Day 17

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. 

In the pre-lab, we calculated the value depends on given value, R1=47, R2=1.1, C=10uF, L=1mH. and according the image of setting up, the R2 and L is series, and C is parallel with R2 and L, then R1 series with all of them.

The circuit is set up on breadboard. 

The picture is about input power, This is a square wave with 100Hz, its peak is 2V with offset 0. 

The picture is Vout.

                                 R1                      R2                       C                          L

experiment value       49Ohm                4.3Ohm              9.92uF                 cannot measure

In the lab, we calculated value of α is 3588.7, but the theoretical value is 1063.8. Its precent difference is -237.4%. 

Summary:

Today, we learned how to calculate the resistor, capactior and inductor in parallel.

Day 16

Series RLC Circuit Step Response

This lab will emphasize 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.

 The image of circuit for setting up.

In the beginning of the lab, we are going to calculate the circuit belongs to overdamped, underdamped, or critically damped. 

In the pre-lab. we find a and w, then we find the circuit is underdamped

Then we use underdamped way to calculte.

The circuit is set up on breadboard,

 theoretical value:    the capacitor is 470 nF, the inductor is 1uH,             and the resistor is 1.1Ohm.

experiment value:    the capacitor is 420 nF, the inductor may be correct, and the resistor is 1.4Ohm.

 

The graph is about input voltage, the frequency is at 500Hz 

This is the graph about Vout

by the calculating, its w is 5.15*10^4. compare to our theoretical value 1.351*10^6, the percent different is 96.2%.

 put some data into excel, find the function is y=3.5006e^-8811x, the value a is 8811, but our theoretical value is 5.5*10^5., the experiment value is 1.6% of theoretical value.

Summary:

Today, we learned how to find solve source free RLC circuit and find boundary value. and we know there three different types of circuit, over damped, underdamped, and critically damped. 

Day 15

In this lab assignment, we will examine the forced response of a circuit which performs a differentiation  that is, the circuit output is derivative with respect to time of the input to the circuit. We will apply sinusoids of various frequencies to the circuit and compare the output with our expectations based on analysis.

In the pre-lab, we calculate Vout with different frequency 1KHz, 2KHz, and 500Hz.

Calculate the percent different about Vout with different frequency.

The circuit is set up on breadboard.

This picture shows the graph when the f=1KHz. 

In the picture, we know its peak voltage is 1.1544V, but our theoretical value is 1.388V, the percent different is 16.8%.

This picture shows the graph when the f=2KHz. 

In the picture, we know its peak voltage is 2.217V, but our theoretical value is 2.776V, the percent different is 20.1%.

This picture shows the graph when the f=500Hz. 

In the picture, we know its peak voltage is 0.603V, but our theoretical value is 0.649V, the percent different is 7.1%.

From the pictures, the percent different increase with frequency. so the error may be caused by frequency increasing.

Day 14

In this lab assignment, we will examine the natural response of a simple RC 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.

In this lab, we calculate the constant time when there is no power is applied. We calculate the constant time is 15.125ms.

      
When we apply 5V power to the circuit, we got a graph, which is a linear part combined with a exponential part
In the lab, we find that the one period is 0.3678s because T=e^-1. And after one period, the value becomes 36.78% of its original value. and we find V0=3.432V, so after one period, the voltage is 1.260V. but When we find period, it is 49.5ms. its difference is -227%.
In part b, a 2.5 V square wave with low frequency is apply into circuit.
In part B, the max Voltage is 3.432V, and in same way, the experiment constant time is 15.25ms, find the different percent is -0.826%
Passive RL Circuit Natural Response
the circuit is set up.
Professor Mason did this lab for us.
Summary:Today we go over Capacitors and Inductors, and learn how to solve RC and RL circuit problems.

before we calculated the R1 and R2, and find C:

                                                                   R1                       R2

                            theoretical value:        1K ohm             2.2K ohm

                            experimental value:  0.98K ohm         2.14K ohm

the Value of C is 22uF.

                                

Breakboard is set up for circuit.

.

The graph about 2.5V square wave is applied.

.

Analysis:

when we apply a square wave with a low frequency, its difference percent is low because during we apply a square wave, we do not to plug and unplug ourselves. It causes the high percent different 227%.

