Operational Amplifier based Integrator Circuit

The description contains information such as design details, equations and practical ideas for the operational amplifier or op amp integrator circuit.


  • Operational amplifier circuits
  • Inverting op-amp
  • Non-inverting op-amp
  • High pass filter
  • Low pass filter
  • Bandpass filter
  • Variable gain amplifier
  • Notch filter
  • Multivibrator
  • Bistable
  • Comparator
  • Schmitt trigger
  • Integrator
  • Differentiator

Operational amplifiers form an ideal basis for creating an integrator circuit.

While it is possible to develop a simple integrator circuit using just a resistor and capacitor, the operational amplifier with its high level of gain enables a far higher performance circuit to be created.

The op amp based circuit is able to give much higher levels of linearity, and a much greater elvel of integrity in its output than one using just the minimum number of components.

Electronic integrator basics

In most op amp circuits, the feedback that is used is mainly resistive in nature with a direct resistive path forming at least part of the network. However for the integrator this is not the case – the component providing the feedback between the output and input of the op amp is a capacitor.

As the name of the op amp integrator implies, it performs a function that is an electronic equivalent to the mathematical integration function. In fact electronic integrator circuits can be used in analogue computers.

In terms of their operation, the circuit produces an output that is proportional to the integral of its input voltage with respect to time.

This means that the output voltage at any time is determined by the start output voltage, the length of time the input voltage has been present and the value of the input voltage.

The basic idea behind an integrator circuit is shown below. Although there are a few changes for the op amp integrator circuit, this concept is what is behind its operation.

Integration of voltage
Integration of voltage

From the diagram, it can be seen that while the input remains at zero, so does the output. However when a step input voltage is applied to the input, the output rises. When the step input returns to zero, the output remains at the voltage it last attained.

Operational amplifier integrator circuit

The basic operational amplifier integrator circuit consists of an op amp with a capacitor between the output and the inverting input, and a resistor from the inverting input to the overall circuit input as shown.

Op amp integrator circuit
Op amp integrator circuit

One of the first points to note is that as the signal is applied to the inverting input, the output of the circuit is the inverse of a basic CR integrator network.

Output from op amp integrator circuit
Output from op amp integrator circuit


Design calculations

The primary calculation required for the circuit is to determine the output voltage for a given input voltage for a given time.

Op amp integrator design equation

Vout = output voltage from op amp integrator
Vin = input voltage
T = time after start of application of voltage in seconds
R = resistor value in integrator in Ω
C = capacitance of integrator capacitor in Farads
c = constant of integration and in this case is the output starting voltage.

The negative sign in the equation reflects the inversion resulting from the use of the inverting input of the op amp.


It is obvious that the output of the integrator cannot rise indefinitely as the output will be limited.

The output of the op amp integrator will be limited by supply or rail voltage and the saturation of the op amp itself, i.e. how close to the rails the output can swing.

When designing one of these circuits, it may be necessary to limit the gain or increase the rail voltage to accommodate the likely output voltage swings.

While small input voltages and for short times may be acceptable, care must be taken when designing circuits where the input voltages are maintained over longer periods of time.

Op amp integrator saturation point
Op amp integrator saturation point

Reset facility

It is sometimes necessary to have a means whereby the op amp integrator can be reset to zero.

The addition of a reset facility or capability is very easy to achieve. It is accomplished by simply adding a switch across the integrator capacitor. This has the effect of discharging the capacitor and thereby resetting the overall integrator.

Op amp integrator with reset switch
Op amp integrator with reset switch

The reset switch can be implemented in a variety of ways. Obviously a simple mechanical switch can be used, but it is also possible to use semiconductor switches. These are typically FET based switches because they have a very high off resistance and can be controlled as switches in this type of application more easily.


