# Feedback resistor value in op amp integrator

Without the feedback resistor, $$\\frac{v_{in}}{R}\$$ current flows through the capacitor and charges it to $$\v_{out}(t) = -\frac{1}{RC}\int\limits_0^tv_{in}\,dt \$$

Good so far. My textbook says a feedback resistor must be added to prevent the output from saturating when no input signal is present. I get that the input offset current will saturate the output.

What I don't get is why the feedback resistor must be $$\\ge 10R\$$. Based on what the author came up with this factor? Why can't the feedback resistor be, for example, $$\R\$$ too ? Lower gain reduces the output offset voltage right?

• What is DC gain of the circuit widout this 10R resistor?
– G36
Commented Apr 3, 2020 at 9:22
• same as openloop gain 100k for 741 @G36 Commented Apr 3, 2020 at 9:23
• So you should see why adding 10R helps prevent op-amp saturation? Yes?
– G36
Commented Apr 3, 2020 at 9:28
• The feedback resistor will influence the integrator part. Do a simulation in the time domain to see how. Or in frequency domine, this feedback resistor and the capacitor will set the pole frequency at which circuit becomes an integrator again. For Rf = R = 1 and C = 1 we have a pole at 0.6Hz but for Rf = 10R the pole is at 0.06Hz. do you see it?
– G36
Commented Apr 3, 2020 at 9:43
• G36-it is not true that above the pole frequency the circuit "becomes an integrator". In practice, we must be sufficiently far above the lowpass pole. Note that - srtictly spoken - we have for real opamps clean integration (wth -90 deg phase shift ) at ONE SINGLE frequency only!!
– LvW
Commented Apr 3, 2020 at 11:35

"What I don't get is why the feedback resistor must be ≥10R."

(Revised, updated):

This is a rule of thumb - nothing else. In electronics it is necessary to find a trade-off between conflicting requirements, in most cases.

• Without a parallel resistor the integration function would be (theoretically) as good as possible. That means: Lower frequency limits are determined by the opamps finite open-loop gain only (mHz range). The upper frequency limit is set by the opamps gain-bandwidth product (GBW). However, the opamps non-ideal offset properties don`t allow such a configuration without dc feedback - therefore, such a parallel resistor is necessary (finite DC ouput offset); as a consequency, the lower frequency limit will be shifted to higher frequencies.

• If this resistor is too small, the resulting DC output voltage would be fine (small) - however, the integration function would be unnecessarily limited to a smaller frequency region. As a consequence, you would have a lowpasss function with a pretty high cut-off frequency. Note that the integration process needs a magnitude slope of -20dB/ec (and a phase shift of -90 deg). Remember that integration is possible only up to a certain upper frequency limit determined by the opamps open-loop gain characteristics (second pole, transition to -40dB/dek).

• A factor of "10" between both resistors (DC gain of "-10" equivalent to 20 dB) seems to be an acceptable trade-off between both limiting effects.

• Strictly speaking: Ideal integration with a phase shift of exact 90 deg is possible for one single frequency only. For very low frequencies (mHz range) the phase shift is -180 deg (inverting operation). Due to the opamps frequency- dependent gain we have to face unwanted additional phase shift. Therefore, in the integrating region, the total phase crosses the -270deg line (-270=-180-90) at one single frequency only (app. the inverse integrating time constant). These phase deviatons from the nominal value (-270 deg) determine the frequency range where a "good" integration is possible.

• Finally, when the integrator stage is used within an overall (outer) loop with DC feedback, the parallel resistor is not necessary in most cases)

The graph shows a simulation of a Miller integrator (opamp: TL071) with R1=1k, R2=10k, C=1nF). Phase: Upper curve ; Magnitude: Lower curve.

