Week 8 day 1

Lecture 

Calculating the Inductor equivalent, Inductor, Equivalent inductor calculation is identical as calculating equivalent resistor in series and parallel:

Our second activity is to calculate the voltage across a resistor in discharging source free RC circuit:

We then used the equation above to calculate a function of the voltage across Source free RC circuit with 0.2 Farad Capacitor and 10-ohm resistor:

Calculating energy dissipate by resistor in source free RC circuit after t=5𝛕

Calculating Maximum switch frequency of our circuit, frequency equal to the reciprocal of 5-time constants (5𝛕):

Calculating the current function over time on a source free RL circuit:

Passive RC Circuit Natural Response

First, we calculated our theoretical time needed for  RC circuit to charge and discharge.

To precisely Measured the time it took to charge and discharge we used triggered option on our oscilloscope. 

In our first graph, we captured the time it took for the capacitor to be fully charged, the horizontal line shows that the capacitor is fully charged. our measured time for RC circuit to be fully charged is equal to 74.03 ms, really close to our theoretical value of 76.5 ms which is about  3% difference.

In our second graph, we captured the time it took for the capacitor is fully discharged.  our measured value of discharging time is equal to 273 ms, compared to our theoretical time of 242 ms.



we then use a square graph with a 2.5 V offset that means the voltage input is alternating between 5V and 0V. 

Passive RL Circuit Response 

 PRE-Lab
we are trying to calculate theoretical values of the inductor

Our RL circuit 

We measured the time it took for inductance to discharge equal to 148.6 ms. using the equation that we derived from pre-lab we calculated an inductance of 20 mH.

SUMMARY

Today in class we learnt how to find equivalence inductance, Passive source RC circuit and passive source RL circuit. Finding equivalent inductor uses the same method as finding resistor equivalent. we also learnt how to derive the function of potential across the resistor of RC circuit and current across resistor across RL circuit. We learnt that the time it took the capacitor to completely discharge is roughly about 5 times constant. in RC circuit time constant is equal to the product of resistor and capacitor, while in inductor the time constant is equal to inductance divided by the resistance. In our first lab, passive source RC circuit we tried to measure the time needed to charge and discharge we calculated the theoretical time of 242 ms to discharge and 76.5 ms to charge. Our measured time discharging time is equal 74.03 and charging time of 273 ms. Our theoretical and experimental are 3% difference for charging time and 11 % difference of discharging time.This % difference is caused by our approximation of 5 tau. In our second experiment, Passive RL circuit response, we are trying to calculate the theoretical value of Inductance (L) 

Week 7 day 2

Lecture

Rage Bridge Discussion, why large sparks occurs when the circuit is initially plugged in in dc circuit

Second activity calculating capacitance and unit of permittivity constant

Best part of the lecture: blowing up electrolytic Capacitor by plugging the wrong polarity demonstration

Calculating potential difference across RC circuit with constant DC current source, assuming it have been plugged in for a long time

Next activity is to calculate the equivalent capacitor of a circuit, capacitor on series is calculated like resistors in parallel, and vice versa :

Calculating current and voltage relations on Inductor, potential difference across inductor is equal to the product of the time derivative of current through the inductor and its inductance (L)

Capacitor Voltage- current Relations Lab

Pre-Lab :

We have to predict what happen to the voltage versus of time and current versus time on an RC circuit:

 WE measured our actual value of capacitor and resistor:

The view of our RC circuit, to measure the current across capacitor, we scope the voltage difference across resistor and divide it by the value of resistance :

Our first input we use 1KHz frequency of cosine wave:

Scope output blue line represent the voltage across the capacitor, the red line represents  current on the resistor and capacitor,  and the yellow line is equal to the voltage across the 100-ohm resistance:

here we observe a phase shift of pi/2 between the current that flows through and the voltage across the capacitor.

we redid our calculation with a 2KHz cosine wave:

Here we still notice phase shift of pi/2 between the potential difference across and the current flowing through the capacitor. The 2K hertz occurs more frequently since the period it took the sinusoidal current is reduced by half.

Lastly, we change our input voltage with a 100Hz triangular input:

Here in the oscilloscope window, we have 3 graphs: the blue line represent the input voltage, red line represent the current flows through the capacitor and resistor and yellow line represents the voltage across the resistor. Here we observe that the graph of current flows through the capacitor is a square wave. The current have a constant positive value when the slope of voltage input is positive, and a constant negative value when the input voltage is negative.

Inductor Voltage- Current Relations Lab

In this Lab, we replaced our capacitor from our previous circuit with an inductor. Our LC resistor is shown by the picture below:

we inputted a 1KHz Blue line represented our input voltage, the yellow line represented the voltage across the resistor and redline represent the current that flows through the resistor.  We observe a phase shift of negative pi/2  between the input voltage and current through resistor:

We repeated the same procedure with the circuit with 2 KHz input voltage:

Summary

In our lecture, we learn about using capacitor and inductor as part of our circuit. According to the passive sign convention current always flow toward the positive side of inductor and capacitor while leaving the negative side. Capacitor stores energy through Electric field.Capacitor resists potential change.When capacitor acted as a wire when it has no initial charge and acted as an open circuit when it's fully charged  The current flows through the capacitor is equal to the product of its capacitance and its derivative of the voltage across over time. The capacitance of inductor is related to its geometrical shape. The most current used capacitor is parallel plate capacitor that uses a metal plate, mostly aluminium foil separated by an insulator. The most common capacitor, the electrolytic capacitor is polarised, that means the positive terminal have to be hooked up to the positive voltage, or we can risk blowing up our circuit. Inductor stores Energy through its magnetic field. Inductor resists a current change, the potential difference of inductor is equal to the product of its inductance and its derivative of current over time. 

