Integrator saturating
I am trying to simulate a leaky integrator in LTspice using an AD8603 op‑amp. My long‑term goal is to model slow‑changing outputs from MOX gas sensors, but the gas sensor itself is not important for this problem — the issue happens even with a simple test signal.
To design the integrator, I followed the calculation steps from the application note (Design Steps, Page 2). I set Rin = 1 MΩ, calculated Cf using the recommended formula, and chose Rf = 10 × Rin, as suggested.
Input Signal
For testing, I am feeding the integrator with a triangular waveform with a frequency of:
f=8.33 mHz(one full cycle every 120 seconds)
Problem
In simulation, the integrator saturates (runs into the supply rails) instead of producing the expected slow rising and falling output. This happens immediately after the simulation starts.
My Question
Why is my integrator saturating even though the input frequency is extremely low (8.33 mHz) and the slope of the waveform is very small?
Could this be caused by my choice of Rin/Rf/C, the AD8603 biasing, or something about how LTspice handles very slow PWL waveforms?
EDIT:
I did a single supply design added a capacitor for blocking the DC content of the trangular input. Now the output is coming as expected.
@Olin, thank you very much for the detailed explanation. That was very helpful.
2 answers
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| newbie | (no comment) | Feb 27, 2026 at 14:12 |
There are so many problems here, it's hard to decide where to begin.
I followed the calculation steps from ...
Blindly following a recipe is no way to design anything. You need to understand the circuit. Once you understand it, there is little reason to follow dumbed-down formulas. You must be able to figure out on your own how to make the circuit do what you want.
Your input signal has an average DC component. That will eventually saturate any true integrator.
I see that you have a bleeder resistor across the integrating capacitor. That will allow the integrator to settle back to 0 if you wait long enough. In your case, the time constant for settling towards 0 is 190 seconds. That is significant relative to your input signal with a 120 second period.
With 120 second input signal period, you will see some low pass filter effects, not just a pure integrator output. Your "integrator" is really a low pass filter with a rolloff frequency of 840 µHz. Put another way, it is only an integrator for frequencies somewhat above 840 µHz.
The output you show doesn't make any sense, especially for a simulation. The opamp is powered from ±2.5 V. That means its output is limited to that range. In the real world, the positive supply might be a few mV high, but not in a simulation. It makes no sense for the output to be above 2.50000 V. Something is wrong. Ditch the simulator and think. You can go back to the simulation once you understand what is going on.
Overall this integrator inverts. The output is really the negative of the integral of the input. Since your input is always 0 or positive, the output should always be negative. The output essentially stuck at the positive limit again makes no sense.
The output waveform is not at all the integration of the input. Again, think about what should be happening. The sudden drop in the peak at the output when the input triangle changes direction shows the derivative, not the integral.
The output doesn't start at 0. It seems you forgot to tell the simulator to start C1 with 0 volts on it. Remember that integrators have state. You need to initialize that state before use.
Let's see what should happen. With 1 V fixed input, 1 µA flows thru R1. For now, let's ignore the bleeder resistor R2 and analyze this as a pure integrator. With an ideal opamp, that current can only flow thru C1, which causes it to rise 53 mV/s. The overall output with a fixed 1 V input should therefore be linearly falling at 53 mV/s. It would take 47.5 seconds for it to get to -2.5 V, again assuming an ideal opamp.
A 0 to 1 V rising input, like the first half of the triangle wave you are putting in, has an average DC of 500 mV, and would therefore cause the output to saturate in 90 seconds or so. That means the first half of the first input cycle would be OK, but the integrator will saturate during the second half of the first cycle.
Even considering the bleeder resistor, the integrator should still saturate quickly with your given input. At most, R2 will bleed off 250 nA, which is only half of the input current average value.
I didn't look at the opamp datasheet (your job), but you need to make sure that the currents you intend to operate at are well above the input leakage current. The same goes for the integrating capacitor.
So the overall lesson here is to stop and think what is supposed to happen and about what is actually happening. By blindly following recipes and then dumping the whole mess into a simulator, you have minimized the chance of actually understanding.
0 comment threads
Why is my integrator saturating even though the input frequency is extremely low (8.33 mHz)
The lower the frequency, the greater the chance that an integrator's output will saturate and, this might be what you are referring to. But also, your input (the triangle wave) is wholly above the 0 volt axis hence, it's bound to saturate the output after a period of time.
So, imagine inputting a DC value equivalent to 0.5 volts (half the peak value of the triangle) and watch the output saturate in due course.
Imagine also that the integration capacitor was removed and you fed the same triangle wave into your circuit. The peak output voltage will be 10x the peak input voltage (due to the ratio of the two resistors) so, if you get saturation without the integration capacitor then, you stand a chance of saturation at particularly low frequencies when the capacitor is present.
But, what you might be calling "saturation" is in fact the natural response when using a feedback resistor i.e. you have, in effect, an imperfect integrator that acts like a regular low-pass filter. That low-pass filter produces the rounded edges on your output waveform. This is all very natural.
Note - op-amp integrators are very imperfect when used on their own due to needing a DC path from output to non-inverting input. That DC path is provided by the feedback resistor and, it's the presence of that feedback resistor that warps the shape of the output at lower input frequencies.
0 comment threads
0 comment threads