A "good" integration is possible app. between 10 and 100kHz only

• Ahh is the -20dB/dec requirement because $v_{out} = -\frac{1}{RC}\int A \cos(\omega t) \, dt = -\frac{1}{\omega RC} A\sin \omega t = -\frac{1}{\omega RC} A\cos(\omega t -90)\\ \Rightarrow \left|\frac{v_{out}}{v_{in}}\right| = \frac{1}{\omega RC } = \color{red}{- 20 \log \omega RC}$ Commented Apr 3, 2020 at 13:05
• High LvW, A question if I may? Why does the 2nd OL pole stop the integrator functioning? Doesn't the integrator have a -20dB/dec beyond the 2nd OL pole? Surely the low pass closed loop response hits the OL response where it is -40dB/dec beyond the 2nd OL pole and then the two responses continue on together, the lowpass response also dropping at -40dB/dec. Wouldn't a low pass filter still be acting like a low pass filter even though it is dropping at -40dB/dec? I am having trouble understanding the -20dB/decade requirement for the integrator to function correctly.
– user173271
Commented Apr 4, 2020 at 12:31
• The requirement (-20dB/dec) comes from the phase specification. From mathematics we know that the process of integration causes a phase shift of 90 deg (integral over sin is cos). On the other hand, we know from system theory that a phase shift of -90 deg is caused by a system that has a negative slope of 20dB/des.
– LvW
Commented Apr 4, 2020 at 12:49
• So the perfect integrator would have a closed loop phase shift of -90 degrees (-270 degrees taking into account the inversion.) This occurs at a single frequency somewhere between the closed loop poles because the phase of a 2 pole amp transitions from 0 degrees to -180 over its bandwidth. The quality of the integration deteriorates with phase change (caused by frequency change) from the optimum -90 degrees (-270 degrees). But isn't it the closed loop poles which determine the -90 degrees (-270 degrees) frequency, not the open loop poles.
– user173271
Commented Apr 4, 2020 at 13:58
– LvW
Commented Apr 4, 2020 at 14:45

What I don't get is why the feedback resistor must be ≥10R.

and

Why can't the feedback resistor be, for example, R too ?

It doesn't make a very good integrator if the feedback resistor was equal value to the input resistor. In fact, if it were of equal value then, the whole circuit would be equivalent to an RC low pass filter (with inverted output).

I'm not saying that it isn't a useful circuit any more but, given that the author is trying to explain integrators, it makes sense to have the feedback resistor much higher in value compared to the input resistor.

Well, let's solve and show this mathematically. We are trying to analyze the following circuit (assuming an ideal opamp):

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1+\text{I}_2=0\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

\begin{cases} \begin{alignat*}{1} \text{I}_1=\frac{\displaystyle\text{V}_\text{x}-\text{V}_1}{\displaystyle\text{R}_1}\\ \\ \text{I}_2=\frac{\displaystyle\text{V}_\text{o}-\text{V}_1}{\displaystyle\text{R}_2} \end{alignat*} \end{cases}\tag2

Substitute $$\\displaystyle(2)\$$ into $$\\displaystyle(1)\$$, in order to get:

$$\frac{\displaystyle\text{V}_\text{x}-\text{V}_1}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_\text{o}-\text{V}_1}{\displaystyle\text{R}_2}=0\tag3$$

Now, when we have an ideal opamp we know that $$\\displaystyle\text{V}_+=\text{V}_-=\text{V}_1=0\$$. So we can rewrite equation $$\\displaystyle(3)\$$ as follows:

$$\frac{\displaystyle\text{V}_\text{x}}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_\text{o}}{\displaystyle\text{R}_2}=0\tag4$$

Now, for the output voltage we get:

$$\text{V}_\text{o}=-\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1}\cdot\text{V}_\text{x}\tag{5}$$

So, the transfer function is given by:

$$\mathcal{H}:=\frac{\displaystyle\text{V}_\text{o}}{\displaystyle\text{V}_\text{x}}=\frac{\displaystyle1}{\displaystyle\text{V}_\text{x}}\cdot\left(-\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1}\cdot\text{V}_\text{x}\right)=-\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_1}\tag6$$

Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform) the fact that the resistor $$\\displaystyle\text{R}_2\$$ is replaced by a capacitor, so:

$$\text{R}_2=\frac{\displaystyle1}{\displaystyle\text{sC}}\tag7$$

So, we get as the transfer function:

$$\mathcal{H}\left(\text{s}\right)=\frac{\displaystyle\text{v}_\text{o}\left(\text{s}\right)}{\displaystyle\text{v}_\text{x}\left(\text{s}\right)}=-\frac{\displaystyle1}{\displaystyle\text{sCR}_1}\tag8$$

Transforming back to the time domain, gives:

$$\text{V}_\text{o}\left(t\right)=-\frac{\displaystyle1}{\displaystyle\text{CR}_1}\int\limits_0^t\text{V}_\text{x}\left(t\right)\space\text{d}t\tag9$$

Now, when we replace $$\\displaystyle\text{R}_2\$$ with a resistor $$\\displaystyle\text{R}_3\$$ parallel to a capacitor $$\\displaystyle\text{C}\$$ we get:

$$\text{R}_2=\frac{\displaystyle\text{R}_3\cdot\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{R}_3+\frac{\displaystyle1}{\displaystyle\text{sC}}}=\frac{\displaystyle\text{R}_3}{\displaystyle1+\text{sCR}_3}\tag{10}$$

