Week 9 day 1

Lecture

Our first warm up problem is to find voltage through an inductor given the current in the unit step and delta form.

Our second topic is to derive a circuit that integrates input voltage. To create Integrator circuit we are replacing feedback resistor of an inverting op-amp with a capacitor, by using nodal analysis we can solve for the output voltage:

We then sketched a few voltage output given different variation of input voltage: sinusoidal, step and ramp voltage

Inverting Differentiator Lab

Pre-Lab: we calculated theoretical output voltage with the input voltage :

Next, we measured the value of resistor and capacitor by using DMM:

We measured 682 ohms of reistance and 92 nF of capacitance. Next, we assembled our breadboard and applied 3 different frequencies: 1KHz, 2KHz, and 500 Hz

We then measured the voltage output and input voltage by using oscilloscope:

 Output voltage(yellow curve) and Input voltage(blue curve) of 100 Hz frequency :

 Output voltage(yellow curve) and input voltage(blue curve) of 250Hz frequency

 Output voltage(yellow curve) and input voltage(blue curve) 0f 500 Hz frequency

Below are the table comparing our theoretical and measured  output voltage peak (amplitutude):

Summary

Today we learned to derive and calculate output voltage of an integrator and differentiator. Integrator and differentiator is made by swapping either the feedback resistor or the input resistor of an inverting op-amp with a capacitor or an inductor. we obtained an Integrator by swapping feedback resistor with a capacitor, and a differentiator by swapping input resistance with a capacitor. In our lab, we constructed a derivator op-amp, our output voltage peak inversely proportional to the input frequency of the input voltage. Our output voltage is bigger by a factor of 2𝝅RCf, where R is the value of resistance, C, Capacitance, and f is the input frequency of sinusoidal input voltage. We compared our theoretical values with our measured values in 3 different frequencies. We found that our average percent difference to be under 1%.

week 12 day 2

Today we learnt how to manipulate AC circuits, There are 8 different that we learnt from our DC circuit :

 Our first problem is to solve find Potential difference across capacitor of DC circuit
First we converted all time dependent part into phasor dependent part, then we applied nodal analysis to solve for the potential difference across the node:

 Our second Problem we used Mesh analysis to find current. since the component are already in the phasor mode, we can directly apply mesh analysis on each loop, and uses the matrix to solve for each current.

 On Our third Problem we used Superposition method to solve our previous problem. TO solve using superposition method we analyze the current for only one source at a time by only opening one source at a time  and then added each of the  current together. Current source become an open circuit when we turn it off, while voltage source turns into short circuit when we turned them off

 Our last Lecture was to convert a circuit into it's Thevenin/Norton equivalent:
First we calculated voltage across point B to found our equivalent Thevenin Voltage

 In order to find our equivalent thevenin resistor, we turned off our current source and connected the circuit  with an independent source across B and ground, we then calculated the voltage across point B to find the voltage, equivalent Thevenin resistor is equal to the voltage across B divided by the current

week 12 day 1

Lecture 

ELI the ICE man: In a resistor, current and voltage are always in phase.In Inductor (L) voltage (E) leads current (I). In Capacitor(C), Current (I) leads voltage (E).

Impedance of Resistor (ZA), capacitor (Zc), and Inductor (ZL) when 𝝎→∞ (DC circuit)

 Solving AC circuit using Impedance :

 Impedance Lab

Pre Lab :

Calculating current and voltage across 3 different type pf component (resistor, Capacitor and Inductor) :

View of the Inductive Impedance Circuit, we measure voltage of both the resistor and voltage across Inductor

View of Capacitive Impedance circuit , we scope both the voltage across the capacitor and resistor

To calculate Input voltage, we simply added the 2 measured voltage together, to Measure the input current we divide the voltage of resistor with 47.

