Inverting Differentiator Lab

Objective:
We covered integrators and differentiator op amps which involve resistors and capacitors In order to create an ideal integrator, we replace the resistor feedback with a capacitor for an inverting amplifier. We can find the current  through nodal analysis at the negative terminal which has a voltage of 0v and can calculate an equation based on the voltage input and output (Vout(t) -V(0) =-1/RC int(Vin(t)dt). To create an differentiator, we can replace the input resistor with a capacitor in an inverting amplifier. the equation calculated is Vout = -RCdvi/dt. Differentiators are unstable and rarely used because of exaggerated noise. We also covered the idea of switching functions and others. In the end, we covered step response of RC and RL circuits.

Group Practice:
1. We are required to find the voltage by using the equation as seen in figure 1. In order to acquire the voltage, we will integrate the current. The voltage graph is then plotted.

Figure 1. Finding the voltage given a current impulse

Inverting Differentiator Lab Procedures and Results:
1. For pre lab, we are required to find an equation of an output voltage as a function of input voltage based on the differentiator op amp. The final equation can be seen in figure 2. We are required to find the frequency needed in the experiment by having the voltage gain Vout/Vin = 1. Therefore, we acquire a frequency of f =2344Hz as long as A = 1, R = 680ohms, and C = 1microfarad.

Figure 2. Calculations for the frequency. 

2. Before, the circuit built, we first measure the actual capacitance of the capacitor and the resistance of the resistor. The resistor has a resistance of R=680 +/-.005 ohms and capacitor of capacitance of C= .945+/-.0005 microfarad. The final build op amp differentiator can be seen in figure 3 & 4. We make sure that we create a voltage sine input with frequencies of 100Hz, 250Hz, 500Hz and measure the voltage output with oscilloscope.

Figure 3. Over all set of op amp differentiator.

Figure 4. Close visual of op amp differentiator.

5. We measured the voltage output when the voltage input with a frequency of 100 Hz was set as seen in figure 5. The voltage output at 250 Hz can be seen in figure 6 and 500 Hz in figure 7.

Figure 5. Voltage output at 100Hz

Figure 6. Voltage output at 250 Hz

Figure 7. Voltage output at 500 Hz

6. We then calculate the theoretical voltage output which can be seen in figure 8 and compare them to our experimental voltage output measurements by looking at the amplitude of the sin wave. We can that they close to each other with percent error less that 3%. We also notice that fussiness throughout the curves.

Figure 8. Comparison of theoretical and experimental voltage outputs for a differentiator op amp. 

Summary of Inverting Differentiator Lab and Learning Outcome:
The lab required us to calculate or find the voltage output as a function of the input voltage in the form of a sinusoid. Since we are dealing with an inverting differentiator, we use the applicable equation derived from earlier in class. We are looking for gain of or Vout/Vin = 1. As a result we can find the frequency in which we can acquire this gain. The frequency we calculated is f = 234 Hz. We also noted that the amplitude of the sinusoidal input will be 1 V and an offset of 0. The measurements of resistor and capacitor are R=680 +/-.005 ohms and C= .945+/-.0005 microfarad. The frequency that we will use to will fall within our frequency of f =234Hz. The frequencies used in the experiment are 100Hz, 250 Hz,  and 500 Hz. From this we noticed that our amplitude increased which means that our voltage output is just proportional to the frequency. We do notice some fussiness in our measurements. Overall when we compared our theoretical and experimental measurements, we see that there is 1.09% error at 100Hz,  2.99% at 250Hz, and 1.69% at 500Hz. The equation for an inverting differentiator hold true since the percent error is relatively low.

Passive RC/RL Natural Response Lab

Objective:
We covered the idea that the equivalent inductance of inductors connected in series or parallel is similar to the idea of resistors in which they add in series and are inversely added when they are in parallel. The goal is to understand what is happening in RC and RL circuit and Passive RC/RL Natural Response Lab will help us do that. For RC circuit, the capacitor initially acts like a wire but after a long time, it acts like an open switch. We will use the equations i(t) = Io*e(-t/RC) or v(t)=Vo*e(-t/RC). For an RL circuit, initially, the inductor opposes rapid change so it acts like an open switch but after a long time, its acts like an ordinary wire. The equations that we will use are i(t)=Io*e(-tR/L).

