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Op-Amp Design Lab
Report
EE210 Spring 2016
Section 001
Name : Nur Farzanah Roslan
TAโs name : Alex Chen
Partner Lab : Titus Umandap
Date : 03/15/2016
Station : 9
Task 1 : Designing a Basic Non-Inverting Amplifier
Design Objective :
In this task, we were given an input and output equation. In this case, we were
provided with a dynamic microphone with a mono audio signal output that has a
maximum amplitude of 200 mVpp. Our aim for this experiment is to build a circuit
that can amplify the signal to a maximum of 20 Vpp so that it can drive audio
equipment. The assumption was that this audio equipment will be damaged if the
voltage amplitude at its input exceeds 20 Vpp. Therefore, we need to design and
build a circuit that achieves this maximum possible amplification of the signal
without damaging the audio equipment.
Schematic Diagram of Network :
Figure 1: Schematic diagram of non-inverting amplifier.
Theory of Operation :
The input for the circuit was 200 mVpp and the desired output was 20 Vpp.
To get these values, we constructed a non-inverting op amp circuit with resistors of
1 kฮฉ and 99kฮฉ. We chose these values by firstly calculating the gain of the
function, then we used the equation, ๐บ๐๐๐, ๐บ =๐๐๐ข๐ก
๐๐๐=
๐ 2
๐ 1+ 1. When substituting
input and output voltages in this equation, we get R2 should be 99 and R1 should
be 1. For all these circuit, the op amp needs to be powered with a positive and
negative of 15V.
Derivation and Analysis:
We did some calculations to get the resistor values in order to achieve the
desired voltage amplitude. We were given input voltage, Vin which is 0.2 Vpp and
output voltage, Vout which is 20 Vpp. We know the formula to find the gain of non-
inverting op amp which is:
๐บ๐๐๐, ๐บ =๐๐๐ข๐ก
๐๐๐=
๐ 2
๐ 1+ 1
Substituting the voltage given, we get:
๐บ๐๐๐, ๐บ = 20 ๐๐๐
0.2 ๐๐๐= 100.
Nominal value for the R2 = 99kฮฉ and R1 = 1kฮฉ. Measured value for the R2
= 99.342 kฮฉ and R1 = 0.9856kฮฉ.
Experimental Results :
Figure 2: The capture of oscilloscope for the non-inverting amplifier signals.
Our data can be seen on the oscilloscope in Figure 2. Based on the results we
obtained, when we used resistor values of 99 kฮฉ and 1 kฮฉ, we were able to achieve
a maximum gain of 20.2 Vpp, which is relatively accurate and perfect. Therefore,
we can calculate the gain as below:
๐บ๐๐๐, ๐บ =20.2 ๐
0.214 ๐= 94.39
Percentage error for all measured values we obtained were calculated as below:
Formula : ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |๐๐๐๐ ๐ข๐๐๐โ๐๐๐๐๐๐๐|
๐๐๐๐๐๐๐ร 100%
For R1 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.9856 ๐ฮฉ โ 1๐ฮฉ |
1 ๐ฮฉร 100% = 1.44%
For R2 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |99.342 ๐ฮฉ โ 99๐ฮฉ |
99 ๐ฮฉร 100% = 0.345%
For output voltage :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |20.2 ๐ โ 20๐ |
20๐ร 100% = 1.00%
For gain :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |94.39 โ 100 |
100ร 100% = 5.61%
Conclusion :
We were successful in amplifying the amplitude by using our formula to
calculate the gain and determine the appropriate resistor values. Our schematic
shows that we simply used a basic non-inverting amplifier, which amplifies the input
signal. There was a 1% error. The experimental value for the output voltage was 20.2
V, which was a little higher than value of 20 V. This problem was most likely
happened due to external noise. To reduce the effect of external noise, we attached
bypass capacitors to our circuit and put it as close to op amp. Another more
significant influence could be that the resistors were not exactly the value written on
them. The resistor also experienced some changes in its value when connected to
circuit.
