ECSE 1010 Proof of Concepts - Omega Lab02

0. Lab Document

1. Prove That the Slope of an $IV$ Curve Corresponds with Ohm’s Law for Two Different Resistor Values

Building Block

Let’s pick two resistor. The first one is

4-Band Color Code: Orange, Orange, Brown, Gold

$$ \begin{align*} 33 \times (1\times10^1) = 330 \Omega \pm 5% \end{align*} $$

have a check

The second one is

4-Band Color Code: Brown, Brown, Red, Gold

$$ \begin{align*} 11 \times (1\times10^2) = 1100 \Omega \pm 5% \end{align*} $$

have a check

Analysis

We know that $IV$ curve means $I$ on the y-axis and $V$ on the x-axis of the plot. Then, it must be a linear function, because both $IV$ don’t have powers.

Using the idea of linear function, we know the slope is $\frac{\Delta X}{\Delta Y}$. Back to our case, it becomes $\frac{\Delta V}{\Delta I}$. Also, we knows the Ohm’s Law, which $\frac{V}{I} = R$. So, the slope is very likely to be the resistance $R$.

If we take $R_1 = 10 \Omega$, $R_2 = 100 \Omega$ (As the simulation set). We should got.

If we plot them together, we got

Here is the data table

Simulation

Measurement

First we built a circuit like this

this is based on the diagram from the lab document

We only changed the $R1, R2$ values. Also, it’s hard to plug multimeter on the breadboard. So, we intersect the $V+$ circuit at the front

This method is not ideal, but works.

Let’s begin

For $V+ = 0.5V$, we got

To save some space and work, we just will not show each result. But here is the data

With this MATLAB code,

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% Step 1: Enter the data
V_plus = [0, 0.5, 1, 1.5, 2, 2.5, 3]; % V+ values
V_R1 = [0, 0.142, 0.238, 0.358, 0.463, 0.572, 0.632]; % V(R1) values
V_R2 = [0, 0.396, 0.724, 1.126, 1.492, 1.831, 1.994]; % V(R2) values
I = [0, 0.3, 0.6, 1.0, 1.3, 1.6, 1.9] * 1e-3; % I values in A (converted from mA)

% Step 2: Plot the data
figure;

% Plot for Resistor R1
subplot(2, 1, 1);
plot(V_R1, I, '-o');
xlabel('Voltage V(R1) (V)');
ylabel('Current I (A)');
title('Resistor R1: Current vs Voltage');
grid on;

% Plot for Resistor R2
subplot(2, 1, 2);
plot(V_R2, I, '-o');
xlabel('Voltage V(R2) (V)');
ylabel('Current I (A)');
title('Resistor R2: Current vs Voltage');
grid on;

we got the plot of $R1$ and $R2$

Now, let’s create a fit line for both. It’s needed to find out the slope ($R=V/I$). To do that, we changed the code a bit into

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% Step 1: Enter the data
V_plus = [0, 0.5, 1, 1.5, 2, 2.5, 3]; % V+ values
V_R1 = [0, 0.142, 0.238, 0.358, 0.463, 0.572, 0.632]; % V(R1) values
V_R2 = [0, 0.396, 0.724, 1.126, 1.492, 1.831, 1.994]; % V(R2) values
I = [0, 0.3, 0.6, 1.0, 1.3, 1.6, 1.9] * 1e-3; % I values in A (converted from mA)

% Step 2: Fit linear regression curves
% Fit for Resistor R1
p_R1 = polyfit(I, V_R1, 1);
slope_R1 = p_R1(1);
R_R1 = slope_R1; % Resistance of R1

% Fit for Resistor R2
p_R2 = polyfit(I, V_R2, 1);
slope_R2 = p_R2(1);
R_R2 = slope_R2; % Resistance of R2

% Step 3: Display the resistances
fprintf('Resistance of R1: %.3f ohms\n', R_R1);
fprintf('Resistance of R2: %.3f ohms\n', R_R2);

% Step 4: Plot the data and fitted curves
figure;

% Plot for Resistor R1
subplot(2, 1, 1);
plot(V_R1, I, 'o');
hold on;
plot(polyval(p_R1, I), I, '-');
xlabel('Voltage V(R1) (V)');
ylabel('Current I (A)');
title('Resistor R1: Current vs Voltage with Linear Fit');
legend('Data', 'Linear Fit');
grid on;