The phase shift between the input voltage and input on the resistor at 1 KHz, 5KHz, and 10KHz:

The phase shift between the input voltage and input current with Capacitive Impedance at 1 kHz, 5kHz, and 10 kHz:

The phase shift between the input voltage and input current with Inductive Impedance at 1 kHz, 5kHz, and 10 kHz:

Values of phase shift and the voltage across 100-ohm resistor, 1 µH, and 100 nF : 

Summary 

In a purely resistive AC circuit, voltage and current are always in phase. In Inductive Impedance (A+jB)  Voltage is leading Current. In Capacitive Impedance (A-jB) Current is leading Voltage.

In our Lab compared 3 different type of  Impedance: purely resistive (100 Ohm ), Inductive (.001 mH Inductor) and capacitive (100 nF capacitor). In purely resistive circuits, the impedance is independent of its frequency. voltage are always in phase with current and their magnitude of voltage divided current is equal to the value of the resistor. In Inductive AC circuit, the frequency is directly proportional to its phase shift, the higher the frequency, the larger the phase shift difference between input current and voltage current. The impedance of inductive circuit also directly proportional to the frequency of the input voltage. from our data, we observe that higher frequency provides a higher drop across the inductor that was resulted from the higher impedance. In a capacitive circuit, the frequency is Inversely proportional to the phase shift and impedance. higher frequency resulted in smaller difference across voltage and current and a smaller voltage drop across the inductor.

week 11 day 2

Lecture

Solving addition of cos and sin using a graphical representation:

Leading and lagging between 2 sinusoidal functions:

 Converting complex number to its polar form:

 Using complex identities to solve complex number operation:

 Changing current from time-dependent form to phasor form :

 Voltage conversion from exponent form to it's phasor form, the bottom part of the board, changing the current from rectangular form to it's time dependent form :

Adding two voltage with different phase angles:

 Solving current by using factor and Impedance:

Passive RL Circuit Response Lab

Pre-Lab : 

Calculating theoretical phase shift  between input  current and voltage with 𝝎= cutoff frequency,  𝝎=1/10 cutoff frequency, and 𝝎=10 cutoff frequency

View of our circuit :

 Input voltage versus input current at 𝝎= cutoff frequency

Input voltage versus input current at 𝝎= 1/10 cutoff frequency

Input voltage versus input current at 𝝎= 10 cutoff frequency

we measure the distance between each peak and dividing over the period of the sinusoidal function to measure the phase shift between the current and voltage. Below are the calculated theoretical value and measured the experimental value of our trials:

Summary 

In AC circuit, It's easier to analyze circuits in phasor form rather than it's time dependent form. Addition and subtraction of two phasors are faster to be done in their rectangular form. Multiplication, division, roots, and power are easier to be done in their polar form. A complex number can be expressed using polar for, rectangular form and exponential form. In our Passive RL circuit lab, we observe that voltage is leading the current (φ-θ) is negative. We observe that frequency is directly proportional to its phase difference and inversely proportional to the voltage across the inductor ( gain). In our experiment, we calculated an average 6.69 % difference. Most of the % difference is from the measurement of 1/10 cutoff frequency because the difference between two peaks are relatively small.

week 11 day 1*

Lecture 

Calculating current across capacitor by adding the transient voltage and steady state voltage 

 Calculating unit step of Parallel RLC circuit  :

RLC circuit Response Lab

Pre Lab : 

using nodal analysis we are trying to calculate the voltage across the resistor across resistor R2

We calculated our alpha,  natural frequency, and damping frequency  of the circuit, according to our calculation the current response across the resistor is an underdamped reaction .
We build our board  :

Then we scope the voltage across the 1-ohm capacitor we obtain a curve that looks slightly underdamped or close to critically damped oscillation instead.

Summary 

To calculate a current or voltage across RLC circuit we can add the transient response and steady state voltage. In our Lab, we obtain a result that is totally off from our expectation, We suspect this result might result from circuit, to be sure we will redo this experiment when we have an extra time.