Task 2 : Designing a Difference Amplifier (Karaoke Circuit)
Design Objective :
In this task, we were given an input and output equation. In this case, we were
given a stereo music signal with left and right channels, each with a maximum
voltage of 200 mVpp. Our aim for this experiment is to build a circuit that will mix
the L and R channels to remove the vocals, and give a resulting signal of maximum
of 20 Vpp by amplifying the difference between the two channels. There are three
assumptions that were made for this experiment. First assumption was that the signal
amplitude on each channel is equally distributed between vocals and instrumental
(i.e vocal amplitude is 100 mVpp and instrumental amplitude is also 100 mVpp).
Second assumption was that the identical vocal signal appears on both L and R
channel. Third assumption was that the instrumental part of the music does not
appear equally in the L and R channels. These assumptions are not always the case,
but usually is the case.
Schematic Diagram of Network :
Figure 3: Schematic diagram of difference amplifier.
Theory of Operation :
The input for this circuit was 200 mVpp and the desired output was 20 Vpp.
To obtain these values, we constructed a circuit with resistors of two 1 kฮฉ, and two
100 kฮฉ which were obtained from gain equation. We assigned the same resistance
to R2 and R4, and, to R1 and R3, respectively, is because we want to simplify the
output voltage equation. To get the desired voltage, we would need to set one of the
inputs to zero by short-circuiting it or remove it from our circuit. To ensure that the
amplitude is 200 mVpp, we would need to adjust the volume of the mp3 signal prior
to connecting our circuit. For all these circuit, the op amp needs to be powered with
a positive and negative of 15V.
Derivation and Analysis:
We did some calculations to get the resistor values in order to achieve the
desired voltage amplitude. We were given input voltage, Vin which is 0.2 Vpp and
output voltage, Vout which is 20 Vpp. We know the formula to find the gain of
difference op amp which is,
๐บ๐๐๐, ๐บ =๐๐๐ข๐ก
๐๐๐=
๐ 2
๐ 1.
Substituting the voltage given we get:
๐บ๐๐๐, ๐บ = 20 ๐๐๐
0.2 ๐๐๐= 100.
We chose R2 = 100 kฮฉ and R1 = 1 kฮฉ. Since R2 = R4, therefore R4 = 100 kฮฉ.
Same goes with R1 and R3. Since R1 = R3 = 1 kฮฉ, therefore R3 = 1 kฮฉ.
Nominal value for the R3 = R1 = 1kฮฉ and R4 = R2 = 100kฮฉ. Measured value
for the R1 = 0.985kฮฉ, R2 = 99.563 kฮฉ, R3 = 0.98313 kฮฉ, and R4 = 99.278 kฮฉ.
Experimental Results :
Figure 4: The capture of oscilloscope for difference amplifier.
The amplitude voltage obtained from the capture above is 6.26V which was
too far from the actual value. We showed our TA the graph we obtained. Percentage
error for all measured valued we obtained were calculated as below:
Formula : ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |๐๐๐๐ ๐ข๐๐๐โ๐๐๐๐๐๐๐|
๐๐๐๐๐๐๐ร 100%
For R1 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.985 ๐ฮฉ โ 1๐ฮฉ |
1 ๐ฮฉร 100% = 1.500%
For R2 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |99.563 ๐ฮฉ โ 100๐ฮฉ |
100 ๐ฮฉร 100% = 0.437%
For R3 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.98313 ๐ฮฉ โ 1๐ฮฉ |
1๐ฮฉร 100% = 1.687%
For R4 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |99.278 ๐ฮฉ โ 100๐ฮฉ |
100 ๐ฮฉร 100% = 0.722%
For output voltage :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |6.26 ๐ โ 20๐ |
20๐ร 100% = 68.70%
For gain :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |31.61 โ 100 |
100ร 100% = 68.39%
Conclusion :
Unfortunately, we were not able to get a desired result from our
design circuit. The percentage error obtained was too high that it was not even
to actual values. This happened due to a few reasons. One reason is that our
selection resistance values. Since the output voltage was much lower than 20Vpp,
it was either, R2 is too low, or R1 is too high. Therefore, next time, we should
choose a higher R2 and lower R1.
