OpAmp Integrator Circuit

[drkidd22]

Hello,

I'm working on an op Amp integrator circuit. I know the basic stuff of how op amps work but never worked with integrator and was hopping that someone here could give me a hand.

The input signal to the opamp is shown in image attached. In this circuit will my gain be the same as if C14 wasn't there? -R8/R18?

The pulse is about 3ms long as shown in the image. Any help analysis this circuit will be appreciated. I've tried to simulate it with LTSpice but had no luck.

Circuit


Input Wave


Output Wave

 

[crutschow]

Post your LTspice .asc file and I'll try to simulate it.

It's not clear what your question is.

C14 is what makes the circuit an integrator. Remove that and the circuit is just an amplifier.

R18 is likely to prevent the integrator from saturating with no input, due to op amp offsets.

 

[ronv]

It looks right. The impedance of the cap is quite low for a 3ms pulse and is in parallel with the 1 meg.

 

[drkidd22]

So I've attached the files. Not getting what I was expecting at the output. So I'm a little confused as of how this circuit should be working. The first post it what I'm getting on the circuit I got. Really just trying to understand how it works. I've done the calculations and stuff, but not sure if I did it right.

 

[MrAl]

Hello,

I'm working on an op Amp integrator circuit. I know the basic stuff of how op amps work but never worked with integrator and was hopping that someone here could give me a hand.

The input signal to the opamp is shown in image attached. In this circuit will my gain be the same as if C14 wasn't there? -R8/R18?

The pulse is about 3ms long as shown in the image. Any help analysis this circuit will be appreciated. I've tried to simulate it with LTSpice but had no luck.

Circuit
View attachment 65997

Input Wave
View attachment 65998

Output Wave
View attachment 65999

Hi,


When you talk about gain for this circuit it has to be frequency dependent. The DC gain is in fact R8/R18 but usually you wont be using that fact for anything if you are using this circuit as a real integrator which it appears you are.

The gain is closer to 1/sRC which shows the frequency dependency by the lower case 's' which is complex frequency.

The resistor R8 is usually included in an integrator to reduce the effects of the op amps input offset voltage which is a by product of the design and is present in every op amp. We dont want it there, but it is hard to design an op amp with no input offset.
Because of this input offset, when the input voltage to the circuit is zero there still exists a small current equal to the input offset voltage divided by R18:
I=Vio/R18

and this current develops a voltage at the output:
Vo=I*R8

but this voltage is less than if we didnt use any resistor at all for R8 because then it would be infinite which would mean the output would ramp up (or down) continuously even with no signal applied to the input and that would saturate the op amp and prevent it from working properly in the application.

So lets see what we end up with here...

With an input offset of 0.001 volts and an input resistor of 8k (rounding) we have an input current of:
I=0.001/8000=0.000000125
and multiplying that times the feedback resistor of 1M we get:
Vo=I*1000000=0.000000125*1000000=0.125 volts

so this could have as much as 0.125 volts output with no input signal.

You'll want to check the input offset of the actual op amp being used however to see what it is (max) and do the math. It may come out less depending on the actual part number of the op amp.

So you see the 1M resistor limits the output with no signal on the input. If it was not there it would saturate and would not work.

There is a catch here though with this actual circuit. That is that the circuit also includes a 'discharge' resistor 47 ohms. That resistor is used to discharge the cap before the start of another integration cycle. If the integration cycle starts soon enough after that 47 ohm resistor is switched off and the output is sampled soon enough after the integration cycle is over and the integration cycle is short, you may not need the 1M ohm resistor. That's because the integration with zero input requires time to charge anyway which could take much longer than the actual integration period.
The ramp period is about 0.6 volts per second, so it would reach 0.125 volts in about 0.2 seconds (without the 1M resistor, with it it would ramp up in about 1 second).

 

[crutschow]

I found several problems with your simulation circuit--

First you used two different port types for the +/- 15V connections which LTspice apparently interprets as two different ports so the op amp wasn't connected to the voltages. To make them the same, delete the power ports on the op amp, then copy them from the source and move them to the op amp.

Second, R1 is shown as a 1m (1 milliohm) resistor, not 1meg (1 megohm) as it should be.

Third, the LT1037 op amp is oscillating at a low level since it is not stable for an AC gain less than 5 (read the data sheet). If you try a LT1007, which is unity gain stable, it works fine.