02/25 - Lab 2: Resistors and Ohm's Law and Lab 3: Dependent Sources and MOSFETs

Lab 2 - Resistors and Ohm's Law: Voltage-Current Characteristics

In this lab, we explored the nature of resistors and how they affect voltage and current. We connected a resistor in series with the power supply with a multimeter in parallel to read the voltage drop across the resistor.

Equation: y = 9.216x -0.0217, R^2 = 1

The data shows that voltage and amperage share a linear relationship in a circuit containing only a resistor. This result is expected due to the equation for voltage V = IR.

Lab 3 - Dependent Sources and MOSFETs

In this lab, we explored the topic of dependent sources by examining how voltage and current behave with a MOSFET added to the circuit from the previous lab

MOSFETs are voltage-dependent current sources, which means they control the current in a circuit based on the voltage entering the gate G. Similar to a diode, MOSFETs have a threshold voltage that must be met in order to pass current through. According to the data collected, the gate of the MOSFET "cracks open" at 1V, and from 2V up to 2.5V the gate really opens up, allowing a flood of current though. From 2.5 V to ~4.5 V, the current plateaus at ~48 mA. Above ~4.5V, the MOSFET severs the circuit. The slope of the line obtained through this period is 68.5, which represents the gate g of the MOSFET. Thus, g = 68.5 A/V. 

In Class Examples

1) Given an incandescent light bulb rated at 75 watts and 120 volts, find the “hot” resistance and “cold” resistance of the light bulb. The filament is made of tungsten (52 × 10-8 Ωm)

2) Determine the number of branches, loops and nodes in the circuit shown below. Check Fundamental Network Thm.

3) Find I and Vab in the circuit for all four cases of polarity in the voltage supplies.

02/23 - Lab 1: Solder-less Breadboards, Open and Short Circuits

02/23

Lab 1 - Solderless Breadboards, Open-circuits and Short-circuits

In this lab, we acquainted ourselves with the solder-less bread boards using a multimeter and various connection setups.

1) Check the resistance between two holes in the same row on the breadboard:

Multimeter indicates a resistance of 1.4 Ohms between the two holes. 

This means the circuit is closed.

2) Check the resistance between two rows of holes on opposite sides of the central channel:

The multimeter reads a resistance of infinity, which means there is no connection between ports on opposite sides of the central channel of the breadboard. This means the circuit is open.

3) Check the resistance between two arbitrary holes of the breadboard:

Again, the multimeter is showing a resistance of infinity, which indicates that there is no connection and therefore the circuit is open.

4) Use a jumper wire to connect the the two rows of the arbitrary holes:

The multimeter displays a resistance of 3.1 Ohms, which makes sense because there is a small resistance added due to the extra wire. This means the circuit is closed.

After running this series of tests, we were instructed to draw a schematic of the breadboards and indicate how each of the nodes are connected internally.

As shown in the diagram, each row is connected horizontally accross all 5 nodes of a column, but the two colums are separated by a central channel. The + and - columns on the outside of the breadboard are connected vertically along the entire length of the breadboard.

In-Class Examples

1) In the first exercise, we were shown a simple circuit with two power sources and three lightbulbs. With the switch off, both light bulbs on the outside of the circuit are equally bright, and the lightbulb in the middle along the switch wire is off. We were asked to determine what would happen to the lights as soon as the switch is turned on, connecting the middle wire to the rest of the circuit.

We took a guess that the two lit bulbs would dim, and the middle bulb would light up to the same brightness as the other two.

In reality, when the switch is flipped, nothing happens. This is because there is no difference in potential between the left and right junctions. As a result, the flow of the current does not change when the switch is turned on, and the lights stay the same.

2) The charge transferred in time is given as shown below where the peak value is 10 C and the time at the maxima is 0.785s. Determine the current, I(t), flowing through the wire.

3) Determine the total charge flowing through an element for 0 < t < 2 seconds when the current entering its positive terminal is i(t) = e^(-2t)

Note: The numbers are good, but we forgot to convert our answer to the appropirate magnitude (mC).

4) Given v = 10V and i(t) as shown below, sketch the power and energy as a function of time

5) The figure shows the current through and voltage across a device. Find the total energy absorbed by the device for the period 0 < t < 4 seconds.

6) Given the circuit below, find V0. Hint: Be sure to use conservation of energy or that the sum of the power in all the elements is 0.