Op-Amp Integrator Circuit (w subtitles)

Op-Amp Integrator Circuit (w subtitles)


In this video, I am going to show you another linear application of operational amplifier. Linear application, that means the operational amplifier will operate in the linear region, and the operational amplifier will be in negative feedback. Now, I will show you the operational amplifier integrator circuit. What is an integrator circuit?? Integrator circuit performs integration in the input waveform. Output is directly proportional to the integration of input waveform. that means Vo ∞ ∫ Vi dt. This is the operational amplifier integrator circuit. In this operational amplifier integrator, output terminal is directly connected to inverting terminal through a feedback capacitor Cf. As Cf is also a passive element, this feedback will be a negative feedback. In this circuit, we are applying input voltage Vi through series resistor R1 in the inverting terminal. The non-inverting terminal is directly grounded, so Vp=0V. The circuit is in negative feedback, so we can write Vp=Vn=0V. The current through the inverting and non-inverting terminal will be equal to zero, We have two currents, I1 is the current through R1, I2 is the charging current of the capacitor Apply KCL at the inverting terminal, I1=I2 + In I1=(Vi – Vn)/R1, As Vn=0V, I1=Vi /R1.. I2 is the current flowing to charge or discharge the capacitor. I2=Cf * d(Vc)/dt Vc is the voltage across the capacitor. As the current is flowing from inverting terminal to output terminal, we can assume Vn>Vo. I2=Cf*d(Vn – Vo)/dt As inverting terminal voltage Vn=0V, We can write I2= – Cf *( dVo/dt) Now, Apply KCL at inverting terminal I1=I2 + 0 Our final job is to find Vo. So, we will keep dVo at one side, and rest of the terms at another side. To find out the Vo we have to do integration on both sides with respect to their integral operator. See this is dVo, that means here the operator is voltage. This is dt, that means the integral operator is time. Let’s assume the change of output voltage is from Vo(0) to Vo(t) for the time t=0 to t=t. Therefore, if we perform the integration, we will get (equation written here) (One last thing) Before applying any input at the inverting terminal, our capacitor must be discharged. That will make the voltage across the capacitor equal to zero. As Vi=0, Vc=0, at that instant Vo(0)=0V. Now, if we put the value of Vo(0)=0V , we will get, Vo (t)=(-1/R1*Cf) * ∫ Vi*dt [limit t=0 to t=t] The output is directly proportional to the negative integral of input voltage. If we apply a rectangular waveform at input, we will get a triangular waveform. But, as the output is directly proportional to the negative integral of input voltage, we will get a negative triangular waveform. If we apply a sinusoidal voltage, in the output we will get a negative cosine voltage. If we use this configuration as op-amp integrator, at very small offset voltage at input, output will be saturated quickly. To avoid this unwanted saturation, we use a feedback resistor Rf in parallel with the feedback capacitor Cf. So our practical op-amp integrator circuit will look like this circuit.

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