Week 6 day 1 Inverting Voltage Amplifier

Lecture

First Problem calculating non-ideal negative feedback operational Amplifier with open loop gain of 200,000, the input resistance of 2 mega ohms and output resistance of 50 ohm:

We calculated again (v out over v source) of 2. Second Problem of our lecture, calculating non-ideal positive feedback operational amplifier. Here we have an identical amplifier but  our Voltage out is connected to the non-inverting voltage input:

We calculated a gain of 9 on our non-inverting op Amp:

Inverting Op Amp Lab

Pre-Lab: 

we calculated Theoretical saturation and Theoretical resistor needed to obtain a gain of 2 from inverting op Amp:

We measured our Actual 2K resistor and 4 K resistor :

Below are the theoretical and measured resistor : 

 Our OP 27 Amp, note that first terminal is marked with a circle

Zoom in Set up of out inverting amplifier circuit

 We Measured the voltage out put of our circuit with different value of input voltage :

We then Sketch the graph of V input and output :

Lastly, we use the waveform generator to scope our output voltage when we have a linear input :

 Summary :

Today we learn about operational Amplifier.  Operational Amplifier is a voltage controlled voltage source that can produce Voltage out using 6 mathematical operations (add, subtract, multiply, divide, integrate and differentiate) from Voltage input. The ability of an amplifier to produce amplified voltage is limited by its outside Voltage. The condition, where voltage out is equal to its external voltage, is called saturated voltage. Open loop Amplifier saturates really easy and therefore impractical.In order to actually have a significant voltage gain, but doesn't saturate really quickly we will use used feedback. In this class, we will use either negative feedback when voltage input is connected to the inverted voltage input, or positive feedback when voltage input is connected to the non-inverted voltage input.

Week 6 Day 2 : SUMMER OP AMP and Difference OP AMP

Lecture

Our First Activity is to sketch the output voltage of an inverting amplifier versus time when we use time-varying voltage input.We use 9V to power our positive external power supply and grounded our negative external power supply terminal. Given that Our voltage input values are -100 mV to 100 mV:

Because this inverting amplifier with a gain of 2, our output should change between -200 mV and 200 mV, but because our amplifier only has a range of 0-9V( grounded negative power supply terminal), therefore our graph will change between 0 and 200 mV.

Our second Activity is to sketch the same graph,, change our input voltage to 0 and 200 mV:

our theoretical output voltage ranges between 0 and -400 mV, but because our inverting amp only has a range of 0-9 V, our output voltage will saturate to 0V the entire time.

Our third Practice is to find Output Voltage of an inverting amplifier using nodal analysis

we investigate two nodes, node A where 2 4K Resistor and 8K resistor branched and Vo, where voltage output is connected to the inverting terminal through 2 K ohm resistor. BY using nodal analysis method we calculated a potential difference of -1.6 V in output terminal.

SUMMING AMPLIFIER LAB

 Pre-Lab:
 Our task os to design a Summing amplifier that uses input resistor greater than 1 K ohm,
In Designing our circuit we use 2- 10 K ohm Resistors as our Input resistor and another 10 K resistor as our feedback Resistor.

Next, we start measuring our input Resistor.

Below is the sketch of our circuit and measured value, with our assembled breadboard:

We then measured The V out of our breadboard with our Digital Multimeter :

Below are the values of output voltages when we set one of our input voltage to 1 and

Our Next Activity is to drive voltage gain across a difference op Amp:

 Difference Amplifier

Pre Lab : 

We calculated the relationship between voltage from inverting input and non inverting input and the voltage output. WE then calculated theoretical resistor needed to obtain gain of 2 from the difference opAMP.  In our theoretical calculation we uses 20K ohm and 10 K Ohm resistor to obtain gain of 2 :

WE measure the actual resistance of 20 K resistor 

Our difference op amp circuit :

 we measured the voltage output by changing the value of V1 (inverting input) and V2(non inverting input)

Below are the graph of V input versus v Output  :

Summary :

Today we learn about 5 Type of OP AMP: unity gain OP AMP, Inverting OP AMP,  non-inverting OP AMP, Summing OP AMP,  Difference OP AMP. Unity gain Amplifier gives a  gain of one and works to isolate two sides of the circuit. To Obtain Unity gain buffer, we shorted (Rf=0) the output voltage with inverting input. Inverting amp, amplify input voltage and change the polarity of the input voltage. To obtain inverting amplifier we connected the input voltage to inverting terminal and ground the non-inverting terminal.