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.

nicely done

Excellent video!

the explanation is quite clear but what will be the output waveform when we have triangular waveform

Excellent video!

thank you for the video. I am finding it difficult to solve for Vout when Vn is not zero. Thus I have a differential integrator.

Good job….thankyou:)

Hi, you're doing a great job.

Thank you very much for putting in efforts to make videos as you do, it's very simple to understand.

Lucid explanation

keep it up.Your videos will be the only one for getting good marks in exam.

awesome video jazak Allah khair please make more videos like this

well done

god bless u for doing such a great job

good one!🖒🖒

video is awesome but the music at the starting make us to feel simpthasie on us to study electronics

if the Vin is a square wave with certain period, how would the amplitude of the triangle wave relate to Vin?

that was precise and elaborate. thank you'

Thank you, this helped me a lot, i'm quiet certain i'm going to pass my exam if i watch more of your videos.

Nice job, one of the better descriptions I've seen. Subtitles helpful

Sir your lecturs are good

Hi, Nice explanation!

Can you tell what are the recommended values for Cf, R1 and Rf?

In my practical application I get a kind of a square wave from 4046 IC (although it looks not like square, just sounds very alike). I've tried to implement the integrator using LM358 and TL074 ICs but no luck 🙁 Also I've inserted R from + to GND (that is in many schematics in Internet).

Now I'll try to use LM833 that seems to be a very good op.amp chip but I guess this will be no luck as well… I mean that I can't get a triangle waveform for my little synth 🙂 PLL chip is nice because it can be frequency-modulated just by a voltage input. This makes it a great thing to create tonal oscillators (like real VCO).

I must note that I use only one 9v battery for this circuit. It's dictated by my target: to make a really portable and small "machine" with a good filter. I've assemblied "MS-20" filter for single supply + AR env. generator and these modules gives really good sound. I've used LM13700 + 2 * LM358 (for AR env.). One battery supports these ICs without problems.

nice.. very simple and easy.. thank you.

Thank you very much, quite helpful

6:00 is that a negative cosine at figure 2 ??

Thanks,,,,, nice video….. Very useful 2 me

Great help.

Thank you, this video was much appreciated 🙂

No current flows to opamp, Voltage at opamp input is zero. Then what is the need of opamp?

WHAT IS AT 2:00

For the first time somebody draws the op amp in and out lines. That's very helpful, thank you.

Thank for making this video.

This is a text book circuit in real world integration of VOS will cause op-amp to go to rail as soon as it's turned on

guys you have to change the intro music ……its scary

Great Allah pak aap ko Kamyabia ataa farmayee

Bahut acche

Thank you for all videos