% Plot for Resistor R2
subplot(2, 1, 2);
plot(V_R2, I, 'o');
hold on;
plot(polyval(p_R2, I), I, '-');
xlabel('Voltage V(R2) (V)');
ylabel('Current I (A)');
title('Resistor R2: Current vs Voltage with Linear Fit');
legend('Data', 'Linear Fit');
grid on;

we got a result of

1
2

Resistance of R1: 331.144 ohms
Resistance of R2: 1069.374 ohms

and the plots

Check this result from multimeter’s reading of resistance

Great! The actual reading is very close to the resistances we determined from your IV measurement data and the linear regression. The average $%$ error is less than $1%$

Discussion

We did a lot of discussion in each session instead of in one. This is just to make the document more logical and follows the flow. So, we will only summarize and add something not appear above.

First, we used LTSpecie to determine $IV$ curve of two resistor $R_1 = 10\Omega$ and $R_2 = 100\Omega$. (This is just for prove our Analysis, so it doesn’t match the $R_1 = 330\Omega$ and $R_2 = 1100\Omega$ we used later). And it matches our Analysis. Both the plot created by Excel and the values.

Then, we built a series circuit, and we know they have the same current across all components. And, the $R$ is only related to $IV$. As long as we got some reading pairs, we can plot the curve. The result matches our expectation with less than $1%$ error. Consider our multimeter can only measure down to $0.1 mV$. This accuracy is amazing!

Thus, we proved That the Slope of an $IV$ Curve Corresponds with Ohm’s Law for Two Different Resistor Values.

2. Prove the non linear $IV$ curve for a light emitting diode

Building Block

Analysis

To plot a $IV$ curve of a diode. We need to find out a few important data.

• Forward Voltage ($V_F$)
• Reverse Breakdown Voltage ($V_{BR}$)
• Reverse Leakage Current ($I_S$)

As the datasheet of QED123 said

• $V_F = 1.7V$
• $I_F = 100 mA$
• $V_{BR} = 5V$
• $I_S = 10 \mu A$

We just plot them into a standard diode $IV$ characteristic diagram and get

Simulation

The turn on voltage of 1N914 is about $0.7V$

Measurement

We create a trig wave like

with amplitude to 5V (10 volts peak to peak), frequency to 200 Hz, and phase to 90 degrees.

Then, we use channel 1 to find out the current using the math function in scope

1

C1/330*1000

and the $IV$ Curve

with this MATLAB Code,

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% Step 1: Import the CSV file
data = readmatrix('P2-4-c.csv');

% Step 2: Extract the columns
voltage = data(:, 2);  % Second column is voltage (V)
current = data(:, 1);  % Third column is current (I)


% Step 3: Plot the I-V curve
figure;
plot(currentvoltage, current, 'k-', 'LineWidth', 1.5);
xlabel('Voltage (V)');
ylabel('Current (I)');
title('I-V Curve');
grid on;

we got

Discussion

Our experimental matches the datasheet. Consider the datasheet said

• $V_F = 1.7V$
• $I_F = 100 mA$

and we got $1.7V$ on $10 mA$ this matches the datasheet curve.

3. Show / demonstrate that the differential resistance changes in different regions in the diode $IV$ curve

Building Block

Analysis

To plot a $IV$ curve of a diode. We need to find out a few important data.

• Forward Voltage ($V_F$)
• Reverse Breakdown Voltage ($V_{BR}$)
• Reverse Leakage Current ($I_S$)

As the datasheet of QED123 said

• $V_F = 1.7V$
• $I_F = 100 mA$
• $V_{BR} = 5V$
• $I_S = 10 \mu A$

We just plot them into a standard diode $IV$ characteristic diagram and get

Simulation

The turn on voltage of 1N914 is about $0.7V$

Measurement

We create a trig wave like

with amplitude to 5V (10 volts peak to peak), frequency to 200 Hz, and phase to 90 degrees.

Then, we use channel 1 to find out the current using the math function in scope

1

C1/330*1000

and the $IV$ Curve

with this MATLAB Code,

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% Step 1: Import the CSV file
data = readmatrix('P2-4-c.csv');

% Step 2: Extract the columns
voltage = data(:, 2);  % Second column is voltage (V)
current = data(:, 1);  % Third column is current (I)


% Step 3: Plot the I-V curve
figure;
plot(voltage, current, 'k-', 'LineWidth', 1.5);
xlabel('Voltage (V)');
ylabel('Current (I)');
title('I-V Curve');
grid on;

we got

Discussion

To find out at least 2 locations on the curve to show that the differential resistance changes along the I-V characteristic. We modified the code a bit to let it find out 2 random point on the plot and its slope.

