op amps 2

The purpose of this lab is to explore further the effect of op amps on a voltage source using an inverting amplifier.

this is the circuit we are dealing with, as you can see we had to change the voltage sources to 12V sources because of availability. and the resistors used are as follows.

Resistor	Nominal Value	Measured Value
R1	10k	9.86k
Rf	100k	98.3k

we predicted the value of the current leaving the op amp to be .1mA and then built the circuit.

Vin Desired	Vin Actual	Vout Measured	Vrf Measured	Iop Calculated
0.25	0.25	-2.67	2.66	.0266mA
0.5	0.5	-4.95	4.92	.0492mA
1	1	-9.9	9.82	.0982mA

We measured

Icc=.933mA

Iee=-1.034mA

Therefore Iop=Icc+Iee

=.933-1.034

=-1.01mA

our power supplied by each of the 12V sources is

P=.0124W

P=.0112W

then we added a 1Kohm resistor

and measured

Vin Desired	Vout Measured	Vrf Measured	Iop Calculated	Icc Measured	IEE Measured
1V	-10	9.96	.1mA	.73mA	-11.09mA

KCL for the op amp continues to hold because the currents are at different values, which means it was divided through the circuit.

the new power supplied by each voltage source is

P=.133W

P=.0087W

Extra Credit

Rf=50Kohms

Vin Desired	Vout Measured	Vrf Measured	Iop Calculated	Icc Measured	IEE Measured
1V	-5.07V	5.05V	1.01mA	.932mA	-1.033mA

these measurements do behave as expected because of the change in resistor it cuts the factor of 10 down to a factor of only 5. because the resistor is half as much as before.

Pspice Tutorial

 In this lab we learned how to use the pspice program and learned how to put circuits into a schematic and check values and run simulations in the program.

1st we set up the first circuit and check the voltages and currents.

then we set up the 2nd circuit on our own and find the voltages and currents across the nodes

then we set up the 3rd circuit and find the voltages and currents across those nodes as well

we also graphed voltage with respect to amps.

then we set up a thevenin equivalent circuit and show the voltages and amps. 

i attempted to show a graph of the max power but was unable to figure out how to do it as i was not in class when we went over it. But i will do my my best to find out how and post it up before the end of the year. thank you.

op amps 1

objective: we have a sensor whose output varies from 0 to +1V and we need to have the corresponding output range from 0 to -10V, and it must draw no more than 1mA

we decided to use 3000ohms and 33,000ohms as our resistors.
we calculate R to be 1152ohms
and Ry to be 105ohms
which makes Rth=230.4ohms

after measuring and recording everything we get

Component	Nominal Value	Measured Value	Power or Current Rating
Ri	1000	976	8th
Rf	10k	9.88k	8th
Rx	1152	1.292k	8th
Ry	104.72	129	8th
V1	12	12.27	2amp
V2	12	12.27	2amp

then we set up the circuit and check the values

Vin	Vout	GAIN	Vri	Iri	Vrf
0	0	0	0	0	0
0.25	-2.79	11.16	0.25	0.08	2.79
0.5	-5.55	10.98	0.49	0.17	5.49
0.75	-8.29	11.05	0.75	0.25	8.29
1	-10.71	10.77	0.97	0.32	10.77

our gain should be by a factor of 10 everytime, however due to the fact that we didnt have access to every resistor possible we had to use a 33000ohm resistor which gave us a gain of about 11

We also measured Iv1 and Iv2 to be .96mA and -1.29mA

We then calculated
Pv1=(.96mA)(12V)=11.52mW
Pv2(1.29mA)(12V)=15mW

Iri=1V/3000ohms=.33mA
and
Irf=10.77V/33000ohms=.33mA
which shows through KCL that our measurements satisfy the condition of using Vin=1V

11.52mW and 15mW are less than 30mW so we did satisfy the power suply constraint. we could increase the values of the resistors to reduce the power drawn from the power supplies.

FreeMAT lab #2

freemat lab #2

these are my freemat and theoretical solutions to #1

this is my formula for converting rectangular and polar

these are the conversions for all of the examples.

this is number 3 with its theoretical proof

solution on freemat for the matrix problem in #4

2nd order lab online

This is our online work we did for the 2nd order lab.

Max Power Lab

objective: to prove that the maximum power transfer occurs when the load resistance is equal to the Thevenin resistance of the circuit as stated in the Maximum Power Transfer Theorem.


The theoretical Values for Voltage and Power at the potentiometer are
V=2.26
P=.234

Voltage 0	Resistance 0	Power 0
0.15	0.21	0.03
0.3	0.42	0.05
0.46	0.69	0.08
0.61	0.95	0.1
0.92	1.59	0.13
1.22	2.46	0.15
1.48	2.72	0.2
1.75	3.54	0.22
2.22	5.39	0.23
2.25	5.56	0.23
2.39	6.29	0.23
2.52	7.02	0.23
2.63	7.8	0.22
2.74	8.58	0.22

I am having trouble pasting my graphs into the blog from excel. The measure of potentiometer resistance that results in max power transfer is 5.56 ohms the theoretical value of resistance is 5.50 ohms, which yields a max power of .234, so percentage error is 1.7% which is an acceptable value! Theoretical Values Measured Values R1=.987 KohmsR2=9.970 KohmsR3=9.959 KohmsR4=.989 KohmsR5=.998 Kohms 4.5V=4.479V=8.89 

 From the picture it is hard to tell but the Highest power input is at , , which is when the Load resistance from the potentiometer is equal to the Thevinin resistance,Therefore we can conclude that max power transfer will always occur when these two values are equal to one another.