Group Practice:
1. We are given a first order differential equation in which we must solve the Voltage of the capacitor in terms of time. We see that tau is the resistance times the capacitance so it can be replaced by tao in equation as seen below in Figure 1. We show that the voltage of the capacitor is 1% of the initial voltage in order to see that the voltage of the capacitor drops close to 0. We will use the idea that
t =5*RC which is the time that the capacitor discharges and we use it as an approximation of time discharge.

Figure 1. Using 1st order linear differentiation to solve for voltage across a capacitor

2. Figure 2 tells us to find the max switch frequency and we can find this by finding the time in which the capacitor discharges which is t=5*RC = .005sec. the Frequency= 1/t = 200Hz which is a low pass filter.

Figure 2. Find the max switch frequency of an RC circuit.

Passive RC Natural Response Lab Procedures and Results:

1. The pre lab requires us to calculate the initial voltage of the capacitor and we do this by showing that Vc = Vo(R2/R1+R2) = 3.438V assuming that capacitor acts as open circuit when the capacitor is fully charged. The voltage across the R2 is the voltage of the capacitor since they are parallel to each other.

Figure 3. Initial voltage for the capacitor calculations.

2. We then find the time constant where tau= RC = .0484sec for circuit a as seen in figure 4. The time it takes to discharge is t = 5*.0484 = .242sec  based on figure 4 circuit a. We find the time constant were tau = C*(R1*R2/R1+R2) = .01513sec for circuit b in figure 4. We find the time when discharging from figure 4b by multiplying by 5 since we assume the time of discharging is t = 5*tau = .0756sec.

Figure 4. Schematics for circuits that will be used in the lab. 

3. We construct the circuit as seen in figure 5 and measure the actual resistance of the resistors which are R1= .95K+/-.05 and R2= 2.13+/-.05. The analog discovery will measure the voltage across the capacitor when quickly disconnecting the power supply from the circuit, thus following the procedure from figure 4 a.

Figure 5. Actual circuit set up

Figure 6. The visual discharge of a capacitor when the circuit is suddenly disconnected from the voltage source.

4. The next step is to turn off the voltage source which will act as a wire. The procedure schematic can be seen in figure 4 b. The visual result can be seen below.

Figure 7. The visual discharge of the capacitor when the circuit is turned off instead of suddenly being disconnected.

5. We also measured the time of the voltage charge of the capacitor which can be seen in figure 7.

Figure 8. The visual charge of the capacitor. 

6. When suddenly disconnecting the voltage source, we measured a time of discharge t =.270 sec. When the voltage source was turned off, we acquired a time discharge of t = .07sec. There is a percent error of %error=(.270-.242)/.242 *100 = 11.5 % when suddenly disconnecting the voltage source. There is a percent error of %error = (.07-.0756)/.0756 *100 = 7.41%

7. The next step is to use a square wave with an amplitude of 2.5V and an offset of 2.5V that oscillates between 0V and 5V and a frequency of 1Hz. 

Passive RL Natural Response Lab Procedures and Results:
1. We are given an inductor and we must calculate the value of the inductor L. In order to calculate the the value of the inductor, we must use the equations i(t) = Io*e(-tR/L). The first step is to use the same setup as the passive RC Natural Response Lab and calculate the voltages across R1 and R2. With these measurements we can find the current that flow in R1. We can then use the equation Vr= Io*R*e(-tR/L) and manipulate it so that we can solve for L

Summary of Passive RC/RL Natural Response Lab and Learning Outcome:
The main focus of the Passive RC Natural Response Lab is to calculate and compare the time it takes to discharge a capacitor when the circuit is suddenly disconnected and turned off. We can assume that when the circuit is suddenly disconnected, the resistor which is connected to the voltage source is neglected since there is no current flowing through the wire. Therefore, we can assume that the time it takes to discharge the capacitor will be higher than when the voltage source is simply turned off. Based on our theoretical calculations, we acquire a time of discharge of t = 5*.0484 = .242sec when the voltage source is suddenly turned off. We acquire a time of discharge of t = 5*tau = .0756sec when the voltage source is simply turned off. We approximated the experimental time as seen in figure 5 and 6. The time of discharge when voltage source is suddenly tuned off is t = .270sec and when voltage source is turned off, t = .07sec. We calculated the percent error when disconnecting the voltage source of 11.5% error and when the voltage was simply turned off 7.41% error. Our theoretical time falls within our experimental results since the percent error is low. We were unable to to calculate the experimental time constant since we did not measure the capacitance of the capacitor. However, out theoretical time constant is Tau = .0484 sec for sudden disconnection and Tau = .01513sec for turned off power source. We were unable to finish the RL natural response Lab but have a sense on how to do it. The idea was to measure the voltage across the both the resistors and use those voltages to acquire the current and finally use the equationVr= Io*R*e(-tR/L).

