May 12, Phasors: Passive RL Circuit Response Phasors: Passive RL Circuit Response

Pre-Lab: 
We did a pre-lab analysis to predict the amplitude gain and phase difference with 3 different frequency: Wc, Wc/10, and 10Wc (Wc = 2.2*10^5 Hz). When W= Wc, expected amplitude gain = 0.032 and expected phase difference = -45degree. When frequency = Wc/10 expected amplitude gain = 0.045 and expected phase difference = -5.7 degree. When frequency = 10Wc expected amplitude gain =24.5*10^-3 and expected phase difference = -84 degree

When W= Wc

When W= Wc/10

When W= 10Wc

The circuit as illustrated previously in the schematic

the data for exp value. The amplitude gain is different, becaese we used  I/V to calculate the  amplitude gain in pre-lab, and we use the Vout/Vin in the final calculation. The results make sense through as f increase the V/V gain decrease. 
The percentage difference for the angle shifted is calculated as 5.6%, 6.3%, 9.2% respectively. 

Summary:

In this lab, we analyze the steady-state response from sinusoidal inputs. We can find the amplitude gain by comparing the Vin and Vout. We also learn how to analyze AC circuits using different methods of analysis including nodal analysis, mesh analysis, superposition, Thevenin and norton analysis. 

The results is good since the percentage difference for angle is less than 10%, the source of error may due to the inaccurate measurement of time from the graph, and the uncertainty of frequency that analogy discovery provided. 

May 7. Impedance

Above is important formula to calculate impedance for each circuit elements.
Prelab
We determined the resistor impedance = 47 + R , the real resistor values came out 48.7 ohm and 100.1 ohm, and therefore the expected impedance for resistors = 148.8 ohm  Inductor impedance = 48.7 + 0.00628j  Capacitor impedance = 48.7 - 1711.34j 

Measurements for part A.

Circuit

This is the RR circuit at 1k frequency. Vt= 1.354 V, I=13.2 mA. 


This is the RR circuit at 5k frequency. Vt= 1.354 V, I=13.2 mA. 

This is the RR circuit at 10k frequency. Vt= 1.354 V, I=13.2 mA. 

Part2 predictions



circuit

This is the RL circuit at 1k frequency. Vt= 0.2652 V, I=40.05 mA. 

This is the RL circuit at 5k frequency. Vt= 1.0584 V, I=34.2 mA. 

This is the RL circuit at 10k frequency. Vt= 1.546 V, I =25.4 mA. 

Part3 predictions


circuit

This is the RC circuit at 1k frequency. Vt= 1.999 V, I =1.05 mA. 


This is the RC circuit at 5k frequency. Vt= 1.9774 V, I =4.955 mA. 


This is the RC circuit at 10k frequency. Vt= 1.924 V, I =9.475 mA. 

Summary:

Today, we analyze AC circuits. We find that a resistor AC circuit has no phase change between the voltage and current. The voltage leads the current by 90° in a circuit with an inductor and the current leads the voltage by 90° in a circuit with a capacitor. We learn that circuit elements can be represented as impedance.  

May5 , No lab today

NO LAB

Today we have an interesting DEMO that how an oscilloscope works. 

On the oscilloscope, the screen is connected to a tube that is far back.

Inside the oscilloscope includes an electron gun. It contains wires and when we apply a voltage across it, the wire gets hot and emits electron in all directions. When we apply source with high intensity, some of the charges bounce around, some of the charges get absorbed, and some of the charges get shoot out the tiny hole and becomes electronic beams. The monitor projects the motion and position of the electronic beams. In the tunnel, we have a positive charged plate on top, and negative charged plate on the bottom, positive charged on the left, and negative charged on the right. When we apply voltage on the circuit, we can see the beam on the display screen. 

April 16. Passive RC/RL Circuit Natural Response

n this lab assignment, we examine the natural response of a simple RC circuit when we open and close the switch. 
Pre lab.
Following is the schematic of our RC circuit. We did calculated time constant since we were given R and C values. We estimated intital capacitor voltage and time constant for the circuit as shown below. R1=100. R2=47
. 
oscilloscope window showing the capacitor voltage response for the circuit shown in previous image where V+ is used as the voltage source. 

