Problems with low frequency integrator.

Hello everyone.

I am using a opamp active integrator (with schmitt trigger) at a very low frequency (500mHz).

The matter is that the integrator R-C values are 56K and 10uF. I build the circuit in a breadboard and measure the capacitor voltage. It's a contiunous voltage of 5.4V (half the value of the opamp VOH. I am using LM358 which reachs the negative rail). I don't know why it gets stuck in that value.

The capacitor used is a 10uF electrolytic one, I am sure it's not broken or burned out. I simulated the circuit with LTspice and it didn't work. I changed its value to 1uF and it worked well.

Then I changed the value to 1uF in the breadboard and it started to work (with a frequency variation but the triangle wave was there). Then I tried decreasing the power soure voltage and it worked. I changed to the 10uF capacitor again and supplying with 8V it worked. Then I put a 10uF TANTLUM one and worked. I raised the power supply to 12V again and was still working. I turned again to the 10uF electrolytic one and didn't work. I tried again with the tantalum one but didn't worked. I tried to repeat the process but it didn't work any more.

What I find strange is that the calcules are right and it should work, but neither in the simulation and in the breadboard it does. I changed the capacitor value to another and it worked. I thought about the capacitor type being wrong (electrolytic) but it worked with the 1uF one.

I've been reading and I found something about capacitors dielectric absorption, which was bad for capacitors used on integrators. The other thing I thought about is the current loss, making it unable to charge at all, but then I realised that it should charge slower but it would reach the schmitt trigger threshold value. The last option possible is that the capacitor is charged with 10.5V and the threshold is about 9V. Is it so high that it doesn't charge at all?

I honestly don't know what to think about, it's so strange, it's the very first time it happens to me.

Thank you in advance.

Post your LTspice schematic.

Is yours an old electrolytic? Can it still hold a charge?

Electrolytics are occasionally known to become leaky as years go by. They develop reduced internal resistance across the plates. This means it will self-discharge if left by itself.

Furthermore it will not charge as quickly. If charging current is slight, it will not charge at all.

I think I bought lots of basic components like 4 or 5 years ago, but I don't know when it was manufactured.

I knew that electrolytic capacitors reduce their capacitor qualities as time goes by. The charging-discharging currents I am working with are of about 45uA minimum, do you think that the leakage is so high that the current isn't enough to charge the cap?

It gets stuck in a value, I don't know if this is right, but does the leakage current increases as the voltage or total charge (Q) does the same? If that's true, then I am sure that the reason for that is the leakage current, it reachs a value that cancels the constant charging current so the voltage variation is null.

I will try adding a resistor in paralel to the capacitor to see if the contiunous voltage decreases. If it does then I am sure of what I said before, just making conclusions.

BrunoARG said:

I will try adding a resistor in paralel to the capacitor to see if the contiunous voltage decreases. If it does then I am sure of what I said before, just making conclusions.

Click to expand...

Of course, you need a feedback resistor providing a small portion of dc feedback (operating point stability) - unless the integrator is part of an overall stabilizing loop.
(Why didn`t you show the circuit?)

Hello. There's no reason for not having shown the circuit, everyone knows the basic active integrator circuit so I thought it was not necessary to put it in the post.

By the way, I finally managed to make it work. It seems that I was connecting the capacitor backwards (with the + terminal to the non-inverting op amp input and the another to its output. I changed the polarity and it started to work fine.

But wasn't the current flowing from the non-inverting to the output through the capacitor when the voltage decreases? I don't know why it should be connected that way. That's all I want to know.

BrunoARG said:

Hello. There's no reason for not having shown the circuit, everyone knows the basic active integrator circuit so I thought it was not necessary to put it in the post.
.

Click to expand...

OK - agreed (in principle), however (read the following):

BrunoARG said:

.. It seems that I was connecting the capacitor backwards (with the + terminal to the non-inverting op amp input and the another to its output. I changed the polarity and it started to work fine.
But wasn't the current flowing from the non-inverting to the output through the capacitor when the voltage decreases? I don't know why it should be connected that way. That's all I want to know.

Click to expand...

To me, this sounds rather confusing (..+ terminal to the non-inv. input..)
Which polarity did you change ?
I think, in this situation a drawing would be very helpful.

Without posting your schematic we must GUESS that your integrator opamp uses a dual polarity supply which causes its integration capacitor to have positive and negative voltage sweeps. Then a non-polarized (film?) capacitor must be used.

Yes, actually having put it would allow you to tell me to invert the capacitor polarity.



I'm not simulating in livewire, I just used it to draw. Some values are not shown because I don't remember them but those ones are right since it's working.

Nothing else to say, I am just curious about why the capacitor is connected that way.

You should always use a non-polar plastic cap for an integrator.

You can achieve longer time constants with small values using 1 to 50M for a 0.5sec time constant with bias current devices and a dump switch, if required.

Yes, but I've been told not to use resistor values higher than 1M since it acts as "open circuit". I think that if a PCB is correctly made and the device is not such an accurate one higher values could be selected.

All I want to know is the current flow of that circuit and why I must connect the capacitor like that.

The 1st output starts with 0 V across the cap then integrates positively after a step voltage determined by the R ratios on 12V since the input is fed into the non-inverting input.

The output gain is 1+ |inverting gain| on Vin+.

Thus the output is more + than input.

Why use the old LM358 that has a low input resistance? You should use a more modern opamp that has Jfet inputs that have an extremely high input resistance.
If the circuit is built on an epoxy fiberglass compact pcb instead of on a messy solderless breadboard or on a cheap electrically leaky phenolic pcb then a high value resistor can be used with a non-polarized film capacitor.

BrunoARG - I am afraid, you have mixed the input nodes of the second opamp ?

BrunoARG - I am afraid, you have mixed the input nodes of the second opamp ?

Click to expand...

Yes, if the circuit is intended to work as an oscillator. Has this been stated somewhere in the thread?

If we assume:

1. R4 = R5
2. R6 = R7
3. The inputs of the second opamp are swapped so the circuit behaves as an oscillator

then the capacitor will be polarized one way during half of the waveform, and polarised the opposite way during the other half of the waveform. Thus a non-polarised capacitor should be used. Tantalum and aluminium electrolytics are not suitable for the circuit.

If those assumptions are wrong then it would help if you show a complete schematic with all the component values, and explain what you actually want the circuit to do.

Yes, in fact, it's an oscillator which generates a low frequency triangle wave as output, which is going to modulate another audible frequency generator. This sound is used as alarm, I was asked to generate a variable frequency sound.

I finally understood why the capacitor is connected like that. As SunnySkyGuy said, the cap will first charge from the op amp output to the non-inverting output, so the polarity of the circuit I posted would be wrong. And I know it's like that because it started to work once I changed it.

I used the old LM358 because the application is a very low frequency one and it's a op amp which reachs the negative rail (I didn't want to use split supply if it was not really necessary). I know there are several and wide better op amps models but those may be expensive and hard to acquire (Once you read where I am living you'll understand why...).

Below is your circuit modified to use a smaller (film or ceramic) capacitor. The feedback from the second op amp is reduced by R6 and R7, allowing a corresponding reduction in the capacitor size without going to large resistor values.

The LM324 is essentially the same as the LM358.