# Integrator again....

Discussion in 'General Electronics Chat' started by forbi, Nov 7, 2014.

1. ### forbi Thread Starter Member

Sep 11, 2012
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0
Hi,

Firstly, How helpful if the datasheet mention the unity gain bandwidth is 3Mhz ?
Will i be able to do a bode plot using this information, it seems that the datasheet for 071 doesn't have a bode plot .....

Secondly, And is there such thing as cut-off frequency in integrator op amp ?

Thirdly, i saw this equation of 0.35/Bandwidth give me the rise time.

Fourthly, on my previous thread http://forum.allaboutcircuits.com/t...tior-for-integrator.103189/page-2#post-779465 . I found out that by adding a resistor parallel to the capacitor all DC offset will be "nulled". And in many reading from searching in google and reading wiki, they suggest for pratical op amp integrator to have a resistor to be parallel with the capacitor.
However, even without the additional resistor connecting parallel to the capactior, a normal op amp integrator circuit should be working normally ?

Lastly, with this couple of information like bode plot , slew rate, RC time constant. How can i piece all this information together to come out a range of frequency so that my output will not have any undesirable or abnormal effect. With a Fixed input resistor of 10Kohm and 1 uF ceramic capacitor.
Example is that with an input square wave of +-5V or at +-1V. my output should be a triangle wave without any DC offset and does not saturate.

*nulling of dc offset in the op amp is already done using a variable resistor. However DC offset occurring again with much higher value is what i mention in the previous thread. http://forum.allaboutcircuits.com/t...tior-for-integrator.103189/page-2#post-779465

Thanks !!!

2. ### crutschow Expert

Mar 14, 2008
16,566
4,475
The gain-bandwidth is a guide as to how high the frequency response of the circuit can be. For a simple non-inverting amplifier, for example, the output -3dB point would be the gain-bandwidth divided by the closed-loop gain.

Since an integrator has a continuing rolloff with frequency it does not have a defined cut-off frequency, other that that determined by the op amp bandwidth.

.35/BW gives you the approximate risetime for a system with a first order rolloff. It does not really apply to an integrator.

Determining what op amp gain-bandwidth is required to generate an integrator output depends upon your definition of "undesirable or abnormal effects". For example if you want a very sharp corner at the top and bottom of a triangle-wave then the required bandwidth is higher than if you can tolerate a little rounding. The easiest way to get a good approximation of what the waveform will look like is to do a Spice simulation. The calculations for that otherwise can be somewhat complex and I'm allergic to math so I let Spice do the calculations.

Adding a resistor across the capacitor will minimize output DC offset but it will limit the low frequency response of the integrator as determined by the RC time-constant (causing bowing of a sawtooth, for example). My simulations show that you don't want to go below about 2 megohm with 10 kohm input and 1 uF feedback.

3. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Woo thanks for the reply.
So having a bode plot . I would be pretty much able to determine the frequency and gain to get a nice sharp triangle wave ? Since my gain is pretty much fix at input reisistor of 10kohm and 1uf feedback capacitor.

If so how could I get a bode plot? Which I don't understand why the data sheet doesn't provide but only the gain bandwidth at unity gain ?

How can I caculate a close loop gain of an integrator ? Cos u mention that to get the -3db point (which I think is the frequency it can go to at the specific gain ? ) will be gain bandwidth divide by closed loop gain.
But for close loop gain of an integrator depends on my input frequency. So I don't really understand how does those 2 relate to each other .

Thanks

4. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Hmm after some reading, when datasheet mention a unity gain bandwidth of 3Mhz. Does it means specifically to the test circuit ?

So when i change the parameters like input resistor of 10Kohm and feedback capacitor of 1uf. the unity gain bandwidth changes to 15.19Hz ?
(1/2*pi*RC). Does this means my op amp integrator will only perform to frequency of 15.19Hz and the gain limited looking at the plotted bode plot for my configuration ?

And i wouldn't have any corner frequency because i didn't place a resistor parallel to the capacitor. Does that means i have infinite gain at low frequency ?
Or how could i determine how the shape of my bode plot will look like since i can't determine the close loop gain starting point at 0Hz ?

5. ### MrAl Distinguished Member

Jun 17, 2014
3,754
791
Hi,

The gain at low frequency is very high because it is the open loop gain. There will be a cutoff frequency because the gain of a real op amp is not infinite, it is limited to values like 10000 to 100000 or about that, and since the integrator is a low pass filter it will have a cutoff frequency anyway. But i am not sure you need to be concerned about the theoretical cutoff frequency here.

