Download Task 1: Basic Non-Inverting Amplifier

Operational Amplifier Design Lab
By Denny Jung
NO lab partner (Missed the lab)
All labs done on Multisim
TA – Freddy Galindo
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Task 1: Basic Non-Inverting Amplifier
Design Objective:
The objective of this task is to amplify the mono audio signal with maximum amplitude of
200mVpp to a maximum of 20Vpp, using a non-inverting operational amplifier.
Schematic:
Schematic #1… Task 1 Schematic
Theory of Operation:
Positive and negative charges are connected to the operational amplifier to supply power to it.
The bypass capacitors are always used with operational amplifiers to remove any additional
sound made. The mono audio signal with amplitude 100 mVp (200 mVpp) goes through the noninverting amplifier to achieve the gain of 100. Using the relationship of a non-inverting operator,
the values of the resistors are then set to achieve the gain of 100.
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Derivations and Analysis:
Gain = 100
R1 = 1kΩ, R2 = 99kΩ
Experimental Results:
**No percent error available because everything was done on Multisim**
Capture #1… Capture of Task 1
Discussion of Results / Concluding Thoughts:
I first attempted to do this during the lab, however, was not able to finish it due to the faulty
wires and complications with the computers. In addition, I had extra wire when setting up the
power supply, which only hampered with the progress of the task. When the schematic was set
up in the Multisim, as done in the lab, the design worked as expected, yielding almost no
experimental error.
One mistake that I have made from the start till the end, which forced me to reconfigure all the
circuits once again, was the difference between the function generator in VirtualBench and
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Multisim. In VirtualBench, the system asks for peak-to-peak amplitude, while in Multisim, only
the amplitude is asked for. By using 200mVpp in place of 100mVp, the output saturated at
+/- 15V. The only way I thought to fix this was to use a higher power source. However, in real
life, the operational amplifier type LF412CN only accepts a power source value of 15.
Jung 4
Task 2: Difference Amplifier (Karaoke Circuit)
Design Objective:
The objective of this task is to mix the Left and Right channels, essentially by subtracting the
Right channel from the Left channel or vice versa. This in result will remove the identical vocal
signal that appears on both the Left and Right channels, while keeping the instrumental signals
from the channels, in addition to amplifying the difference to get the desired output amplitude
and inverting the Right channel signal.
Schematic:
Schematic #2… Task 2 Schematic
Theory of Operation:
From this task on, the bypass capacitors and the power supplies are used for the same reason as
stated in task 1. In task 2, two different function generators are used for each channel because
each channel consists of vocal and instrumental signals. The two function generators are also put
in series with one another so that the signals are summed. In the inverting input, the resistors
need a ratio of 100 in order to provide a gain of 100. However, when this is set to 100, the gain
of non-inverting input will be 101 due to the relationship of a non-inverting operator. Therefore,
two more resistors are added before the non-inverting input to create voltage division.
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Derivations and Analysis:
Gain- = -100, Gain+ = 101
R1 = 1k, R2 = 100k, R3 = 1k, R4 = 100k
𝑅
𝑉𝑥 = 𝑅𝑥 𝑉𝑠 ; Rx is the selected resistor value, RT is the total series circuit resistance
𝑇
Experimental Results:
**No percent error available because everything was done on Multisim**
Capture #2… Capture of the Right channel and the output
Discussion of Results / Concluding Thoughts:
When the task was first done, the schematic was missing the Resistor 3 and 4. The output was
fairly close to the expected with an experimental error of 1%. Nonetheless, Schematic #2 is the
ideal schematic that should be used to get the exact gain of 100 for each channels. In addition to
this problem, another problem was that because we were told to measure the input of the Right
channel and the output, 440Hz vocal signal was still part of the input. Therefore, it was difficult
to spot a single point in which the output had an approximate gain of -100.
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Task 3: Variable Two Channel Mixer with Unbalanced Inputs
Design Objective:
The purpose of this task is to mix an unbalanced stereo signal in which the left and right channels
have different voltage amplitudes. The circuit requires the left channel to have a fixed gain of -20
while the right channel has a varying gain between -20 and -50.