In this lab assignment, we will 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.

On pre-lab, we assume apply  a square wave with amplitude 2.5V and offset 2.5V for the circuit. two constant time, one is 10ms and other one is 30ms.

the graph is made by professor Mason.

Day 13

In this assignment, we will measure the relationship between the voltage difference across a capacitor and the current passing through it. We will apply several types of time-varying signals to a series combination of a resistor and a 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.

Pre-lab: understand circuit and draw pictures. 

Here is our predicted graph.

before the lab, we calculate the R and C

The Rand C are measured:  

                                         R                          C

theoretical values:       100 ohm                1 uF

experiment value:       100.1 ohm            1.096 uF

breadboard is set up.

Measure the wave after we apply three different frequency AC.

First: A sine and amplitude 2V with 1kHz is applied.

Second: A sine and amplitude 2V with 2kHz is applied.

Third: A sine and amplitude 4V with 100Hz is applied.

Analysis: 

after we got picture, our analysis are almost right. the formal I=V/R is used to get current. Then we found, the capacitor has a right angle phase shift. And we can know the V is between middle line to top or bot value.

Summary:

Today we learned capacitors and inductor. When we calculate the capacitor with series, its calculation way is same with parallel about resistance. When the capacitor is parallel, we calculate it with adding.
We calculate inductor, it is same way with resistance.

DAY 12

In this lab assignment, we will design and implement a measurement system which outputs a voltage which is indicative of temperature. A thermistor will be used to measure temperature. The resistance of the thermistor changes with temperature.

The picture about lab setting.

In this lab, we hope to get output voltage at 0V at room temperature, and it increase 2V around human being' temperature 37 degrees.

theoretical value:              True Value:

R1=R2=10K Ohm            R1=R2=9.89K Ohm

Rt=12.3K Ohm                 R3=Rt=12.3K Ohm

R1/R2=R3/Rt

The picture about lab:

Find R5=R6=7.9K Ohm

        R7=R8=100k Ohm

The lab is set up. the designed circuit amplify Vout, the thermistor can be changed in temperature.

By calculating, we find the Vout is amplify 12.5 times.

This shows how the Vout changes without amplifying.(The temperature is droping)

Shows us Vin and Vout, Before the saturation, its ratio is 12.5.

                                   Vin                Vout
room temperature:       0mV              0mV
body temperature:     579mV           3.52V

Summary:
We learn the Vin can be magnify by using op-amp. at the same time, when we are going to change temperature, the Vout and Vin can be changed.

Day 11

In this lab, we implement a simple operational amplifier-based circuit. Since operational amplifiers are used commonly in circuits used to implement mathematical operations we implement the processes of summing two voltages.

Design an inverting summing circuit which performs an addition of two signals. The input resistance seen by the two voltage sources(Va and Vb) should be at least1kQ.

Find threes resistances, their theory values are two 1k ohm and one 100 ohm.

Experiment set up;

Apply an input voltage when Vb = 1V, and measure the output voltages.

Record all data from the measuring output voltages

Calculate all percent different  when each different Va is applied.

In this lab, we implement a simple operational amplifier-based are used commonly in circuits used to implement mathematical operations, we implement the process of taking the difference between two voltages.

In this lab we find four resistances of same theory ohm. and their true values all are 9.7k ohm.
Experiment set up:

Same with the lab Summing Amplifier. Record all the output voltages data.

1) When Vb = 1V; calculate the all theory output voltages, record the all true output voltages, and calculate all percent different.

2)When Vb = -1V; calculate the all theory output voltages, record the all true output voltages, and calculate all percent different.

When we use Vb as 1 V， we find V out, and calculate the percent difference. 

Graph of Vout vs. Va
  

When we use Vb as -1 V, we find V out, and calculate the percent difference.

Graph of Vour vs. Va.

When two graphs are combined together, a graph Vout VS. Vin is gotten. the positive saturation is around 3.5V, the negative saturation is around -4V.

Conclusion: when there is above 3.5V or -4V are supplied, the ideally the voltage op-amp should show out 5V. However, this op amp is cheap, so it causes some energy lose.

Summary:

Today, I learn about op-amp, and how to use it when we apply a voltage. although the op-amp is used as ideal, even there are an input resistance and an output resistance.