950 Words4 Pages

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 +90⁰ 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*…show more content…*

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) 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 180⁰ 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

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) 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 180⁰ 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

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