Schematic:
Schematic #3a… Task 3 Schematic before short-circuiting a channel
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Schematic #3b… Task 3 Schematic with right channel short-circuited
Schematic #3c… Task 3 Schematic with left channel short-circuited
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Theory of Operation:
Similar to Task 2, this task requires two channels, but both of them are amplified and inverted.
This means the non-inverting input is grounded. Left channel, on one hand, will have a fixed
gain of -20, while the Right channel has a varying gain between -20 and -50. The Right channel
requires a potentiometer, in order to vary the gain. Thus, using a potentiometer, the total resistor
value of the Right channel varies from 13.3k to 33.3k, which results in ratio of 50 and 20,
respectively. This would result in the Left channel to have a resistor value of 33.3k and the Rf to
have a value of 665k.
Derivations and Analysis:
Gain-Left = -20, Gain-Right = -20 ~ -50
R1 = 1k, R2 = 100k, R3 = 1k, R4 = 100k
Experimental Results:
**No percent error available because everything was done on Multisim**
Capture #3a… Capture showing a gain of -20 from the Left channel (at 0%)
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Capture #3b… Capture showing a gain of -50 from the Right channel (at 100%)
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Capture #3c… Capture showing a gain of -20 from the Right channel
Discussion of Results / Concluding Thoughts:
Without the prior knowledge that we are not allowed to change the resistance value of the
potentiometer, I first changed the value of the potentiometer to provide the gain of -20 ~ 50. At
this point, no resistor values are unique, as we can easily change the values along with the
potentiometer. Using the potentiometer resistance value of 20k, all the resistor values used in this
schematic become unique. Moreover, I did not short circuit each side when first measuring; this
resulted in a convoluted result that made it impossible to figure out if the operational amplifier
was providing the expected gain.
Jung 11
Task 4: Level-Shifting Amplifier
Design Objective:
The purpose of this task is to invert and amplify a mono audio signal that is an output of an
electret condenser microphone, while removing the DC offset voltage that is always included in
an electret condenser microphone.
Schematic:
Schematic #4… Task 4 Schematic
Theory of Operation:
An electret condenser microphone has an output signal, along with a DC voltage offset value of
5V, as provided by the lab handout. In order to amplify and invert the signal, while removing the
DC offset from the microphone, a level-shifting amplifier is needed. Using the positive power
supply source and the rule of voltage division, I found the ratio of resistors to be 1 to 2.
Therefore, the non-inverting input of the operational amplifier is supplied with 5V. From looking
at the figure provided in the lab handout, the gain is -100 and by using the relation of inverting
operator, I found the values of the resistors.
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Derivations and Analysis:
Gain-Left = -100, Gain-Right = -20 ~ -50
R1 = 1k, R2 = 100k, R3 = 50k, R4 = 1.5k, R5 = 100k
𝑅
𝑅
Vout = − 𝑅2 𝑉𝑎 + 𝑉𝑏 (1 + 𝑅2 ); Va is the input from the inverting side, Vb is the input from the
1
1
non-inverting side
Experimental Results:
**No percent error available because everything was done on Multisim**
Capture #4… Capture of Task 4
Discussion of Results / Concluding Thoughts:
In the Theory of Operation section, I purposely left out the reason I included the 1.5kΩ resistor.
This is because when I first did the experiment, I did not think this value would prove to have
any effect on the circuit. When looking at the oscilloscope captures, that was when the error was
spotted: even though the maximum output was approximately 10V, the minimum output was -
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𝑅
𝑅
5V. By doing the math, this error was resolved. Vout = − 𝑅2 𝑉𝑎 + 𝑉𝑏 (1 + 𝑅2 ), where Va is the
1
1
input signal from the inverting terminal and Vb is from the non-inverting terminal. Therefore,
when the 1.5kΩ was not used, it had an “amplified” output – an excess of 5V. Using a 1.5kΩ
resistor supplies a lower input voltage at the non-inverting terminal of the operational amplifier
but will completely cancel out of the DC offset voltage, without any leftover voltages.