Capacitor/Inductor Voltage-Current Relations Lab

Objective:
The goal for today's class meeting is to review the idea of capacitors in circuits. Capacitors are devices that store electrical energy in the electric field. There are two plates which are usually aluminum since they are cheap to use. Also, the plates are oppositely charged. The voltage between the capacitor plates is V=Q/C  where C = (epsilon)*A/d can be replaced. We can also find the equivalent capacitance which is almost related the equivalent resistance but opposite. In other words, when capacitor are in parallel, they are added together. When the capacitors in series, they are inversely added. We also covered and reviewed inductors which are coils with many loops that oppose rapid changes in current. The voltage difference between an inductor is V = L*di/dt. We will use this idea for the lab by using the oscilloscope and waveform generator.

Group Practice:
1. The image below is an experiment that shows a capacitor being blown up due the having the positive voltage terminal input being connected to the negative side of the capacitor. The positive side  or plate of the capacitor is connected to the negative input terminal. This makes the capacitor expand due the liquid salt  inside the capacitor that makes it expand and explode.

Figure 1. Set up of blowing up a capacitor when the voltage inputs are connected in the opposite direction.

2. We are given a problem where we are told to find the voltages at the capacitors. In order to this, we will assume that the capacitors act like an open switch. We can then use current division and find the current flowing through the 3k,2k and 4k resistor. Figure 2 shows how to find the current. We can then find the voltages at 4k and 2k resistors which are actual voltages of the capacitors since they are parallel to each other. 

Figure 2. A circuit in which there are capacitors. 

3. We are given a voltage equation in terms of t and an initial current at t=0, We are told to find the current in a circuit where there is an inductor. We use integration as seen in figure 3 and solve for current through the conductor.

Figure 3. A problem in which we must solve for the current flowing through the inductor and the energy.

Capacitor Voltage-Current Relations Lab Procedures and Results: 
1. Before we start the lab, we are required to do a pre lab that will give us an idea on what we expect for our capacitor current when when sinusoidal and triangle wave voltage is applied. As seen in figure 4, since i=Cdv/dt, we can calculate the current by taking the derivative of the voltage input.

Figure 4. A plot that shows a voltage signal and the expected current. 

3. In order to test our the theoretical equation where the current is the derivative of voltage with respect to time, we will build a circuit where there is a resistor and capacitor in series. The schematic of our planned building circuit can be seen in figure 5 and the actual circuit can be seen in figure 6.

Figure 5. Schematic for circuit where there is a voltage source connected in series with resistor and capacitor.

Figure 6. Actual circuit build. 

4. We apply a sinusoidal input voltage with f=1kHz, A =2V and offset 0V. We record the voltage difference of the resistor as seen in figure 7 with the wave colored in yellow. The voltage difference of the capacitor can be seen represented by the color yellow and the current through the resistor by the color orange. To measure the current, we had to create a math channel where i = C1/100 where C1 represents the voltage across the resistor. The actual resistance of the resistor is R= 98 ohms+/-0.5

Figure 7.  Results when frequency is 1kHz with a sinusoidal voltage. 

5. The next step is to change the f = 2kHz. The results can be seen in figure 8.

Figure 8. Results when frequency is 2kHz

6. The final step is to apply a triangular input voltage with f = 100Hz, A = 4V, and offset=0V. The result can be seen below. The voltage difference across the capacitor is in yellow and the voltage input is in blue. The current of the resistor is in orange. 

Figure 9. Results when f =100Hz with a triangular voltage input. 

Inductor Voltage-Current Relations Lab Procedures and Results:

1. The image below is similar to the previous lab. However, there is a inductor in this case. The actual circuit set up can be seen in figure 11.

Figure 10. Schematic of a voltage input connected to a resistor and inductor in series.

Figure 11. Actual circuit.

2. We apply a sinusoidal input voltage with f =1kHz, A = 2V, and offset= 0V. The yellow is the voltage across the resistor. The blue wave represent that voltage difference between the inductor and the orange is the current through the resistor.  

