# Why feedback resistor be at least 10 times of the input resistance?

Figure below is an integrator with reduced output offset due to the 10R shunted at the capacitor. It was written on the book that:

One way to reduce the effect of input offset voltage is to decrease the voltage gain at zero frequency(DC) by inserting a resistor in parallel with the capacitor, as shown in figure below. This resistor should be at least 10 times larger than the input resistor Why should the resistor be at least 10 times of the input resistor?

Why not use the same resistance on the input for the feedback resistance so that the input offset voltage will only be amplified by 1 instead of 10 when using 10R?

• It’s a hypothetical integrator with leakage and large R values – Tony Stewart Sunnyskyguy EE75 Aug 7 at 4:40
• @Tony Stewart Sunnyskyguy Why does this integrator needs large R value for feedback? – Iwatani Naofumi Aug 7 at 5:05
• Where did you get that "10 times input resistor" specification? That is not right. R sets the input impedance and Rf sets the gain. C is chosen for the roll-off point. Rf must be larger than the minimum output load, as the (-) acts like a virtual ground. – user105652 Aug 7 at 5:21
• It is on the book. Why is Rf dependent on load? Is it because of the Zout? – Iwatani Naofumi Aug 7 at 5:39
• Rf acts as a load connected between the op-amp output and virtual ground (inverting input); so it must not exceed the minimum output load (Rf >> RLmin). Practically, it is always fulfilled since Rf >> R. When Rf is too high, it can be replaced with a "T resistor network". Another trick, in the case of an AC integrator, is to replace the grounded resistor of the T-network with a capacitor (low-pass filter). – Circuit fantasist Aug 7 at 8:06

It should be much larger than the input resistor. The number 10 is just a nice round number somebody picked.

The feedback resistor is a necessary evil. Ideally you wouldn't want to have it, but without it the output is likely to saturate. If the op-amp saturates then the op-amp will no longer behave like an op-amp and you won't get anything close to the integral at the output.

The feedback resistor takes some current that would otherwise be going into the capacitor to create the integral function. Having it there creates a little bit of error in the output. The larger the feedback resistor is, the less error there will be.

Normally if you are trying to get a pure integral function, then you make a trade-off between having the feedback resistor as large as possible, but not so large that the output saturates.

So how does it prevent the output from saturating?

Imagine you put a sine-wave into the input that is centered around 0V. Theoretically the average value of this waveform is 0V, so the integral should be bounded. But in reality the average won't be exactly 0V, and even if it was, the op-amp input has finite input resistance and will take some bias current.

Without the feedback resistor the output would eventually wander all the way to one of the power supply voltages and the integrator would stop working. What the feedback resistor does is steal a little current from the output in such a way that it tends to very slowly pull the integral back towards 0V. So as long as the average of the input is very close to 0V then the output wont wander too far.

Note that if your capacitor is leaky enough, you may not need to add a separate resistor, since you essentially have one inside the capacitor.

• Ah, if Rf is too small some current will go through the Rf instead of all going through the caps and this will create an output error, AND if RF is too large the input bias current will all go through the capacitor and will saturate opamp through the rails. – Iwatani Naofumi Aug 7 at 5:16

An IDEAL integrator does not exist: It would require an ideal operational amplifier with infinite gain and zero offset.

Therefore, we need a trade-off between integrating properties and DC operational point considerations. As a consequence, each real opamp-based integrator is a lowpass (R parallel to C) with a rather small integrating area: Far above the 3dB-lowpass frequency (that is the reason for a large parallel R), but still below the transit frequency of the opamp.

In this context - note that the circuit will have a 90deg phase shift (ideal integration) at ONE SINGLE FREQUENCY only.

Remark: If the integrator is used within an overall stabilizing feedback loop (filters, opscillators, control systems) there is no need for a damping resistor in parallel to the feedback capacitor.

• LvW, I suggest that we accept, once and for all, that there are no perfect devices in this real world, so that we do not repeat it every time. And if we ever come across an ideal device (maybe in another world), let's note it then. But I suggest that when discussing concepts, we assume that devices are perfect ("ideal"). Any focus on imperfections at this stage hinders the understanding of ideas. This is a basic principle of creative thinking where the term of "ideal final result" is frequently used. Agree? – Circuit fantasist Aug 7 at 8:29
• Partly! Your suggestion surely is helpful and makes a lot of sense when discussing with experienced engineers. But what about answering questions like: Why is the voltage at the inv. terminal identical to the voltage at the non-inv. terminal of an opmap? In such cases, it is absolutely necessary to mention the difference between real and ideal(ized). Here is another example: Rather often I read that the gain of a simple common-emitter stage (with RE-stabilization) would be RC/RE. How would you react? – LvW Aug 7 at 15:18
• Really, there is a problem to explain the operation of circuits with (negative) feedback... there is something like a "magic circle". I imagine that the op-amp acts as a servo… or an integrator (slowly acting device that "analyzes" the input difference and changes its output voltage until it reaches the equilibrium); this is true during the transition. Also I carry out a mental experiment with an amplifier with variable gain (from zero to infinity) and observe how the input voltage difference decreases up to zero. This creates the idea of ​​an "ideal amplifier". – Circuit fantasist Aug 7 at 17:52
• Regarding the CE stage, we can assume the gain is Rc/Re if we imagine an "ideal transistor". In the case of a common-source configuration, the same current (Id = Is) flows through both resistors; so VRd/VRs is exactly equal to Rd/Rs. The only error is the small difference between VRs and Vin (VRs < Vin) because of the limited gm and Vth. – Circuit fantasist Aug 7 at 18:02
• Of course, we can develop several different view or models for such a circuit. But my scepticism remains: What do we gain with such artificial sights? Does it help to better understand the transistor function? Does it help to understand how other cicuits (common-base or common-coll.) will work? I still have some doubts. As a typical example, let me mention the so-called re-model. Sometime ago somebody had the idea to describe such a model - and, yes, it works. However, has this model some advantages in comparison to other descriptions? I think: NO - in contrary !! – LvW Aug 10 at 7:00