Op Amp Lab Report

Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012

1

Task 1: Basic Inverting Amplifier.

Figure 1: Schematic of Op Amp Design 1 & Oscilloscope screen via

Multisim®.

Design Objective: In this task we were given an input and an output

equation. In this case we were given a dynamic microphone whose output

is a mono audio signal with maximum amplitude of 200 mVpp. The goal of


2

this experiment is to invert and amplify the signal to a maximum of 16 Vpp

so that it can drive audio equipment. The assumption was that this audio

equipment would be damaged if the voltage amplitude at its input exceeds

16 Vpp; therefore we had to design and build a circuit that achieves this

maximum possible amplification without damaging the audio equipment.

Theory of Operation: The input for the first function was 200 mVpp and the

output voltage was 16 Vpp. To get these values we constructed an inverting

op amp with resistor values that were 80 KΩ and 1 KΩ. We picked these

values because we used the equation !!"#!!"

= !!!!!

. This equation relates to the

gain of the function. By subbing in the known input and output voltages

you can see that R2 should be 80 and R1 should be 1. For all these circuits

you need to power the op amp with a positive and negative voltage of 15

Volts. In the picture channel 1 represents the input and channel 2

represents the output voltages. A representation of the given circuit is

shown in Figure 2.


3

Figure 2: A simplified schematic of basic Inverting Op Amp circuit.

Calculations: In order to achieve the desired voltage amplitude we had to

do some calculations so that we can choose the resistor values in order for

the circuit to amplify the given input to 16 Vpp.

Formula:

!!"#!!"

= !!!!!

Given:

Vout=16 Vpp, Vin=.2Vpp

!!"#!!"

= !".! = 80 Gain

Percent error Equation: 𝑴𝒆𝒂𝒔𝒖𝒓𝒆𝒅!𝑨𝒄𝒕𝒖𝒂𝒍𝑨𝒄𝒕𝒖𝒂𝒍

* 100 𝟏𝟔.𝟑𝑽!𝟏𝟔𝑽𝟏𝟔𝑽

* 100=

1.875%


4

Experimental Results: Our data can be seen on the oscilloscope in Figure 1.

Using our resistor values of 80kΩ and 1kΩ we were able to achieve a

maximum gain of 15.9 Vpp, which is relatively accurate and nearly perfect.

The schematic consists of our two resistors an input and output and +15V &

-‐15V supplied through the function generator and Channel 1 (in yellow)

shows the input of the circuit and Channel 2 (in blue) shows the output of

the circuit and on the right we can see the Peak-‐to-‐Peak voltage amplitude

and we can see that we have successfully built the desired circuit using a

basic inverting amplifier.

Conclusion: We were successful in amplifying the amplitude by using our

formula to calculate the gain and determine our resistor values. Our

schematic shows that we simply used a basic inverting amplifier, which

scales and inverts the input signal. We were not 100% accurate but we

were actually able to build this circuit and got it to function and showed it

to our TA.


5

Task 2: Weighted Summing Amplifier


Multisim®.

Design Objective: In this task we used an input from an audio signal that

we got from the computer. The output of the right channel is 500 mVpp

and the output of the left channel is 200 mVpp. The objective of this task

was to produce a 16 Vpp output that is balanced and that is an inverted

sum of the two outputs of the audio signal. A simplified circuit is shown in

Figure 4 below.


6

Theory of Operation: To find the resistor values we broke the design up

into two parts. Each of the inputs would have to have half of the total

desirable gain. To find this gain you can use the equation from the first task.

You would need to set one of the inputs to zero (also known as super

position) to determine the resistor value for that input. Once you found

that resistor you would use the same feedback resistor to solve for the

second unknown resistance. You can also use nodal analysis to figure out

what the resistors should be.

Figure 4: A simplified schematic of Weighted Summing Amplifier.


7

Calculations:

Formula:

!!"#!!"

= !!!!!

Vout= 16 Vpp Vin= .5 Vpp or .2 Vpp

Break the problem up using super position.

!.! = 16 !