Figure 12. Results when f=1kHz

3. The frequency is then changed to 2kHz to acquire new results as seen in figure 13.

Figure 13. Results when F =2Khz

4. A triangular voltage is applied to the inductor and resistor and acquire new results. 

Figure 14. Results when triangular wave is introduced. 

Summary of Labs and Learning Objective: 
Our assumed results in figure 4 relates to our results from figure 7 and 9 in terms of voltage input and the voltage difference across the capacitor. Out theory shows accurate based on our results for the voltage but not the current and the voltage across the capacitor. Unfortunately, there was a mistake entering our mathematical input for the current across the capacitor. Instead, we measured the current across the resistor which was not necessary which is represented by the orange wave line. The same procedure is produced to inductor voltage-current relations lab. We came to the conclusion that it looks similar to the results from the capacitor voltage-current relations lab. The main problem in both lab is that we did not measure the current across the capacitor and inductor which will show whether or not out assumed results for voltage- current relations hold true for a capacitor. 

Temperature Measurement System Design Lab

Objective:
We covered the idea of cascading operational amplifiers which are op amps that are connected in cascade or heat to tail to get a larger gain. We also discussed out Wheatstone bridges are which are circuits that are used to convert variations in resistance to variations in voltage and are used in measurements systems where there are sensors that have a variation of resistance due to external influence such as thermistors. We apply this idea into our temperature Measurements system design lab.

Group Practice:
1. The circuit below shows two operational amplifiers where we are told to find the voltage output and the current as seen in Figure . We will find the voltage output of the first non inverting amplifier in which we calculated to be Vo = 100mV. We then use that voltage as the voltage input for second non inverting amplifier. We would then calculate the voltage output using the equations Vout = (1+(Rf/R1)Vin. The result is Vout =350mV . The current is found by treating the current input in the positive terminal to be 0 and treating the resistors in series. We use the equations I=Vout/(R1+R2). The current is I = .025E-3mA.

Figure 1. 

2. The circuit in Figure 2. tells us to find Va and Vb. Since we have Vs which is the input voltage. We will calculate the voltage at a by using voltage division as seen below. We then calculate the difference between the points a and b.

Figure 2. Using voltage division to find the voltage at point a and b. 

Lab Procedures and Results:
1. The purpose of this lab is to build a simple temperature measurements system which outputs a DC voltage which indicates temperature. We will use a thermistor, potentiometer, and the idea of a Wheatstone bridge circuit as seen in figure 4 and 5 as its equivalent circuit.

Figure 3. Schematic of circuit which is planned to be built.

Figure 4. Wheatstone bridge circuit

2. We are considering a thermistor where the resistance is 10K when the temperature is at room temperature or 20C. When the temperature increases to 37C, the resistance increases to Rhot= 37K+/-.05. We measure the actual resistance of the thermistor Rth = 10.35K+/-.01 at room temperature.

3. When we connect all the components to the circuit, we notice that we must connect the potentiometer in series with the resistor at the bottom right of the circuit as seen in Figure 5. This will help us acquire a voltage difference between point a to b with 0 and measure the voltage Vab easier when the temperature changes for the thermistor.

Figure 5. Actual build for the Wheatstone bridge circuit.

3. We measure the resistance of the resistors, potentiometer, and thermistors that will be used in the Wheatstone Bridge circuit: Rth = 10.35K+/-.05 at room temp, R1 = 10K+/-.05, R2 = 10K+/-.05, R3 = 9.98K+/-.05, R(Potentiometer) = 9.07K+/-.05.  The potentiometer has the ability to change the resistance with a knob but in this case we are trying create a voltage difference between point a and b of 0 which can be seen in figure 6. This will help us measure the voltage when the thermistor increases in temperature. The voltage at body temperature is Vab= .25V+/.05.

Figure 6. Taking necessary steps to make the Vab equal to 0 at room temperature

4. We will use the voltage Vab for the input voltage for the operational amplifier. as seen in Figure 8. However, there is more work to be done since we need to find the necessary resistors for the operational difference amplifier as well as the voltage output. We note that the we must have a voltage output of 2V. In this case we will use the equation Vout = -(R2/R1)Vin. Our voltage input must be Vab=.25V+/-.05. Our calculation was wrong and acquired a theoretical resistors of resistance of R1= 11.87K+/-.05, R2 = 2.93K+/-.05, R3 = 9.98K+/-.05, R4 = 12K+/-.05. This means, based on our resistors used, that the voltage output should be 1V at body temperature which is shown as a video in Figure 9. The video shows that the voltage output increases as temperature increases. The voltage output increases to 1v in this case based on our error of calculations.

