4/12/16 - Capacitors and Inductors for RLC Circuits.

Today, we start off the class on review about capacitors. We are asked to design a variable capacitor and came up with a design of moving the plates back and fourth to change the capacitance, another design we had was moving the dielectric in and out of the gap between the plates. I personally like my design of just having a large amount of capacitors and hooking up switches to each of them.

We are then asked to derisive the formula for the energy stored in a capacitor. We know the the current is the capacitance times the time derivative of the voltage and since power is the voltage times current, we can calculate the energy stored in a capacitor as the integral of the capacitance times voltage times differential voltage. We find that calculate the energy stored in a capacitor is just (1/2)Cv^2.

Now given a current graph, we are asked to sketch the voltage graph based on the current graph, since we know that the current is a time derivative of voltage, we can just integrate and find a piece wise voltage graph.

Now we are asked to find energy stored in each capacitor of this circuit. 

Since we know that capacitors act as a open circuit in a DC loop, we can simply remove them and treat this as a single loop with resistors in series to find the voltages across the resistors. And the capacitors should have the same voltage. We use this to find the energy stored in each capacitor.

Now given two voltage graphs, we are asked to find the current graphs of each signal. We just derive and turn the sinusoidal wave into a co-sinusoidal wave and turn the linear signal into a constant current.

We also see some similarities between Ohm's law and the equation for current with a capacitor.

There are also a few leaves on the left side which are the result of some boredom. They will mature into something soon.

While waiting around for the next activity, we can see that the leaves matured into a set of berries, but someone ate most of them. So now we can safely say that it is Eric's Berry.

Next, we set up the analog discovery and hook it up to a resistor in series with a capacitor.

We hook channel 1 and 2 up to measure the voltage across the capacitor and resistor.

We input some voltage in the form of a sinusoidal wave to produce this graph on the scope.

We can see that the voltages of each are off by a phase shift.

Here we employ the math function to add an additional channel to calculate the voltage across the capacitor divided by the capacitance.

More pictures of the same graph,

Next, we test our experiential function by adding a trigger to the scope, we tell it to only track the difference past 1V. And we can see when we manually plug and unplug the voltage source, we can see the voltage across the resistor falls linearly (?) and the voltage across the capacitor is experiential as we expected.

Last problem, we are asked to find the equivalence capacitance of this circuit. We know that capacitors work opposite of resistors and series equivalence adds while parallel capacitors add normally.

4/07/16 - Wheatstone Bridges and DACs

Today, we continue the discussion of all the different op amp circuits we can create, and if we were to connect multiple op amps up in a circuit, Here are some calculations showing what would the output be based on different configurations.

Here we have an example of the output of one non-inverting amplifier being the input to another non-imputing amplifier. We solve for each circuits' gains and then multiply them to find our final output voltage.

Next up, we have an application for the difference amplifier. We are told that a strain gauge is designed with a varying reisstance based on how far the item has changed in shape, since it won't have the same resistance.

Here we hook up that small difference in resistance to a circuit known as the Wheatstone bridge, we have the same resistor ratio to be out positive input terminal and hook up a varying resistor and the thermistor to the negative input terminal.

Here are all the recorded values of the elements used in our circuit, we also measured the difference in voltage when we touched the thermistor and found that we had a desired gain of 4 to give us the 0-2V range that Professor Mason wanted.

We set up the circuit with the Wheatstone Bridge on the right side and the difference amplifier on the right side. We notice that the potentiometer does not have the resistance range to match the thermistor at room temperature, so we hooked up a resistor in series to give it an offset.

Please ignore the mini-inductor in the circuit.

We begin by calibrating the Wheatstone bridge to give us an initial difference of 0V by turning the potentiometer.

Then we connected the two terminals of the op amp top find that we were not getting what we expected as our output, we were even getting a voltage though our difference should be 0V.

After double checking everything, we notice that we were producing a non-zero output voltage even though our Wheatstone Bridge was inputting a 0V difference.

We could not produce results that confirmed our discussion of the strain gauge and instead, blamed it on the lousy op amp.

Now we move on to the topic of Digital to Analog Converters, DAC for short. We are asked how would we produce a analog output given a digital signal. Based on the most and least significant digit, we can convert this 4bit digital number into a analog siganl with a summing amplifier and different resistors for the inputs.

An example problem dealing with the DAC and asked to turn the 4bit 1001 into its analog equivalent given that our range is 0-1V and our feedback resistance is 500 Ohms.

4/05/16 - Signal Transformations

Today we learn about signal transformations, and we start off by asking what it would require to turn a sinusoidal wave with a amplitude of 0.1 to get the expected graph on the right. We find that there is a scalar and offset and find it. We say we can implement this using a DC offset and a Op Amp.

Now, given an Op Amp circuit shown on the right, we are asked to find what would the expected output signal be if we were to input a square wave. We calculate the gain to be -3 and say it has a range of 0-9V. Therefore, we have a half a period phase shift and don't have a negative portion of the graph due to the supply being ground.

