PSpice Review

Given problem

Place a DC current source between terminals and sweep the current from 0 to 3 amps. We can find Vth=10V when I terminal = 0A. Rth is the slope which is equal to 10/3 =3.33 ohms.

Then, assume RL=R4=10 ohms. The Thevenin Equivalent circuit is shown as following

Norton Equivalent circuit is 

Set RL is from 1 to 10 as following

Then graph power (V6*V6)/RL vs RL, and then RL should be 3.33 ohms in order to get the maximum power which is 7.5 W.

PSPICE Tutorial

Purpose


The purpose of this lab is to learn how to PSpice to simulate a complex circuit.

Given circuit

By using PSpice, the voltage at each node and the current on each branch are easy obtained as following,

Thevenin Equivalents

Purpose

       The purpose of this lab is to simplify a complex circuit by using Thevenin Equivalent method. 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, 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,  RL2 = 825 Ω.

Results

Config          Theoretical Value          Measured Value           % error

RL2                   Vload2 = 8V                   7.99V                       0.13%

RL2 infi.            Vload2 =8.64V               8.63V                       0.12%

Extra

Pmax = 8.64^2/(4*66)=0.28W

Overall, values were fairly close with both circuits. Percent error may have been the result of using 9.07V for power sources instead of an ideal 8V.

Beta of a BJT

Purpose

The purpose of this lab is to figure our the beta(gain) of a BJT by finding the linear relationship between Ie and Ib. For a NPN typed BJT, Ie = (1+beta)*Ib

The built circuit looks like as following

Figure 1.

Measured V         Rb(Theoretical)         Rb(actual)            Ib                   Ie

6.05 V                       530 k ohm           529 k ohm          0.011 mA      1.36 mA

6.05 V                       265 k ohm           263 k ohm          0.021 mA      2.81 mA

6.05 V                       176 k ohm           176.6 k ohm       0.033 mA      4.18 mA

6.05 V                       132 k ohm           132.5 k ohm       0.046 mA      5.56 mA

6.05 V                       106 k ohm           106.4 k ohm       0.058 mA      6.98 mA

Using EXCEL to graph Ie v.s Ib 

Therefore, 1+beta=117.3 ========> beta=116.3

Nodal Analysis

Purpose

The purpose of this lab to analyze a circuit with nodal analysis since the given circuit combines elements in series or parallel. Therefore, choosing a node and identifying a ground are important steps before solving the circuit.

Given Problem

Figure 1. 

We can write down equation on V2 and V3 as following

(V1-V2)/100=V2/1000+(V2-V3)/220------->1

(V2-V3)/220+(V4-V3)/220=V3/1000------->2

Solve for equation 1&2 we can get V2=10.26V and V3=8.67V

Ibat1=(V1-V2)/100=0.0174A

Ibat2=(V4-V3)/220=0.0015A

Pbat1=-0.0174*12=-0.2088W

Pbat2=-0.0015*9=-0.0135W

Next we build the circuit up 

Figure 2. 

Connect it with two voltage supplies, and measure the battery current

Figure 3.

The results come out as following

Variable    Theoretical Value     Measured Value         Percent Error 

Ibat1         17.4  mA                 17.12  mA                  1.61 %

Ibat2         1.5    mA                 1.77   mA                    18  %

V2            10.26  V                  10.34  V                     0.78 %

V3            8.67  V                    8.77    V                    1.15 %

Next, we design that we want V2=V3=9V, we need to solve the desired battery voltages

Set the equations up

(V1-9)/100=9/1000+9/220======>V1=9.9 V

(V4-9)/220=9/1000======>V2=10.98 V

We implement the new battery voltage in the circuit and measure the new current again

V2=8.92 V, V3=8.80 , I bat1 = 9.29 mA, I bat2 = 8.03 mA

P.S. This guy allows us to adjust our voltage we want in a range between 0V to 20 V

FreeMat Activity (Assignment 4)

Purpose
The purpose of this activity is to solve a circuit problem using nodal method and matrix technique in FreeMat. 
Problem

KCL:
i1+i2=i3


V3=10i3
V2=5i2
V1=20i1


KVL:
15-V3-V1=0  
V3+7+V2=0  
==> solve for
20i1+10i3=15
5i2+10i3=-7

i1+i2-i3=0
by using FreeMat, i3 can be solved as following 

Therefore, i3 = -0.1875A

Introduction to Biasing

Purpose


The purpose of this lab is to design a proper bias circuit to light up two LEDs for preventing them from burning out.


List of materials

• 9 V power supply
• LED 1 (5V 22.75mA)
• LED 2 (2V 20mA)
• Resistors
• Multimeter
• BreadBoard

Process

1. We determine the value of resistor by using given current 22.75 mA and 5 V across LED 1, and 20 mA, 5 V across LED 2. R(LED 1)=5/22.5*1000 = 220 ohm, and R(LED 2)= 2/20*1000= 100 ohm.
2. Next decide each resistor R1$ R2 need to be used in series with each LED. R1= (9-5)/22.75*1000=176 ohm. R2= 7/20*1000=350 ohm.
3. In order to get those values of resistance, we have to use two 150 ohm and 100 ohm to get 175 phi, and two 100 ohm and one 150 ohm to get 350 ohm. So, we pick up three "Brown, Green, Brown Gold" resistors and three "Brown, Black, Brown, Gold" resistors. The actual value resistance are 151.1 ohm and 98.5 ohm, and their wattage are 1/4 W. 
4. Build the circuit as following Figure 1.

