op-amp design lab report ee210 spring 2016 section 001

Op-Amp Design Lab

Report

EE210 Spring 2016

Section 001

Name : Nur Farzanah Roslan

TA’s 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 kΩ and 99kΩ. We chose these values by firstly calculating the gain of the

function, then we used the equation, 𝐺𝑎𝑖𝑛, 𝐺 =𝑉𝑜𝑢𝑡

𝑉𝑖𝑛=

𝑅2

𝑅1+ 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:

𝐺𝑎𝑖𝑛, 𝐺 =𝑉𝑜𝑢𝑡

𝑉𝑖𝑛=

𝑅2

𝑅1+ 1

Substituting the voltage given, we get:

𝐺𝑎𝑖𝑛, 𝐺 = 20 𝑉𝑝𝑝

0.2 𝑉𝑝𝑝= 100.

Nominal value for the R2 = 99kΩ and R1 = 1kΩ. Measured value for the R2

= 99.342 kΩ and R1 = 0.9856kΩ.

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 kΩ and 1 kΩ, 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:

𝐺𝑎𝑖𝑛, 𝐺 =20.2 𝑉

0.214 𝑉= 94.39

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.


Figure 3: Schematic diagram of difference amplifier.


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 kΩ, and two

100 kΩ 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.




output voltage, Vout which is 20 Vpp. We know the formula to find the gain of

difference op amp which is,


𝑉𝑖𝑛=

𝑅2

𝑅1.

Substituting the voltage given we get:


0.2 𝑉𝑝𝑝= 100.

We chose R2 = 100 kΩ and R1 = 1 kΩ. Since R2 = R4, therefore R4 = 100 kΩ.

Same goes with R1 and R3. Since R1 = R3 = 1 kΩ, therefore R3 = 1 kΩ.

Nominal value for the R3 = R1 = 1kΩ and R4 = R2 = 100kΩ. Measured value

for the R1 = 0.985kΩ, R2 = 99.563 kΩ, R3 = 0.98313 kΩ, and R4 = 99.278 kΩ.


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:



For R1 :


1 𝑘Ω× 100% = 1.500%

For R2 :


100 𝑘Ω× 100% = 0.437%

For R3 :


1𝑘Ω× 100% = 1.687%

For R4 :


100 𝑘Ω× 100% = 0.722%



20𝑉× 100% = 68.70%

For gain :


100× 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 – 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.


Figure 3: Schematic diagram of a variable two channel mixer with unbalanced input.


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.



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:

𝐺𝑎𝑖𝑛, 𝐺 = −𝑉𝑜𝑢𝑡

𝑉𝑖𝑛= −

𝑅2

𝑅1

We calculated gain for each channel by substituting the voltage given, we get:

𝐺𝑎𝑖𝑛 𝑎𝑡 𝐿 𝑐ℎ𝑎𝑛𝑛𝑒𝑙 = −𝑅2

𝑅1= −20

Hence, we know R2 = 20R1. We chose R2 = 500 kΩ, then R1 = 25 kΩ (with

measured values of R2 = 502.76 kΩ and R1 = 25.052 kΩ). 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 kΩ,

−𝑅2

𝑅3 + 20kΩ= −20

For R4 = 0 kΩ,

−𝑅2

𝑅3= −50

From these, we got the value of R3 = 5 kΩ (with measured value of 5.026 kΩ)


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:

𝐺𝑎𝑖𝑛, 𝐺 =0.469 𝑉

0.325 𝑉= 1.44

Percentage error for all measured values we obtained were calculated as

below:



For R1 :


25 𝑘Ω× 100% = 0.208%

For R2 :


500 𝑘Ω× 100% = 0.552%

For R3 :


5𝑘Ω× 100% = 0.52%



20𝑉× 100% = 97.66%

For gain :


100× 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.


Figure 7: Schematic diagram of level-shifting amplifier.



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.







