# What does it mean that an integrator has an infinite DC gain?

I mean, I get that, in the ideal case, $$\G\rightarrow \infty \text{ as } s \rightarrow 0\$$, but it doesn't make sense for a physical device to be able to amplify a signal to an arbitrarily large value, as that would seemingly violate energy conservation. So what does it mean in the case of an actual integrator circuit? I mean, I know the transfer function would look different than $$\\frac{1}{s}\$$ if we aren't approximating it as ideal and so wouldn't actually have infinite gain, but that's not what I mean. What I mean is, what does the "ideal" version having infinite gain tell us about the real version? Or does it not tell us anything except that the model is imperfect?

• It tells you that you need to stop integrating at some point. This can be done with switches. For an example, see the ACF2101 which is a classic integrator system. (I've used them some decades ago.) Jun 8 at 0:12

What happens depends on the details of how the circuit is designed.

If we're talking about an op-amp integrator, ultimately the required output voltage will approach the power supply provided, and the op-amp won't be able to continue increasing (or decreasing) the output. Then the op-amp will enter saturated operating mode and the output voltage will stop changing.

In RF circuits or passive filters, an "integrator" might be implemented with a simple capacitor. In those cases it's usually the source that fails to provide a sufficient signal to continue the integrating behavior rather than the filter itself. In an extreme case it's conceivable that the capacitor fails destructively, and the circuit no longer works as intended ever again.

I mean, I know the transfer function would look different than $$\\frac{1}{s}\$$ if we aren't approximating it as ideal

This is a nonlinear behavior. Analysis with the Laplace transform assumes the circuit behavior is linear. Therefore this behavior can't be described in the Laplace domain.

• "If we're talking about an op-amp integrator, ultimately the required output voltage will approach the power supply provided, and the op-amp won't be able to continue increasing (or decreasing) the output. Then the op-amp will enter saturated operating mode and the output voltage will stop changing." Ah, so basically just replace infinity with the maximum power output of the source? Or, in other words, an integrator will have the maximum gain that's physically possible given the details of the power source? Jun 8 at 0:44
• @MikaylaEckelCifrese not a maximum gain, but a maximum output voltage. When it hits this limit it stops integrating and then you have to think about whether the word "integrator" means a mathematical function which this circuit stops doing when it hits the limit, or whether it means the actual circuit you built in which case the mathematical theory doesn't apply when it hits the limit. Either way, if you want to use the mathematics to help you with the the circuit then you have to keep the output voltage within the bounds where the circuit does the mathematics properly. Jun 8 at 2:39
• Ah, ok, so, basically, in order for the circuit to actually be an implementation of integration, the gain HAS to keep increasing as the input input frequency keeps increasing? So once the input voltage frequency and magnitude are high enough that the required gain would increase the voltage magnitude beyond what the power source can provide, it's no longer accurately implementing integration? Jun 8 at 3:56
• @MikaylaEckelCifrese, There's all kinds of nonlinear behavior that can make a real circuit not behave ideally when the input amplitude is too high. Again, the details of how or why this happens depends on the specific circuit design in question. For example in the RC integrator in SImon Fitch's answer, the circuit becomes nonlinear (and no longer an ideal integrator) when the output voltage becomes a significant fraction of the input voltage. If you want to ask about some specific example circuit, you could submit another question to do that. Jun 8 at 5:47
• @MikaylaEckelCifrese, in the op-amp integrator, if everything else is well-designed then the saturation of the op-amp output is what keeps the circuit from integrating indefinitely. Input bias current, capacitor nonlinearity, and probably a dozen other things can limit the accuracy of the circuit. Jun 8 at 5:55

This is a great question, I have never considered integration from an energy perspective. There are two typical implementations of an integrator, one passive (left) and the other active (right): simulate this circuit – Schematic created using CircuitLab

For the passive integrator, your concerns about conservation of energy are justified, since all energy stored in the capacitor must come from source $$\V_1\$$, which is also the source of signal potential being integrated. Energy is also lost in $$\R_1\$$. This places a severe constraint on the maximum energy that $$\C_1\$$ is able to receive and store, and thus integrating behaviour will be highly dependent on the input $$\V_1\$$ itself. The transfer function is:

\begin{aligned} \frac{V_{OUT1}(s)}{V_{IN1}(s)} &= \frac{1}{sC_1R_1 + 1} \\ \\ &= \frac{1}{C_1R_1}\times \frac{1}{s+\frac{1}{C_1R_1}} \end{aligned}

The right-hand term approaches $$\\frac{1}{s}\$$ for $$\s >> \frac{1}{C_1R_1}\$$. We can interpret this to mean that it functions as an integrator only for signals of frequency much higher than $$\\omega = \frac{1}{C_1R_1}\$$.

