Oscilloscope 101

        Using an oscilloscope, which is an electric device able to display time varying currents, we connected a function generator and practiced using different functions. The frequency was set to 5kHz and the voltage 5V. With the knobs we could adjust the position and # of period of the graph. By counting the grid we could find that the zero to peak amplitude was 5V, and peak to peak amplitude was 10V. We measured anticipated RMS by dividing 5V by square root 2, which was 3.54. Using DMM, the value was 3.33V relatively close to measured value.

        Afterwards, a +2.5V DC offset was added, and the sinusoidal graph needed to be readjusted. The VAC was 3.30V, similar to part 1, and the VDC was 2.51V, just as expected when adding 2.5V DC offset.

       In the end, we connected the scope to two mystery signals, and using our acquired skills we used the graphs to approximate DC voltage, Frequency, and Pk-Pk Amp of Channel 1 and 2:

Operational Amplifiers

In this lab we had the problems of wanting a voltage source varying  0-1V while having 0-10V of output. In this case we used an inverting amplifier to obtain a gain of -10. 6V battery supplies powered the op amp.

We first calc. the smallest Ri value for input resistance using Ohm's law (V = IR)         Ri = 1000Ω.

Afterwards, nodal analysis was used to calculate a value of Rf, the resistor between the inverting input and output of the op amp., that would satisfy a gain of -10.          Rf = 10000Ω.

For the  variable source (Rx) to vary 0-1V, a resistor Rx was placed, acting as a voltage divider. Assuming worst case scenario, where Ry = 0Ω, we calc. Rx = 283Ω. However, closest resistor available was 360Ω.

We determined Ry using KCL, measuring  5.95V from power supply and asuming V=1V for Ry.    
Ry = 72.72Ω.   Actual R used: 75Ω.

After calculations, we built the breadboard, switching the +/- terminals of one battery to have both
+ and - 5.95V power supplies. We then measured different Voltage values at 0-1Vin, as followed:

VIN          Vout  (V            Gain              VRi (V)            IRi  (mA)            VRf  (V)
                                      (Calculated)                              (Calculated) 
0.0V            0.02                  0                       0                       0.00                  0.03       
0.25V          2.62                  10.08                0.26                  0.26                  2.62
0.50V          4.06                  9.90                  0.41                  0.41                  4.17
0.75V          4.06                  9.44                  0.43                  0.43                  4.39
1.00V          4.06                  9.02                  0.45                  0.45                  4.62

With VIN = 1.00V, we measured the current from the power supplies:

(+Vcc) IV1 = 0.03 mA
(-Vcc)  IV2 = 0.76 mA

The gain was consistent, although it decreased as the voltage increased, and the current did not pass 1mA.

Webassign Blog

        Wanting the Thevenin Eq. of the circuit external of R3 (= 2kΩ), the R3 is removed from the circuit. VTh becomes the open circuit voltage at the terminals, which is 75V. 
        For RTh, we can treat current sources as open circuits, voltage sources as short circuits, and add Rs in parallel and series. Adding R1 and R2 in parallel gives RTh = 5.33kΩ.

        Alternatively, we could use Norton's theory to treat the open circuit as a short circuit and find the current through it. Using Kirchoff''s current law, I3= 11.81 + 2.25 ~ 14mA. Using IN(I3) = VTh/RTh, RTh = 5.33Ω.

              VTH and RTH is found the same way as part 1. The voltage across the open circuit approximates to 61.2V, and adding R1 and R2 in parallel gives the RTh = 42.4Ω.
              If the R of the load is to dissipate as much power as possible, it must be equivalent to RTh; therefore, RL = 42.4. Using PL = VTh/4RTh, PL = 22.1W.

 DC sweep was introduced in this Pspice lab, where node voltages and currents can be obtained with source that changes through a range of values, rather than staying constant as in nodal analysis.

 

In this circuit V1 had a DC sweep from 0-20V in 1-V increments. The graph above has its traces with the voltages with respect to current.

For more complicated circuits, Thevenin and Norton equivalents can be found for elements. In this case, a DC sweep graph was done at branch I2. With it being Voltage over Current, the resistance is the slope of the graph (around 6 ohms) and VTh is the zero intercept, which is 20V.

 The current source is then replaced with a voltage source to find the Norton equivalent. With the graph being Current vs Voltage, the slope of graph represent the inverse of resistance, conductance, with is around 0.17S. Zero y-intercept gives IN = 3.34A.

In the last circuit we graphed a chart of power. As expected, the power is maximum when the load resistance = the Thevenin Resistance, or in this case, when R is 1k.

Thevenin Equivalents

           We were given the engineering problem of finding the smallest equivalent load resistance that can be used in a system of multiple sources and loads. Using Thevenin Equivalent, we wanted to simplify everything into a circuit with only VTh, RTh, and Rload 2. Therefore, using nodal analysis, we calculated Vth = 8.64V and I load 2 = 0.131 A, and found RTh with RTh = VTh/ I load 2 = 66 Ω. With a min. Voltage of 8V across load 2, we used Ohm's Law to find RL2 = 825 Ω.

 

 

              We first built the thevenin equivalent circuit on a breadboard. After measuring values, we treated the RL2 to have infiinite resistance and measured the voltage. The values were as followed:

Config                     Theoretical Value        Measured Value         Percent Error

RL2 = RL2, min          VLoad 2 = 8V                 VLoad 2 = 7.77V            2.9%                

RL2 = ∞Ω                VLoad 2 = 0V                 VLoad 2 = 0V                  0%

 

 

 

       

          Afterwards, we built the original circuit, consisting two power sources and more Rs, and calculated:

Config                     Theoretical Value        Measured Value         Percent Error

RL2 = RL2, min          VLoad 2 = 8V                 VLoad 2 = 8.04V            0.5%              

 

Values were fairly close with both circuits. Percent error may have been the result of using 9.11V for power sources instead of an ideal 8V, and using R's slightly different from nominal values.

