Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
1
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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. 2
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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!𝑨𝒄𝒕𝒖𝒂𝒍 𝑨𝒄𝒕𝒖𝒂𝒍
* 100
𝟏𝟔.𝟑𝑽!𝟏𝟔𝑽 𝟏𝟔𝑽
* 100=
1.875%
3
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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. 4
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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. 5
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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!𝟏𝟔𝐕 𝟏𝟔𝐕
* 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
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
Task 3: Two Channel Mixer with Balanced Inputs.
Figure 5: Schematic of Op Amp Design 3 & Oscilloscope screen via Multisim®(.4 Vpp). 9
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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. 11
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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. 12
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
14
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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 19
Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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Ahmed Al Haddad TA: Jared Price Op Amp Design Lab 11/04/2012
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.
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