Task 3 : Designing a Variable Two Channel Mixer with Unbalanced
Input
Design Objective :
In this task, we were given input equation and output equation. In this case,
we were provided an unbalanced stereo signal in which the left and right channels
have different voltage amplitude โ the left channel is 500 mVpp and the right channel
is 200 mVpp. Our aim for this experiment is to build an op amp circuit with
potentiometers so that the left channel is amplified with a fixed gain of -20 and the
right channel has a gain that can vary between -20 and -50. Besides that, we want
our circuit able to mix L and R channels into a single amplified inverted output while
varying the gain of the right channel only.
Schematic Diagram of Network :
Figure 3: Schematic diagram of a variable two channel mixer with unbalanced input.
Theory of Operation :
To produce the circuit that could vary the gain of only one channel, we need
to include a potentiometer at one of the inputs. Because we used the potentiometers
as variable resistors as opposed to voltage dividers, we had to disconnect one of the
terminals. There was one problem with this approach, however; if the potentiometers
were set to their minimum values (zero), then the resistance in front of the input
values would be zero. According to the gain formula given in circuit two, this would
cause an infinite gain. To rectify this dilemma, we included a resistor in addition to
the potentiometer in front of the input signals. This way there would always be input
resistance even if the potentiometer was turned all the way down.
Derivation and Analysis:
We did some calculations to get the resistor values in order to achieve the
desired voltage amplitude. We were given left input voltage, Vleft which is 0.5 Vpp
and right input voltage, Vright which is 0.2 Vpp. We were also given gain at L
channel, Gleft which is -20, and gain at R channel, Gright which is between -20 and -
50. We know the formula to find the gain at inverting op amp which is:
๐บ๐๐๐, ๐บ = โ๐๐๐ข๐ก
๐๐๐= โ
๐ 2
๐ 1
We calculated gain for each channel by substituting the voltage given, we get:
๐บ๐๐๐ ๐๐ก ๐ฟ ๐โ๐๐๐๐๐ = โ๐ 2
๐ 1= โ20
Hence, we know R2 = 20R1. We chose R2 = 500 kฮฉ, then R1 = 25 kฮฉ (with
measured values of R2 = 502.76 kฮฉ and R1 = 25.052 kฮฉ). Now, we calculated
gain at R channel. Since there is potentiometer at R channel, the calculation is
slightly different than one we did. Here is the calculation:
For R4 = 20 kฮฉ,
โ๐ 2
๐ 3 + 20kฮฉ= โ20
For R4 = 0 kฮฉ,
โ๐ 2
๐ 3= โ50
From these, we got the value of R3 = 5 kฮฉ (with measured value of 5.026 kฮฉ)
Experimental Results :
Figure 6: The capture of oscilloscope for a variable two channel mixer with unbalanced input.
From the oscilloscope capture, we obtained output voltage equals to 0.469 V
which was not as we expected, and gain as calculated below:
๐บ๐๐๐, ๐บ =0.469 ๐
0.325 ๐= 1.44
Percentage error for all measured values we obtained were calculated as
below:
Formula : ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |๐๐๐๐ ๐ข๐๐๐โ๐๐๐๐๐๐๐|
๐๐๐๐๐๐๐ร 100%
For R1 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |25.052 ๐ฮฉ โ 25๐ฮฉ |
25 ๐ฮฉร 100% = 0.208%
For R2 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |502.76 ๐ฮฉ โ 500๐ฮฉ |
500 ๐ฮฉร 100% = 0.552%
For R3 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |5.026 ๐ฮฉ โ 5๐ฮฉ |
5๐ฮฉร 100% = 0.52%
For output voltage :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.469 ๐ โ 20๐ |
20๐ร 100% = 97.66%
For gain :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |1.44 โ 100 |
100ร 100% = 98.56%
Conclusion :
Unfortunately, we were not able to get a desired result from our
design circuit. The percentage error obtained was too high that it was not even
to actual values. This happened due to a few reasons. One reason is that our
selection resistance values. Since the output voltage was much lower than 20Vpp,
it was either, R2 is too low, or R1 is too high. Therefore, next time, we should
choose a higher R2 and lower R1.