Pspice Introduction

          This lab offered an introduction to using the program Pspice in order to create circuits, thus obtaining values of current and voltage electronically.The program is quite user friendly, where elements and wires could easily be placed, rotated, or adjusted as commanded. Values of elements can be changed by clicking on the specified elements, and having to add ground makes it practical to real world appliances. In the end, the circuit could be run as a simulation to obtain values of current and voltages wherever specified, as shown above.

         For the circuit below,we had to find the voltage drop across R1 and the total power being produced by VS. Using V=IR, the voltage drop through R1 was 8.66 Voltage. For VS, taking the difference of the two V values on both sides gives the V drop. Using P= VI, (P = (30)(5.118)), the power for VS was around 153.5V.

Nodal Analysis

 

 In this lab we constructed a circuit designed to be 'reliable,' where damage to one part of the circuit may not destroy the entire circuit, leaving other parts functional. The idea is to have two loads supplied by two different batteries, so that if one load fails,  circuit breakers (not included in this circuit) would separate the damaged circuit and the other battery can continue to support the functional parts of the circuit.

       Given resistor and battery voltage values, the currents can be found using nodal analysis by finding the voltages of the nodes and choosing a ground. The 3 upper resistors act as cable resistance (100, 220, and 220 ohm  Rs from left to right) and the 2 vertical 1000 ohm Rs are the loads. With Bat1(nearer to 100 R) = 12V and Bat2 = 9V, the values for I bat 1and2 and Voltages of the nodes were calculated and measured.

Variable    Theoretical Value     Measured Value         Percent Error 

Ibat1         17.5  mA                 17.52  mA                 6.3%

Ibat2         1.5    mA                 1.62    mA                 8.0%

V2            10.3  V                    10.23  V                    0.7%

V3            8.67  V                    8.70    V                    0.3%

(V2 and V3 being the nodes above first and second 1000 ohm Rs, respectively).

Surprisingly, there were no large percent errors when dealing with a small current such as 1.5 mA. It was to be expected that with a 12V Ibat1 would have lots of current since it has a small 100 ohm R.

Voltage Dividers

An unregulated power supply was tested with three resistors (1 kOhm each) in parallel to the circuit. In this lab, we wanted to maintain 4.75-5.25V (5% variation about 5V) while varying the # of Resistances in parallel. Simplification of calculations was made by adding R's and solving Req, and using the Req and max. and min. of the voltage of the load (Vbus) gave the voltage & resistance of the battery source (Vs and Rs).

Vs = 5.54 V          Rs = 54.9 Ohm

Using V=IR, the calculated min. and max. Ibus can be found, which were 5.25mA and 14.25A respectively. For the lab, the power supply had only intervals of 2V, so 6V was used instead of 5V, and a Resistor box measured around 55.4 for the Rs. Measured values of circuits with different # of resisters (loads) were as followed:

Config     Req (Ohm)     Vbus (V)     Ibus (mA)     Pload (W)

1 Load    1000               5.72              5.81 -min      0.033

2 Load    500                 5.43              10.97            0.060

3 Load    333                 5.18              15.6 -max     0.081

All Vbus were within the 5% variation except for the 5.72V (circuit with 1 load). Naturally, the use of 6V would increase all the values of Vbus. The higher Vs (6V, not 5V) also explains the higher min. and max. Ibus, as more current flows as Vs increases and R stays constant. This lab demonstrates that with less Req, there is more current to flow, and using P=VI, shows that more power is applied to the load as well.



Introduction to Biasing

  

The goal here was to have a 9V Alkaline battery light up two LEDs with the following requirements:
                       LED1 rated 5V & 22.75mA            LED2 rated 2V & 20mA.
 LEDs were placed 'in parallel' to regulated their individual currents, with a resistor 'in series' to each LED to control their voltages. Using Kirchoff's current and voltage law we determined the ideal current, voltage, and resistors necessary, and used resistors closest to the calculated values.
                      Calculated Rs:   R1 = 176ohm     R2 = 350ohm
                      Actual Rs:         R1 = 150ohm     R2 = 360ohm
Using an ammeter and voltmeter we measured the current, voltage, and resistance of the circuit with different configuration as followed:
                      Config                    ILED1       VLED1       ILED2        VLED2      ISupply
                      Both LEDs             14.5mA   6.86V       21.0mA    1.65V      35.4mA
                      No LED2 branch    14.5mA   6.81V       X              X             14.5mA
                      No LED1 branch    X             X             20.8mA    1.65V       20.8mA
The bulbs lit up and did not explode because of too much power or current. There was a 26.4% error between the achieved LED current and the desired value, which was most likely the result of using different R values than calculated.

Simple DC Circuit

The objective of this lab was to approximate the max. distance between a battery and load (the 1000 ohm resistor) that would still allow the circuit’s proper functioning, as a 12V battery would work accurately if the voltage is greater than 11V. Rather than use AWG #30 wires, we used a resistor box to act as the resistance of the cables. The box reached its peak at 87 ohm by the time the battery lowered to 11V, and finding the power used by the load (0.1232 W) and the power lost to the cable (0.011 W), the efficiency (= Pout /(Pout + Plost ) was 92%. Assuming the resistance of an AWG #30 wire to be 0.3451 ohm/m, the max. distance rounded to about 252m. With the cable looping back to the battery, the displacement from the battery to the load would actually be 126m, which is still a long distance. Don’t worry anymore about connecting to your neighbor’s battery supply during days of short circuits!