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.