Next up, we are given a slightly more complicated circuit with an ideal op amp. We use nodal analysis to solve for the output voltage.

Solving for an output voltage of a non-inverting op amp, we can see that the gain is simply the ratio of the sum of the feedback resistance and input resistance over the input resistance.

Next, we are shown the example circuit for a summing apliffer, and given an non ideal op amp, we can see that we have some current passing though into the op amp. We can calcualte this current with Ohm's law and find that the total voltage is the negative sum of all the voltages over their resistance times the feedback resistance.

We then set up the circuit shown in the diagram to test our calculations of an summing amplifier.

Here is the circuit diagram for our amplifier and the measured resistances of the resistors used. Due to saturation of the op amp voltage supply, we have to use halving resistances in order to see the full range of its output.

A picture of the OP27, analog discovery, and resistors used. We have two waveform generators hooked up and have one supplying a constant 1V while the other we are going to vary in the graph.

Here we have our chart of V_o versus V_in. We can see that we have negative gain due to it being an inverting amplifier and a slope of 0.5 due to the resistances halving the total input into the op amp.

Afterwards, we are asked to take the difference of two signals, we find that if we subtract a DC signal from a sinusoidal signal, we end up with an offset-ed sinusoidal signal.

So next up after that was the difference amplifer. We tkae the difference of two input signals and apply some gain to it to get our output voltage. We are also told that for the difference aplifer to work, we need to have the ration of R_1 to R_2 equal to the ratio of R_3 to R_4.

Here is Professor Mason's example of an difference amplifier circuit, we use nodal analysis to find out the output voltage.

So given a square wave with an amplitude of 5V, if we feed it into the circuit, we should expect an output square wave with no negative due to the ground being the negative supply and a amplitude of 5/3V.

Lastly, we are shown an unknown circuit in class and are asked to figure out whether it is a non-inverting amplifier or a difference amplifier. We are assigned to solve for it as a non-inverting amplifier and find that we are wrong. We're told that if we are shown an unknown circuit that looks like it could be either a non-inverting amplifier or a difference amplifier, we should just treat it as a difference amplifier or use nodal analysis.

3/29/16 - Operational Amplifiers

Today we are introduced to a new circuit element, the operational amplifier, Op Amp for short. We're told that its equivalent circuit is a resistor between its two input terminal and it outputs a voltage controlled voltage source.

We draw the circuit for a inverting amplifier and use KCL to solve for the output voltage. We found that the output voltage is amplified by a factor of -2 (gain), hence the inverting amplifier. There is also a small depiction of negative gain.

We can also find the gain by taking the negative ratio of the feedback resistor and input resistor.

Here we have a similar set up, except that this is a non-inverting amplifier. We do KCL again and find the gain, now positive. We can also find the the gain by taking the ratio of the sum of input resistance and feedback resistance over  the input resistance.

Here we set up a circuit with the Analog discovery, two resistors, and an OP27 circuit element to test our calculations.

We shove a DC voltage though it and measure the values of the output voltage, we used resistors that should give us an expected gain of 2.

We measured the output voltage and found that it has a gain of roughly 2.

Here is a chart of our input voltages versus the output of the op amp.

3/22/16 - Max Power Theorem

After learning about Thevenin's and Norton's equivalent circuits, we went over the topic of the max power being delivered to the circuit. We plotted power delivered to the circuit versus the load resistance and found that it has a peak value. We found this peak value to be the same as the Thevenin resistance.

So given this circuit, we first do our source transformations to find the Thevenin resistance and voltage.

We then plotted difference load resistances and found the power delivered to the circuit with each resistance value. We see that the relationship appears to be a inverse quadratic as we expected with the peak value being the Thevenin resistance.

Now we have a design problem, given three motors that have a defined load resistance and their operating voltage, we are told to find the correct Thevenin resistance to get the max power out of the motors, we took our resistance value incorrectly and failed to solve the problem.

Next, we are given a circuit and asked to find R_L to get maximum power out of it.

We source transform the circuit and find the Thevenin voltage and resistance. With that, we find the load resistance and max power. There is also a nice banana drawing showing itself off.

Next, we set up a circuit with a potentiometer to test the max power theorem.

We measure the voltage across the load resistance, the potentiometer, to find our power.

We were also told to make a table of power versus resistance.

I seem to have forgotten to take a picture of the circuit, so here's Chris' version of the table of P vs. R.

Here is the circuit diagram, we find beforehand that our max power graph should have its peak value at 4.78 kOhms.

After measuring the voltages across the potentiometer, we find that we have our peak value of 1.35W just around the 4-5 kOhm mark as we expected.

Lastly, we are told that a battery supplies exactly 12 voltage to a load resistor of 2Ohms, and we are expected to find the internal resistance of the battery. We are also given that the battery measures 12.4V on an open circuit, we use these two values to solve for the internal resistance and then find the voltage across an 8 Ohm load resistance instead.