Lab Report Day Seventeen - RLC Circuit Response

RLC Circuit Response

This lab will emphasize modeling and testing of a second order circuit containing two resistors, a capacitor, and a n inductor. In this assignment, the step response of the given circuit is analyzed and tested. The measured response of the circuit is compared with expectations based on the damping ratio and natural frequency of the circuit. 

We built the circuit with a 1.1Ω resistor, a 47 Ω resistor, a 10uF capacitor and a 1 uH inductor. 47 ohm resistor and the inductor are in series while they are in parallel with the other components in the circuit.

 

This is the set up for the circuit. 

 

This is the input of our power. It is a square wave at frequency at 100 Hz, and amplitude of 2V at offset 0

This is the graph for V.

In the lab, the 47 ohm resistor has a real value of 49 ohm. The 1.1 ohm resistor has a real value of 4.3 ohm. The 10 uF capacitor has a real value of 9.92 uF, and we cannot measure the 1 uH inductor.  (In the picture, the 1 uF should be 1 uH.)

We calculated our experimental value of α to be 3588.7, while our theoretical value is 1063.8. It has a precent difference of -237.4%. 

 

Summary:

Today we learn how to analyze circuits containing a resistor, capacitor and an inductor in parallel. There are three possible cases that may occur in a RLC circuit which include over damped, underdamped, or critically damped. We also do a lab about RLC circuit response. In this lab we get large percent error between theoretical and experimental values. It may result from not measuring the data correctly and not correctly inputting the values. 

Lab Report Day Sixteen - Series RLC Circuit Step Response

 

Series RLC Circuit Step Response

This lab emphasizes modeling and testing of a series RLC second order circuit. This lab assignment will consist of two parts:

•In Part I of this assignment, the step response of a given circuit is analyzed and tested. The measured response of the circuit is compared with expectations based on the damping ratio and natural frequency of the circuit.

•Part II of this assignment consists of a simple design problem: the circuit of Part I is to be re-designed to make it critically damped, without changing either the natural frequency or the DC gain. Again, the circuit step response is measured and compared to expectations.

Since we do not have time, Professor Mason only ask us to do the first part and record the data. 

 

This is the circuit we are going to build. First, we are going to calculate whether the circuit is overdamped, underdamped, or critically damped. 

This is our pre-lab. Based on the value given, we can see that our circuit is underdamped. We also calculated the value we need to make it critically damped. 

This is our set up for the circuit. In the circuit, we use a 470 nF capacitor, a 1 uH inductor, and a 1.1 ohm resistor.

 

Our experiment value for capacitor is 420 nF.

Our experiment value for resist is 1.4 ohm. 

We are not able to measure the inductance. We assume it is correct. 

This is the input graph. It is a square wave at 500 Hz with 0 V offset. 

This is our output function. 

We calculate our experimental omega to be 5.15*10^4. Compared to our theoretical value 1.351*10^6, we get a percent difference 96.2%. That is really a big percent difference. 

 

When we record some data points form the graph, and have an offset by 2V, I get this graph in Excel. The plotted function is Y=3.5*e^(-8811x). We can rewrite it as V=3.5*e^(-8811t). This means that our experimental α is 8811. Compared to our theoretical α, 5.5*10^5, it is only 1.6% of the theoretical α. Possible causes of error could be the way we measure the circuit, or could from our input value. 

To calculate our experimental damping ratio, we get α/omega, and we get 0.171. 

Summary:

Today, we learn how to find boundary value, how to solve source free RLC circuit. We also learn how to determine  whether the circuit is over damped, underdamped, or critically damped. We do a lab of a underdamped circuit, and we get a big percent difference which is almost 100%. Possible causes of error could be the way we measure the circuit, or could from our input value. 

 

 

Lab Report Day Fifteen - Inverting Differentiator

In this lab, we are going to prove that the output signal is the derivative of the input signal. We do some derivation, and get Vo = A*R*2*Pi*f*sin (omega*t). We can see that the output voltage is related to the frequency.  

When calculating the theoretical output voltage, there is one trick there. At first, we take f as omega. Then, we figure out that omega=2*pi*f.

We calculate the Vout at frequency of 1k Hz, 2k Hz, and 500 Hz. 

 

This is the circuit we build. We use 470 ohm resistor, and 470 nF capacitor. 

 

We measure our real resistance to be 461 ohm, and real capacitance to be 415 nF. 

 

This is the output when it is at 1K Hz. We get our experimental value to be 1.1544V. Compared to our theoretical value 1.388 V, we have a 16.8% percent difference. 

 

This is the output when it is at 2K Hz. We get our experimental value to be 2.217 V. Compared to our theoretical value 2.776 V, we have a 20.1% percent difference.

 

This is the output when it is at 1K Hz. We get our experimental value to be 0.603V. Compared to our theoretical value 0.649 V, we have a 7.1% percent difference. 

 

This is a summary table of our measured value and our calculation of % difference. The error could be caused by the inaccuracy of the value we use for capacity.

