# Basic Op-amp Integrator circuit question

• Jdo300
In summary: JasonO,I would need to know the capacitance of the cap and the frequency of the piezo element. I could try to measure those values and then calculate the resistor.In summary, adding a larger resistor to the feedback capacitor seems to cure the problem of the floating offset.
Jdo300
Hello All,

Today, I was experimenting with a basic op-amp integrator circuit (I have a piezo element that I ripped from a buzzer, which I would like to use as a mechanical pressure sensor). So I'm using an integrator circuit to integrate the voltage signal produced by the piezo to get a pressure magnitude reading. The circuit is put together using an LM358 dual op-amp IC with a 1uF capacitor and 1k resistor for the feedback network (see the attached schematic).

The circuit works great for the most part except for one small problem. That is that the zero level of the circuit slowly starts to float more and more negative (rather the input is connected or not). If I press on the piezo, it gives me a nice positive hump and then returns exactly back to the zero point, but if I just leave it sitting there, the line floats down more and more until the output signal hits the negative rail (in my case, I'm powering it with +- 5V).

What I'm wondering is if there is a way to compensate for the floating problem to keep the signal stable? I'm guessing that it has something to do with the finite offset difference between the + and - input signals but I'm not sure what to do about it.

Any input greatly appreciated, thanks!

- Jason O

P.S. I should mention that this floating effect occurs over several seconds to minutes, changing by about -1V every 30-40 seconds on average.

#### Attachments

• Integrator.PNG
1.1 KB · Views: 528
Looks like you need a large value bleeder resistor around the cap...

Yep. Berkeman is correct. There is NO DC negative feedback.

Hi guys,

Thank you very much for the reply. Is there any good way of knowing/determining the size bleeder resistor to use across the cap?

Thanks,
Jason O

You could put a 100 K in instead of the capacitor.

This would give a stable output, a gain of 100 and improve the responsiveness of the circuit.

Not sure why you would want an integrator for this.

vk6kro said:
You could put a 100 K in instead of the capacitor.

This would give a stable output, a gain of 100 and improve the responsiveness of the circuit.

Not sure why you would want an integrator for this.

I think the cap is the piezo sensor itself that he's trying to monitor. I could be wrong, though.

Jdo300 said:
Hi guys,

Thank you very much for the reply. Is there any good way of knowing/determining the size bleeder resistor to use across the cap?

Thanks,
Jason O

Calculate the bleeder resistor value based on the bandwidth you want from the circuit. You know the capacitance value, so set the resistor to give you an RC constant that works for the measurements that you want to make.

Tell us what you come up with and why...

You should first try putting a 1k resistor from the non-inverting pin to ground in place of the short. This will help eliminate input offset current differences. Then you will be able to get by with a larger value for the bleeder.

Hint:

What is the value of the integral:

$$\int_{0}^{t}{(-\varepsilon) \, dt'}$$

I think the cap is the piezo sensor itself that he's trying to monitor. I could be wrong, though

I suspect that a piezo sounder like this would be a bit like a crystal microphone. So, it would be equivalent to a voltage generator, in series with fairly large resistor, in series with a small capacitance but with a large output.

If so, it would probably be better off driving the non-inverting input of the opamp, which would at least be high impedance. Put a 10 Megohm resistor from the noninverting input to ground.

I would make the first stage a voltage follower by joining the output to the inverting input.

Then another stage of amplification.

The output will still be AC, though, so you would have to rectify that to drive a meter.

vk6kro said:
You could put a 100 K in instead of the capacitor.

This would give a stable output, a gain of 100 and improve the responsiveness of the circuit.

Not sure why you would want an integrator for this.

Hi vk6kro,

I need to use an integrator with the piezo element because it's voltage output is proportional to the change in pressure applied and not according to the absolute magnitude of the applied pressure. Also, the output voltage from the piezo element is large enough (on the order of a few volts), so I don't need to directly amplify the input voltage itself.

