Title: Operational Amplifier, OP-AMP Abstract: In the experiment, the application involving the operational amplifier was investigated. A ±15volts voltage is supplied to the op-amp. The 15volts also the working ranges for the op-amp. There are two basic op-amp circuit, inverting and non-inverting amplifier. For an ideal op-amp, the voltage gain of the circuit is the function of ratio of resistor. There are also other configuration investigated like the buffer amplifier which has the same input and output voltage and the summing amplifier.
When a capacitor is connected as a feedback element, the output voltage is the integral of the input voltage. The summing and difference amplifier is basically the same concept with the basic circuit of inverting amplifier Introduction: Operational amplifier, op-amp is a voltage amplifier electronic device. There are several kind of op-amp like linear, non-linear and frequency dependant. The op-amp is normally made up of 20-30 transistors. Below are the examples of the most common op-amp. The op-amp is three terminal devices, which is inverting terminal, non-inverting terminal and the output terminal.
In the ideal case, op-amp has a infinite input impedance, infinite open-loop voltage gain and zero output impedance. The invention of op-amp is still new which less than 100years. However, the op-amp is now widely used in electronic device or even the musical instrument. Using different connection to the input resistance and feedback resistance, the op-amp can bring out different kind of application. The most common inverting amplifier is actually the same concept and function with the RC circuit. Op-amp can use in low pass and delay function. Equipments and Components a) µ741 Op-Amp. 2x (100 k? 20 k? , 10 k? ), 2k? , 1. 5 k? , 100? , 4. 7 nF b) Signal Function Generator, Oscilloscope, DC power supply Procedure: +15V and -15V was connected to pin 7 and pin 4 respectively for the op-amp. Pin 1,5,8 remain no connection. 4. 1 Inverting Scaling Amplifier a) The inverting amplifier circuit was connected as shown in above. b) The inverting amplifier circuit was connected with R1 = 20k? and Rf = 100k?. The resistors, R1 and Rf were measured. c) The input was applied with sine wave, 1 kHz and amplitude 0. 5 V d) The input and output waveforms were sketched and the phase inversion was noted. ) Rf/R1 was calculated. Vout peak amplitude was measured. f) Multisim was used to simulated. 4. 2 Non-inverting Scaling Amplifier a) The non-inverting amplifier circuit was connected as shown in above. b) The non-inverting amplifier circuit was connected with R1=20k? and Rf=100k?. c) The input was applied with sine wave, 1 kHz and amplitude 0. 5 V d) Input and output waveforms ware sketched and the same phase was noted. e) (1 + Rf/R1) was calculated. Vout amplitude was measured. f) Multisim was used to simulate. 4. 3 Buffer amplifier (unity gain voltage follower) ) The buffer amplifier circuit was connected as shown in above. b) The input was applied with sine wave, 1 kHz and amplitude 0. 5 V c) The input and output waveform were sketched. d) Vout amplitude was measured. e) The purpose of the buffer amplifier was illustrated. 4. 4 Summing Amplifier a) The summing amplifier circuit was connected as shown in above. b) The circuit was connected with R1 = 20k? , R2 = 100k? and Rf = 100k?. c) The input was applied with sine wave, 1 kHz and amplitude 0. 5 V d) The input and output waveforms were sketched. e) Vout amplitude was calculated.
The measure value was noted. 4. 5 Integrator Amplifier Figure 4. 5 a) The integrator amplifier circuit was connected as shown in above. b) The circuit was connected with R1 = 1. 5 k? and Cf = 4. 7 nF . c) Both input were applied with square wave, 1 kHz and amplitude 0. 5 V. d) The input and output waveforms were sketched. e) Sine wave and cosine was repeated for Vin. 4. 6 Difference Amplifier a) The difference amplifier circuit was connected as shown in above. b) The circuit was connected with R1 = 10 k? and Rf = 100 k?. c) The input was applied with Sine waves, 1 kHz and amplitude of v1 = 1 V and v2 = 0. 5 V. d) Vout amplitude was calculated. Vout was measured. e) (b) and (c) for v2 = 1 V and v1 = 0. 95 V were repeated. Result: Part 4. 1: R1= 19. 60 k? Rf= 98. 6 k? Rf / R1= 5. 04 Measured Vout peak amplitude = 2. 6V Input waveform: Output waveform: Part 4. 2: R1= 19. 61 k? Rf= 98. 8 k? (1+ Rf / R1)= 6. 04 Measured Vout peak amplitude = 3V Part 4. 2: Input waveform: Output waveform: Part 4. 3: Measured Vout amplitude= 0. 5V Input and output waveform: Part 4. 4: R1=19. 62 k? R2=98. 8 k? Rf=98. 6 k? Calculated Vout amplitude= 3. 012V Measured Vout amplitude = 3. 2V Input waveform:
Output waveform: Part 4. 5: Square wave input and the output waveform: Sine wave input (left) and it output (right): Cosine wave input and it output: Part 4. 6: Case V1 = 1V and V2=0. 95V Calculated Vout amplitude = -0. 5V Measured Vout = -0. 5V Case V1 = 0. 95V and V2=1V Calculated Vout amplitude = 0. 5V Measured Vout = 0. 5V Multisim Solution: Part 4. 1 Input waveform: Output waveform: Circuit: Part 4. 2: Input waveform: Output waveform: Circuit: Part 4. 3: Input waveform: Output waveform: Circuit: Part 4. 4: Input waveform: Output waveform: Circuit: Output waveform Part 4. 5 quare wave Input and output waveform: Input waveform Sine wave input and output: Input waveform Output waveform Circuit: Part 4. 6: Case1: Input waveform: Output waveform: Case 2: Input waveform: Output waveform: Circuit: Discussion: Part 4. 1 During the experiment, one of the simplest inverting op-amp was build. In the experiment, the non-inverting terminal is connected to the ground result in the terminal to be 0V. There are no current or very little current to be passing into the op-amp. In the deal case, the inverting and non-inverting terminal is connected with an infinity high resistor.
