week 13 day 2

Lecture 

Calculating Effective Current (I RMS) and Effective Voltage (V RMS):

 Calculating V RMS of  voltage graph with an offset of amplitude (Vp)

 Calculating Apparent Power and power factor using V RMS and I-RMS:

Calculating Complex Power: 

Apparent power and Power Factor Lab

Calculation of Vrms, I rms, Apparent power, Power factor, and Average power : 

Graph between input Voltage( Blue line), load voltage (yellow) and current across the load ( red )

RL=10 Ω

RL=47Ω

Rl= 100Ω

Below are the values of our calculated  and measured values:

RL 10 Ω

Rl=47Ω

Rl=100Ω

Summary 

Because voltage and current of AC circuits are sinusoids, it's more effective to describe voltage and current as it's effective voltage (Vrms) and effective current ( Irms) we can represent power as it's complex power Phasor. Complex power consists of average power as it's real part and reactive power as it's imaginary part. Average power is the product of apparent power( S) and power factor. apparent power can be calculated by finding the product of the magnitude of effective current and the effective voltage. In our, we obtain a low percent difference between calculated and measured effective voltage and current. by comparing the Power dissipated by resistor and load and taking the ratio,  we observe that the average power comes purely from the real part of the load. The ratio between the 2 power is identical with the ratio between two resistors.  therefore we can assume that all the power dissipated through the load is dissipated only by the two resistors. we also observe that power factor( P avg load divided by S load) is equal to the cosine of phase shift  angle between voltage and current)

Week 13 day 1

Lecture 

Using Nodal Analysis to calculate the output voltage of inverting op-amp, note that we have to convert all the capacitor into their phasor form first :

An op-amp with no feedback: when the voltage across the inverting and non-inverting input are equal, then the output voltage is equal to zero

 Instantaneous Power equation, given  sinusoidal current and voltage, applying Sum Trig identities  we can separate power to time independent average power

Op-Amp Relaxation Oscillator 

Pre Lab: 

Designing Op AMP relaxation oscillator  to produce 777 Hz  frequency  using 100 nF capacitance

we calculated a theoretical resistance of  7.32

We measured the  capacitance of 90 nF acorss the capacitor 

TO obtain resistance close to 7.5 K connected 5.49 K and  1K resistor in series

To check our circuit we run the model in every circuit, here we get a frequency of  769 Hz, which is really close to our desired frequency.  Don't forget to shake the op amp inorder to jump start !!

The final step we run our circuit with a +/- 5 V saturation:

view of our Circuit:

first, we scope the feedback resistor to compare our graph with every circuit graph:

Our frequency across the feedback resistor is 812 Hz, that's 4.5% difference from our desired Frequency 

Lastly, we scoped the voltage across the resistor and the output voltage of the OP AMP :

The result is identical with what our theoretical graph predicted, the output voltage of op amp is a step function, while voltage across the capacitor is a combination between an inverse exponential and exponential graph 

Summary 

In an AC OP  AMP circuit, we can apply nodal analysis to calculate output voltage by using phasor and impedance.  In our OP AMP Relaxation oscillation Lab, we use saturation of op amp as a way to create a periodical function. We designed a circuit that gives 777 Hz frequency, we then tested our design using both every circuit and oscilloscope. Our final design are able to produce oscillator with 813 Hz frequency. We obtained a 4.5% difference, these difference in values is caused by our inability to find a precise resistor. 

Week 10 day 1

Lecture

Finding Initial Value Problem of an RLC circuit:

Our Initial Value Problem Solution: After switch is turned on for a long time Capacitor acted as an open circuit and Inductor acted as a short circuit.

Second order RLC formula derivation  in series :

Overdamped RLC Problem:

Parallel RLC problem :

Series RLC Step Response Lab

Pre Lab: Calculating damping ratio (1/𝛼) , natural frequency (𝝎o), damped frequency (𝝎d) and overshoot (

We calculated a damping ratio of 0.002 and a damped frequency 99999 with an overshoot of 4 V.
We measured the value of capacitor and resistor :

Since the value of the resistor is relatively small, our DMM couldn't measure the proper resistance :

View of our Breadboard :

WE used a unit step response of 2 V with zero offset

scope of  Voltage across the capacitor:

Graph of voltage across the capacitor in response with the unit step function we can measure the damping period by dividing 2𝝿 with the distance across peak to peak of the sinusoidal function.

We can measure the overshoot of the graph by finding the distance between the highest peak and the equilibrium line . we obtain an overshoot of 5.75 V and damping frequency of 101341.

Summary 

 To Determine the  total step response of a second order circuit we need to calculate both a steady state response and transient response of the second order circuit. Steady state response of a second order circuit is the value of either voltage or current across a component after a long time (t approaching infinity). To calculate transient response of a second order circuit we turned off independent sources and uses KCL and KVL to calculate our characteristic root. In our lab today we tried to measure the response of voltage across capacitor in response of a unit step function with 2 V amplitude with zero offset( the input voltage is switching between 2 V and -2V). Our calculated damoing frequency is 9999, 1/s while  our measured damping frequency is 101341 1/s  We obtained a 1.34 %  difference in theoretical and our measured value of damping frequency. 

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.