Task 4 : Designing a Level-Shifting Amplifier
In this task, the input signal was a mono audio signal that is the output of an
electret condenser microphone. For this experiment, we were using electret
microphone, which is a type of condenser microphone, which its output always
includes a DC offset in addition to the signal itself. Assuming that this microphone
gives an output signal with a voltage swing of 200 mVpp plus a DC offset of 5V.
This 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. Therefore, we need to design and build a circuit that will cancel out
the DC offset and achieve maximum signal amplification without exceeding the 20
Vpp input limit of the audio equipment. Another assumption was that the only DC
voltage source available is the +/- 15 V used to power the op amp IC.
Schematic Diagram of Network :
Figure 7: Schematic diagram of level-shifting amplifier.
Theory of Operation :
The input for the circuit was 200 mVpp and the desired output was 20 Vpp.
In this design problem, there is a DC offset of 5V within the input signal of the audio
signal. We need to counter this DC offset so that there is no offset in the output of
the function. To do this, we can start by constructing an inverting op amp so that we
get a voltage of 5V at the positive terminal of op amp to counter the offset. We must
use the positive 15V power supply that is powering the op amp so we used a voltage
divider at positive terminal of op amp to get our desired voltage of 5V. Therefore,
we can find R4 and R3 values from the voltage division at non-inverting terminal
and R2 and R1 from gain equation at inverting terminal.
Derivation and Analysis:
We did some calculations to get the resistor values in order to achieve the
desired voltage amplitude. We were given input voltage, Vin which is 0.2 Vpp and
output voltage, Vout which is 20 Vpp. We know the formula to find the gain of
inverting op amp which is:
๐บ๐๐๐, ๐บ =๐๐๐ข๐ก
๐๐๐=
๐ 2
๐ 1
Substituting the voltage given we get:
๐บ๐๐๐, ๐บ = 20 ๐๐๐
0.2 ๐๐๐= 100
Therefore, we chose R2 = 100 kฮฉ and R1 = 1 kฮฉ. (with measured values of R2 =
99.546 kฮฉ and R1 = 0.9819 kฮฉ).
We chose R3 and R4 by performing voltage division at node voltage at non-
inverting terminal, Vx. Here is the calculation:
๐ 4
๐ 4 + ๐ 3ร ๐๐๐ข๐ก = ๐๐ฅ
๐ 4
๐ 4 + ๐ 3ร (15๐) = 5๐
We solved for R4 and R3, we obtained:
๐ 4
๐ 4 + ๐ 3=
1
3
We chose R4 = 1 kฮฉ, then R3 = 2 kฮฉ. (with measured values of R4 = 0.9822 kฮฉ
and R3 = 1.9713 kฮฉ).
Experimental Results :
Figure 8: The capture of oscilloscope for level-shifting amplifier.
From the oscilloscope capture, we obtained output voltage equals to 19.1 V
and gain of:
๐บ๐๐๐, ๐บ = 19.1๐
0.189๐= 101
Percentage error for all measured values we obtained were calculated as
below:
Formula : ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |๐๐๐๐ ๐ข๐๐๐โ๐๐๐๐๐๐๐|
๐๐๐๐๐๐๐ร 100%
For R1 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.9818 ๐ฮฉ โ 1๐ฮฉ |
1 ๐ฮฉร 100% = 1.82%
For R2 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |99.546 ๐ฮฉ โ 100๐ฮฉ |
100 ๐ฮฉร 100% = 0.454%
For R3 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |1.9713 ๐ฮฉ โ 2๐ฮฉ |
2๐ฮฉร 100% = 1.435%
For R4 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.9833 ๐ฮฉ โ 1๐ฮฉ |
1 ๐ฮฉร 100% = 1.67%
For output voltage :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |19.1๐ โ 20๐ |
20๐ร 100% = 4.50%
For gain :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |101 โ 100 |
100ร 100% = 1.06%
We obtained a percent error of 4.5% for output which is not too bad. We also get
1.06% error for the gain which is good.