Figure 7. Difference operational amplifier

Figure 8. Schematic on how to connect the Wheatstone bridge circuit and difference operational amplifier.

Figure. 9. Video of successful experiment

Summary of Labs and Learning Objective:

The lab takes two components into consideration and they the use of building a Wheatstone bridge circuit and an operation difference amplifier. We also take thermistor and potentiometer into consideration since we are building a simple temperature measurements system. The reason why we build a Wheatstone bridge circuit is because its resistance change will convert to a voltage change. The Wheatstone bridge has a difference output that will be close to zero according to our set up. As a result we used an operation amplifier in order to increase the output voltage of the whole system to 1V. The lab require us to create a voltage output of 2V. However, we make a miscalculation when find the necessary resistor values for the operation difference amplifier. The resistance used and the resistance of the thermistor are Rth = 10.35K+/-.05 at room temp, R1 = 10K+/-.05, R2 = 10K+/-.05, R3 = 9.98K+/-.05, R(Potentiometer) = 9.07K+/-.05. This values helped us acquire a difference voltage between the point a and b of circuit from figure # to zero. We connect this Vab which is considered as out voltage input for the operational difference amplifier to the proper terminals. Yet we calculated the resistor needed in order to create a voltage output of 1 V while the voltage difference is in fact Vab= .25K+/-.05 when there is a temperature change. We use the equation Vout = (Rf/R1)Vin where Vin = Va-Vb. The resistors that were used in order to acquire a output voltage of 1v are R1= 11.87K+/-.05, R2 = 2.93K+/-.05, R3 = 9.98K+/-.05, R4 = 12K+/-.05. We tested our circuit by raising the temperature of the thermistor by squeezing it in order to represent a change of temperature based on our body temperature. The lab was a success and saw that the voltage output increases as temperature increased as seen in out video in Figure 9.

Summing and Difference Amplifier Lab

Objective:
The goal for today's class meeting is to understand buffet, non inverting, summing, and difference amplifiers. The lab will be based on summing and difference operation amplifiers. The equations for some of the operational amplifiers based on type are:
Inverse: -Rf/Ri * Vinput
Summation: Voutput = -Rf(V1/R1 + V2/R2)
Non-inverting: Voutput = (1+Rf/Ri)* Vinput
Difference: Voutput = R2/R1(V2-V1)

Group Practice:
1. The first in class problem that we worked on was based on plotting a block digram that represents the voltage input as seen in Figure 1. We are given two resistors with resistance of 2.5K and 7.5K. We are then told to plot the output voltage which can also be seen in figure 1. We note that the the lower limit is 0 volts so the bottom portion of the output block diagram is erased.

Figure 1. Plotting the input and output voltage. 

2. We are then told that the voltage input is between 0 to 200mv. Using the equation of an inverting operational amplifier where Vout = -(R2/R1)*Vin, we acquire a new voltage output. The plot can be seen in Figure 2. We note that the voltage output can not be below 0. Therefore it voltage output is 0.

Figure 2. Plotting the voltage output when voltage input is from 0 to 200mv.

3. We are given a circuit which can be seen in Figure 3. We are told to solve for voltage output. In order to this, we assume the circuit to be an ideal operational amplifier. Therefore, no currents flow flowing into the inverting or non inverting terminals. We see that voltage difference between is zero, meaning they are equal V1=V2. We then use nodal analysis at two points which can be seen in the figure below and acquire a Vo = -18/11.

Figure 3. Nodal analysis calculations for voltage output in an inverting operation amplifier.

Summing Amplifier Lab Procedures and Results:
1. The pre lab involves designing a summing circuit with two voltage inputs which can be seen in Figure 4. We note that the circuit is an inverting operational amplifier. As we do the calculations using nodal analysis, we acquire an equation Vout = -(R3/R1)(Va+Vb)

Figure 4. Schematic build for op amp

2. The circuit require us to find R3 as seen in figure 5. The way in which we find the resistance of the resistor is to note that Vout = -(R3/R1)(Va+Vb). In order to get R3, we will note that Vo = Va +VB. In order for this to true, then R3 and R1 must be equal. However, we must consider that we may not have this resistor, so we acquire the resistors that are available in class that are at least 1K . We use resistors of R1 = 2.15K+/-.01, R2 =2.17K+/-.01, R3 = 2.17K+/-.01.

