# analogue integration, using opamps

Discussion in 'General Electronics Chat' started by sponge_bob, Dec 14, 2002.

1. ### sponge_bobNew Member

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as i am still writing my extended essay about analogue integration vs. digital, another problem has emerged.
My collected data is really not, what it should be. An amplified sin-curve should look kinda streched in the y -axis, shouldn't be? (i.e. y=a sin(x), where a is the amlification.) Well, what i got, is really y=sin(x) + a ??? :? . is that normal? HELLLP!!! The curve is just shifted up/down on the x-axis, that doesnt seem right to me at all.
Well, I hope, someone will be able to help me out of this mess, thanx a lot guys!

Jon

2. ### kinjalgpActive Member

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Seems you are trying to say The curve is just shifted up/down on the "Y" axis

I think its a DC offset problem. Can you explain me your experiment setup?
:arrow: I mean to ask, are you simulating the circuit or doing it in the laboratory on breadboard or something?
:arrow: From where are you feeding the signal? If it is from signal generator check for DC offset of it. If it is ON turn the pot to zero volts.
:arrow: Another thing is are you using dual rail supply i.e. +-15V or so for the op-amp or just 0-15V? If it is the later case the wave will be shifted up to around 7.5V.
:arrow: And finally if possible I would like to see your circuit.

3. ### sponge_bobNew Member

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uhhhm, ya, i meant on the x-axis,

it is not simulated, i built it on a "plaque montage rapide", dunno what the english equivalent is.... a board, where u can put electronic components on and take 'em off without soldering. - i calibrated everything using an oscilloscope and recorded it then, using a data logger, so i have it all on the computer for further data processing.

i am using +-18V (supply voltage for that opam) and a common earth. it is shifted about 14-15V. (on y axis)

http://de.photos.yahoo.com/bc/nonic...bc/nonicknameleft/lst?&.dir=/&.src=ph&.view=t
sorry, for the bad quality, i am in a little hurry, hope the diagram helps. The two circles there represent PD- metres, for input and output signal.

oh, one more thing, i am using a signal generator, but it doesnt have any DC offset regulation....

alright, that is all i think. i am just totally confused, as it worked b4, unfortunately, i didnt collect all the data needed at that point...

thanx a lot

Jon

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5. ### kinjalgpActive Member

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Hi, I saw your circuit but it is not very clear because you have not marked the ground point. You may try the circuit given below. The only differenec between your circuit and this is that the Op-amp is in inverting mode and in the output side there is a capacitor in series to remove any DC content from the signal.

6. ### sponge_bobNew Member

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hmm, okay, thanx for that circuit, my lab time is run out anyway, so i have to come up with a quiet good evaluation...

concerning the circuit:
i had a common earth, like on the diagram u posted, the only difference would be the second capacitor. would that one have done the deal, to change that problem? 'cause all the diagrams i found in my research never had that capacitor.

well,... i probably just messed up building the whole thing, or blew the opamp sometime . If u have any idea, how to get out of that mess, i'd appreciate it a lot.

Jon

7. ### kinjalgpActive Member

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Capacitor has a property of blocking DC signals and passing only AC signals. This is made use of over here to remove any DC content from the output of the integrator so that only pure AC is present at the output. Using this capacitor your level shifting problem will be solved.

8. ### sponge_bobNew Member

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sorry, to bother again....
so, ya i got the whole capacitor, seems logical,
but even though i didnt have it in my circuit before, i should have had and amplified output signal of -1/RC, right?
Now, all i had was a non-amplified output signal, shifted..... and before, when i didnt get that shifted curve, the signal was not amplified as well (i used various capacitor - resitor arrangements, all givin the same :? .

Or is that problem resolved by the capacitor as well?

Jon

9. ### pebeMember

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Your original circuit wouldn't work as an integrator because you were applying feedback from the output to the non-inverting input. That is regenerative and would have given you a good square wave oscillator.

Use kinjalgp's circuit but leave off the output cap and resistor - that's a differentiator circuit and will negate the effect you want. You need to be able to see the varying DC level.

To provide offset, connect the op-amp's non-inv input to a potential divider across the + and - rails.

10. ### sponge_bobNew Member

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@ pebe:
hmm, the results i had from that arrangement are not very encouraging either. There is no shift on the y-axis, however, no amplification is observable. Am I stupid?? The amplitude of the output should be higher than the one of the input, right?
Anyway, for my purposes it should be enough,
as it doesnt really matter for that comparison, as long i got two curves and some numbers to work with.
Just not very scientific, huh....? I'll have a major job writing an evaluation then - or talk to my physics teacher to get some labtime b4 xmas.

11. ### kinjalgpActive Member

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Agreed Pebe that its an diffrrentiator but look at the R and C values. The differential RC time constant is larger than Integral Time constant and hence differentiator will just act as DC blocker and nothing else. If you think 0.1uF and 10k make small difference with 0.1uF and 1k you may use larger value for output capacitor like 100uF or so.

12. ### mechieNew Member

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Integrator ?

