I've been trying to figure this out for a while to no avail. Could someone help me out?

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So far I've tried computing KCL around a couple of different nodes but wasn't able to get a model describing the entire circuit. I also tried working backwards from the given equation but wasn't able to get anything out of it.

  • \$\begingroup\$ Look at the parallels between this circuit and a differential amp configuration: electronics-tutorials.ws/opamp/opamp_5.html \$\endgroup\$
    – John D
    Commented Oct 7, 2014 at 3:49
  • \$\begingroup\$ Hint : treat the circuit as a combination of 2 complex impedances ; the left RC form a parallel branch, while top RC are in series. \$\endgroup\$ Commented Oct 7, 2014 at 4:57

3 Answers 3


With the Laplace transform the capacitors have impedance

$$Z = \frac{1}{sC}$$

Use this impedance for the capacitors and analyze the circuit the same way as you would with regular resistors.

For an ideal op amp the inputs \$v_{+}\$ and \$v_{-}\$ are at the same voltage and have infinite impedance. Consequently there is a voltage divider at the input:

$$v_{+}(s) = \frac{\frac{1}{sC}}{\frac{1}{sC} + R}v_s(s) = \frac{1}{1 + sRC}v_s(s)$$

Similarly there is a voltage divider with the output:

$$v_{-}(s) = \frac{R}{R + \frac{1}{sC}}v_{o}(s) = \frac{sRC}{sRC + 1}v_{o}(s)$$

Since \$v_{+}(s) = v_{-}(s)\$ set the two equations equal to each other:

$$\frac{1}{1 + sRC}v_s(s) = \frac{sRC}{sRC + 1}v_{o}(s)$$

Re-arrange to find \$v_{o}(s)\$:

$$v_{o}(s) = \frac{1}{sRC}v_{s}(s)$$

Take the inverse Laplace transform to get back to the time domain (remember that dividing by \$s\$ is equivalent to integration):

$$v_{o}(s) = \frac{1}{sRC}v_{s}(s) \longleftrightarrow v_{o}(t) = \frac{1}{RC}\int_{0}^{t}v_{s}(x)dx + v_{o}(0)$$


This circuit solution of a non-inverting integrator stands out for its originality from many straight-forward solutions. It was invented a long time ago; I saw it for the first time sometime in the 80s. I was amazed that it contained a simple passive RC circuit that was not connected in the negative feedback circuit and yet the integration was perfect.

This circuit piqued my curiosity again by stumbling upon this old question a few days ago. Now, with the help of CircuitLab, I was able to reach some intuitive explanation, which I share below. To illustrate the point of this circuit solution, I have added a few preceding steps to this inventing scenario.

Passive C integrator

A humble capacitor C is the simplest yet perfect integrator with current input. When we supply it with a constant current source...


simulate this circuit – Schematic created using CircuitLab

...the (output) voltage across it changes linearly through time.


Passive RC integrator

But we need an integrator with a voltage input. So, we connect a resistor R to convert the input voltage into an input current. However, as the capacitor is charged, the voltage drop across it increases, the resulting voltage across the resistor and the current decrease.


simulate this circuit

As a result, the famous exponent is obtained.


So we have to somehow compensate for the reduction in current.

Inverting integrator

One possible way (and the most commonly used) is to insert an additional voltage source (op-amp output) in series. Its voltage Vout = -Vc is added to the input voltage and the current (I = (Vin - Vc + Vc)/R = Vin/R) is constant. This idea is implemented in the op-amp inverting integrator. Vout is inverted to be in the same direction with Vin and added to it; but for this reason the circuit requires one more (negative) source.


simulate this circuit


Non-inverting integrator

So to supply the circuit with only one (positive) voltage source, it must be non-inverting.

Floating non-inverting integrator

The first thing that comes to mind is to replace the simple resistor in the RC circuit acting as an imperfect voltage-to-current converter with the more complex non-inverting op-amp converter. The only problem is that the capacitor (and the output) is "floating", and the output voltage includes, besides Vc, the input voltage as well.


simulate this circuit

So when we apply a 1V step input voltage, the voltage across the capacitor changes linearly through time but appears at the output "lifted" by 1V. What do we do?


Grounded non-inverting integrator

When we apply a 1 V step input voltage in the circuit above, the op-amp begins increasing its output voltage, accordingly the current through the RC circuit, until equalizes the voltage drop across the resistor R. Thus it creates a copy of the constant Vin across the constant R, and a constant current I = Vin/R flows through the capacitor C. So the voltage drop Vc across C is a perfect integral of Vin. It remains only to do something with VR so that it does not "jump" but begins to change smoothly from zero.

The simple RC circuit has such a behavior, so let's connect another R1C1 circuit to the input and to see what will happen.


simulate this circuit

Unbelievable but fact - VC2 and VR2 change non-linearly through time, but their sum (Vout) changes linearly!


Let's zoom in on the graph and try to understand how this "magic" happens.


Intuitive explanation. The input voltage is exponentially formed by the input R1C1 circuit acting as a log converter. Then it is converted to current by the op-amp and R2. This exponentially increasing current passes through C2. It is perfectly integrated but its rate of change is slowed in the beginning; as a result, the curve is bulging down. Summed with the bulging up curve of the voltage VR2 (VC1) it gives a straight line. Of course, math has the last word for a precise quantitative explanation.

There is another famous circuit of a non-inverting integrator - the Deboo integrator... but this is another story...


if you want to solve within Laplace,

let input voltage of opamp is 'V?'.

using KCL on - input of opamp

: V?/R + (CV?)'= (CVout)'

same with + input.

: V?/R + (CV?)'= Vs/R

left hand side are same. so you can get the answer easily.


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