Conclusion :
The percent error of the amplitude was 4.5%, and the measured value was
lower than the expected value. This problem was caused by the AC voltage source
and inverting input gain resistors. Since the value was too small, either R2, the
resistor from Vout to the inverting input, was too small, or R1, the resistor from the
inverting input to the AC voltage source, was too large. Next time, use appropriate
resistance value so that more accurate data will be obtained.
Task 5 : Designing a Variable Level-Shifting Amplifier
In this design problem, the input signal was from an electric condenser
microphone as in previous task except that the DC offset now is unknown. But, we
do know the offset is between 4V and 6V. This signal need to be inverted, amplified,
and the DC offset needs to be removed before being input to the same sensitive audio
circuit as before. Our aim for this experiment is to build a level-shifting amplifier
circuit that can be adjusted to cancel any DC offset between 4V dc and 6V dc. Again,
we were asked to design a circuit that will achieve maximum amplification of the
signal without exceeding the 20 Vpp limits of the audio equipment.
Schematic Diagram of Network :
Figure 9: Schematic diagram of a variable level-shifting amplifier.
Theory of Operation :
The input for the circuit was 200 mVpp and the desired output was 20 Vpp.
In this design problem, we know the DC offset is between 4V and 6V. We need to
counter this DC offset so that there is no offset in the output of the function. This
task is similar to Task 4. The only difference is that the offset is varied and in order
to vary the voltage obtained at non-inverting terminal to cancel the offset, we need
to add potentiometer, and choose resistor values that will give the desired voltage.
We must use the positive 15V power supply that is powering the op amp so we used
a voltage divider at positive terminal of op amp to get our desired voltage of between
4V and 6V. Therefore, we can find R4 and R3 values from the voltage division at
non-inverting terminal and R2 and R1 from gain equation at inverting terminal.
Derivation and Analysis:
We did some calculations to get the resistor values in order to achieve the
desired voltage amplitude. We were given input voltage, Vin which is 0.2 Vpp and
output voltage, Vout which is 20 Vpp. We know the formula to find the gain of
inverting op amp which is:
๐บ๐๐๐, ๐บ =๐๐๐ข๐ก
๐๐๐=
๐ 2
๐ 1
Substituting the voltage given we get:
๐บ๐๐๐, ๐บ = 20 ๐๐๐
0.2 ๐๐๐= 100
Therefore, we chose R2 = 100 kฮฉ and R1 = 1 kฮฉ. (with measured values of R2 =
99.546 kฮฉ and R1 = 0.9819 kฮฉ).
We chose R3 and R4 by performing voltage division at node voltage at non-
inverting terminal, Vx. Here is the calculation:
๐ 4
๐ 4 + ๐ 3 + ๐ ๐๐๐กร ๐๐๐ข๐ก = ๐๐ฅ
For Rpot = 0 kฮฉ:
๐ 4
๐ 4 + ๐ 3ร (15๐) = 5๐
We solved for R4 and R3, we obtained:
๐ 4
๐ 4 + ๐ 3=
1
3
For Rpot = 20 kฮฉ,
๐ 4
๐ 4 + ๐ 3 + 20๐ฮฉร (15๐) = 5๐
We solved for R4 and R3, we obtained:
๐ 4
๐ 4 + ๐ 3 + ๐ ๐๐๐ก=
1
3
We chose R4 = 1 kฮฉ, then R3 = 2 kฮฉ. (with measured values of R4 = 0.9822 kฮฉ
and R3 = 1.9713 kฮฉ).
Experimental Results :
Figure 10: The capture of oscilloscope for a variable level-shifting oscilloscope.
Based on the capture of oscilloscope above, the output amplitude obtained is 18.9 V
which is close to actual value, 20V. Notice that the mean (the two circled arrows) is
exactly in the center of the graph (on the horizontal axis). This means that the DC
offset was successfully cancelled out by the voltage at the non-inverting input.