Figure 1. Biasing LED circuit

• I(LED1)=13.7mA, V(LED1)=6.55V, I(LED 2)= 19.3mA, V(LED2)=2.18V, I(supply)=3.29V

     5. Next we remove LED 2 from the circuit

Figure 2. Remove LED 2

•    I(LED1)=18.93mA, V(LED1)=6.56V, I(supply)=3.29V

     6.  Remove LED 1

Figure 3. Remove LED 1

• I(LED 2)= 19.25mA, V(LED2)=2.19V, I(supply)=3.29V

Calculation

1. Useful 9V battery life is 0.2 A-hr, so we want to calculate how long can the circuit operate before the battery goes to low. 

• I(total)=13.7+19.3=33mA=>0.2A*Hr=33mA*t  =>  t=6.1 hr

    2.  We figure out the percent error between LED current and the desired value for both LEDs.

• % error LED1 = (13.7-22.75)/22.75*100% = -39.8%
• %error LED 2 = (19.3-20)/20*100% = -3.5%

    3.  the power efficiency is pout/(Pin+pout)*100 % = 0.079/(0.079+0.297)*100% = 21.01%

    4. pin= 6 *(22.75+20)*10^-3= 0.255W, pout = 0.154W. power efficiency=    
        0.154/(0.154+0.225)=40.62 goes up.

         

Introduction to DC Circuit

Purpose


The purpose of this lab is to determine the maximum permissible cable resistance required for the load to still function normally, and the maximum distance the battery and load can be separated if use AWG # 30 cable, the distribution efficiency, and the approximate time before the battery discharge. The simple system can be illustrated as following

Figure 1. A Simple Cabling System

1. The theoretical value for R(load) can be determined by using R=V^2/(P)= 12^2/0.144=1 Kohm.
2. Since we need to measure the current through the circuit and the voltage across R(load), ammeter and voltmeter need to be inserted into circuit in series and parallel respectively.
3. The resistor box power rating is 0.3 W and the max supply voltage and current are 12 V and 2 A.
4. Based on the value of the resistor, we choose "Brown black red gold" resistor to act as R(load) which provides us 1000 ohm within 5 % error, and its real measured value is 980 ohm. Also, its wattage is 0.25 W. For R(cable), the measured value is 87.3 ohm and its theoretical value is 88 ohm. Its wattage is 0.3 W from the label.
5. We measure the actual voltage which is 12.09 V that a 12 V power supply can really provide.
6. Next, we build the circuit as following
7. Then we measured V(load)=11.1 V, I(Bat)=11.3 mA, and R(cable total)= 87.3 ohm.
8. We use the Ahr capacity of the battery  to estimate the amount of time the battery can supply this load before being discharged. The capacity of the battery is 8 Ahr. By calculation, 8 Ahr= 11.3mA*t therefore, t = 708 hr.
9.  Moreover, we want to know distribution efficiency of this circuit, so we have to calculate power to   the load and to the cable. P(load)=11.1^2/980=0.126 W, and P(cable) =(12.0911.1)^2/87.3=0.0112 W. The efficiency is p(out)/(P(out)+P(lost))=0.126/(0.126+0.0112)*100 = 91.8 %
10.  If we have a AWG #30 wire which the resistance is 0.3451 ohm/m. we can determine the maximin  distance between the battery and the load. That is 1/2 * 87.3 ohm/0.3451 ohm/m = 126.5 m

Figure 2. Actual simple circuit

Review Exercise: Using a Multimeter

Purpose:

The purpose of this lab is to learn how to prevent a multimeter from burning out its fuse by doing resistance, voltage, and current test.

• Resistance Test

1. First we set the multimeter to read resistance when its two probe do not touch anything, the reading from the multimeter is zero ohm.
2. Touching the two probes together, the reading of the multimeter is 1 ohm since the metal tips are conductors.
3. Then we need to grab a set of four different resistors and record the measured and theoretical values. Here is the way to figure out the color:

   
       original source above is from  http://www.afiata.com/how-to-read-resistor-color-codes/.

   First resistor we choose is "Red orange brown", so resistance value from colors is 230 ohm, and
its measured value from multimeter is 218 ohm. Next, the color of resistor is Orange orange red", so the resistance value from its colors is 3300 ohm, and its measured value from multimeter is 3290 ohm.  Third, the color are "Orange, orange, brown"; therefore, the resistance value from its colors is 330 ohm, and its measured value from multimeter is 329 ohm. Finally, we choose "Red, black, red", so the resistance value from colors is 2200 ohm, and its measured value from multimeter is 2010 ohm. Notice that there is a color code at rightmost location, and the color of all tested resistors are gold, which tell us that the error for the certain resistor is within five percent.

• Voltage Test

1. Set the multimeter to read DC voltage.
2. Use two probes of the multimeter to touch a 9 volts battery, the measured value is 9.10 V.
3. Testing wall wart adapter, we get 23.0 V. (Output from label is 18V) If we use two alligator clips to attach a 150 ohm resistor across two multimeter terminal the voltage we measured is still 9 V.
4. Make a circuit as shown in figure 1. 

Figure 1. An LED circuit

The measured voltage across LED is 2.13 V. The voltage of the battery to the sum of the voltage drops across the resistor and the LED is the same.

• Current Test

1. Set the multimeter to read DC current and make a circuit as figure 1.
2. To measure the current of the circuit, the multimeter has to insert the circuit as in series. The measured current is 15.9 mA.