 

Summary:

Today, we talk about how op-amp can act as an integrator and differentiator. We talk more about 1st order linear circuit. We talk about the singularity functions. We do a lab on inverting differentiator.  We get a max % difference of 20.1% and a minimum of 7.1%. I guess it is due to the capacitance we use is 415 nF instead of 470 nF which is its theoretical value. 

Lab Report Day Fourteen - Passive RC/RL Circuit Natural Response

Passive RC Circuit Natural Response Lab

In this lab assignment, we examine the natural response of a simple RC circuit. We use both a manual switching operation and a square wave voltage source to create our circuit’s natural response. We see that the method used to create the response affects the circuit being measured.







This is our RC circuit. In the pre-lab, we calculate the time constant for the circuit when there is no power source. We calculate our time constant to be 15.125 ms. In addition, we find that time constant has the same unit as time, which is s.



We build our circuit as shown. We measure our experimental value for R1 to be 0.98k, and R2 to be 2.14k Ohm, which is close to their theoretical value. I get our C value to be 22 uF. We assume our capacitance is accurate.




This is the set up of our circuit.




We use analog discovery to apply a 5V source ot the circuit. We get our oscilloscope graph as a linear part combined with a exponential part.





We know that e^-1 is about 0.3678, which means that after 1 time constant period, the value will be 36.78% of its original value. We have our initial value of 3.426. We calculate the voltage after one time constant is 1.260 V. We find the time difference is 49.5 ms, which is the experimental time constant. We have a % difference of -227%, which is very big.



In part b, we apply a 2.5 V square wave, and with a offset of 2.5 V at a low frequency. This way, we do not need to plug and unplug the power supply ourselves.

This is the graph we get when we apply a square wave.




In part B, we get maximum voltage of 3.432 V. Using the same method, we find experimental time constant to be 15.25 ms, with a % difference of -0.826 %, which is much accurate. This could be cause by the small difference of the theoretical and experimental value of resistor.

Analysis:
As we can see, when we apply a square wave at a low frequency, it has a low percent difference. I think it is because when we apply a square wave, we do not to plug and unplug ourselves. By plugging and unplugging, some current may loss and cause our part one has a high percent different (227%).

Passive RL Circuit Natural Response

In this lab assignment, we examine the natural response of a simple RL circuit. We will use both a manual switching operation and a square wave voltage source to create our circuit’s natural response. We will see that the method used to create the response affects the circuit being measured.




This is our pre-lab. We predict the graph when we apply a square wave with amplitude 2.5V and offset 2.5V to the circuit. We calculate two possible time constant, one is 10ns, and another one is 30 ns.



This is the set-up of this circuit,

Since we do not have time, Professor Mason does this lab for us.


This is his output graph. We can see that the graphs match. Our prediction is correct.

Summary:
Today we go over Capacitors and Inductors, and do labs on RC and RL circuits and learn how to solve RC and RL circuit problems.

Lab Report Day Thirteen - Capacitor Voltage-Current Relations

In this assignment, we measure the relationship between the voltage difference across a capacitor and the current passing through it. We apply several types of time-varying signals to a series combination of a 100 ohm resistor and a 1u (micro) capacitor. The voltage difference across the resistor, in conjunction with Ohm’s law, will provide an estimate of the current through the capacitor. This current can be related to the voltage difference across the capacitor.

We apply three waves with different frequency across the circuit using Analog Discovery. We apply two sinusoidal waves with magnitudes of 2 V and frequencies of 1k Hz and 2k Hz, and one triangular wave with magnitude 4V and a frequency of 100 Hz.We are asked to predict what would happen to the current.

Here is our predicted graph.

Here are our measured R and C. Our theoretical values are 100 ohm and 1 u F, and our experiment value is 100.1 ohm and 1.096 u F, which are very close. 

This is the set up of our circuit.

The first time varying signal sent into the circuit was a sine function with an amplitude of 2V and a frequency of 1kHz.

The second time varying signal sent into the circuit was another since function but the time with a frequency of 2kHz.

The last time varying signal sent into the circuit was a triangular function with an amplitude of 4V and frequency of 100Hz.

Analysis: 

As what we see from the screenshot,  our predictions are almost. We use Ohm's law, I=V/R to get the measurement if current in mA.  We can see that the current through the capacitor has a 90 degree phase shift (Max or min to the middle point of V).

Summary:

Today we learn about capacitors and inductors. When calculating capacitor's capacitance, they add together when they are in parallel, and add inversely while in series. We can draw non-ideal capacitor  a resistor in parallel with it. Inductors are mostly coils of wire. When calculating the inductance of inductors, they just act like resistors. We can draw non-ideal inductor with a resistor in series. In DC, capacitor acts as open circuit, and inductor acts as short circuit. 

Lab Report Day Twelve - Temperature Measurement System Design Lab

Temperature Measurement System Design Lab 

In this lab assignment, we need to designed a simple temperature system which outputs a DC voltage that indicates temperature. Our system will use a thermistor to indicate the temperature. We use a Wheatstone bridge circuit to convert this resistance change into a voltage change. The voltage output of the Wheatstone bridge circuit is small relative to the amount of temperature change, so we use a difference amplifier to increase the overall sensitivity of the temperature measurement system.

This is a circuit diagram for a Wheatstone bridge. It can detect small changes in temperature and amplify it in a circuit.

