# What is the purpose of an opamp in an integrator circuit?

What is exact role of opamp in integrator circuit....?

• It does the integration. – DerStrom8 Oct 15 '15 at 13:40
• There are multiple reasons, you can easily turn a square wave into a ramp signal. Commonly used for charging purposes due to its gradual changes. – Josh Jobin Oct 15 '15 at 13:41
• No, the integration is done by the capacitor. – LvW Oct 15 '15 at 14:54

Here comes another explanation: For integrating purposes we need a passive RC lowpass with a very low cut-off frequency. This would require excessive large capacitor values.

Therefore - why not using the capacitance multiplication offered by the MILLER effect? Do you remember the MILLER effect in a simple transistor stage? Hence, we connect the integrating resistor to the inverting input of an amplifier which has a capacitor between inverting input and output. Now - the capacitor value is amplified by the open-loop gain of the opamp (as seen from the input).

Example: 1nF*1E5=100µF. Together with an integrating resistor of R=10kOhm we would have a time constant of 1sec (cut-off at f=1/6.28 Hz). As another advantage - the output is available at the low-resistive opamp output.

By the way: That is the reason the shown circuit is called "MILLER integrator".

An opamp in such a negative feedback configuration will try to keep the voltage between the - and + input zero. The + input is grounded (0 Volt) so the opamp will control it's output such that the - input will also be 0 Volt. Let's assume that Vin is 1 Volt then there will be 1 V across resistor R. This means that a current of 1/R will flow through the resistor. This current cannot flow into the - input of the opamp (the inputs are high-impedance) so the current must flow to the capacitor, charging it. Since the - input stays at 0 Volt and the capacitor is charging, the other plate of the capacitor (connected to the output of the opamp) must linearly rise in voltage. So at the output there would be a linearly increasing voltage.

To make the output voltage decrease you would have to apply a negative voltage (for example -1 Volt) to the input. If Vin = 0 then the circuit enters a static situation as the capacitor is not charged nor discharged.

• As per the figure given by Mr.Nuts, if there was a positive 1 V input, the output would linearly decrease becoming more and more negative. – Ryan Jensen Oct 16 '15 at 21:47

By virtue of negative feedback, the inverting input is kept at the same voltage as the non-inverting input i.e. ground or 0V. This is called a virtual earth.

It has the effect of this - any voltage applied to the input creates a current in R that is Vin/R. There is no option here - the inverting input, to a massive extent is held at 0V by negative feedback.

Given that the input impedance of the op-amp can be regarded as being so high as to be regarded as infinite, the input current can only flow through the capacitor.

Q = CV or $\dfrac{dQ}{dt} = C \dfrac{dV}{dt}$ = input current

Because the voltage on the capacitor (V) in the above equation is one-side tied to the same virtual earth, the op-amp output must actually be -V.

Can you see why this is an integrator?

$\dfrac{V_{IN}}{R} = -C\dfrac{d(V_{OUT})}{dt}$

Integrate both sides to solve for Vout.

I like to this of it as a voltage-to-current converter. The op amp converts the input voltage to a feed-back current.

The current that gets fed back through the capacitor is given by ohms law: Icap = Iresistor = Vin/R This is true because the op amp strives to keep the voltage the same at its two input terminals, so the voltage at the inverting terminal should always be 0 V. So the current flowing through the resistor should be the input voltage divided by the input resistance.

The voltage across the capacitor (Vout) is then given by the capacitor equation:

Vcap = integration(Icap)/C Vout = -Vcap Vout = -integration(Vin/R)/C Vout = -integration(Vin)/(RC)

• Well, I guess, strictly speaking, the resistor is the voltage-to-current converter... The op amp keeps one leg of the resistor at 0 V and forces the same current from the resistor to flow through the capacitor. – Ryan Jensen Oct 19 '15 at 23:42

My inference,

Opamp dont care about the capacitor…it does its work tirelessly to equate both input terminals…it our work how to basically exploit this property to our need of integration.

So what our requirement

To linearly discharge the capacitor for true integration.

How will us achieve step 1…and the answer is to charge capacitor via constant currents

Now what are the problem that comes in generating constant current ….and those are —> i) If input voltage is varying for e.g sine cos etc and ii) the capacitor voltage Vc

Now with some voltage drops over R due to Vin at inverting terminal of opamp …the opamp according to its nature will try to equate the voltages of both terminal to the same value thus it output such a voltage which will make same current to flow across the capacitor and the same current to flow across R to cancel out….so literally its the same thing that means a same and opposite amount of current flows across the capacitor which is also flowing across R….hence the two currents are same ..

Now considering fluctuation of Vin..first we should know what does constant current means..it means that the value of current should remain same no matter what Vin and R value is..it should be fixed at some value always.Suppose Vin rise positively to some extent(any atributary value)..as current starts flowing through R… voltage is developed at the inverting terminal…now in order to maintain constant current across the branch(resistor + capacitor) the opamp will output the same voltage in opposite direction..which literally maintains same potential in the branch ..no matter what value Vin goes ..Vout will go down or up the same extent hence maintaining the same current in the branch despite of Vin fluctuations.

Also one should not be confused about the direction of the current that is flowing in the branches ,there are no opposite currents..since the left end of cap is at zero volts due to virtual ground which is still at greater potential than the negative Vout potential of the cap..hence the current will maintain the same direction…(ie…Vin—>R—->C——>Vout..)….

Am I right...