a single power source and some equivalent resistor.

Along with the diagram, we are given the resistance of the lower motor which is brushless. For the upper two motors, CIM, a spec chart is provided and we are to calculate the resistance of each motor.

Below is a better image of the calculations. Once the CIM motor's internal resistance is determined, the three motors may be treated as resistors and added accordingly. A maximum power may now be calculated by stating that the Thevenin resistance is equal to the already calculated load resistance and solving using a 24 Volt source.

Exercise 4:

Max Power Lab

We have learned through Thevenin's theorem that maximum power is delivered to the load when the load resistance is equal to that of the Thevenin resistance of the circuit. The purpose behind wanting to utilize maximum power is because it results in minimum loss of power as heat dissipation.

Above we constructed a simple circuit where we have a controlled voltage source, feeding into the circuit, and a variable resistor (i.e. potentiometer) that we will control at 1 kilo-Ohm  increments.

Below, is a table of voltage out, the voltage across the variable resistor, so we may calculate power at the load.

Our good friend MATLAB has come to the rescue again, providing us with the following plots.

Blue: Ideal plot
Orange: Measured plot
Light Blue: 4th Polynomial Fit
Red Dot: Max Power

Table of max values.

m1 = Voltage Out

m2 = Thevenin Resistance (constant)

m3 = Load Resistance

m4 = Power

As we can see, the load and Thevenin resistance are not exactly equal, but rather "close enough". Looking back at the graph and our data, it is possible that smaller increments of load resistance could have improved our values.

17-Mar-2016: Day 8

Exercise 1:

The opening topic of the day is Thevenin's Theorem which states that any combination of voltage sources and resistors may be replaced with a single equivalent voltage and a single resistor connected in series with the ends open. The voltage across the open ends of the circuit is the same as the new voltage source value, the Thevenin voltage.

Below we ran a practice example to use the theorem. The variable resistor on the far right may be removed temporarily and the circuit left open. Voltage sources may be replaced with short circuits and current sources may be removed all together. This will help in solving the Thevenin resistance.

Below is what the new circuit can be drawn as, where any load resistor can be simply added in series in order to solve for voltage drops due to the loader.

Exercise 2:

A trickier problem, as seen below can require source transformation to reach a central element.

Exercise 3:

Lab:
"Thevenin Circuit Lab"
Given the circuit below, we are to solve for the Thevenin resistance and voltage in order to create a Thevenin circuit. After only 4 failed attempts at solving for the voltage, nodal analysis made our lives much more happy.

Theoretical

Rth = 7.4 kilo-Ohms 

Vth = 0.46 V

Once solved for, we will want to build and measure for the same values in order to compare them with our theoretical ones. To measure the equivalent resistance we must short circuit the voltage sources and measure across the open ends of the circuit.

Measured

Rth = 7.63 kilo-Ohms

Vth = 0.437 V

Now, what if we add the variable resistor, potentiometer in this case? Well, we hypothesized that if a resistance equaling the Thevenin resistance, 7.4 kilo-Ohms, were added to the circuit, then the voltage drop across that resistor would be exactly half of our Thevenin voltage.

Haza! we were correct.

Exercise 4:

Last extra topic of discussion was the Norton equivalent circuit... essentially the source transformation of the Thevenin equivalent circuit.

15-Mar-2016: Day 7

Exercise 1:

Lab:
Nothing like starting the day off with a simple circuit lab.
By using the waveform generator feature of the Analog Discovery we may generate time-varying signals and use the oscilloscope to measure them.

Here we will predict the sine, triangular, and square graphs for the lab setup.

Exercise 2:

Problem:
Continuing to solve system of equations.

MATLAB doing what it does best.

Exercise 3:

Problem:
But what if there was another way to solve for unknown currents or voltages, superposition. We will treat a voltage source as a single wire and a current source as an open circuit to solve for a single element. The summation of the two answers is the value of the unknown element.

More practice with superposition.

Exercise 4:

Problem:
... perhaps a method that reminds us of transformers... robots in disguise... after all, there is more than meets the eye. Source transformation, the greatest thing since pancakes for dinner. Below is a diagram of the circuit we will be analyzing, and in the upper left is a 4 ohm resister with some current flowing through it. The task is to use our newly found technique to solve the unknown current through the 4 ohm resister.

Exercise 5:

Lab:
Learning about the source transformation technique we can now use it as part of this lab procedure in order to first predict the unknown voltage across the 6.8 kilo-ohm resister. Below is a sketch of the circuit we will build and a table with our measured and predicted values.

The percent variance between the measured and predicted values for the voltage are minimal and may be passed off to rounding error as well as the calibration in the resistances.