So, we get as the transfer function:

\begin{alignat*}{1} \mathcal{H}\left(\text{s}\right)&=-\frac{\displaystyle1}{\displaystyle\text{R}_1}\cdot\frac{\displaystyle\text{R}_3}{\displaystyle1+\text{sCR}_3}\\ \\ &=-\frac{\displaystyle\text{R}_3}{\displaystyle\text{R}_1+\text{sCR}_1\text{R}_3}\\ \\ &=-\frac{\displaystyle\frac{\displaystyle\text{R}_3}{\displaystyle\text{R}_3}}{\displaystyle\frac{\displaystyle\text{R}_1}{\displaystyle\text{R}_3}+\frac{\displaystyle\text{sCR}_1\text{R}_3}{\displaystyle\text{R}_3}}\\ \\ &=-\frac{\displaystyle1}{\displaystyle\frac{\displaystyle\text{R}_1}{\displaystyle\text{R}_3}+\text{sCR}_1} \end{alignat*} \tag{11}

Now, when we have:

$$\frac{\displaystyle\text{R}_1}{\displaystyle\text{R}_3}\to0\tag{12}$$

We get a pure integrator back. This implies that $$\\displaystyle\text{R}_3\to\infty\$$.

In your case we have $$\\displaystyle\text{R}_3=10\text{R}_1\$$, which gives:

$$\mathcal{H}\left(\text{s}\right)=-\frac{\displaystyle1}{\displaystyle\frac{\displaystyle\text{R}_1}{\displaystyle10\text{R}_1}+\text{sCR}_1}=-\frac{\displaystyle1}{\displaystyle\frac{\displaystyle1}{\displaystyle10}+\text{sCR}_1}\tag{13}$$

In order to guarantee feedback, even at very low frequencies, there is placed a resistor parallel to the capacitor. Because this circuit does not represent a pure (theoretical) integrator, in order to get as close as it can be to an integrator we need to choose the resistor as 'big' as possible, as shown.

• "as big as possible"? What does this mean? Why not R3=100R1? More than that, you should finalize your calculation and multiply the last expression by a factor of 10. In this case, the resulting lowpass cut-off is at wo=s(10R1C).....increased by a factor of 10.
– LvW
Commented Apr 3, 2020 at 13:32
• @LvW Theoretically speaking the value of the resistor $\text{R}_3$ should be infinite to go back to only an integrator. Of course, it is not possible to get a resistor that big, but I assume that the OP understands that. So I do not see what you're looking for? Commented Apr 3, 2020 at 14:13
• Jan...it was not clear to the OP why we require a factor of 10. Hence, he did not know which impact on the integrator function such a resistor may have. When you say "as big as possible", the question - of course - arises: Why not factor 100? Factor "10" is certainly not "as big as possible". The answer must contain the information that the choice of such a resistor is always a TRADE-OFF between the wanted function (integrate over a large frequency range)) and the unwanted parasitic effects (output offset). Do you see, what I mean?
– LvW
Commented Apr 3, 2020 at 14:51

Let's call the feedback resistor R2 and input resistor R1. Ideally you don't want the resistor R2. In that case your frequency response looks like the red curve in the figure.

Now, for practical reasons you mentioned, you have to put some feedback resistor. In that case, for low frequencies your gain will be set by the feedback resistor (since cap is a effectively open) but at high frequencies the cap will set the gain as its impedance goes down. The overall curve will be as shown above, where your feedback resistor levels off low frequency gain. So for these low frequencies you don't have an integrator. To maximize the range of frequencies where your circuit behaves as integrator you need to put this cutoff at some low frequency. What is this cutoff frequency? It is given by: $$\frac{1}{sR_1C} = \frac{R_2}{R_1} \implies \omega_{cutoff} = \frac{1}{R_2C}$$ Clearly you want high R2 to have low cutoff frequency, so you would make it as high as possible. Possibly 10R1.

• or 100 R1. Or more. It all depends on how good an integrator you want. Commented Apr 4, 2020 at 14:42

If the integrator operating as a low pass filter, in order to get good integration, is the crucial factor then for a more slowly changing input signal the cut-off frequency must be quite low. But surely this means that Rf must be large when considering the size of C as the low pass cut-off frequency is 1/(2*piRfC).

The size of Rf compared with Rin seems unimportant when considering whether the integrator will actually act as an integrator instead of an amplifier.

