# Basic Op-amp Integrator circuit question

1. Jul 8, 2011

### 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.

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2. Jul 8, 2011

### Staff: Mentor

Looks like you need a large value bleeder resistor around the cap...

3. Jul 8, 2011

### Averagesupernova

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

4. Jul 8, 2011

### Jdo300

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

5. Jul 8, 2011

### vk6kro

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.

6. Jul 9, 2011

### Staff: Mentor

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

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....

7. Jul 9, 2011

### Antiphon

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.

8. Jul 9, 2011

### Dickfore

Hint:

What is the value of the integral:

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

9. Jul 9, 2011

### vk6kro

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.

10. Jul 10, 2011

### Jdo300

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: Jul 10, 2011
11. Jul 10, 2011

### Carl Pugh

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.

12. Jul 11, 2011

### Averagesupernova

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.

13. Jul 11, 2011

### Staff: Mentor

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

14. Jul 12, 2011

### Jdo300

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