We want the output of the circuit to be 0 V +/- 20 mV at initial room temperature at 25 degrees and have minimum increase of 2V when it reaches 37 degrees, which is approximately body temperature.

R1 and R2 are two identical 10K ohm resistor.

R3 is where we put in series a potentiometer with a 10K resistor.

Rt in this case is where we put our thermistor. Our calculated value is 12.30 K ohm.

We build Wheatstone bridge with R1/R2 = R3/Rt

This is our Wheatstone bridge

Our data table. We measure that R1=R2=9.89k Ohm, and R3=RT= 12.30 k Ohm.

This is our circuit for our designed system.

We have R5=R6=7.9k Ohm, R7=R8=100k Ohm. And the gain is 12.5

This is our designed circuit that can amplify the output voltage.The thermistor can detect changes in temperature and as a result the voltage will drop when measuring Vout from the difference amplifier.

 

We calculate that our gain is 12.5. So the output voltage should be amplified for 12.5 times. 

This shows how the Vout changes without amplifying. When the thermistor detects a change in temperature, the output voltage starts to drop.

This shows us the original output value and the amplified output voltage. Before it reaches saturation, it has a ratio of 12.5. The saturation voltage is about 3.52V. 

Summary Data:

Vab measured at room temperature: 0 mV

Vab measured at body temperature 579 mV

Vout measured at room temperature: 0 mV

Vout measured at 37 degrees: 3.52 V (It should be bigger, but it has reached its saturation)

Summary:

Today, we learn cascaded amplifiers, which includes multiple amplifiers. We also design a temperature measurement circuit, and learn how to build a balanced Wheatstone bridge. We learn that in a Wheatstone bridge, the resistors in the circuit has to have the ratio R1/R2=R3/R4, thus the two related resistor need to be identical. Since we cannot get this, this could cause some errors in this lab. 

Lab Report Day Eleven - Summing Amplifier, Difference Amplifier

We talk about different types of amplifiers in class, such as Buffer Amplifiers, which doesn’t have much change in voltage when there is a big change in output voltage, therefore Vin = Vout. And we talk about Inverting Amplifier, which produces an output with opposite sign, Vout = -Rf/Ri * Vi. And we do experiment on Summing Amplifier and Difference Amplifier. 

 

1. We first do an experiment on Summing Amplifier. 

By doing this lab, we can know and prove how a Summing Amplifiers combines two input signals. For any amplifier shown in the circuit, the Vout = - R3/R1 * (Va+Vb). 

(on the graph, R3 should be 0.1K, like the data in the right)

In the pre-lab, we need to figure out which resistors to use to avoid saturation. We make a mistake at first. We want to make the calculation simple, so we first use all 1K resistors. Then we figure out that it will be at saturation if we use all 1K resistor. Thus, we use 0.1 K for R3 instead. 

We have R1=R2= 984 ohm, and R3=98.2 ohm

We find out the theoretical value for Vout should be

Vout= - 1/10 * (Va+Vb). 

This is the set up of the experiment.  (from different views)

To build the circuit, we used two 1K resistors, one 0.1K resistor, one Op Amp 27, an Analog Discovery to provide +/- 5V and two input voltages. We use a DMM to measure the output voltage.

 

This is the result table of our data, with Va varying from -4V to +5V, and Vb keeps constant at +1V. We do fine on this lab. We get a minimum % difference of 0% and a maximum of 5%, which is not too bad. 

 

2. We then do the lab for Difference Amplifier.

The purpose of this lab is to find and prove output voltage to be the difference between two input voltages. It is called Difference Amplifier because the output voltage is the difference between two input voltage. For the pre-lab section, we figure out that if we use four identical resistors, the out put voltage is simply Vout = Va - Vb. We choose several 10K ohm resistor, and choose 4 that has same experimental and theoretical value to reduce uncertainty. We get our four resistor to have true value of 9.7 k ohm. 

 

 

This is the set up of this experiment. It looks very similar to the first experiment. We build the circuit with four 10k resistors, one Op Amp 27, an Analog Discovery device to provide +/-5V and two input voltage. We use a DMM to measure the output voltage.

 

The right half of this white board is our result table. 

 

When we use Vb as 1 V, we get the following output results.

This is a graph of Vout vs. Va
As the graph shown, we can see that it reaches the positive saturation at V+ = 3.48 (about 3.5V). 

 

When we use Vb as -1 V, we get the following output results.

This is a graph of Vour vs. Va.

As the graph shown, we can see that it reaches the negative saturation at V- = -4.26 V(about -4V). 

Or if we combine them together, and plot a Vout vs. Difference graph. We can also clearly see that it reaches positive saturation at about 3.5V, and negative saturation at about -4 V. 

 

Conclusion: We find that that the voltage begins saturating when voltages above 3.5V or -4V are supplied. Ideally the voltage coming out of the op-amp should be 5V but because the op-amp is inexpensive so it could cause some voltage loss.

 

 

Summary:

Today, we learn more about operational amplifiers and analysis few amplifiers. Operational amplifiers include input resistance and output resistance, but we can usually treat them as ideal Op Amps.