AC Signals #1

The above circuit was analyzed with professor's computer.
Anticipated Vrms = 2.12 V
Actual Vrms = 1.97 V
Zcap = -717j Ohms
Vcap = ~2V
Vcap,rms = 1.6 V
t_x = 90 microseconds

Oscilloscopic Skeleton

The point of this lab was to become acquainted with oscilloscopes and how to use them for determining the frequency and amplitude (voltage) of an oscillating current.
Part 1 - Displaying and Measuring a Sinusoid
The FG was set to have f = 5kHz and peak to peak amplitude to have 5V. We measured the period with the O-scope to be 200 microseconds. The the peak to peak amplitude measurement was off by a factor of 4.: 20 V. Same with the zero to peak amplitude: 10 V.we got the Vrms to be 7.07V.

Using the DMM on the leads: VDC = 0.06 V, VAC = 4.07 V

Part 2 - Including a DC offset
The difference between the DC and the AC coupled is a change along the y axis. (voltage)
Using the DMM again: VDC = 2.5 V, VAC = 3.28 V

Part 3 - Square Wave
Again with DMM:
VDC = 2.5 V
VAC = 5.17 V

Vrms = 3.66V
Part 4- Measuring two Mystery Sources

Channel 1:
DC V= 0 V
f = 100 Hz
Pk -to- pk Amp: 0.6 V

Channel 2:
DC V=0.5 V
f = 6.25 kHz which is about 6.06 kHz (the actual frequency)
pk -to-pk Amp: 0.3 V

Thus we learned how to use oscilloscopes to learn things about alternating/time dependent currents.

Op Amp Lab

In this lab, a 741 op-amp was used to create inverting amplifier circuit under the pretense of creating a signal conditioning circuit for a sensor.

A voltage divider was devised to get a voltage of one volt for the V_IN and mimic a sensor's output.
The following data was collected on the components we used.

 Above is the layout of our circuits.

I_V1 = 8.95 mA
I_V2 = 0.95 mA
Because these aren't equal to each other, we couldn't go on to the last page of the lab because he lab was written in anticipation of each group getting a 3rd voltage source, but only 2 were available for each.

Equivalents HW using PSpice

Here's some screenshots for the evidence.

On the last one, after selecting the trace cursor to display, I used the Trace>Cursor>Peak to get teh peak wattage.

Thevenin and Norton Equivalents in PSpice

We modeled circuits in PSpice and used DC sweeps to o determine their Thevenin and Norton equivalents.

In order to do this in the first circuit, we placed a current source I2 across the terminals that we were interested in and then used a DC Sweep to change it's value from 0A to 1.0A. After running the simulation, a trace of "V(I2:-)" was added to plot the voltage across the current source I2.

 The y-intercept gave the Vth and the slope of the line gave the Rth because V=I*Rth.
Thus, the Vth = 20V and Rth = 6 Ohms.

Then we worked out the Norton equivalent. We did this by replacing the current source I2 with a Voltage source that was varied from 0 volts to 1.0V by increments of 0.1V.

Here the y-intercept gave us the I_N = 3.335 A and the slope gave G_N = 0.17 S.
In the last problem, I found the load that maximizes the power dissipation.
I put a resistor across the terminals that had values of { RL } and I set RL's value with a Global Parameter. This global parameter was swept from 100Ohms to 5kOhms with 100Ohm increments. I ran the simulation and got the following graphs. The top one was arrived at through a linear sweep, and the bottom through a octave sweep. The peaks show at 250microW with RL = 1kOhm, as expected.

Thevenin Equivalents' Equity

The purpose of this lab was to find a Thevenin equivalent of a circuit and model it to see how accurate it was.

The V_Th = 8.65V,
the R_Th = 66.0Ω,
and the I_Th = 131mA.

Here's a picture of our circuit.

Here's a the table of the components we used.

 Here's a table of the results we got from those components followed by another table of components we used.

 

However, the components in the seoncd table didn't really get the time needed to be used. We had started the lab too late and had to leave. Instead of making the whole class come back on Friday to finish the last page of the lab, Professor Mason allowed us to call it quits.

PSpice Fun

After installing PSpice, I had to run Capture Student as administrator to get the PSpice menu to show up in it.
I had to import some libraries to get the parts I need to use, and then it was simply place parts in a sensible pattern and connecting them with wire.

Below is my PSpice for working out the last problem of the 8th WebAssign assignment.

More pictures from Node Analysis

This shows a reading for our V₁ during the last problem when we were trying to make it and V₂ equal to 9V.

Nodal Analysis is for Knowing Nerd Noggins

We (James Dunn and I) used nodal analysis to determine the theoretical voltage between two points and a reference node AKA "ground" and the currents coming from two voltage sources.

Circuit Diagram

I'll get a better picture from James soon.

Our theorectical values for V₂ and V₃ were 10.26V and 8.67V respectively.
We determined that I_{Bat 1} and I_{Bat 2} were 17.4mA and 1.48mA, which would give power supplies of 208.8mW and 13.35mW respectively.



In the last part, for extra fun, we figured out what voltages we would need to load across the circuit to make both V₂ and V₃ equal 9V. We worked out that V_L1 would be 9.9V and V_L2 would be 10.98V. To get these particular load voltages, we would need to use a voltage divider to bring 12V down to these voltages.