@All,

I did try adding several different resistor values in parallel with the feedback capacitor. It appeared that any value less than 1 MOhm caused the feedback capacitor to discharge too fast (though it did correct the floating offset problem). I didn't have any larger resistors at that moment, but I it clear that a much larger resistor could cure the problem.

@ berkeman,

As for predicting the bandwidth, if I used, say a 10 MOhm resistor in parallel with the 1uF cap and model the pair as a low-pass filter, I get 15.7 Hz as the cutoff frequency, is this correct?

- Jason O

Last edited:
A piezo element can only output AC.
If you want to measure pressure over a long period of time, try a different approach.

If you only want to measure the change in pressure over a period of a few minutes, use National Semiconductor LMC**** (or equivalent in other brands) operational amplifier and very high value resistors.

Here's a trick that will probably take care of your problem. Install two diodes in series cathode to cathode in the feedback path. There is probably enough leakage current to fix the problem. You may want to be choosy about your diodes, but I have seen this trick work.

Averagesupernova said:
Here's a trick that will probably take care of your problem. Install two diodes in series cathode to cathode in the feedback path. There is probably enough leakage current to fix the problem. You may want to be choosy about your diodes, but I have seen this trick work.

That's a clever trick -- hadn't seen that one before.

Carl Pugh said:
A piezo element can only output AC.
If you want to measure pressure over a long period of time, try a different approach.

If you only want to measure the change in pressure over a period of a few minutes, use National Semiconductor LMC**** (or equivalent in other brands) operational amplifier and very high value resistors.

Hi Carl,

Actually, I'm not looking to take a pressure measurement over a long period of time. After the integrator circuit, the output goes into a peak detector which measures and holds the peak output voltage value from the integrator. So if the pressure sensor value decays, it won't affect my measurement. I'm just hacking this circuit together to do some basic experiments. If I need to do anything really fancy, I would just purchase one of those piezoresistive sensors to do it properly.

@Averagesupernova,

Thanks for the tip, I'll have to try that.

Thanks,
Jason O

## What is a basic op-amp integrator circuit?

A basic op-amp integrator circuit is an electronic circuit that uses an operational amplifier (op-amp) to perform mathematical integration of an input signal. It is commonly used in analog signal processing applications.

## How does a basic op-amp integrator circuit work?

The op-amp integrator circuit works by using the feedback loop of the op-amp to integrate the input signal. The input signal is fed into the inverting input of the op-amp, and the output of the op-amp is fed back into the inverting input through a capacitor. This creates a low-pass filter, which integrates the input signal over time.

## What are the advantages of using a basic op-amp integrator circuit?

One advantage of using a basic op-amp integrator circuit is that it can perform mathematical integration without the need for complex mathematical calculations. This makes it a useful tool in analog signal processing and control applications. Additionally, it is a simple and low-cost circuit to implement.

## What are the limitations of a basic op-amp integrator circuit?

One limitation of a basic op-amp integrator circuit is that it is highly sensitive to noise and can introduce noise into the output signal. Additionally, it can only integrate signals up to a certain frequency due to the limitations of the op-amp and the capacitor used in the circuit. It is also prone to stability issues if not designed correctly.

## How can I design a basic op-amp integrator circuit?

To design a basic op-amp integrator circuit, you will need to choose an appropriate op-amp and capacitor based on the desired input and output signals. You will also need to consider the frequency response and stability of the circuit. There are many online resources and tutorials available that can guide you through the design process.

• Electrical Engineering
Replies
6
Views
3K
• Electrical Engineering
Replies
3
Views
961
• Electrical Engineering
Replies
9
Views
4K
• Electrical Engineering
Replies
20
Views
1K
• Electrical Engineering
Replies
6
Views
2K
• Electrical Engineering
Replies
2
Views
794
• Electrical Engineering
Replies
9
Views
3K
• Electrical Engineering
Replies
6
Views
2K
• Electrical Engineering
Replies
12
Views
3K
• Electrical Engineering
Replies
6
Views
1K