# 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 Apr 3 at 9:22
• same as openloop gain 100k for 741 @G36 – across Apr 3 at 9:23
• So you should see why adding 10R helps prevent op-amp saturation? Yes? – G36 Apr 3 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 Apr 3 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 Apr 3 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 $v_{out} = -\frac{1}{RC}\int A \cos(\omega t) \, dt = -\frac{1}{\omega RC} A\sin \omega t = -\frac{1}{\omega RC} A\cos(\omega t -90)\\ \Rightarrow \left|\frac{v_{out}}{v_{in}}\right| = \frac{1}{\omega RC } = \color{red}{- 20 \log \omega RC}$ – across Apr 3 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. – James Apr 4 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 Apr 4 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. – James Apr 4 at 13:58

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.

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. – WhatRoughBeast Apr 4 at 14:42

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:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{x}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_\text{o}-\text{V}_1}{\text{R}_2} \end{cases}\tag2$$

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

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

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

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

Now, for the output voltage we get:

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

So, the transfer function is given by:

$$\mathcal{H}:=\frac{\text{V}_\text{o}}{\text{V}_\text{x}}=\frac{1}{\text{V}_\text{x}}\cdot\left(-\frac{\text{R}_2}{\text{R}_1}\cdot\text{V}_\text{x}\right)=-\frac{\text{R}_2}{\text{R}_1}\tag6$$

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 $$\\text{R}_2\$$ is replaced by a capacitor, so:

$$\text{R}_2=\frac{1}{\text{sC}}\tag7$$

So, we get as the transfer function:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{v}_\text{o}\left(\text{s}\right)}{\text{v}_\text{x}\left(\text{s}\right)}=-\frac{1}{\text{sCR}_1}\tag8$$

Transforming back to the time domain, gives:

$$\text{V}_\text{o}\left(t\right)=-\frac{1}{\text{CR}_1}\int_0^t\text{V}_\text{x}\left(t\right)\space\text{d}t\tag9$$

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

$$\text{R}_2=\frac{\text{R}_3\cdot\frac{1}{\text{sC}}}{\text{R}_3+\frac{1}{\text{sC}}}=\frac{\text{R}_3}{1+\text{sCR}_3}\tag{10}$$

So, we get as the transfer function:

$$\mathcal{H}\left(\text{s}\right)=-\frac{1}{\text{R}_1}\cdot\frac{\text{R}_3}{1+\text{sCR}_3}=-\frac{\text{R}_3}{\text{R}_1+\text{sCR}_1\text{R}_3}=$$ $$-\frac{\frac{\text{R}_3}{\text{R}_3}}{\frac{\text{R}_1}{\text{R}_3}+\frac{\text{sCR}_1\text{R}_3}{\text{R}_3}}=-\frac{1}{\frac{\text{R}_1}{\text{R}_3}+\text{sCR}_1}\tag{11}$$

Now, when we have:

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

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

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

$$\mathcal{H}\left(\text{s}\right)=-\frac{1}{\frac{\text{R}_1}{10\text{R}_1}+\text{sCR}_1}=-\frac{1}{\frac{1}{10}+\text{sCR}_1}\tag{13}$$

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 Apr 3 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? – Jan Apr 3 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 Apr 3 at 14:51

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.