We did this with a resistor box of 213Ω in series with V_bat1 and 103Ω in series with V_bat2. We got as close as 9.09V for V₂ and 8.99V for V₃, after fiddling with the resistances a bit. :) Our currents became 9.90mA and 8.61mA for I_bat1 and I_bat2 respectively.

Voltage Dividers

Given 1 to 3 loads of parallel 1kΩ resistors, we calculated the Vsource and Rsource of an unregulated power supply to make the voltage variations on the loads + or - 5% of 5V.

We recognized that the R_EQ,max and the R_EQ,min were 333Ω and 1kΩ, respectively, and that the larger resistance corresponded with the larger V_BUS. I worked out then the following:

• V_S = 5.54V
• R_S = 55.5Ω
• I_BUS,max = 14.26mA
• I_BUS,min = 5.25 mA

Below is collected data. (I'll fix the format later.)

Color Code	Nominal Value	Measured Value	Wattage (W)
Brwn,Blck,Red,Ag	1 kΩ	0.995	kΩ	0.25
Brwn,Blck,Red,Ag	1 kΩ	1.001	kΩ	0.25
Brwn,Blck,Red,Ag	1 kΩ	0.989	kΩ	0.25
Resistor Box	55 Ω	56.4	Ω	0.3
Power Supply	6 V	6.06	V	NA

We made sure nothing would be broken by the circuit we were planning to make, so we put it together. Here's a picture with all 3 loads connected.

These are the results:

Configuration	REQ (Ω)	VBUS (V)	I BUS(mA)	Pload(mW)
1 Load	1001	5.75	5.74	33.0
2 Loads	499	6.74	10.9	66.3
3 Loads	332	5.20	15.6	81.2

The power of 2 loads was found by P = V^2/R = (5.46 V)^2 / 499Ω = 66.3 mW

The actual % variations were quite different from the theoretical because we used 6.06 V instead of 5.54V.

There would be theoretically a -9.4% variation from 5V if we were using a 5.54V power supply and added a fourth 1kΩ load.

If we redesign it for a ±1% variation of V_BUS from 5V, then we would have a R_S = 10.2Ω and V_S= 5.10V.  

LED circuit

On Monday we built a circuit to light two LEDs (light emitting diodes), a red (LED1) and a yellow (LED2), maximally without burning them out with too much wattage.
The red LED's given voltage and current ratings were 5V and 22.75mA respectively, yellow's were 2V and 20mA.

The LEDs were placed in a circuit in parallel with a resistor (R1 and R2) in series on each. A 9V supply was given by a power supply.
Ohm's Law was used to calculate the equivalent resistances of the LEDs.
RLED1 = 219.8Ω, RLED2 = 100Ω

Using the current  and voltage ratings, theoretical voltages and resistances values were found for R1 and R2.
V1 = 4V, V2 = 7V
R1 = 176Ω, R2 = 350Ω

We only had access to 100, 150, 220, 360, and 470-Ohm resistors, so we approximated R1 with a 150Ω and R2 with a 360Ω. Their measured values were 148.2Ω and 360Ω respectively. Both were quarter-watt resistors.

Three configurations were used:

1. Both LEDs in the circuit
2. Without LED2
3. Without LED1

These are the results:

Configuration	I of LED1 (mA)	V of LED1 (V)	I of LED2 (mA)	V of LED2 (V)	I of supply (mA)
1	14.5	6.68	19.71	19.71	35.1
2	14.8	6.74	NA	NA	15.1
3	NA	NA	20.2	1.65	20.3

Using this data, I calculated a few things.
a) If the supply was just a 9V battery with a 0.2 A-hr capacity of useful voltage, the circuit in config1 would last 4.68 hours.
b) The % error between experimental and desired values of LED current were 36.3% for LED1 and 1.45% for LED2. This was caused by resistor restrictions.
c) The efficiency of the circuit in configuration 1 was found to be 40.8%, which is pretty poor.
d) Assuming current is the same, if a 6V battery was used instead of a 9V battery, ... I really don't understand this problem.

Oh, yeah, and we blew up a red LED by overloading it with voltage.

Modelling Power lines

In our first official lab, we worked out the maximum length of cable between a load and its power source using a resistor box to model the length of cable.

The ideal power source of 12V was modeled with a power supply. The load required 11 V across it and it was worked to have a resistance of 1 kΩ so it was modeled with the appropriate resistor (brown black red).

This is the setup of another group. Gerrek (I have no idea about the spelling), my lab partner, took pictures of our set up, but I haven't received a message from him (I gave him my card with my contact info) and didn't get his contact info, so I may have to upload the picture he took after class on Monday.

We used two multimeters to simultaneously measure the current in the circuit and the voltage across the load resistor. We varied the resistance of the resistor box until the voltage across the resistor was 11.00V.
At this point: current = 10.93mA and R of total cable = 97Ω.

I worked out the time the hypothetical battery rated at 0.8Ahr would discharge in 73.2 hours, that the power to the load was 0.1195W and the lost in the cable was 0.01159W, and thus the efficiency was 91.2%.
The resistor box's rating of 0.3W was not exceeded in this experiment. Finally, I worked out that if the cable were AWG #30 wire, the load could be as far as 142m away from the power source.