There is no gain on the integrator, this should be provided by a second stage.

I agree with Pebe that your original circuit is not an integrator - I see an unstable differentiator here. Also with Kinjal's circuit the output RC bit will remove DC but it is an differentiator ! swap the cap for a 10uf or 100uf or use the op-amp output directly (that would be more honest).

For kinjal's integrator the input R and feedback C are giving you the integration, the op-amp supplies the drive (allowing you to ramp as far as a power rail). No gain per-se.

Any apparent gain will be an illusion, a result of the integrator time being small compared to the sine period, as the sine frequency increases so the intergrator will appear to lose gain.

The original DC offset problem could be caused by the (differentiator) cap having a DC charge which can't discharge as the sine signal has a net DC content of zero. It thus simply adds the offset to the input ... there is still something weird with the circuit though - no gain control, square waves out ??? or uV signal in ???
.

13. ### kinjalgpActive Member

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I have simulated the circuit which gives a DC level shifht of arond -7V which was removed with the capacitor without affecting the shape/phase of the integrated wave. Have a look at the simulated output.

The Blue color waves are after capacitor and purple are before capacitor. See the DC shift.

14. ### sponge_bobNew Member

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okay, now, to summarize:

yeah, i have noticed, i messed up the circuit, i actually had it the other way around (the feedback on the inverting input), i'll try to include the whole removal of DC from my output and get a common ground.

however,.....
should the output signal be amplified?? So far, from what i know it pretty much should be, right? So, this suggests i really should do the wholething again,......

- Jon -

15. ### pebeMember

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Hi all,

It might help to consider the circuit operation under differing input conditions.

Referring to kinjalgp's circuit, I said to put a potential divider to the + i/p. That implies a fixed voltage. What I should have said was connect a pot between the two supply rails and feed its slider to the + i/p. This was to cancel out any offset between the two inputs, because in an ideal op-amp the o/p = 1/2 supply rail when the two inputs are equal - but they almost never are.

In the following, it is assumed that the above status is the starting point and the cap is discharged. By follower action the o/p will then be the same as set by the pot at the + i/p. I'll call the non inverting input +in, and the inverting -in.

1. Steady DC input signal. I have seen this circuit used as a linear ramp generator. If a +ve input voltage is applied, current flows through R into the C at -in. As it tries to lift -in the o/p give go -ve to counteract it. It will continue to fall until limited by the -ve supply rail as mechie said and the amp will no longer act as an amp. The fall time will be directly proportional to the RC time constant.

2. Square wave. If a series of +ve pulses is applied to the input (eg. a square wave), then the circuit will integrate those pulses. The o/p will start to fall while the pulse is high, then remain constant while the pulse is at zero, then fall further at the next pulse. The result is the familiar staircase with flat steps and sloping risers. Again, the o/p will be limited by the supply rail, as before.

3. Sine wave. Here the o/p will be a sine wave as shown by kinjalgp, because the result of integrating a sine wave, is a sine wave (ie. it will give the same reponse as a passive RC circuit, but inverted). The amplitude of the o/p depends on the time constant RC as mechie says, but you can have gain if the time constant is very short (think of the current going through a 1K resistor into a 100pF cap at 50Hz)!

Note that under all conditions the o/p will drift with time because there is no DC feedback from o/p to -in. So if tests are undertaken then perhaps the C should be shorted with a switch until you are ready to carry out the test.

16. ### sponge_bobNew Member

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okay, from all this I conclude,
i should have gotten some sort of amplification,so.... i better get my butt out in the lab and do the whole thing again, paying close attention to everything :? .

merry christmas!

I'll tell ya guys if it worked then or not...

thanks a lot

Jon

17. ### kinjalgpActive Member

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Inetgration of Sin A = -Cos A. And Cos A = Sin (90-A). Therefore if a sine wave is applied, the output will be a CoSine Wave with 90 degree phase shift and inverted. But the op-amp is already in inverting mode and so the output will only be +Cos A. :wink:

18. ### pebeMember

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I stand corrected! I tend to think of all sinusoidal waves as sine waves.

19. ### sponge_bobNew Member

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okay, here's the whole solution:

the y- axis shift was due to offset, that was fine, however, why was ther no amplification, even though it is

-1/(RC)integralof V dt

well, when considering sin(x), it is important to include the frequency f!!!
therefor:

sin(2pi*f*t)

just integrating this term will give:

-1/(2pi*f)cos(2pi*f*t) + c

considering f as a constant in the integration
when now the amplification of -1/(RC) is applied, we get A=1/(RC2pif)
which is at the frequency corresponding to the RC - value approximately one. It gave me a 95%-99% yield, considering high school lab equipment not too bad
so, the output curve should really look something like: (1/(RC2pif))cos(2pift)
which fits my results, except that offset, which can b negleted comparing phase relations and amplitudes
so, that was just really to finish off. I gonna post, my finished essay sometime, as soon i finish it and sent it off.
thanx a lot for your help guys, might need it again, as soon i stumble over the next thing -

Jon