07-Apr-2016: Day 14

Exercise 1:

Cascading operational amplifiers may look frighting until we realize that each piece is to be attacked as a small individual block. The total gain is then just the product of the individual gains. Below is an example.

Exercise 2:

A strain gauge is a device used for indicating the strain of a material or structure at some point of attachment. Students have probably worked with these in the form of a force sensor. As the material is deformed, its resistivity changes and thus resulting is a correlation between strain and resistance.

Exercise 3:

Building our first multi-component circuit...

Temperature Measurement System Design Lab:
(Long Name = Long/Tedious Lab)

In this lab we were to design and implement a measurement system that outputs a voltage which is related to the temperature. A thermistor, resistor that decreases as temperature increases, will be used in a Wheatstone bridge and then connected to a difference amplifier to give us some voltage response on the output as a function of the temperature.

The Wheatstone is depicted below.

The voltage drops at point 'a' and 'b' are depicted below.

With the above being true, then the difference in potential energy between points 'a' and 'b' may be written as follows.

Notice above that if the dR << R then it can as shown below.

If R >> dR then 4R >>>> dR, therefore dR may be ignored in the denominator but not in the numerator and the above is true.

Next, we build and become frustrated... ideally because the resistors in the Wheatstone bridge are all "equivalent" to the thermistor at room temperature then we should be getting a voltage reading of zero. Unfortunately in the real world we must work with what we have and what we have is a variance of almost 1 kilo-Ohm between the room temperature thermistor and the other resistors in the bridge.

We fought and scratched our heads before realizing that if we had simply ignored the fact that we could not start at zero then when we run the experiment the voltage decreases from some small positive voltage, reaches zero, then continues to increase in the negative direction. When the voltage is zero, here is where the thermistor is "equivalent" to the other resistors in the bridge. Still not convinced? Well by now implementing a potentiometer into the bridge connected in line with the thermistor across 'b' then we could dial our resistor value to match that of the thermistor at room temperature and reach a value of "zero" because "zero" is good enough, as seen below.

Below is a video of what happens.

It also turns out that the voltage you put in plays a large role in the interference that is picked up and below you see we used only 20 mV to reach a "pretty" result.

Below is a sketch of the circuit we built and some of the chicken scratch we jotted down.

As it turns out, the issue we had of the mysterious voltage readings even when voltage input was set to zero could be simply solved with an instrumentation amplifier.

Exercise 4:

Our next brief topic of discussion was about digital to analog converters, DACs. They are in many of our pockets without many of us even knowing about them. Video and audio transmissions are often transferred using analog signals while high resolution video is an example of digital signals. Most communication between integrated circuits is in the form of a digital signal so the DACs act as mediums between the integrated circuits and input/output "hardware".

Up next we will be discussing capacitors and inductors... stay tuned.

05-Apr-2016: Day 13

Exercise 1:

We began the class discussing unity gain buffer amplifiers which simply output the same voltage that is put in because the difference between them is so negligibly small.

Continuing with class, we are asked to look at the image below. The sketch on the left can be transformed into the sketch on the right and some simple equations may be written.

It is a an amplified version of the original that is shifted upward.

Exercise 2:

Continuing , we are asked to look at a circuit, sketch below. It is an inverting op amp circuit powered by positive 9 volts and the negative power supply connected directly to ground. What may we predict will happen?

At the bottom of the above image is a sketch of what the graph may look like had it  been powered by both positive and negative terminals. Now, due to the fact that the negative terminal has a zero supply delivered, the actual sketch is depicted to the right of the above image. The function only exists above the horizontal axis, zero.

Exercise 3:

Using nodal analysis, we are asked to solve for the voltages Va and Vo. Once the 2 KCL equations are written in terms of the variables, a few of them may be eliminated because they are equal to zero (i.e. ideal Op-Amp). At this point we end up with 2 equations and 2 unknowns, making this a much simpler problem.

Exercise 4:

If there is a parallel rule for solving circuits, does that imply there is a perpendicular rule? No, not really... But, while we know of inverting op-amps, does that imply that there are non-inverting op-amps? Yes! and to solve the gain we must once again do some nodal analysis.

Below is our analysis of the nodes for the above circuits. Since it is an ideal op-amp, two of the currents are equal and a single KCL equation may be written. Below, the output voltage may be expressed as a function of the gain and the input voltage.

Something that we may notice with non-inverting op-amps is that the gain may never be less that 1... this makes them less useful because of that limitation.

 Exercise 5:

Let us now jump on over to analyzing summing op-amps. If we recall, op-amps are amplifiers that perform mathematical operations, this one will do just that by summing, currents in this case. Nodal analysis is performed and something interesting may be seen. If all the resistors are the same, then the voltage out is simply the sum of the voltages going in.

It may also be noted that by playing with the resistors, we can turn a sum amp into an averaging filter.

Exercise 6:

Summing Amp Lab

Below is the sketch of the circuit we are to design and build. We must chose resistors that will not result in a saturation of our data.

Once we calculate and design the circuit we may build and test it. Below is the simple summing circuit we built.

Data was taken and entered into MATLAB in order to graph and analyze. In a perfect world there are calculated sets of data that are equal to any measured data. In the real world there is wear and tear and that leaves room for error. As shown in the graph below, although we calculated for resistor  values that would not lead to saturation, our sad little operational amplifier did so anyways. Still, we initially had a line almost parallel to that of the predicted result... and for a couple of simple engineering students and Mt. SAC, that is good enough.

Exercise 7:

If there are summing amplifiers then there must also be difference amplifiers, right? Right! Let us first looks at the sketches below and see what is happening. If you solve for the difference between the upper left sketch and the lower left sketch, the result is the lower right sketch. It is the upper right sketch shifted down 1 unit.

Exercise 8:

Continuing with difference amps, below we see a circuit, and by calling it an ideal op-amp and using nodal analysis we may solve for the voltage out in terms of some gain value and the voltage sources. The result is the product of the gain and the difference of the voltage inputs.

Exercise 9:

One more exercise to practice solving circuits with difference op-amps.

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.