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% Step 1: Import the CSV file
data = readmatrix('P2-4-c.csv');

% Step 2: Extract the columns
voltage = data(:, 2);  % Second column is voltage (V)
current = data(:, 1);  % Third column is current (I)

% Step 3: Select two random points
num_points = length(current);
random_indices = randperm(num_points, 2);  % Generate 2 unique random indices

% Step 4: Extract the voltage and current values for the selected points
V1 = voltage(random_indices(1));
V2 = voltage(random_indices(2));
I1 = current(random_indices(1));
I2 = current(random_indices(2));

% Step 5: Calculate the slopes
slope1 = (V2 - V1) / (I2 - I1);
slope2 = (V1 - V2) / (I1 - I2);  % This is the same as slope1 but calculated in reverse

% Step 6: Print the slopes
fprintf('The slope between the randomly selected points (I1 = %.4f, V1 = %.4f) and (I2 = %.4f, V2 = %.4f) is: %.4f\n', I1, V1, I2, V2, slope1);
fprintf('The slope between the randomly selected points (I2 = %.4f, V2 = %.4f) and (I1 = %.4f, V1 = %.4f) is: %.4f\n', I2, V2, I1, V1, slope2);

We got

The slope between the randomly selected points (I1 = 0.0097, V1 = 0.2959) and (I2 = 0.0036, V2 = -2.6254) is: 479.8789

The slope between the randomly selected points (I2 = 7.9784, V2 = 1.2568) and (I1 = 2.8170, V1 = 1.1975) is: 0.0115

We can see they are very different.

4. Prove That Nodal Analysis Solves Unknown Nodal Voltages in a Circuit

Building Block

Analysis

To make our life easier, I rewrite some equation in $\LaTeX$.

Current through a resistor:

$$ I_R = \frac{V_A - V_B}{R} $$

Kirchhoff’s Current Law (KCL) at node B:

$$ I_{R_1} + I_{R_2} + I_{R_3} = 0 $$

KCL at node C:

$$ I_{R_3} + I_{R_4} = 0 $$

Expressing currents in terms of voltages. From the first equation:

$$ \frac{V_B - V_A}{R_1} + \frac{V_B}{R_2} + \frac{V_B - V_C}{R_3} = 0 $$

From the second equation:

$$ \frac{V_C - V_B}{R_3} + \frac{V_C - V_D}{R_4} = 0 $$

Substituting known values. Given $V_A = 5$ and $V_D = 0$, the equations become:

$$ 2.5V_B - V_C = 5 \\ 2V_C - V_B = 0 $$

Matrix form: $\begin{bmatrix} 2.5 & -1 \\ -1 & 2 \end{bmatrix}$ $\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix} 5 \\ 0 \end{bmatrix}$

Solve them “by hand”

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% Define the matrix A and the vector b
A = [2.5, -1; -1, 2];
b = [5; 0];

% Solve the system of linear equations A * x = b
x = A \ b;

% Display the solution
disp('The solution is:');
disp(x);

we got

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The solution is:
    2.5000
    1.2500

Thus, $\begin{bmatrix} V_B \\ V_C \end{bmatrix} =$ $\begin{bmatrix}2.5 \\ 1.25 \end{bmatrix}$

Simulation

Measurement

For $V_C$, we got

For $V_B$, we got

Discussion

Our Analysis matches the Simulation. The Experimental data has less than $2.5%$ error than expect, which is very less. Thus, we proved That Nodal Analysis Solves Unknown Nodal Voltages in a Circuit.

5. Prove / demonstrate your approach to designing a circuit using nodal analysis

Building Block

Analysis

To make our life easier, I rewrite some equation in $\LaTeX$.

Given values:

• $V_A = 3 , \text{V}$
• $V_C = 0 , \text{V}$
• $V_B$ is unknown.