Thevenin Equivalents

Objective: our goal is to take a circuit with multiple voltage sources and resistors and calculate and test the Thevenin Equivalents

First we calculate Vth which is equal to Vx

then we Calculate the following,
Vy=5.11V
Vy/Rc3=Isc
Rth=Voc/Isc
Rth=65.98ohms

1.
then we find the smallest permissible RL2 using a voltage divider which is
Vth/(Rth+RL2)*RL2=VL2
VL2=8V
RL2=819.9ohms


2.
The short circuit current is then found by
8/819.9=.0098A


3.
The open circuit voltage is found to be
8.64V

then we set up the experiment

and come up with the following values

Component	Nominal Value	Measured Value
Rth	66Ω	65.9Ω
Rl2,min	819.9Ω	820Ω
Vth	8.64V	8.66V

Config	Theoretical Value	Measured Value	% Error
Rl2=Rl2,min	Vload2=8V	7.78V	2.75%
Rl2=infiniteΩ	Vload2=8.64V	8.71V	1.00%

Component	Nominal Value	Measured Value
Rc1	100	98
Rc2	39	38
Rc3	39	39
RL1	680	667
Vs1	9	9.1
Vs2	9	9.18

Config	Theoretical Value	Measured Value
Rl2=Rl2,min	Vload2=8V	6.9
Rl2=infiniteΩ	Vload2=8.64V	7.94

Configuration	Vload2
RL2=0.5Rth	1.8
RL2=Rth	2.85
RL2=2Rth	4.16

in conclusion we have found that by calculating the Thevenin Equivalents we can easily find new values for V and R thevenin when sources or resistors are added to the circuit.

Voltage Dividers

Objective: we need to determine the bus voltages so that anywhere between one to three loads can be applied to the voltage and still meet the voltage and current requirements.

first we calculate

Req,max= 1Kohm
Req,min=1/3Kohm

upper and lower voltage bounds are

max=6.25V
min=5.75V

plugging in equations

I bus,max=19.2mA
I bus,min=6.25mA

data collected from the three possible congigurations

Config	Req	Vbus	Ibus	Pload
1 load	1Ω	6.09V	6.1mA	.04W
2 loads	1/2Ω	5.69V	11.4mA	.065W
3 loads	1/3Ω	5.33V	16.1mA	.086W

the actual percentage in load variation goes from 1.5% to 5.1% to 11%, the difference can be found from the imperfections in the tools used in the experiment, the variation of power given, the variation in the resistors and so on, everything factors in and creates problems for experiment.

if a fourth load was added, that would give us an Req of 1/4 a Vbus of about 5.0, an I bus of about 21mA and a Pload of about .11W

the new source parameters would have to be between 5.95-6.05.

in conclusion our tests went fairly well and our percentages almost made it within our requirements.

Intro to Biasing

Our objective is power and light up two separate LEDs that have different Voltage ratings and different current ratings. By doing Biasing we will attempt to establish correct voltages and currents across the LEDs so they will not burn out

we start by calculating the resistances needed to acquire our desired voltages.

R led1

176Ω

R led2

350Ω

we then calculate the following

I R1	22.75mA	I R2	20mA
V R1	4V	V R2	7V
R1	176Ω	R2	350Ω
P R1	.091W	P R2	.14W

and after looking at the available resistors we decide to use

R1

220Ω

R2

360Ω

we choose our resistors and acquire the following measurements

color code (List Colors)	Nominal Value	Measured Value	Wattage
2200	220	213	1/8W
3600	360	352	1/8W
4700	470	459	1/8W

after building the ciruit

we obtain these values

Config	I led1	V led1	I led2	V led2	I supply
1	10mA	5.9V	14mA	2.11V	24mA
2	10mA	5.9V	X	X	10mA
3	X	X	14mA	2.11V	14mA

BONUS*

Both LEDs must be in this circuit for it to work properly because of increased current.

A. 8.3 hrs
B. 44%
C. 93%

introduction to DC circuits

Our objective in this experiment was to determine the max cable resistance required, the max distance the battery and load can be separated with awg#30 cable, the distribution efficiency, and the time it takes for the battery to discharge

we determine the theoretical value for R(load) to be 1000Ω

Vload=11.01V
Ibatt=11.6mA
Rcabletot=73Ω

Time to discharge
.8Ahr/.0116A=68.97 hours

power to the load=.1309 W
power to the cable=.0098W

the efficiency is (.1309/(.1309+.0098))100=93%

we are not exceeding the power capability of the resistor box because we are well below .13w

the max distance of the wire is 73/.3451=105.77m


e. (optional)
the ohm per foot ratio in AWG#28 wire is 0.06490Ω
V=IR
5=(.020)R
R=250Ω
250/.06490=3,852 ft.

however we will probably not be getting ideal values, therefore done with a  voltage of .2
.2=(.020)R
R=10Ω
10/.06490=154 ft.

only being able to drop 12V I calculated R=1.2ohms dividing that by 150 ft. I get .024
therefore,
if the sub is approximately 150 ft. away then the minimum cable gauge would be #22