Using the open loop gain A then the cutoff frequency is:
w=sqrt(2*A^2-1)/((A+1)*R*C)

for low gains like 100 it is about 1.4/RC, and for infinite gain it is sqrt(2)/RC, so we dont see much difference here.

I think the gain you would want to know would be the response with frequency which is:
Vout/Vin=1/(w*R*C)

This tells you the amplitude of a sine wave at the output for a given frequency so you have some idea what to expect.
Since you are dealing with a square wave input, you might want to think about the time equation which is t/RC. In your case this is t/0.001 or simply 1000*t.
So if you input a square wave front that goes from 0 to 1v it takes about 1ms for the output to reach 1v because 1000*t with t=0.001 equals 1.
Now if that wave then shoots downward (as square waves do) at t=1ms, then the output will ramp up to 1v and then start down.
That is equivalent to a 500Hz input square wave. So at 5khz, the output would only reach about 0.1v before it started downward again.
Does this seem like what you are trying to understand?

6. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Yup I understand that. Because the data sheet is giving me the unity gain bandwidth of the op amp as 3MHz. How do they get this reading from and if is specifically to their test circuit for the input resistor is same value of feedback resistor. But that seems to be the case for inverting op amp.

For integrator.
So I would need to recalculate my unity gain bandwidth with input resistor as 10kohm and feedback capacitor as 1uf ? Which will result in 15.19hz

7. ### MrAl Distinguished Member

Jun 17, 2014
3,754
791
Hi,

If that's all you want then yes the response goes through 1 at w=100 which is f=15.915Hz, not really 15.19Hz.

8. ### crutschow Expert

Mar 14, 2008
16,566
4,475
Unity gain bandwidth refers to operating the amp with a resistive feedback at a non-inverting gain of 1, not where an integrator may have a gain of 1. Don't confuse the two as they are unrelated.

9. ### forbi Thread Starter Member

Sep 11, 2012
37
0
Hmm so if I set my input frequency above 15.91Hz . What kind of result would I expect ? Issit some undesirable output signal or is just that my gain is limited by the frequency if input for example at 100Hz.

10. ### crutschow Expert

Mar 14, 2008
16,566
4,475
An integrator rolls off at -6db/octave with frequency so will roll off at that rate as you increase the frequency above 15.91Hz (where it has a gain of 1) giving a gain of about -16dB down or 0.16 at 100Hz.

11. ### MrAl Distinguished Member

Jun 17, 2014
3,754
791
Hello again,

Lets try this one more time.

If you input a square wave to an integrator and assuming it does not saturate, the output is:
Vout/Vin=t/RC

Your RC time constant is 0.01, so we have:
Vout/Vin=t/0.01=100*t

Now when the square wave goes up to 1v the integrator starts to integrate from 0 up. It ramps up such that when t=0.01 the output is:
Vout=t*100=0.01*100=1.

Now assume the square wave drops down to -1 volt. The output starts from 1 and ramps down toward zero again. After 0.01 seconds it will be back to zero volts at the output.

If it is a square wave then it will jump back up to +1 volt, and so the output will ramp up again. Since the half cycle period is 0.01, it reaches 1v again at t=0.01.
The total period is twice that, or:
Tp=2*0.01

Looking at the above, we see that with a period of 2*R*C we got a change that went from 0 to 1 and back to 0 again.

With a square wave of plus and minus 0.5v, the output would rise up to 1 volt in 0.005 seconds, and back down to zero in another 0.005 seconds, so the output woudl change by 1v and back to 0v in 0.01 seconds. So the period is now:
Tp=R*C

So with 1v peak to peak input square wave we get 1v peak to peak output triangle when Tp=R*C.

If we double the square wave frequency, the output only reaches up to 0.5v in 0.005 seconds, and back down in the next 0.005 seconds, so we end up with 1v peak to peak input and only 0.5v peak to peak output.

So you see where this is leading. The output amplitude will match the input amplitude (peak to peak) when the period is Tp, and will be greater when Tp is longer and will be less when Tp is shorter. So we can call the 'gain' 1 when Tp=R*C. Anything above that will result in an amplitude cut, and anything below that will result in an amplitude 'boost'.

Several points tabulated
1Hz, Tp=1, Vout=100::
10Hz, Tp=0.1, Vout=10
100Hz, Tp=0.01, Vout=1
1kHz, Tp=0.001, Vout=0.1
10kHz, Tp=0.0001, Vout=0.01

You can see this resembles a low pass filter, but of course we cant get 100 volts out so anything below about 10Hz will saturate the op amp and lead to great distortion. Anything too high will lead to a tiny tiny output so that probably wont do us much good either.