Complete and easy explanation – Difference amplifier and Voltage follower

An op-amp with no feedback is already a differential amplifier, amplifying the voltage difference between the two inputs. However, its gain cannot be controlled, and it is generally too high to be of any practical use. So far, our application of negative feedback to op-amps has resulting in the practical loss of one of the inputs, the resulting amplifier only good for amplifying a single voltage signal input. With a little ingenuity, however, we can construct an op-amp circuit maintaining both voltage inputs, yet with a controlled gain set by external resistors.

If all the resistor values are equal, this amplifier will have a differential voltage gain of 1. The analysis of this circuit is essentially the same as that of an inverting amplifier, except that the noninverting input (+) of the op-amp is at a voltage equal to a fraction of V2, rather than being connected directly to ground. As would stand to reason, V2 functions as the noninverting input and V1 functions as the inverting input of the final amplifier circuit. Therefore:

If we wanted to provide a differential gain of anything other than 1, we would have to adjust the resistances in both upper and lower voltage dividers, necessitating multiple resistor changes and balancing between the two dividers for symmetrical operation. This is not always practical, for obvious reasons.

Another limitation of this amplifier design is the fact that its input impedances are rather low compared to that of some other op-amp configurations, most notably the noninverting (single-ended input) amplifier. Each input voltage source has to drive current through a resistance, which constitutes far less impedance than the bare input of an op-amp alone. The solution to this problem, fortunately, is quite simple. All we need to do is “buffer” each input voltage signal through a voltage follower like this:

Now the V1 and V2 input lines are connected straight to the inputs of two voltage-follower op-amps, giving very high impedance. The two op-amps on the left now handle the driving of current through the resistors instead of letting the input voltage sources (whatever they may be) do it. The increased complexity to our circuit is minimal for a substantial benefits

Voltage follower

A voltage follower (also called a unity-gain amplifier, a buffer amplifier, and an isolation amplifier) is a op-amp circuit which has a voltage gain of 1.

Voltage Follower

This means that the op amp does not provide any amplification to the signal. The reason it is called a voltage follower is because the output voltage directly follows the input voltage, meaning the output voltage is the same as the input voltage. Thus, for example, if 10V goes into the op amp as input, 10V comes out as output. A voltage follower acts as a buffer, providing no amplification or attenuation to the signal.

Voltage Follower Input and Output

What is the Purpose of a Voltage Follower

One may ask then, what is the purpose of a voltage follower? Since it outputs the same signal it inputs, what is its purpose in a circuit? This will now be explained.

An op amp circuit is a circuit with a very high input impedance. This high input impedance is the reason voltage followers are used. This will now be explained.

Voltage Followers Draw Very Little Current

When a circuit has a very high input impedance, very little current is drawn from the circuit. If you know ohm’s law, you know that current, I=V/R. Thus, the greater the resistance, the less current is drawn from a power source. Thus, the power of the circuit isn’t affected when current is feeding a high impedance load.

Let’s look at both illustrations below:

The below circuit is a circuit in which a power source feeds a low-impedance load.

Power source with low impedance load

In this circuit above, the load demands and draws a huge amount of current, because the load is low impedance. According to ohm’s law, again, current, I=V/R. If a load has very low resistance, it draws huge amounts of current. This causes huge amounts of power to be drawn from the power source and, because of this, causes high disturbances and use of the power source powering the load.

Now let’s look at the circuit below, connected to an op-amp voltage follower:

Power Source with High Input Impedance

This circuit above now draws very little current from the power source above. Because the op amp has such high impedance, it draws very little current. And because an op amp that has no feedback resistors gives the same output, the circuit outputs the same signal that is fed in.

This is one of the reasons voltage followers are used. They draw very little current, not disturbing the original circuit, and give the same voltage signal as output. They act as isolation buffers, isolating a circuit so that the power of the circuit is disturbed very little.

Voltage Followers Are Important in Voltage Divider Circuits

So, current, as explained above, is one of the reasons voltage followers are used. They simply don’t draw a lot of current, so they do not load down the power source.

Another reason voltage followers are used because of their importance in voltage divider circuits. This again deals with ohm’s law. According to ohm’s law, voltage= current x resistance (V=IR).

