Op Amp Circuits

The problem that we are tasked to solve for in this lab is to make a sensor work with a micro-controller by inserting an operational amplifier in between the two elements. The output voltage of the sensor is given to be from 0 V to 1 V.  For the micro-controller to function it needs an input of about 10 V. Given these constraints we are to create an op amp circuit. While creating the circuit we found that changing the feedback resistor will determine the output of the operational amplifier.

 

 

PSpice Thevenin and Max Power (Homework)

To solve for Thevenin Voltage for the first problem we must place a current source where the Thevenin Voltage is to be found, like the figure below at the far right.

 

 When we simulate the DC Sweep a graph is given to analyze.

When my current source is at zero (x-axis) my voltage is 112.50 volts. This is my Thevenin Voltage. For the slope of this graph I get (695.833V-112.500V)/.1A, which is equal to 5833.33 V/A, this is my Thevenin Resistance.

To find my Thevenin Equivalents in my next problem I will perform the same procedures like before.

After Running DC Sweep I get my Thevenin Voltage to be 75.65 Volts. The slope of this graph is (105.065 V - 75.645 V)/.4 A= 73.55 ohms.

 

 

 The second part of this problem asks to find the value of the resistor load that will dissipate as much power as possible and to find that max power. The first thing to do is to transform the above circuit into the Thevenin Equivalent circuit. Then we set up the DC Sweep to give us the graph that we want

When the Circuit is simulated, the graph is traced to find the max power and the value of the resistor load. The graph below shows that in order to dissipate maximum power the resistor has to have the value of 73.55 ohms. At that point the graph states that the value of max power will be 19.453 Watts.

PSpice for Thevenin and Norton Equivalents and Max Power

In this assignment we learned how to use PSpice to find Thevenin and Norton Equivalents. But to be familiarize with the program we first created a simple circuit to find the voltages on each nodes. To accomplish this, we used DC Sweep, a feature used in the Schematics.
 

 

Now to find Thevenin Equivalent, we draw out our new circuit and then place a current source where the Thevenin Voltage is to be found. just like the figure below.

 

Running the DC Sweep will give us the graph that looks like this:

  

The zero x axis of the graph will point out the Voltage Thevenin and the slope will give us Thevenin Resistance.

To find the Norton Equivalence we must remove the current source on the far right of the circuit and place a voltage source.

 

 The zero intercept of this graph will give us the Norton Equivalence and the slope of this graph will be the Norton Inverse Resistance

 

 In this Thevenin Equivalent circuit, we are going to find the maximum power that the resistance load will dissipate in the circuit.

With the help of PSpice, this figure shows that the max peak of the graph is the max power, which is 250 microwatts.

  

Thevenin Equivalents

In this lab we analyzed a circuit that we wanted to turn into a Thevenin equivalent. We were first asked to compute the Voltage Thevenin using nodal analysis and we found our value to be 8.643 Volts.

To find the Thevenin resistance we first needed to find the current found by short circuiting the open circuit. Once we found the current with nodal analysis, we divide the V_th with I_sc to find the Thevenin resistance. The resistance was 65.964 ohms. To check our value we can also short circuit the two only voltage sources and find the equivalent resistance, which was 65.946 ohms.

When we created our Thevenin equivalent circuit, our elements measured were: R_th = 66.9 ohms, R_L2,min = 828 ohms, V_th = 8.64 Volts.

Component    Nominal Value    Measured Value
R_th             66 ohms             66.9 ohms
R_L2,min      820.479 ohms     828 ohms
V_th             8.64 Volts           8.64 Volts

Config                    Theoretical Value    Measured Value    Percent Error
R_L2=R_L2,min     8 Volts                   7.81 Volts              2.4%
R_L2= inf Ohms    0                            0                            0

Pspice Tutorial and Homework 5

To start off a DC analysis, you must first launch the Capture Student application. Then click File>New>Project from the menu.Click on Analog or Mixed. The next page will ask if you want to open an existing or a blank project, click blank project. To start on the circuit, all wires, resistors, current sources, and voltage sources will be found on Place on the menu bar. Once the circuit is finished go to PSpice>new simulation profile. Type Bias in the name field and click create. Select Bias Point under the Analysis type and click OK. Finally, run the simulation by selecting PSpice from the menu and selecting run.


 This an example done for one of the Homework problems from assignment 7

Nodal Analysis

For this lab, we were asked analyze a "reliable" circuit that contains multiple loads and two power supplies. We were given the values for all resistors and both power supplies. With these we calculated what our voltages were supposed to be using nodal analysis. When we finished, we needed to then make a circuit to be exactly like the one we used to calculated our voltages. With the multi-meters, we measured the voltage and current values and compared them with our calculated values.

Variable    Theoretical value    Measured Value    Percent Error
I_Batt1    17.45 mA               17.43 mA              0.11%
I_Batt2    14.82 mA               15.24 mA              2.79%
V_1         10.255 V               10.17 V                 0.88%
V_2         8.674 V                 8.58 V                  1.09%

Voltage Dividers

In a house where there are multiple loads in a circuit, there is going to be some kind of variation in the voltage source. We are going to create a circuit that will help us understand what our voltage source should be. With the given values for the resistance loads and voltage bus max and min, we can calculate what our voltage source and our bus resistance should be.

V_source= 5.54 volts, R_s=55.56 ohms
Voltage bus max = 5.25 volts, Voltage bus min = 4.75 volts. (A 5 percent variation about 5 volts).

This is what our circuit was like when all resistance loads where connected to each other.

Calculated:
Voltage bus min= 5.18 volts, Voltage bus max= 5.72 volts

Because our Voltage source could not have been set to 5.54 volts, we were strictly forced to make it 6 volts. This new voltage changed our voltage variation by 9.91 percent.