Can anyone explain the behaviour of this integrator simulation - it's not as I expected.
There is an active and a passive integrator, they both use a +/-5V square wave @ 100Hz as the input.
The output of the passive integrator is doing what I expected - but the output of the active integrator swings down between the negative rail and 0V - why is this?
I also notice that if I use a 100Hz Sine as the input - the signal goes straight to 0V to -12V - why is this?
Putting a 1V PP square wave through has exactly the same effect except the waveform is much lower amplitude and it takes longer to fall to the negative rail.
I don't know what "UniversalOpamp2" is but I would suggest using something with known characteristics because the input bias current will affect the performance of an integrator.
Hello,
Can anyone explain the behaviour of this integrator simulation - it's not as I expected.
There is an active and a passive integrator, they both use a +/-5V square wave @ 100Hz as the input.
The output of the passive integrator is doing what I expected - but the output of the active integrator swings down between the negative rail and 0V - why is this?
You shouldn`t forget that the active integrator circuit is an INVERTING integrator. Thus, everything looks good and as expected (as far as I can see).
Moreover, the integrating function (active circuit) is limited due to the supply rail (maximum possible negative output voltage).
Please note, in addition, that the passive circuit shows the expected step response for a LOWPASS. this circuit can be regarded as an integrating device for very high signals only (very far beyond the 3-dB edge)
There are two points you apparently didn't consider:
- initial solution in transient analysis
- symmetry of generator waveforms
The sine input integrator is showing effectively ideal integrator behaviour according to the characteristic of the universal OP model. You get exactly the shown out1 waveform for an initial value of zero. Integrate the sine input by pencil-and-paper method and see, why the output has a negative bias.
With the square generator, the integrator is initialized in positive saturation because the initial input voltage is negative. But the square wave isn't exactly symmetrical (due to a small finite rise and fall time), causing a slow drift and negative clipping. You can change the behaviour by adjusting VPULSE and .tran parameters.
A real integrator would show arbitrary output DC bias without suitable means to control the level. e.g. implementation as a lossy integrator, overall DC feedback or a periodical reset.
Presumed, you managed to generate a relative symmetrical input, then the integrator will keep the bias set by the initial transient solution.
In the present case, it's a starting point for V(out1) in saturation. The output waveform simply starts from this point. You can activate the "skip initial transient solution"-option to make the output voltage start from zero. Or apply arbitrary .IC statements, review the LTSpice help how to.
As FvM noted the sine wave output is offset due to the starting point of the AC waveform. For example if you add a 90° phase shift (Phi) to the sinewave then the output will be centered around 0V.
You need to activate the .uic option or apply an initial condition, too. Otherwise the initial voltage would cause an integrator starting point in saturation.
To modify the circuit towards real integrator hardware, you can e.g. add a reset switch across the capacitor that's opened at t=0.
You need to activate the .uic option or apply an initial condition, too. Otherwise the initial voltage would cause an integrator starting point in saturation.