Using Kirchhoff’s Current Law (KCL) at node B:

$$ \frac{V_B - V_A}{R_1} + \frac{V_B - V_C}{R_2} + \frac{V_B - V_C}{R_3} = 0 $$

Substituting the given values and resistances:

$$ \frac{V_B - 3}{1} + \frac{V_B - 0}{4} + \frac{V_B - 0}{4} = 0 $$

Simplifying the equation:

$$ (V_B - 3) + \frac{V_B}{4} + \frac{V_B}{4} = 0 $$

Combine terms:

$$ V_B - 3 + \frac{V_B}{2} = 0 $$

Multiply through by 2 to clear the fraction:

$$ 2V_B - 6 + V_B = 0 $$

Combine terms:

$$ 3V_B = 6 $$

Solve for $V_B$:

$$ V_B = 2 $$

Simulation

Measurement

For $V_B$, we got

Discussion

Our Analysis matches the Simulation. The Experimental data has less than $1.2%$ error than expect, which is very less. Thus, we proved That Nodal Analysis Solves Unknown Nodal Voltages in a Circuit.

6. Prove the function of an op amp comparator

Building Block

Analysis

A non-inverted comparator has a transfer function of

$$ \begin{equation*} V_{out}=\begin{cases} \text{if} ; V_{in} < V_{ref}, V_{out} = V_s - \\ \text{if} ; V_{in} > V_{ref}, V_{out} = V_s + \\ \end{cases} \end{equation*} $$

In our case, we got

$$ \begin{equation*} V_{out}=\begin{cases} \text{if} ; V_{in} < 0V, V_{out} = -5V \\ \text{if} ; V_{in} > 0V, V_{out} = 5V \\ \end{cases} \end{equation*} $$

Our supply voltage are $5V$ and $-5V$, and the input is a SINE wave with amplitude of $1V$, and the reference voltage is $GND$ which is $0V$

Simulation

Measurement

Discussion

Comparing our simulation to our experiment, we see that both of them are square waves with the same periods and similar amplitudes. They are fluctuating between 5 and -5, which are our supply voltages. This makes sense, because the supply voltages are the outputs of op amp comparators.

This proves our concept of an op amp comparator.

7. Prove the function of a mathematical op amp

Building Block

Analysis

Summing amplifier circuit has a transfer function like

$$ V_{out} = - \frac{Rf}{R1} \cdot V1 - \frac{Rf}{R2} \cdot V2 $$

In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it can be adjusted according to our demand. Then, we got

$$ \begin{align*} V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\ V_{out} &= - V1 - V2 \\ \end{align*} $$

Simulation

We just used two SINE waves with different frequencies ($500 ; \text{Hz}$ and $1K ; \text{Hz}$) in our simulation.

Measurement

Then, we setup the circuit. We just connected scope channel 1 to the $V_{out}$ to check it works or not

We have $V_s + = 5V$ and $V_s - = -5V$

We used wave generator to create to SINE waves of $500 ; \text{Hz}$ and $1K ; \text{Hz}$

And we checked the output wave using scope channel 1+

Discussion

As we see, the shape of the output wave is exactly the same as our simulation. Both the amplitude of the output wave in the simulation and measurement is around $1.75V$, and the period is the same.

Since the shape and all features of our experimental wave matches our simulation, we know this op-amp works in different voltage ranges.

This proved our concept of summer amp which is a mathematical op-amp.

8. Prove the concept of transfer functions of Two-Channel Audio Mixer

Building Block

Analysis

Summing amplifier circuit has a transfer function like

$$ V_{out} = - \frac{Rf}{R1} \cdot V1 - \frac{Rf}{R2} \cdot V2 $$

In our case, we want to use $50K \Omega$ potentiometer as the resistors, so it can be adjusted according to our demand. Then, we got

$$ \begin{align*} V_{out} &= - \frac{\cancel{50K}}{\cancel{50K}} \cdot V1 - \frac{\cancel{\cancel{50K}}}{\cancel{50K}} \cdot V2 \\ V_{out} &= - V1 - V2 \\ \end{align*} $$

Simulation

We just used to SINE wave with different frequency ($500 ; \text{Hz}$ and $1K ; \text{Hz}$) to test what we expect.

Measurement

Then, we setup the circuit. We just connect scope channel 1 to the $V_{out}$ to check it works or not

We supply $V_s + = 5V$ and $V_s - = -5V$

And use wave generator to create to SINE wave of $500 ; \text{Hz}$ and $1K ; \text{Hz}$

And we checked the output wave using scope channel 1+

Discussion

As we see, the shape of the output wave is exactly the same as what we simulated. Both the amplitude of the output wave in the simulation and measurement is around $1.75V$, and the period is the same.

This proved our concept of summer amp.