Op Amp 1

  

In this lab we are to have a sensor whose output ranges from 0 to +1V and whose Thevenin output resistance is negligible. the output range must be between 0 and -10V and must draw no more than 1mA from the senseor and the op amp power supplies should be no more than 30 mW.
From our calculations we determined Ri = 1 k ohms and Rf = 10 k ohms
We proceed by determining Rx = 8 * V^2 = 1152 ohms      RY / (RY + RY) * 12 =1     RY = 288 ohms
From this we find that RTH = 230.4 ohms which is at least 20 times smaller than the resistance value of R1

Applying our theoretical results we attain the following

Component Nominal Value Measured Value Power Rating

        Ri                1k ohm      984 ohm            0.25 W

        Rf                10k ohm      9.74 ohm            0.25 W

        Rx                1k ohm      982 ohm            0.25 W

        Ry                     x                    x                         x

        V1                  12 V      12.06 ohm               2 A

        V2                  12 V       12.1 ohm               2 A

Vin           Vout Gain          VRi         IRi           VRf

0.0V          0V             0           0V          0 mA      0

0.25V -2.48V -9.92 .249V .253 mA -2.48 V

0.5V -       4.97V -9.94 .5V         .508 mA -5.01 V

0.76V -7.48V -9.97 .751V .763 mA -7.52 V

1.004V -10.03V -9.98 1.005V 1.02 mA -10.01 V

IV1 = 1.737 mA                            IV2 = 1.276 mA

PV1 = IV1 * V1 = 20.9 mW            PV2 = IV2 * V2 = 15.4 mW

I_Ri = V_in / R_i =1.004 / 984 = 1.02 mA

I_Rf =  V_Rf / R_f  = 10.01 / 9740 = 1.02 mA

From our calculations we see that the power supply constraint of no more than 30 mW is maintianed. In order to reduce the power drawn we should increase the resistor values in a way that the ration Rf/Ri = 10.

PSpice

In this lab we practicing using the program PSpice.

First Circuit

Second Circuit

VTH = 10 V    RTH = 3.33 ohms    IN = 3 A

Maximum Power Transfer

In this lab we verified the Maximum Power Transfer Theorem in a simple series circuit and use the theorem to determine the Thevinin Resistance.

We then build the circuit using a potentiometer and recorded the values of V, R and P from the lowest to highest resistance of the potentiometer.

we then built the following circuit

we followed by using the data we obtained in LoggerPro

Thevenin Equivalents

The purpose of this lab is to determine the smallest equivalent load #2 resistance that can be succesfully used. We are to reduce the network to an equivalent source and series resistance. We are given that RC1 = 100 ohms RC2 = RC3 = 39 ohms RL1 = 680 ohms VS1 = VS2 =9V VLoad 2,min = 8V.

Through the use of nodal analysis we establish that VLoad 2 = 8.643V VY = 5.11V RTh = 65.98 ohms. The smallest permissible RL2 = 819 ohms short circuit current I = 9.76 mA and open circuit voltage V = 8.64V

Transistor Switching

In this lab we are to learn how to use a transistor. We first create the following circuit.

We then demonstrated that the led still glows after removing R2 and completing the circuit with our finger tip. This demonstrates that the transistor responds by amplifying the current applied to the base.

We then add a potentiometer and remove the LED

we collect the following data

A1      0     .1     .2      .31     .44     .58     .75

                                                A2   .  1    34   42.3   47.3   51.6   53.5   54.7

As we can see the transistor amplifies the current but eventually becomes saturated.

FreeMat

In this lab we were to learn how to use basic function in freemat in order to give us another tool to solving multiple equations.

Exercise 1: Create a 2-D plot

Exercise 2: Given a circuit find Current

Exercise 3: Plot 2e^(-t/T) and 2(1-e^(-t/T))

Voltage Divider

In this lab we are to determine the characteristics of the Unregulated Power Supply given an acceptable range of variation of the bus voltage. Given the model below we are to calculate the required values that will function if only one load is ON.

 We are given that R1 = R2 = R3 = 1K ohms and that anywhere from one to all three loads may be on at any time. We determined REQ,max = 1K ohms and REQ,min = 1/3 K ohms. VBUS,max = 5.25V VBUS,min = 4.75V VS = 6.53V RS = 45.45 ohms. IBUS,max = 6.25 mA IBUS,min = 17.25 mA. We follow through by implementing our design and got the following results.

There was a 5.5% and an 0.8% voltage variation which is different from our design goal because of our use of an unregulated power supply.

Adding another 1k ohm load will have no affect on the load voltage because its in parallel.

Introduction to Biasing

In this lab we had to determine which circuit to use in order to light up two LEDs. We must establish the correct voltage across and the current through a component so it operates properly. We are to use a 9V battery, LED1 rated for 5V and 22.75 mA, LED2 rated for 2V and 20 mA, and resistors R1 and R2. We determined the Resistance of the LEDs to be RLED1 = 219.8 ohms and RLED2 = 100 ohms. Using KCby the resistors PR1 = 0.091 W and PR2 = 0.14 W. Resistors are only available in certain discrete values based on these values we chose R1 = 220 and R2 = 470 as they are closest to our values and will ensure that our LEDs are not destroyed. After constructing our experiment we determined the following values.

We determined the Resistance of the LEDs to be RLED1 = 219.8 ohms and RLED2 = 100 ohms. Using KCL, KVL, and Ohm's Law we determined IR1 = 22.75 mA IR2 = 20 mA VR1 = 4V VR2 = 7V R1 = 175.8 ohms R2 = 350 ohms. To ensure we do not destroy our equipment we determined the power consumed by the resistors PR1 = 0.091 W and PR2 = 0.14 W. Resistors are only available in certain discrete values based on these values we chose R1 = 220 and R2 = 470 as they are closest to our values and will ensure that our LEDs are not destroyed. After constructing our experiment we determined the following values.

a. I= ILED1 + ILED2 = 22.75 mA + 20 mA = 42.75 mA

.2 A -hr / .04275 A = 4.678 hr

b. (40.5 - 42.75) / 42.75 * 100 % = -5.3% error

The error in our experiment is attributed to the resistors not being exact to our theoretical resistors.

c. PLED1 = 9.1 mA * 4.73 V = 43.043 mW

PLED2 = 31.4 mA * 2.24 V = 70.336 mW

Pout = PLED1 + PLED2 = 113.379 mW

Pin = 9 V * 39.2 mA = 352.8 mW

113.379/352.8 * 100% = 32.1% efficiency