Effect of OP-AMP bandwidth!!!

[samiran_dam]

Hi all,

What is effect of finite op-amp bandwidth on low-pass filter's cutoff frequency using Sallen-Key architecture? I mean what happens if op-amp's b.w. < cut-off frequency?

Thanks&Regards
Samiran

 

[keith1200rs]

The opamp bandwidth needs to be considerably higher than the cut-off frequency otherwise you will not get the characteristic you designed.

Texas Instruments free filter design software shows the recommended opamp bandwidth for each stage of a filter:

**broken link removed**

Keith.

 

[FvM]

Strictly spoken, additional poles are introduced to the filter characteristic by the OP bandwidth limitation in any active filter. The
question is, if they are neglectible. You may either follow rules of thumb about required surplus bandwitdh or calculate the exact
transfer function. Some filter tools, e.g. Nuhertz Filter Solution have an option to consider OP bandwidth in calculation.

 

[LvW]

One additional remark:
There are many different filter topologies besides S&K structures. And you should be aware that some are less and some other are more sensible to the limited opamp bandwidth. In this respect, S&K topologies have pretty good properties, but this topology is rather sensible to passive parameter tolerances. Thus, filter design is always a compromize between conflicting requirements.

 

[keith1200rs]

The Filter Pro software shows the required bandwidth of the opamps and you will soon see how the bandwidth requirements of Sallen Key become crazy with high order filters.

Nuhertz software is good but you need to be careful with the opamp modelling. They assume a simple, single dominant pole characteristic but most real opamps aren't like that. You need to take the Nuhertz design and re-simulate in Spice with manufacturers' models. You can also simulate with ideal opamps and then correct small deviations from ideal with the real opamps by tweaking the component values.

Keith

 

[LvW]

May I add one point?
As samiran_dam would like to know something about the "effect" of limited opamp bandwidth:
Normally (in most cases) the pole frequency is lower if compared with ideal opams and the pole Q is larger! (Example: Butterworth turns over into Chebyshev style and 1-dB-Chebyshev perhaps to 2 dB ripple or something else).

 

[samiran_dam]

@LvW:

That is you mean to say - if I simulate a 2nd order S&K architecture with an ideal op-amp behavioral model and set other passive component values according to that to obtain the intended cut-off frequency, it will definitely fail to achieve the same while simulated using a real op-amp having finite bandwidth. Is the hand-analysis only way to obtain the effect of op-amp's non-ideality like finite b.w.on filter characteristics, I mean by simulations will it not be cumbersome to track?

@all:

And, one more question - what I know is that low-pass filter system Q = cut-off frequency (undamped natural frequency or magnitude of the complex conjugate pole-pair) "divided by" [absolute sum of the negative real parts of complex conjugate pole-pair]. How is it related to individual pole-Q?

Thanks & Regards
Samiran

 

[FvM]

I don't see that the said "system Q" is a general accepted or commonly known quantity. I also wonder about it's purpose.
Where did you get the definition?

 

[samiran_dam]

The defn. is available in any text-book on analog filter. I got this from the book of Continuous Time Filter Design by Deliyannis T. Anyhow, for second-order low-pass system, the standard form of the denominator of transfer function is:

s² + (ω/Q)s + ω² = 0;

where, ω = natural frequency of the second-order system,
Q = quality-factor of the second-order system.

Solving this eqn., we get-

s1, s2 = -(ω/2Q) ± 0.5*√[(ω/Q)² - 4ω²]

from the above solution -

Q = ω/(|real(s1)|+|real(s2)|);

So, from this relation we can conclude to that definition. Am I making any mistake?

Actually my question was how dos the pole get calculated by the SPICE simulator and does it have any relationship with the Q defined in the standard second-order system's denominator?

Thanks & Regards
Samiran

Added after 1 minutes:

sorry a typo in the prev. reply:

It wud be:

Actually my question was how dose Q of a pole get calculated by the SPICE simulator and does it have any relationship with the Q defined in the standard second-order system's denominator?

 

[LvW]

samiran_dam,

For low pass functions there is only one definition for a quality factor - and that is the "quality" of a single pole pair (Qp). That means, the parameter Qp is applied only for a second order function and it is defined as

Qp=wp/2σ (with wp: pole frequency and σ: real part of the complex pole).

 

[samiran_dam]

Yes, I got it. Thank you all for the co-operation!

 

[LvW]

Hi samiran_dam,

Accidently, I found something in my documents (notices), which completes my answer from April 18th:
If you like you can define for a 2nd order low pass a "pole frequency wp" as well as a "quality factor Qp" also for the case that both poles sn1 and sn2 are real,
which means
sn1=σn+wn=rn1 and sn2=σn-wn=rn2

Then:
* wp=sqrt(rn1*rn2)
* Qp=0.5*cosφ
with φ=angle between the neg. real axis and the tangent between the origin and the circle which crosses rn1 and rn2 at the neg. real axis. Of course, the center of this circle is σn (σn is a negative number).
However, I know that this is a rare case and does not happen very often.