Figure 5. Finding a resistance value for R3

3. We are told that input Va will change from -6,-5,-4,-2,-1,0,1,2,3,5V and the Vb will stay at 1V. We measure the first voltage output  of -2.98V+/-.01 when Va = -4, Vb = 1. The actual circuit can be seen in figure 6.

Figure 6. Output voltage measurement

4. We measure the rest of the voltage outputs as we change the the Va. The experimental and theoretical measurements can be seen in Figure 7. Percentage difference between theoretical and experimental is low between the voltage output ranging from 4<Vout<-2.97 based on what was inputed by Va and Vb. There is saturation at above a voltage output above 4 and -3.37V.

Figure 7. Table that includes experimental and theoretical as well as the percent difference.

Difference Lab Procedures and Results:
1. This lab includes the idea of difference operational amplifiers which is a device that amplifies the difference between two inputs but rejects any signals common to the the two inputs. The first step is to derive an equation which we can apply to the difference operational amplifier as seen in Figure 8 based on the schematic as seen in figure 7.

Figure 7. Schematic for a difference operational amplifier.

Figure 8. Calculating the voltage output based on the resistors in the circuit and the voltage inputs delivered to the positive and negative terminal of the operational amplifiers.

2. We draw out the schematic and plan on building our circuit. We acquire Resistors that will be used in the circuit. The resistance of the resistors are R1=9.99K+/-.01, R2=9.87K+/-.01, R3=19.98K+/-.01, R4=19.82K+/-.01.

Figure 9.

3. We measure the voltage output when the Va ranges from -4V to 5V and Vb stays at 1V. The measurements can be seen in figure 10. The difference is low between the theoretical and experimental voltage output.

Figure 10. Measurements for voltage output when Vb is 1 and Va ranges from -4 to 5V

Figure 11. Vout Vs Vin plot when Vb = 1V

4. We then measure the voltage output when the Va ranges from -4v to 5v and the Vb stays at -1V. The measurements can be seen in figure 12. The difference is low between the theoretical and experimental voltage output.

Figure 12. Measurements for voltage output when the Vb is -1 and Va ranges from -4 to 5V.

Figure 13. Vout vs Vin plot when Vb=-1V

Summary of Labs and Learning Outcome:

In the Summing Amplifier Lab, we notice that the operational amplifier is an inverting amplifier since the input voltage is connected to the negative terminal of the amplifier. We can assume that there will be a relationship in the equations when there are multiple voltage inputs connected to the negative terminal of the amplifier. We can use nodal analysis  for the summing amplifier in the first part of lab in order to find the voltage output which we found it to be Vo= -((Rf/R1)V1+(Rf/R2)V2). This is similar to inverting amplifier lab where Vo=-(Rf/R1) *Vin. We built the circuit and measured the resistance of the resistors that we plan to use. The resistors must be the same so that Vout = -(Va+Vb). The resistors in use are R1 = 2.15K+/-.01, R2 =2.17K+/-.01, R3 = 2.17K+/-.01. We measure the voltage output which can be seen in figure 7 and we notice saturation when the voltage output is above 4 and -4 volts. Anything above that, we see a high percent error between the theoretical and experimental voltage out of 20% when Va=-6 and Vb = 1. Anything below the output voltage of -4, the percent error become high as well. 

The difference amplifier works differently in which the device amplifies the difference between two inputs and rejects any signals common to the two inputs. For the lab we, we are require to find the Voltage output equation by using nodal analysis. Thus, we acquire the equation Vo=V2-V1. We measure the resistance of the resistors of R1=9.99K+/-.01, R2=9.87K+/-.01, R3=19.98K+/-.01, R4=19.82K+/-.01. We then measure the output voltage when Va varies from -4 to 5V and Vb stays the same. We can see based on the graphs and measurements in figures 10 and 11 that there is saturation around 4v and below -3.5V. We see that the theoretical and experimental out voltages have small percentage error below 1% between the saturation points. In other words our theoretical equation satisfies with our experimental measurements as long as the voltage output is within the linear region of Vout vs Vinplot. When the Vb is switched to -1V and Va varies from -4v to 5v. The plot is then reversed but the saturation is the same.