PSpice

The following is a demonstration on using PSpice.

The first exercise was to simulate this circuit:

Once the circuit was built, PSpice immediately gave us values for the current through each element and voltages through each node. 

The next part was to determine the Thevenin equivalent for the circuit:

Instead of just simply analyzing the circuit, we had to perform a DC sweep on the bottom current source (I3) . This puts I3 onto the x-axis of the graph. We wanted to observe the voltage as a function of current and obtained the following graph:

The graph tells us that when the current is zero (open circuit), the Thevenin voltage is 10 V. The slope of the graph gives us the Thevenin resistance, which is 3.33 Ω. The graph also tells us that when the voltage is zero (short circuit), the Norton current is 3 A.

The last part of the activity was to replace I3 with a variable resistor to show that maximum power transfer occurs when the variable resistor equals to the Thevenin resistance. This was our result:

As expected, our peak occurs when the variable resistance is 3.3 Ω.

As opposed to performing tedious calculations by hand, PSpice is a convenient tool in analyzing circuits and other properties of a circuit.

Maximum Power Transfer

The experiment was to verify the maximum power transfer theory of the following circuit:

We expected to see that when the potentiometer (load) is of the same value as the 5.6 kΩ, the maximum power of the load would be achieved. The first experimental setup was simple:

Our experimental data came out as follows:

We came across some bad data and simply omitted those points, giving us the linear function seen on the graph. Despite the encountered errors, we were able to determine the maximum power when the load resistance is 5120 Ω, thus supporting our hypothesis.

The next part of the experiment had us build the following circuit:

We ran the circuit through a software called Logger Pro in order to determine maximum power across the potentiometer, similar to the first part of the experiment. We were, however, unsuccessful because of the noise in the current graph. This prevented us from giving us an accurate reading of the power graph.

Perhaps the biggest cause of our problem were the adapters used to power the Logger Pro instruments. The AC source could attribute to the noises we see in our graphs because of the changes in current.

Thevenin Equivalents

Thevenin's theorem allows us to take a complicated system and simplify it down to a voltage source and an equivalent resistance.

We first found the Thevenin resistance and Thevenin voltage of a seemingly complex circuit. Our analysis yielded us values of

V_Th = 8.64 V
R_Th = 65.95 Ω

We then calculated the smallest value of R_L2 (the load resistance) when V_Th = 8 V.

R_L2 = 824.3 Ω

The next step was to take measurements of V_{Load 2}, the load voltage outside of the Thevenin circuit, for the original circuit and its Thevenin equivalent. We took load voltages for the smallest permissible resistance and for an open circuit. The Thevenin equivalent data are as follows:

Config	Theoretical Value	Measured Value	Percent Error
RL2 = RL2,min	VLoad 2 = 8.00 V	8.37 V	4.63 %
RL2 = ∞	VLoad 2 = 8.64 V	9.06 V	4.86 %

For the original circuit:

Config	Theoretical Value	Measured Value	Percent Error
RL2 = RL2,min	VLoad 2 = 8.00 V	8.55 V	6.88 %
RL2 = ∞	VLoad 2 = 8.64 V	9.06 V	4.86 %

At the end, we calculated some values of power supplied to the load resistor:

Config.	VLoad 2	PLoad 2
RL2 = 0.5RTh	2.88 V	0.252 W
RL2 = RTh	4.32 V	0.283 W
RL2 = 2RTh	5.76 V	0.252 W

The table shows that when R_L2 = R_Th, the maximum power is supplied.

Transistor Switching

The transistor is a widely used component in many modern electronic devices. It consists of three leads, the collector, base, and emitter. Current flows from the collector and flows out of the emitter when active. However, current must also flow into the base in order for the transistor to be operational. Our experiment shows how transistors work.

We began by constructing the following circuit:

When the switch is closed, the base is provided with current and thus, current flows through the 2N3904 transistor and lights up the LED shown in the picture above.

The next demonstration is of the same circuit except it shows that the tip of our finger can be used to close the switch and power the transistor 

The next part of the lab was to investigate the amount of current going into the base and coming out of the emitter. The circuit is built as follows:

We found the relationship of the current at A2 and at A1 by turning the linear potentiometer (P1). The results were:

The beta (gain) of the transistor is the ratio of the current coming out of the emitter and the current going into the base. The slope of this graph corresponds to the gain, whose value is 87.362.

FreeMat

This is a demonstration of the software FreeMat. Its uses are similar to that of the more widely known MatLab. Our sample includes:

1. A graph of the functions sin x and cos x between 0 and 2π.

2. Finding the current through the 10 Ω resistor in the following circuit:

We set i₁ and i₂ to the left and right loop, respectively, with both loops moving clockwise. By setting the systems of equations in FreeMat, shown below,

we find that the current is 0.1857 A in the direction of i₂ .