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Task 5: Variable Level-Shifting Amplifier
Design Objective:
The objective of this task is to invert and amplify a signal from an electret condenser microphone
that has a varying DC offset voltage between 4 and 6. The circuit requires a potentiometer so that
it can be adjusted to cancel any DC offset between 4V and 6V.
Schematic:
Schematic #5a… Task 5 Schematic with 4V DC offset
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Schematic #5b… Task 5 Schematic with 6V DC offset
Theory of Operation:
Task 4 and 5 are very similar to each other; the only difference is the potentiometer allowing the
circuit to adjust its voltage at the non-inverting input of the operational amplifier. Therefore, a
potentiometer arrow is used, in place of where the wire is used for voltage division in Task 4. In
doing so, the non-inverting input voltage can be varied between 4 and 6. Using the potentiometer
resistance value of 20k and the rule of voltage division, the resistor values needed for offset were
found. The 1.5kΩ resistor is put in there for the same reason stated in Task 4.
Derivations and Analysis:
Gain = -100
R1 = 1k, R2 = 100k, R3 = Rpot = 20k, R4 = 40k, R5 = 90k, R6 = 1.5k
𝑅
𝑅
Vout = − 𝑅2 𝑉𝑎 + 𝑉𝑏 (1 + 𝑅2 ); Va is the input from the inverting side, Vb is the input from the
1
non-inverting side
1
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Experimental Results:
**No percent error available because everything was done on Multisim**
Capture #5a
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Capture #5b
Discussion of Results / Concluding Thoughts:
Channel A of the oscilloscope, which measures the input signal, varies between 5.9V and 6.1V
and also between 3.9V and 4.1V. The values are centered at 6V and 4V because of the DC offset
voltage. Also, they vary by 100mV because of the sine function that is part of the input signal.
The expected outcome should be the amplified signal of the sine function, which peaks at +/10V, without any DC offset voltage, that should have been cancelled out by the varying noninverting input.
Jung 18
Post-Lab Activities
1. An output signal in the voltage range of 5Vpp ~ 20Vpp actually means 2.5Vp~10Vp. To
create a variable-gain amplifier with an output signal as given, we could possibly place a
potentiometer in series with the Rf.
100 = 1 +
𝑅𝑓 + 𝑅𝑝𝑜𝑡
𝑅𝑓 + 20𝑘
= 1+
𝑅𝑠
𝑅𝑠
25 = 1 +
𝑅𝑓
𝑅𝑠
Substituting Rf/Rs, which is 24, in the first equation, we can find the value of Rs, which comes
out to 0.26667kΩ and therefore, the Rf value comes out to be 6.4kΩ. Using these values, we can
achieve the variable-gain amplifier with the desired output voltage range.
2. If the amplitude of the vocals on the 2 channels that I wanted to cancel out was not identical,
we can use voltage division to drop off the difference. For example, if one has 0.5Vp and another
has 0.1Vp, we could set up a two-resistor system to drop the voltage on the 0.5Vp signal. One
resistor will be connected to the ground, while the other is in series with the signal input. The
values of the resistor can be figured out by using the rule of the voltage division.
3. Considering how we are already varying the Right channel gain, the only thing we would need
to change in this circuit design is the Left channel. Similar to how the Right channel is designed,
we add another potentiometer in series with the previous resistor and change the value of the
resistor depending on the value of the gains.
4. The additional complication to wanting a variable signal amplification is having to change the
input signal without having to change the gain. If the input was varying in Task 4, which has an
input going into the inverting input of the operational amplifier, we would need to use a
potentiometer right after the signal input but before the input goes through the op amp. This is
because in doing so, we can vary the signal before it goes through the operational amplifier
without changing the gain. If the potentiometer is placed on any other spot on the circuit, it will
change the gain or the voltage going into the non-inverting input of the operational amplifier.
5. If the DC offset voltage could be either positive or negative, we would need to incorporate
both power sources. The problem with this is that in Task 5, the potentiometer is only connected
to the positive source, which would never supply any values below 0. I would address this
problem by using a switch that will flip between the positive and the negative end of the power
sources.