opamp integrator losing signal

[BkraM]

Hi All,

I've build a circuit in which a 20~40 us signal is fed to an opamp integrator.
The output of the integrator is used as a measure of the length of the 20~40 us signal.
After 200 us the integrator is reset and the process starts again.

Unfortunately the capacitor in the opamp circuit (ceramic 100 nF) loses its charge to fast for the circuit to produce decent results (significant decay in 200us).
Using a larger capacitor with a smaller resistor (electrolytic 1 uF) did not help a lot.

How can I make the capacitor hold it's charge longer? Or would i be better of to try a circuit without an integrator?

Thanks,

edit : error in header, dignal = signal

 

[JimB]

edit : error in header, dignal = signal

Fixed that for you.
JimB

 

[alec_t]

Can you post the schematic of your integrator?

 

[MikeMl]

Get an opamp with tiny input bias current... They are available...

 

[steveB]

Without more details it is hard to be sure of the answer. Capacitor Leakage, and bandwidth limits on the opamp are possible concerns. As mentioned input bias is possible.

 

[KeepItSimpleStupid]

Polystyrene caps are usually used combined with a low Ib OP AMP. See: https://www.mouser.com/ProductDetail/Xicon/23PW310

 

[crutschow]

How are you resetting the integrator?
Post your circuit diagram.

 

[BkraM]

I think my problem might be a bit more to the basics than i thought.

I'm using the attached circuit, which apparently doesn't do what i expected it to do in the simulation.
Resetting it with switch B, doesn't seem to work.



**broken link removed**

In my tested circuit, on a breadboard this reset issue could be (part of) the problem. In the circuit i did not add anything for a automatic reset as the decay of the cap was to high for the circuit to work anyway.

I found that a 100 nF cap(100 uF in the circuit), without any circuitry loses its 5V load within a second, this would interfere with the correct working of the Integrator i presume.

Thanks,

 

[MikeMl]

LM324 is anything but a low input bias current opamp; it is one of the worst ones you could select.
Why is R1 (1meg) there?
What is the input impedance of the stupid green box in the upper right hand corner? (Not the fake one; the real one?)

 

[BkraM]

I put the 1meg there as a test, wasnt Sure if i needed a pull down resistor.

The real green box is a oscilloscope, not sure what the input resistance is, but should be high

 

[jjw]

Are you just simulating this or do you have a real circuit?
If real, what are the switches, how are they controlled, the supply voltage and the control signal levels for the switches?
Show your whole circuit diagram.

 

[crutschow]

You need to use a ceramic or film cap and a low input bias current op amp.

 

[KeepItSimpleStupid]

A typical scope has an input Z of ~1 meg ohm shunted by about 20 pf which isn't high in this circuit (Channel A). If your using a probe, the input Z jumps to about 10 M which still isn't high in this circuit for channel A.

 

[BkraM]

I begin to understand that the charge I'm putting in the resistor is very low (+/- 500nJ)
My multimeter, oscilloscope and opamp all dissipate some, adding up to a significant amount compared to this small value.

Removing my multimeter from the circuit, keeping only the scope already helped to get things more acceptable.
What would be a better opamp to use for this purpose?

Thanks for the help guys

 

[BkraM]

double post

 

[crutschow]

Pick an op amp with a CMOS or JFET input.

 

[tomizett]

TL071 (or 072, 074) would be a good start. If you're still using an electrolytic then try to replace it with a polyester or (better) polypropylene film type.
As mentioned in #13, also bear in mind that this circuit won't work properly woth your scope in the inverting input of the opamp, due to the scope's input impedance (this node should be at ground anyway).

[edit for bad typing]

 

[Tony Stewart]

Integrators follow the basic time constant rules for Capacitors Ic=C dv/dt, T=RC, shown below re-arranged

(1) dv/dt= Ic/C or 5 V/1 s = Ic / 100 nF ... thus Ic = 500 nA = 0.5μA
or V/I=R = 5V/0.5μA = 10MΩ ( e.g. 10:1 scope probe)
(2) T = RC where T= decay of 60% = 1 second for your example from 5V on 100 nF cap
so R=T/C = 1 s /0.1 μF = 10MΩ

Ceramic and Electrolytic Caps have a self-leakage time constant on the order of magnitude of 1 second +/- 1 order of magnitude depending on vendor, and model

Plastic Film Caps have a self-leakage time constant on the order of 100 seconds to 1000 seconds and have lower density so tend to be much bigger and more expensive. Rank of quality and cost in descending order; Teflon, Polystyrene, Polyester,.

Anywhere a very high resistance is required, equally low input bias current Op Amps using FET inputs are needed. These requirements can be easily computed from basic formula given above.

To make an Integrate and Dump ADC converter, or Sample and Hold circuit , the prudent designer will also avoid using ceramic due to other non-linear behaviors like memory and voltage sensitivity. Plastic film is always the good choice.

 

[MrAl]

Hi,

There are also ready made sample and hold circuits in an IC package. With the right (low droop) capacitor they work pretty well. Check them out too, part numbs LF198/298/398 (see edit note).

{EDIT: As per Tony Stewart's note "LM198/298/398" should have read "LF198/298/398" but it is fixed now, thanks to Tony for that.}

 

[Tony Stewart]

LM398 error should read LF398 for a FET input S&H.

Regardless, the point is awareness of all sources of capacitor leakage ( internal R, external Bias, and scope probe)
and learning the reasons for better choices for low sag. on storage cap..