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
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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%
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
5
Task 2: Weighted Summing Amplifier
Figure 3: Schematic of Op Amp Design 2 & Oscilloscope screen via
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
9
Task 3: Two Channel Mixer with Balanced Inputs.
Figure 5: Schematic of Op Amp Design 3 & Oscilloscope screen via
Multisim®(.4 Vpp).
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
10
Figure 6: Schematic of Op Amp Design 3 & Oscilloscope screen via
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
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
14
Task 4: Level-‐shifting Amplifier
Figure 8: Schematic of Op Amp Design 4 & Oscilloscope screen via
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
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
18
Figure 10: Schematic of Op Amp Design 5 & Oscilloscope screen via
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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