Thus DC is not integrated, nor are any frequencies lower than some arbitrary value above the break point of this filter.

I'm not sure how to relate this mathematically to energy, except to repeat that this passive system cannot possibly exhibit precise "integration behaviour" due (as you said) to its inability to accumulate energy beyond what's available from signal source $$\V_1\$$, and after $$\R_1\$$ has disposed of some.

This constraint is addressed in the active, op-amp implementation. Limitations of the op-amp itself notwithstanding (such as bandwidth and input offset), this design is close to ideal, having the transfer function:

\begin{aligned} \frac{V_{OUT2}(s)}{V_{IN2}(s)} &= -\frac{\left(\frac{1}{sC_2}\right)}{R_2} \\ \\ &= -\frac{1}{C_2R_2}\times \frac{1}{s} \end{aligned}

This is precise integration, at any frequency (including DC), as you can see from the $$\\frac{1}{s}\$$ term. It is able to achieve this by drawing energy to charge $$\C_2\$$ from a separate, and presumably inexhaustible source $$\V_3\$$, derived via the op-amp's output.

The source of potential being integrated is $$\V_2\$$, but the only power required from that source is $$\\frac{{V_2}^2}{R_2}\$$, due to the virtual ground on the far side of $$\R_2\$$. The signal itself is decoupled from the integrating capacitor $$\C_2\$$ by that virtual ground, and $$\V_2\$$ is relieved of any duty to charge the integrating element $$\C_2\$$.

The only constraint on maximum energy that $$\C_2\$$ can accumulate is the potential difference of the auxiliary supply $$\V_3\$$, but until that limit is reached, this arrangement is able to integrate DC with the expected linear rise/fall of output potential.

At high frequency the op-amp's limited bandwidth will cause behaviour to deviate from the $$\\frac{1}{s}\$$ ideal. From a practical standpoint, if the signal being integrated were for example some velocity, the goal being to derive position, this would certainly limit the circuit's ability to accurately track position if the velocity signal were to contain high frequency components, corresponding to sudden acceleration or deceleration.

Other op-amp idiosyncrasies, such as input bias current and offset voltage will also contribute to inaccuracy, since they will introduce asymmetry. A zero-input potential will not necessarily result in zero current through the integrating capacitor $$\C_2\$$. For this reason there's usually a large resistance placed in parallel with $$\C_2\$$, to cap gain at DC and prevent the output from slowly migrating away from zero.

Output error due to finite open-loop gain could be of concern, but it is dwarfed by other imperfections, like input offset voltage. I think open-loop gain would be of concern only if it is unusually small, far less than 105. For most op-amps, gain is sufficiently high that there are far worse problems to contend with first.

How to understand infinite gain at DC in an integrator?

There are two typical cases

• Where the integrator is being used as part of a feedback loop
• Where the integrator is open loop

If it's in a feedback loop, and if the circuit is operating without the signals saturating, or 'hitting the rails' as we often say, then it means the input to the integrator will be zero in the long term average. The integrator can have any output value with an input value of zero, which you can regard as the result of $$\0 \times \infty \$$.

This technique is used in a PID control loop, the 'I' being the integrator, so the loop can generate whatever control signal the process needs to get to the set point, without having to have a constant error from the set point, which would be the case if the loop only had finite gain.

If the integrator is used open loop, then it means it is going to saturate at some point in the future, it will hit the rails and stop operating as an ideal integrator.

Circuits like the PID can sometimes have their loop broken, if the sensor or actuator fails, that sort of thing. If they operate for a while open loop, then the integrator can hit one rail or the other, with possible bad effects for the process when control is restored. Software PIDs often take steps to detect this, and correct it with an integrator reset, before slamming the process with a saturated control value.

Obviously in a real integrator, the DC gain will not be infinite, but 'only' 106 or so, representing the op-amp's open loop DC gain. This is usually close enough to infinity to meet either definition of an integrator above, to make the input error 'zero', or to hit the rails in the long term.