It appears to me that the main reason for making Rf = 10*Rin is to give the output some headroom because with Rf equal to 10 times the size of Rin then the integrator will function well up to an output amplitude of 10 times Vin at which point the output will be limited. When I say "the integrator will function well" I am assuming that the input signal freq is above the cut-off freq of the low pass filter effect caused by Rf and C.

# Some history

At the beginning of the 80s, I was doing my thesis on Vibration measurement system. Its input part contained two cascaded AC integrators through which vibration acceleration was converted into velocity and displacement. Then I met for the first time with this "R2 trick". Since the frequency was too low (1 - 2 Hz), the R2 resistance had to be very high. I remember using some exotic 1 GΩ resistors sealed in glass ampoules (apparently because of the leaks).

# The power of low-level explanations

Then I was trying to understand the idea and choose the right resistance value by reading complex explanations with many formulas like most here. Intuitively, I felt that the idea was simple, but I was never able to find that simple explanation that I needed.

# Miracle on Christmas Eve

They say that miracles happen on Christmas night. I generally do not believe in miracles, but now I am probably going to start believing because a "miracle" happened - I managed to find the simplest possible explanation of this trick!

This was after I realized that for the purpose of this problem (finding the value of R2):

• the input voltage does not play a role and can be considered equal to zero,

• the input resistor R2 also plays no role because no current flows through it (see below) so it can be excluded,

• the ratio R2/R1 between the two resistors does not play a role either.

# Basic idea

## Current-supplied capacitor

With these considerations in mind, the problem can be represented by the simplest possible setup of two elements - a current source producing the op-amp input bias current Ib that charges the capacitor C. To make the effect more noticeable, I have chosen a much higher op-amp input bias current (1 μA).

simulate this circuit – Schematic created using CircuitLab

Theoretically, the voltage across the capacitor should grow to infinity (run Time-Domain Simulation).

## Capacitor shunted by a voltmeter

But when we connect a voltmeter with 1 MΩ internal resistance...

simulate this circuit

... we notice that the voltage stops changing at some, albeit very high (1 kV), value.

## Capacitor shunted by a resistor

Aha... this is how we can limit the voltage across the capacitor to a minimum value - by connecting a resistor R in parallel. Let's assume 1 V for starters; so the resistance is R2 = 1 V / 1 μA = 1 MΩ (Ohm's law). It can be said that with the help of Ib and R we have made a Norton voltage source, and somewhere after the fifth second the voltage across the capacitor becomes equal to the voltage of this source.

So the basic idea can be represented by a simple setup of only three elements - an Ib current source charging the capacitor C shunted by a resistor R.

simulate this circuit

Now the voltage stops changing much sooner - at 1 V.

# Building an inverting integrator

## Conceptual inverting integrator

Before the practical circuit, let's look at the conceptual diagram below in which the op-amp is represented by the so-called behavioral voltage source VOA controlled by the voltage V- of the middle point between the two resistors (the so-called virtual ground). Its voltage is set by an expression - VOA = 1000000*V- (i.e., this is an "op-amp" with a gain of 10E6). The role of this source (op-amp below) is to add a voltage VOA = VR = Ib.R in series to the input voltage Vin with the purpose to compensate for VR (that is why this voltage is negative). This is the idea of such a perfect inverting integrator.

As you can see, the bias current Ib does not flow through R1 and Vin because there is no voltage difference between the real and virtual ground. It flows through C and eventually through R towards the negative VOA terminal (this corresponds to Schematic 1.3 above). Then it passes through the VOA source and returns to the input current source.

simulate this circuit

As you can see in the graph below, the voltage Vout (VOA) is a mirror copy of Vc, and the sum of the two voltages is zero.

## Op-amp inverting integrator

Now, let's replace the VOA voltage source with an op-amp to obtain a real op-amp inverting integrator. To match the experiments above, I have set the op-amp input bias currents to 1 μA and to be exiting.

simulate this circuit

The graphical results are the same.

## Conventional circuit diagram

Finally, let's draw the schematic in the usual "textbook" form and supply the circuit by a 1 V AC input voltage. The AC input current Iin = Vin/R1 and the constant 1 μA op-amp input bias current are summed, and the total current charges the capacitor.

simulate this circuit

Here are the graphical results after running the Time-Domain Simulation for three values of R2 - 100 kΩ, 1 MΩ and 2 MΩ. The AC output voltage is respectively shifted with 100 mV, 1 V and 2 V.