.! = 40 Since .5Vpp has a lower gain that that of the .2Vpp

input you can say that you will need a higher resistor value at that input

source so that they would be equalized and have the same output of 8 Vpp

and then summed together at the end to get a total of 16 Vpp.

The resistor values that we chose were 400k for the feedback resistor, 25k

right after the .5 Vpp input, and 10k right after the .2 Vpp input.

Percent Error:

𝟏𝟔𝐕!𝟏𝟔𝐕𝟏𝟔𝐕

* 100= 0%

Experimental Results: Our data can be seen on the oscilloscope in Figure 3.

Using the values we got from our calculations of 400k, 25k and 10k for


8

resistors R1, R2 and R3 respectively we were able to get an input of 200

mVpp in Channel 1 and 499 mVpp in Channel 2. Channel 3 (in purple) is the

sum of both of the frequencies and we can see that we got amplitude of

15.8 Vpp.

Conclusion: In conclusion, we did not have any %error and were 100%

accurate, and we were able to construct this circuit on the breadboard in

class using our Gain formula and the concept of superposition to calculate

our resistor values and successfully build a weighted summing amplifier.


9

Task 3: Two Channel Mixer with Balanced Inputs.


Multisim®(.4 Vpp).


10


Multisim®(16 Vpp for Vout).

Design Objective: For this circuit we have a balanced stereo signal in which

both the right and left channels have equivalent amplitude of 0.5 Vpp. But

in this scenario we had to mix these two channels into a single inverted

output while independently varying the gain of the two channels. Therefore


11

we were asked to use potentiometers so that we could modify the gain. A

simplified representation of the circuit can be seen in Figure 7.

Theory of Operation: To get a variable gain we used 20k potentiometers

that can change resistances with ease. You would do the same process as

task 2 but now when you measure the gain from either the right or the left

channel you will need to use half of the variable gain to determine your

other resistor values. When your gain is at its maximum you will have all of

the 20k potentiometer in your equation and when the gain is at its

minimum value you will turn the potentiometer so that it will have no

resistance in the system.

We can see that Channel 1 represents the right channel with a frequency of

440 Hz and Channel 2 is the output of the function. The frequencies are

different but it doesn’t change your calculations for determining your

resistor values it just alters the output graph on the oscilloscope when you

sum the two input functions together.


12

Figure 7: A simplified schematic of Two Channel Mixer.

Calculations:

For .4 Vpp output with .5 Vpp input for at both the left and right channels

we will use 100% of the 20kΩ potentiometer.

!!"#!!"

= !!!!!

Vin = .5 Vpp Vout = .2 just for half the circuit using

super position

.!

.! = 0.4 for both !!!

!!!!"! R2= 8kΩ R1= 500Ω !"""

!""!!""""= 0.39024

For 16 Vpp output with .5 Vpp input for both the left and right channels we

will use 0% of the potentiometer.


13

Conclusion: Unfortunately we were not able to successfully build this circuit

in class but were able to simulate it on Multisim and we successfully got our

varying gains as shown in Figures 5 & 6. We reached our resistor values

using the gain formula and super positionThis circuit has many applications

in the real world and could be used to change the volume of an audio

device.


14

Task 4: Level-‐shifting Amplifier


Multisim®.

Design Objective: In this design problem, the input signal was a mono

audio signal that is the output of an electret condenser microphone. An

electret microphone is a type of condenser microphone, which eliminates

the need for a polarizing power supply by using a permanently charged

material. An electret condenser microphone output always includes a DC

offset in addition to the signal itself. Assuming that this microphone


15

outputs a signal with a voltage swing of 600 mVpp plus a DC offset of 2.5 V.

The microphone output signal needs to be changed before being input to a

sensitive audio circuit. Specifically, it needs to be inverted, amplified, and

the DC offset needs to be removed. We were asked to design and build a

circuit that will cancel out the DC offset and achieve maximum signal

amplification without exceeding the 16 Vpp input limit of the audio

equipment. A simplified representation of the circuit is shown in Figure 9.

