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Jeffrey Chen Section 001H Oct 29 2012 Op Amp Design Lab Task 1: Basic Inverting Amplifier Objective: Design an op amp circuit that inverts and amplifies a mono signal from a dynamic microphone. Schematic of Networks: Comments on Implementation: Voltage sources of ±15V are implemented using the 20V output of an Agilent Power Supply. V1 is implemented using an Agilent Function Generator in “high Z” mode. All calculations assume that the op amp is operating in linear mode and that it is ideal. Resistor values can be any multiple of the ones chosen while maintaining the ratio between the two. Theory of Operation: The op amp linearly amplifies a signal within its operating specifications. When input is on the negative terminal, it tries to match the two terminal voltages by outputting and amplified signal and uses a negative feedback loop to control the amount of amplification. The amount of amplification is called the gain and is negative for input on the negative terminal and positive for input on the positive terminal. Measurements: None Calculations: πΊπππ = ππ 8 = − = −80 ππ .1 πΊπππ = − π 2 = −80 π 1 π 2 = 80π 1 π 1 = 3πΩ ∴ π 2 = 240πΩ % πππππ = = (ππ π‘ππππ‘π − πππ‘π’ππ) πππ‘π’ππ (16 − 15.78) 15.78 × 100%. × 100% = 1.39% Oscilloscope Captures: Observations/Conclusions: The op amp circuit is fairly easy to design and build. With minimal parts, an input signal can be amplified linearly to a certain degree specified by the voltage inputs on the op amp as well as the op amp specifications. Although no op amp is truly ideal, realistically, most come fairly close which is why the implementation comes so close to the theoretical calculations. The weak point in the circuit is definitely the resistor values. Because the gain is so high in this implementation, small changes in resistor values due to manufacturing tolerance can make a significant difference in the operation of the circuit. Task 2: Weighted Summing Amplifier Objective: Design an op amp circuit that takes an unbalanced stereo signal and sums, inverts, balances, and amplifies it. Schematic of Networks: Comments on Implementation: See previous task. Theory of Operation: The output is a product of constructive interference and varied gain on the two inputs. The principle of superposition applies because the circuit is assumed to be operating linearly. Therefore, the two signals are added together and amplified according to the resistors directly in series with them. Measurements: None Calculations: πΊππππ ππβπ‘ = ππ ππ πΊππππ ππβπ‘ = − 4 = − .1 = −40 πΊππππΏπππ‘ = π 2 π 1 = −40 ππ ππ πΊππππΏπππ‘ = − 4 = − .25 = −16 π 2 π 3 = −16 π 2 = 40π 1 = 16π 3 π 2 = 80πΩ ∴ π 1 = 2πΩ AND R 3 = 5πΩ Oscilloscope Captures: Observations/Conclusions: Using the principles of superposition, it is relatively simple to calculate resistor values needed to achieve the desired output. As with the previous task, the resistor values are arbitrary as long as the ratios remain the same and the operation is very close to ideal. This task illustrates the flexibility of operation of op amps. This type of summing amplifier can be used in a number of audio applications for instance the amplifier on a subwoofer which requires a mono signal or a down-mixer for editing purposes. Task 3: Two Channel Mixer with Balanced Inputs Objective: Design and build an op amp circuit that mixes two inputs with variable gain to have a set voltage swing. Schematic of Networks: Comments on Implementation: The potentiometers have one end disconnected. The wiper terminal is connected to the op amp input to allow for a variable resistance. Measurements: None Calculations: Gain of the op amp in lower limit: Gain = Gain = −R 7 R1 + R 3 Vout 0.4V =− = − − 0.4Vpkpk 2 ∗ Vin 1V Gain of the op amp in higher limit: Gain = −R 7 R3 Vout 16V =− = −16Vpkpk 2 ∗ Vin 1V Assuming 20kΩ potentiometers and solving for the resistors when R 3 = R 4 yieldsR 3 = R 4 = .513kΩ and R 7 = 8.205 kΩ. Gain = Oscilloscope Captures: Wiper set to 0% Wiper set to 100% Observations/Conclusions: Unlike the previous circuits, there is only one correct resistor value for each resistor used in the circuit. While only showing the maximum and minimum voltage, the potentiometer allows for virtually infinite adjustment within the designed range. Introduction of another resistor value also introduces another source of error. The voltages were slightly off even though the calculations were absolute indicating that potentiometers are not as accurate as fixed value resistors. Again a very obvious application is in audio such as on a DJ turntable. Two circuits could be used in conjunction to mix stereo signals. Task 4: Level-shifting Amplifier Objective: Design a circuit that amplifies a microphone signal while removing an inherent DC offset simultaneously. Schematic of Networks: R4 80kΩ XSC1 VEE -15V C2 Tektronix P G 1 2 3 4 T 10nF V1 4 R1 U1A 2 3kΩ 1 0.3 Vpk 440 Hz 0° 3 8 LF412CN C3 R2 R3 84kΩ 16kΩ 10nF VCC 15V Comments on Implementation: Voltage division is used to achieve the correct level shift voltage on the positive input terminal of the op amp because there is only a fixed 15V power line. Measurements: None Calculations: Gain = −R4 R1 Gain = 1 + for the inverting terminal R4 R1 for the noninverting termnal Oscilloscope Captures: Observations/Conclusions: Due to the nature of the gain, the resistors used in voltage division are dictated by the ones chosen for the negative terminal gain. Because of this, it is very hard to design the circuit so that all the numbers work out in rational whole numbers. There is a slight difference in the projected values and the implemented values for this very reason. There are many reasons for removing a DC offset on an input signal. It can be used to prevent clipping and to level match the signal with the reference signal level. Task 5