From the oscilloscope capture, we obtained output voltage equals to 18.9 V
and gain of:
๐บ๐๐๐, ๐บ =๐๐๐ข๐ก
๐๐๐
๐บ๐๐๐, ๐บ = 18.9๐
0.206๐= 91.75
Percentage error for all measured values we obtained were calculated as
below:
Formula : ๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |๐๐๐๐ ๐ข๐๐๐โ๐๐๐๐๐๐๐|
๐๐๐๐๐๐๐ร 100%
For R1 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.9818 ๐ฮฉ โ 1๐ฮฉ |
1 ๐ฮฉร 100% = 1.82%
For R2 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |99.546 ๐ฮฉ โ 100๐ฮฉ |
100 ๐ฮฉร 100% = 0.454%
For R3 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |1.9713 ๐ฮฉ โ 2๐ฮฉ |
2๐ฮฉร 100% = 1.435%
For R4 :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |0.9833 ๐ฮฉ โ 1๐ฮฉ |
1 ๐ฮฉร 100% = 1.67%
For output voltage :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |18.9๐ โ 20๐ |
20๐ร 100% = 5.50%
For gain :
๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐๐ = |91.75 โ 100 |
100ร 100% = 8.25%
Conclusion :
A variable level shifting can be used to remove DC noise, especially when the
exact amount of DC is unknown. Although our percent error is considered small, it
still was non-zero. There were some errors in the circuit. Since the Vout value was
small than actual one, it means that the inverting input was high. The inverting input
gain could be too high if either R2 was too high or R1 was too low. Therefore, we
should use no resistors lower than 10k, so it would be a good idea to change R2 to
1Mฮฉ and R1 to 10kฮฉ.
Post Lab Questions
1. To create a variable gain amplifier, one way to achieve desired circuit would
be to connect a potentiometer from R2 to output voltage. Then, there would
be two new equations that need to be calculated to find the new resistor values
that non inverting input will output voltage in the range of 5 Vpp โ 20 Vpp.
The gain will vary from 25 to 100.
Letโs assume potentiometer resistance is 20kฮฉ.
For max Rpot = 20kฮฉ,
๐บ๐๐๐ = ๐ 2 + 20๐ฮฉ
๐ 1ฮฉ+ 1 = 100
๐ 2 = 99๐ 1 โ 20
For min Rpot = 0kฮฉ,
๐บ๐๐๐ = ๐ 2
๐ 1+ 1 = 25
๐ 2 = 24๐ 1
We got two equations and then by solving these, we get R1 = 0.27 kฮฉ and R2
= 6.4kฮฉ.
2.
3. To be able to produce the circuit that vary the gain of both channels, we need
to include potentiometer at both the inputs. Because we used the
potentiometers as variable resistors as opposed to voltage dividers, we had to
disconnect one of the terminals. There was one problem with this approach,
however; if the potentiometers were set to their minimum values (zero), then
the resistance in front of the input values would be zero. According to the gain
formula given in circuit two, this would cause an infinite gain. To rectify this
dilemma, we included a resistor in addition to the potentiometer in front of
the input signals. This way there would always be input resistance even if the
potentiometer was turned all the way down.
4. What would happen if I wanted a variable signal amplification is that I would
need to connect a potentiometer at R2. If I connect it, the inverting input gain
will change, and this will cause a change in non-inverting gain as well as
resulting to change in DC offset which will not to be as expected. We can fix
this problem by connecting the third terminal from the potentiometer at R2 to
the non-inverting input and attenuated properly so the DC offset would not be
affected by the altered gain.
5. The problem is that we assumed the DC offset was positive, so after going
through the inverting input it would become negative. For this reason, we
connected the positive side of the power supply to the non-inverting input so
that it would cancel the negative DC offset. To fix this problem of negative or
positive, we could connect a switch that would change from the positive side
of the power supply to the negative side based on the polarity of the dc offset.
A user could do this manually, or a comparator could be used to check the
polarity and act accordingly.
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