Theory of Operation: There is a DC offset of 2.5V within the input signal of

the audio signal. To counter this DC offset so that there is no offset in the

output of the function you must first construct an inverting op amp. At the

positive terminal you want to have a voltage of 2.5V to counter the DC

offset. You must use the positive 15V power supply that is powering the op

amp so we used a voltage divider to get our desired voltage of 2.5V. Results

are shown in Figure 8.


16

Figure 9: A simplified schematic of a Level-‐shifting amplifier.

Calculations:

For the resistor values use the inverting op amp formula: !!"#!!"

= !!!!!

Vout=

16Vpp Vin= .6Vpp

!!"#!!"

=!".!=26.6 gain so the resistor values that we chose were

R1=8kΩ and R2=300Ω

To get 2.5V at the positive terminal we used a voltage divider.

�!= !!"!!"!!!"!!!"

*15V=2.5V

Percent Error: No percent error all values were exactly the same as the

intended values.


17

Conclusion: In conclusion we got our desired voltage of 16Vpp. In our

calculations we did not get any percent errors and the oscilloscope caption

via Multisim® can validate this. We got exact values for the input and

output of the circuit. The DC offset can be shown by the mean in the

caption. It starts off as being negative 2.5 and then switches to positive 2.5.

This will get rid of the offset and essentially be zero.

Task 5: Variable Level-‐Shifting Amplifier


18


Multisim®.

Design Objective: For our final design, we have a similar set up as Task 4

however, now we don’t know the DC offset. But we do know that the offset

ranges from 1V-‐3V. The signal needs to be inverted, amplified and the DC

offset needs to be removed before being input into the audio circuit as

before. We were asked to build a level-‐shifting amplifier that can be

adjusted to cancel any DC offset between 1 VDC and 3 VDC. Again, we need

to design our circuit so that it does not exceed the maximum amplitude of

16Vpp. A simplified schematic of the circuit is shown in Figure 11.

Theory of Operation: This circuit is similar to the previous circuit but to

construct this circuit the only thing that we need to change will be the

resistor values. For the feedback resistor we will need to have 13.3kΩ and a

500Ω resistor right after the input. Results are shown in Figure 10.


19

Figure 11: A simplified schematic of Variable Level-‐Shifting Amplifier.

Calculations:

For the resistor values use the inverting op amp formula: !!"#!!"

= !!!!!

Vout=

16Vpp Vin= .6Vpp

!!"#!!"

=!".!=26.6 gain so the resistor values that we chose were

R1=8kΩ and R2=300Ω

To get 2.5V at the positive terminal we used a voltage divider.

𝑉!= !!"!!"!!!"!!!"

*15V=2.5V

Percent Error: No percent error all values were exactly the same as the

intended values.

Conclusion: This circuit was more complex than the other circuits but is

similar to the circuit in task 4 but we used a different voltage divider and


20

were able to build a level-‐shifting amplifier circuit and get a very accurate

amplitude at the output because we ran this on Multisim and were not able

to complete the task in class due to time constraints.

Post lab Questions

1) If you wanted a variable-‐gain between 8 Vpp and 16 Vpp you would

add a 1kΩ potentiometer to the circuit right after the input comes in.

!!!!!!!!

, R2= 80kΩ, R1= 1kΩ, !"!!!!!!

= 40*.2Vpp= 8Vpp

2) You would construct a non-‐inverting amplifier with the ground at the

terminal and the input at the inputs at the positive terminal to get a

circuit that does not invert the signal.

3) You would add a potentiometer value that had a large resistance

value that when active will decrease the voltage so much that

essentially it will appear to be zero.

4) You can add potentiometers in task 4 and 5 just like what was done

in task 3 with variable gain.


21

5) You can tie a voltage divider that is tied to a negative source on one

side and a positive source on the other. You can put a potentiometer

in the middle of the voltage divider to choose whether you would

want a negative or positive voltage to counter the DC offset

whatever it may be.

Op Amp Lab Report

Documents

weighted summing

basic inverting

inverting

5v percent

mono audio

vout 16vpp

shifting amplifier

op amp