𝑉𝑖𝑛=

𝑅2

𝑅1



0.2 𝑉𝑝𝑝= 100

Therefore, we chose R2 = 100 kΩ and R1 = 1 kΩ. (with measured values of R2 =

99.546 kΩ and R1 = 0.9819 kΩ).

We chose R3 and R4 by performing voltage division at node voltage at non-

inverting terminal, Vx. Here is the calculation:

𝑅4

𝑅4 + 𝑅3× 𝑉𝑜𝑢𝑡 = 𝑉𝑥

𝑅4

𝑅4 + 𝑅3× (15𝑉) = 5𝑉

We solved for R4 and R3, we obtained:

𝑅4

𝑅4 + 𝑅3=

1

3

We chose R4 = 1 kΩ, then R3 = 2 kΩ. (with measured values of R4 = 0.9822 kΩ

and R3 = 1.9713 kΩ).


Figure 8: The capture of oscilloscope for level-shifting amplifier.


and gain of:

𝐺𝑎𝑖𝑛, 𝐺 = 19.1𝑉

0.189𝑉= 101


below:



For R1 :


1 𝑘Ω× 100% = 1.82%

For R2 :


100 𝑘Ω× 100% = 0.454%

For R3 :


2𝑘Ω× 100% = 1.435%

For R4 :


1 𝑘Ω× 100% = 1.67%


𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 = |19.1𝑉 − 20𝑉 |

20𝑉× 100% = 4.50%

For gain :

𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 = |101 − 100 |

100× 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.


Figure 9: Schematic diagram of a variable level-shifting amplifier.



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.







𝑉𝑖𝑛=

𝑅2

𝑅1



0.2 𝑉𝑝𝑝= 100

Therefore, we chose R2 = 100 kΩ and R1 = 1 kΩ. (with measured values of R2 =

99.546 kΩ and R1 = 0.9819 kΩ).

We chose R3 and R4 by performing voltage division at node voltage at non-

inverting terminal, Vx. Here is the calculation:

𝑅4

𝑅4 + 𝑅3 + 𝑅𝑝𝑜𝑡× 𝑉𝑜𝑢𝑡 = 𝑉𝑥

For Rpot = 0 kΩ:

𝑅4

𝑅4 + 𝑅3× (15𝑉) = 5𝑉


𝑅4

𝑅4 + 𝑅3=

1

3

For Rpot = 20 kΩ,

𝑅4

𝑅4 + 𝑅3 + 20𝑘Ω× (15𝑉) = 5𝑉


𝑅4

𝑅4 + 𝑅3 + 𝑅𝑝𝑜𝑡=

1

3

We chose R4 = 1 kΩ, then R3 = 2 kΩ. (with measured values of R4 = 0.9822 kΩ

and R3 = 1.9713 kΩ).


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.


and gain of:


𝑉𝑖𝑛

𝐺𝑎𝑖𝑛, 𝐺 = 18.9𝑉

0.206𝑉= 91.75


below:



For R1 :


1 𝑘Ω× 100% = 1.82%

For R2 :


100 𝑘Ω× 100% = 0.454%

For R3 :


2𝑘Ω× 100% = 1.435%

For R4 :


1 𝑘Ω× 100% = 1.67%


𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 = |18.9𝑉 − 20𝑉 |

20𝑉× 100% = 5.50%

For gain :


100× 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

1MΩ and R1 to 10kΩ.

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 – 20 Vpp.

The gain will vary from 25 to 100.

Let’s assume potentiometer resistance is 20kΩ.

For max Rpot = 20kΩ,

𝐺𝑎𝑖𝑛 = 𝑅2 + 20𝑘Ω

𝑅1Ω+ 1 = 100

𝑅2 = 99𝑅1 − 20

For min Rpot = 0kΩ,

𝐺𝑎𝑖𝑛 = 𝑅2

𝑅1+ 1 = 25

𝑅2 = 24𝑅1

We got two equations and then by solving these, we get R1 = 0.27 kΩ and R2

= 6.4kΩ.

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.

op-amp design lab report ee210 spring 2016 section 001

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