Superposition

Superposition

In the lecture, we talk about super position and how this method can be better used over mesh analysis. As shown in the picture below, one circuit is solved using mesh analysis and the other is solved using superposition.  

Circuit solved using mesh analysis

 Solved using Superposition

For the lab, we predict the voltage of the circuit shown:

Then we record the actual resistance of the resistors we got and made the schematic for it: 

Next, we set up the circuit and record the resulting voltage across the 6.8k resistors: 

With only the 3V source:

 With only the 5V source:

With both sources:

Using the data above, we calculate the percent difference between the calculated sum of voltages of the independent 5V and 3V source and compared it to the measured one: 

[(2.685-2.69)/2.69]*100% = 0.19% difference. 

Here is the table of the analysis and experimental results: 

Summary:

We learn about a new method called the superposition where it's a method used for the circuits. It can be shown to be better to use than the mesh analysis. In the lab, we use superposition to find the voltage across the 6.8 KΩ.

Mesh Analysis 2 and Time Varying Signals

Mesh Analysis 2

In this lab, we predict what the current I₁ and V₁ for what this circuit will be.

And doing the calculations as shown below:

 We get that:

I₁= -3.2mA

V₁= 5.0184V

After predicting the current and voltage, we collect the resistors and record the actual resistance as shown below:

Thus, taken from the picture, (Except for the 10KΩ is exactly as measured), the circuit for the actual resistors are drawn as shown below: 

After measuring the current I₁ and voltage V₁ and V₂  we get the data below:

Current I₁

% error:  0.9375%

:

 voltage V2

 Voltage V1

% error: 0.964%

For the lecture, mesh analysis is used here in order to find I_B,I_C, and V_o. Thus, by doing the calculations below: 

We get that:

I_B:  .165mA

I_C: 8.25 mA

V_o: 5.175

After the lecture, we went test the response of the circuits by having images of the input and output voltage sketches: 

Summary

In conclusion, we use mesh analysis to find the hidden current and voltage and apply those calculations onto the lecture and the lab. We record the actual measurements and then measured the voltage and current to get the % error between the predicted data and the actual data. Then we also did mesh analysis to find the currents and voltage involving a transistor current. Afterwards, we use the waveform programs to see varying oscillation responses from the circuit. 

Nodal and Mesh Analysis

Nodal and Mesh Analysis 

In this lab, we use the breadboard to set up the circuit and then we calculate V1 and V2 as shown in the picture below:

Using nodal analysis, we find from the picture above, we find the voltage of the node along with V1 and V2 {Voltages of the 22kΩ and 6.8kΩ resistor}. 

Once we calculated V1 and V2, we then proceed to set up the circuit and then record and measure V1 and V2 as shown below: 

Then calculating the % error for V1 and V2: 

The reason the recording shows positive is because we placed the red and black alligator clips at the opposite ends of the sides. The difference for V2 seems quite large. This may be because the voltage provided is not as much as we calculated.

For the lecture, we learned about mesh analysis and how it's applied to circuits: 

In the picture above, we find out what are the currents i1 and i2. To do so, we use mesh analysis which involves the equation around the first loop and second loop using the sum and difference of the voltages of the resistor or the battery supply. For the 20Ω resistor, we use i1-i2 for the first loop and i2-i1 for the second loop as current for the 20Ω resistor. 

This picture involves the same thing as the previous picture. The equation was set up based around the three loops. With this, we may use matlab to find the currents of the first, second, and third loop. 

Summary

In the lab, we use node analysis to calculate the voltage. Then in lab, we set up the circuit and record the voltage. Afterwards in lecture, the professor taught us mesh analysis which involves finding either current, resistance, or voltage. 

Temperature Measurement System

Temperature Measurement System

In the lab, we design a temperature measurement system with a termistor, a resistor, Vout Voltage, and the 5V power supply. Before doing the lab, we determine Vout as a function of Rth and R. 

From this Picture, Vout is calculated. Then we verify if the Vout temperature increases as temperature increases and by using the quadratic formula in the picture above, it is confirmed that it does. 

When the circuit was set up. We check to see the resistance at room temperature and watch what happens as temperature increases:

From the picture it is shown that as temperature increases (As I touch the Thermistor) the resistance Rth decreases. 

Now we record the voltage of the circuit from when it is at room temperature and from when I touch the thermistor. 

From the picture shown, Voltage increases as temperature increases. when I was taking the picture as I'm touching the thermistor, the Voltage kept increasing. 

From the calculated Voltage: 

10.95kΩ would be calculated to 2.39 V

6.39 kΩ woudl be calculated as 3.05V. 

Doing the % error: 

1.3% error for the 10.95kΩ

and

2.3% error for the 6.39kΩ

At the lecture, we learn about the nodal analysis and how the method was used.

On this problem, we need to calculate the node voltage on the circuit: 

Using KCL and Ohm's law, we get that: 

V1 = 13.33V

and

V2 = 20V

Summary

In the lab, we record the voltage of the thermistor.  After recording the voltage which is at room temperature, we touch the thermistor to record the voltage. When we touch the thermistor, it increases temperature which also increases the voltage. This is because as temperature increases, the resistance decreases. 

Voltage Dividers

Voltage Dividers

In this lecture, we talk about finding the equivalent resistance and in the lab, we used a photocell resistance to power the LED light. 

The Picture above shows how equivalent resistance is calculated. First, the two resistance from the right are parallel. Thus, the two resistance are added with this equation: (1/R1+1/R2)^-1. 

Then, afterwards, the combined Resistor is added to the resistor on top due to the fact that it's connected in series, and then it is added parallel to the 3R resistor. Thus the Req = R. 

This picture above shows how to find the Voltage of the Photocell. From picture for the photocell, the resistance can be controlled from 5kΩ to 20kΩ. Thus, we used the equation of the photocell resistance plus the total resistance and multiplied that by the current to find the Voltage of the Photocell. 

Speaking of the Photocell, in this lab, we make a circuit to power the LED using the photocell. This lab would show why resistance is important. 

From this video, we can see that by preventing the light from entering the photocell. The LED lights up. This would be because decreasing the voltage from the light will also decrease the resistance. This causes the LED to light up because by having too much current in the circuit, the LED is "fried". 

Measuring the Photocell Resistance:

When it's dark, the resistance of the photocell is: 110 kΩ

When light reaches the photocell, its resistance is 1.3 kΩ

The above two pictures represent the voltage from the diode Vd measured when light is prevented from entering the photocell to when light contacts the photocell.

The above two pictures is the same as the next two above except it's measuring the voltage from the base to the ground Vb. 

The calculated values of Vb for photocell resistance of 5kΩ and 20kΩ is 5/3 and 10/3 respectively. 

Summary
In lecture, we calculate the equivalent resistance by combining the resistors that are connected in either parallel or series. Then in the alb, we find the resistance of the photocell. Here, it is shown why resistance or resistors are needed as too much current onto the LED will "Fry" it. This causes the LED not to light up meaning that if the photocell was blocked of light, the LED will light up because resistance will significantly increase.