Hence, the potential difference between the input and out is very low and can be ignored. The input impedance for the inverting op-amp is input resistor. For the output resistor, the impedance is very low and it is 0 for ideal cases. The formula for the inverting op-amp can be derived using Kirchhoff current law at the node connected between inverting terminal, input resistor and the feedback resistor. Vin-0Rin-0-VoutRf=0 VinRin=-VoutRf VoutVin=-RfRin From the formula above, the voltage gain of the op-amp can be seen to be the ratio between the input resistor and the feedback resistor.
As the name implies, the inverting op-amp when invert the sine wave graph to negative sine wave graph. In another word, there is a phase shift of 180? Part 4. 2 There are some difference between the inverting and the non-inverting amplifier. Instead of an inverting amplifier, the non-inverting amplifiers do not have a phase shift. The input impedance of a non-inverting terminal is very high. The input voltage also supplied to non-inverting terminal instead of the inverting terminal. As stated in part 4. 1, the potential difference between the inverting and non-inverting terminal is almost the same.
A voltage divider consists between the feedback resistor and the resistor connected to the ground, R1. The voltage across the R1 is the feedback voltage. The feedback voltage is applied to the inverting terminal. One of the functions of the non-inverting amplifier is to control the output voltage. Following is the derivation of voltage gain of non-inverting amplifier. V+=V-=Vin Vout=Vin(R1? RfR1+Rf) VoutVin=1+RfR1 Part 4. 3: Buffer amplifier or unity gain voltage follower is amplifier that always has a voltage gain of 1. The input voltage is being copied from the non-inverting terminal to the inverting terminal without any load.
In another word, the input voltage is equal to the output voltage. The function of the buffer is important as it can transfer voltage from a network of high impedance to another network of low impedance. In ideal case, the input impedance is infinity high where the output impedance is 0. Part 4. 4: The summing amplifier is very similar to the combination of few inverting amplifier into 1. The summing amplifier can produce an output voltage proportional to the summation of all the input voltage varied with the input resistance. The output voltage can be calculated using the superposition voltage.
The differential inputs are virtually grounded. The output voltage formula is similar to the part 4. 1, VoN=-RfRinNVinN N= the number of input voltage Which the total output voltage is summation of total input voltage Vout=Vo1+Vo2+Vo3+ ……+VoN Vo=-(Rf1Rin1Vin1+Rf2Rin2Vin2+Rf3Rin3Vin3+ ……) There are several application using the summing amplifier. The summing of several signal with equal gain is done in the audio mixer. Besides that, the summing amplifier can convert the signal with binary number voltage, digital-to-analog convertor. Part 4. 5
An integrator amplifier is a circuit which output voltage proportional to the time integrator of the input voltage. As the name implies, the function of integrator amplifier is configure to perform the mathematical calculus, integration. Integration is the process to find the area under the graph. Hence integrator amplifier, the output voltage it designed to work responds to the input voltage respect to the time. By applying KCL VinRin=-C(dVoutdt) Vout=-1RC? 0tVin(t)dt The integrator also always has a phase shift of +90? to the output voltage.
In which, the input voltage of a sine wave will be outputted with a cosine wave and vice versa. The interchanging of capacitor, Cf and resistance, R1 will become a differentiator circuit. ?= +90°-tan-1(2? RC) As an ideal cases, there are no current can flow into the op-amp. This is due to the grounding at the non-inverting terminal and result in the 0V in the inverting terminal. During the experiment, one of the procedures is to use the dc offset to make sure no dc voltage is in Vin. This is because the input offset voltage is not zero or another word, the supplied voltage work together with AC and DC.
Hence, the dc input voltage will be integrated and the output voltage starts to drift. Part 4. 6 An op-amp without a feedback resistance is original a difference amplifier. The difference op-amp produces the algebraic difference between the two input signals. Few cases can be done in the difference op-amp. Example like the two input voltage have the same magnitude and the output voltage will be zero which will be the most common. There are no amplification take places when all the external resistors value is the same. Hence, the amplification of the difference op-amp is depends on the ratio of the resistor value.
V+ V- B A The output voltage equation can be derived using super position theorem and voltage divider. At first, we will ground the V2 causing the non-inverting terminal and inverting terminal to be zero volt. Hence, at node A Vo1=-R2R1V1 When V1 is grounded, V+=R2R1+R2V2 V+=V- Vo2=1+R2R1V- Vo2=(1+R2R1)R2R1+R2V2 V02=R2R1V2 By superposition theorem Vo=Vo1+Vo2 Vo=-R2R1V1+R2R1V2 Vo=R2R1(V2-V1) Conclusion: In the experiment, most of the op-amp have a 180? phase shift except non-inverting amplifier and buffer amplifier. The integrator amplifier is different from the others which is 90? phase shift.
Inverting and non-inverting amplifier are two most basic amplifiers with a voltage gain more than 1. The buffer amplifier has the same input and output voltage. Summing amplifier are similar to the combination of few amplifier. The integrator amplifier can perform the mathematic calculus integration or respond with the input voltage over time. The difference amplifiers are amplifier that will produce an output voltage of the difference between two input voltage multiply with the ratio of resistor. Reference: 1. Neamen, D. A. (2010). Microelectronic (4th ed. ). New York, America: McGraw-Hill.