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.

For part B

Part B prelab predictions. R1=978. R2=2150, C=22 uF

Part B measurements

in part B, we get maximum voltage of 3.4 V. Using the same method, we find experimental time constant to be 0.0172,which is about one fifth of the predicted value. the percentage difference would be 80%. 

Part B graph

We can see the shape of open switch and close the switch. 

signal input data

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.

ciucuit

after class DEMO.

Summery

As we can see, when we apply a square wave at a low frequency, it has a low percent difference. the results are not very good comparing to the expectation. That may due to the uncertainty of measure time from the graph and the inaccurate resistance since the inductor has some resistance which we are not considering in calculations. Overall, we learn the step response of circuit, calculate the time constant and visualize what happened when open and close switch. 

April 9. Temperature measurement system design

In this lab we are going to design a temperature measurement system that could make a voltage difference greater than 2V when temperature changes from 25 to 37. 

we are going to use OPAMP to amplifier the output voltage and use a Wheatstone bridge circuit to calibrate the measurement

Practice in class.

lab circuit

Resistor value in this lab. 

For Wheatstone Bridge, we make R1 = R2 = 10K, We know at room temperature, Rn=11.8K, so we calculated that we need a 4.7K resistor to balance the circuit. Actually, we use a 10 K potentiometers to balance the circuit. 

Below is the measured values for true resistance of those resitors that been used in the experiment. and the measured output range when heating the thermistor to 37 degree. output range is 0 - 0.276V.

The opamp circuit. 

Below is the resistors value we choose for this circuit, the ratio between R2 and R1 is the multiplaction of the circuit which could amplify the signal 10 times. 

after done the experiment, the actual output range is 0 to 1.8V the multiplication is around 9 times.

the real circuit could been seen in the videos. 

Summary:

Today, we learn cascaded amplifiers and how to build a balanced Wheatstone bridge. There are some very important applications of opamp, while we figure out that the opapm we uses is not ideal, so the feedback is very strong,we initially use the around 10K resistor for R2 and R4 and the feedback is too strong that the amplifier not functioning well. Then we switch to high resistance 1M and 100K, the circuit works much better. All in all we kind successfully finished this experiment with the output range is slightly smaller than 2V. if we could use larger resistors for opamp circuit we could get even better results. 

May 14. Inverting Voltage Amplifier

Today  we build an inverting voltage amplifier using an op-amp.  We first calculate the theoretical voltage gain and the phase shift.

Pre-lab
Here we are given the circuit diagram, and used the node to get the relationship between  the Vin and Vout.

theoretical angular frequency, gain and phase angle will be at different frequencies.

the resistance and capacitance of elements are listed on the graph.The built circuit consists of 2 Resistors (10kOhm), 1 Capacitor (1 microF), 1 Op Amp (OP 27)

100Hz, the waveform for Vin(red), and Vout(blue).

 1000Hz, the waveform for Vin(red), and Vout(blue).

5000Hz, the waveform for Vin(red), and Vout(blue)

circuit

post lab calculations
we can see the percentage difference for gain and angle shift are both very small. This experiment is well performed.


Part 2
everycircuit simulating

Prelab calculations.
my last 3 digit is 155, so the resistance need to have f=155 is 29428 ohms.
in order to have this value, we use 2 10K resistor connect in series with a 10K potentialmter.

For the first time, We got the Graphics is not very ideal which is not a perfect square waveform. 

After we reconnect the wires, we get what we want the perfect square waveform. and we can adjust the resistance to change the frequency. Wow

Summary:

Today we go over how OP Amps work in AC circuit, and do labs to see how to see how they can apply in real life. 

For part 2, We learn how op amp relaxation oscillator works and how we can make our desired frequency. 

The main source of error may due to the inaccurate measure of period ( time) from the graph on screen.