3. Comparing exponential graphs with different time constants. The first graph is the function e^-t/τ

Assignment 1-1 is a graph of 2e^-t/τ. One time constant is 100 ms and 200 ms for the other.

Assingment 1-2 is a graph of 2(1-e^-t/τ) with the same time constants.

4. Adding sinusoidal functions. 

Assignment 2-1 is the sum of two sinusoids: 3 sin(2t+0.175) and 5 cos(2t-0.525).

Assignment 2-2 is the same graph with a frequency of 10 Hz. The script for the output goes as follows:

and the graph 

Voltage Dividers

The design of our voltage dividers consists of three load resistors of equal resistances connected parallel and powered by an unregulated power supply. Each load can be switched on or off such that we exhibit an upper and lower bound voltage and resistance. It should be intuitive that closing a switch will give us the maximum equivalent resistance and closing all three switches will give us the minimum resistance. We needed to find the resistance for the shunt resistor (R_S) to prevent the system from short circuiting. Our calculated value for R_S was 45.05 Ω and thus our voltage source needed to run at 6.53 V. From these values, we further determined our current (I_BUS) with the maximum equivalent resistance was 6.25 mA and 17.2 mA for the minimum equivalent resistance.

Our setup consisted of a resistor box as the shunt resistor and the three parallel load resistors at 1000 Ω each, wired to the breadboard, all connected in series. An ammeter was connected in series with the system to measure I_BUS and a voltmeter is connected parallel to the load resistors to measure V_BUS.

Our data goes as follows:

1 Load

R_eq = 9.79 Ω

V_BUS = 5.49 V

I_BUS = 5.65 mA

P_Load = 0.0312 W

2 Loads

R_eq = 489 Ω

V_BUS = 4.75 V

I_BUS = 9.80 mA

P_Load = 0.0470 W

3 Loads

R_eq = 326 Ω

V_BUS = 4.25 V

I_BUS = 13.19 mA

P_Load = 0.0567 W

P_Load can be calculated as

P_Load = I_BUS² R_eq

The actual percentage in load voltage variation was 9.8% and 15.0% up and down, respectively. The variations in the values of the resistance box and the load resistors differed from their true respective values enough to cause a discrepancy.

By adding a fourth 1 kΩ resistor parallel to the circuit, and assuming the shunt resistance and voltage supply remains the same, the new load voltage was calculated to be 5.525 V.

If we wanted to reduce the load voltage variation to ±1% for the 3-load case, our source parameters would have to be V_S = 5.05 V and R_S = 46.2 Ω.

Introduction to Biasing

We are looking at two different LEDs in parallel connected to the same battery source. Each LED has a different rating. In order to put the correct voltage through each LED, we use a technique called biasing to establish these predetermined voltages by wiring resistors in series with each LED.

Our setup consists of a 9 V power supply and LEDs rated at 5 V, drawing 22.75 mA and 2 V, drawing 20 mA. We calculated that we need 175 Ω resistor for the 5 V LED and 350 Ω for the 2 V, however, we only had access to 150 Ω and 330 Ω for the experiment.

The setup

We measured voltage and current for three configurations for the LEDs as well as the current in the power supply.

Configuration 1 - Both LEDs in the circuit

I_LED1 = 12.86 mA

V_LED1 = 6.14 V

I_LED2 = 18.85 mA

V_LED2 = 2.18 V

I_Supply = 31.05 mA

Configuration 2 - Remove LED 2 from the circuit

I_LED1 = 13.06 mA

V_LED1 = 6.19 V

I_Supply = 13.05 mA

Configuration 3 - Remove LED 1 from the circuit

I_LED2 = 18.99 mA

V_LED2 = 2.21 V

I_Supply = 18.96 mA

With a 9 V battery operating at 0.2 A-hr, we calculated that the circuit could operate for 6.4 hours before the battery voltage gets too low.

Our percent error between the achieved current in the LED and the desired value of the two LEDs was 25.63%. The reason for this error was that we did not use the required resistors in our theoretical calculations.

For the efficiency of the circuit with both LEDs, we found the power supplied by the battery (P_in) and the power dissipated by the two LEDs (P_out) to be 279.45 mW and 120.1 mW, respectively. The efficiency is the ratio between the power dissipated and the power supplied which we found to be 43 %. 

If we changed the design to a 6 V battery while keeping the LED parameters, the efficiency would increase because we will use smaller biasing resistors while keeping the current the same, thus reducing the power dissipated in the load. Using a 5 V supply would give the best efficiency because it will provide the minimum voltage needed to run the circuit while demanding the smallest biasing resistors.