In a circuit, voltage divides up or is allocated according to the resistance or impedance of components.

Because an op amp has a very high input impedance, the majority of voltage will fall across it, (since it’s so high impedance). So it’s very valuable when used in a voltage divider circuit because strategically doing so can allow a designer to supply sufficient voltage to a load.

This will now be illustrated so you can see.

So let’s say we have a circuit shown below which represents a voltage divider with a load attached to the output.

Voltage divider circuit that does not work

So the above circuit will not work and it will be explained now why not.

So in the circuit above, we have a voltage divider between the top 10KΩ resistor and the bottom 10KΩ and 100Ω resistors in parallel. So the voltage divider equation is characterized by the following equation, 10KΩ and 10KΩ||100Ω.

Doing the math across the 10KΩ and the 100Ω resistors in parallel gives us, 10KΩ || 100Ω = (10KΩ)(100Ω)/1.1KΩ= 99.01Ω ~ 99Ω.

So we next have a voltage divider between the 10KΩ resistor and the 99Ω resistor.

We now can use the voltage divider formula to see how much voltage will fall across the top 10KΩ resistor and the bottom 10KΩ resistor in parallel with the 100Ω resistor.

The voltage divider formula for the voltage across the top 10KΩ resistor is, V= 10V(10KΩ)/(10KΩ+99Ω)= 9.9V.

The voltage divider formula for the voltage across the bottom 10KΩ resistor and the 100Ω resistor is, V= 10V(99Ω)/(10,099Ω)= 0.098V or 98mV.

Remember, we use 99Ω because this is the equivalent resistance of the 2 resistors (the 10KΩ resistor and the 100Ω resistor in parallel).

Because the resistors are in parallel, they have the same voltage across each other, which is 98mV.

Now let’s say the load needs about 5V to operate. You can see based on the calculation, there will not be sufficient voltage at the output. As we calculated, we had 98mV as our voltage across the load at the output.

The 100Ω resistance (load) carries down the resistance at the output too low. Therefore, in a voltage divider circuit, the load gets very low voltage, since voltage drops across loads in direct proportion to the resistance (V= IR).

However, if we take out the 100Ω load and instead connect an op amp instead (with its high input impedance), the resistance at the output (which ultimately powers the load) doesn’t get drawn down. So the load can receive sufficient voltage.

Let’s see how this circuit changes now with an op amp, with its high input impedance, and the load connected to the output of the op amp.

This is shown below.

Voltage divider circuit that works

So this circuit above now works.

The voltage divider is now between the top 10KΩ resistor and the 10KΩ resistor and op amp at the bottom.

The op amp virtually offers infinite input impedance. Obviously, it’s not really infinite in real life, but it is hundreds of megohms. Let’s assume it’s 100MΩ, though it can be much more.

So the equation that would characterize our voltage divider is between, 10KΩ and 10KΩ || 100MΩ.

Doing the math on the equivalent parallel resistance of the 10KΩ || 100MΩ resistance gives, (10KΩ)(100MΩ)/(10KΩ + 100MΩ)= 9999Ω ~ 10KΩ.

So we have, 10KΩ || 10KΩ.

Any voltage divider composed of the same 2 resistances gives half the voltage of the power supply. But just to show the math, we have the voltage divider formula, 10V * (10KΩ)/(10KΩ + 10KΩ)= 5V.

So 5 volts falls across the top 10KΩ resistor and 5V falls across the bottom 10KΩ resistor and the 100Ω.

Since the 100Ω and 10KΩ resistor are in parallel, they both receive the same 5V.

So you can see how the op amp allowed us to buffer the output of this circuit so that the load receives the voltage it needs.

So these are the 2 chief reasons we use voltage followers. We either don’t want to load down the power supply and/or we want to buffer the output voltage from a circuit so that a load (especially a low-impedance one) can receive the voltage it needs.

So voltage followers are important to either isolate a circuit so that it doesn’t draw down power or buffer a low impedance load so that it receives sufficient voltage.