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.

Day 10

In this lab, we are going to learn a knowledge, and find the relationship between Vin and Vout.

we measure two resistors R1 and R2. the R1 is 1.76k ohm, and the R2 is 3.49k ohm.

Experiment set up:

Every time, when we enter an Vin, we can find a Vout. Recording the data.

We enter the data into Excle, and use data to make a graph.

In the graph, we find the highest Vout is 4.0V, and lowest Vout is -3.5V  In fact, the value should be positive 5V to negative 5V. but depend it is quality, the answer is different.

Day 9

In the beginning of the lab, we go to find the power source when it is ideal and not ideal. When we are going to do ideal, the current 0.0455 A, and the power is 1, so the power should be P=VI = 0.0455W. However, when we are going to calculate, we find the current always be lower than 0.0455, it shows there is a other resistor in the inside of power source. And its power is 22/(22+Rs)^2 W.

In the 22 ohm resistor, its real value is 22.4ohm.

Because there is a inside resistor, when we want to get 1V, it only shows 0.994V. However, we increase it to 1V

measure the output voltage is 0.980V.

 We use a 39ohm resistor, but its true value is 38.8 ohm, and we found its voltage is 0.993V

 

In the lab, we calculate two resistor 22.4ohm and 38.8ohm, and its voltage is 0.980V and 0.993V. after the calculate, we find that the inside resistor is 0.457ohm and 0.273ohm.

In this lab, first when we are doing pre-lab, we calculate power on a 2.2K ohm resistor. And its power 2.84*10^-3W

In the experiment, we choose two 2.2k ohm resistors, the real value is 2.16k ohm.

 We measure that the voltage on the resistor is 2.48 V. So we calculate the power is 2.847*10^-3 W. its percent difference is  -0.246 %.