# Low-pass filter with unideal op-amp

1. Dec 14, 2012

### xortan

Hello,

I am designing an op-amp to be used as a low-pass filter and am using a TLC271. I have attached an image of the topology of the circuit. I found the transfer function to be G(s) = Y(s)/X(s) = 1/(10^-4s + 1).

I plotted the bode plot in MATLAB and have attached the result in the second figure. However when looking at the datasheet there is a graph that shows how the phase and gain will change with frequency. (Also included this image)

How can I determine the frequency dependent gain and phase shifting of TLC271 on the frequency response of the circuit?

Thanks!

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2. Dec 14, 2012

### Staff: Mentor

Do you mean that when built around the ordinary ideal inverting op-amp, that expression would give the gain?
An op-amp with a fixed phase shift of 90 degrees over most of its useful range. Wow! How is that going to enliven amplifier design? http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]
TI do make available the Pspice model, so maybe simulate your filter and see what materialises?
http://www.ti.com/product/tlc271a?CMP=AFC-conv_SF_SEP#doctype1

Last edited by a moderator: May 6, 2017
3. Dec 14, 2012

### Enthalpy

Every op amp exhibits 90° phase lag over its useful frequency range. It's the effect of the compensating capacitor.

To compute the effect of the op amp's frequency response, model its finite gain-bandwidth product as if it were an integrator, that's good enough in this frequency range (in fact, it's good in normal use of an op amp).

So the op amp's input (differential) voltage is τ*s times the output, where τ*2pi*GBW=1.
Recalculate the stage's attenuation function using this finite gain, simplify the result a bit, and usually τ just adds to the RC time constant.

4. Dec 14, 2012

### Staff: Mentor

So it does! (Memo to self: be more observant of datasheets!)

With the particular opamp mentioned, OP will be able to set the corner frequency by choice of the bias level.

5. Dec 15, 2012

### xortan

Thank you for the input! Yes, I mean that when I model the op-amp around ideal characteristics that is the expression that I will have. I will play around with it some more :)

Last edited: Dec 15, 2012
6. Dec 15, 2012

### xortan

So after working on this thing a bit longer I'm getting confused. Looking at the datasheet I see the GBW is 1.7 Mhz. So am I supposed to be recalculating τ or use the τ from my transfer function and calculate the GBW? Also when you say model it as an integrator would I just multiply the current transfer function by GBW / s?

Edit: I calculated the tau to be 93 ns, I added this onto the RC time constant and now my bode plot starts to roll off at approximately 10 KHz, does this seem correct? I'm still confused by what finite gain you are speaking of however.

Last edited: Dec 15, 2012
7. Dec 15, 2012

### AlephZero

See http://en.wikipedia.org/wiki/Closed-loop_transfer_function

Using the notation of that link, H(s) is the transfer function of the passive components in your circuit. G(s) is the gain of the op amp. For a given frequency, You can read the amplitude and phase of G from the plot in the data sheet.

If |G(s)| is several orders of magnitude larger than |H(s)|, the difference between
G(s) / (1 + G(s)H(s)) and 1/H(s) is small. Your "ideal op amp" analysis assumed the G was "infinite" and so the overall transfer function was 1/H(s).

8. Dec 15, 2012

### xortan

So if I use the equation in the link and read the value of G from the datasheet plot this should give me the phase/amplitude I need for a given frequency?

9. Dec 15, 2012

### AlephZero

Yes. In real life, transfer functions (like G here) are often measured data, not defined by a formula.

10. Dec 16, 2012

### xortan

Thank you so much! This helps me a lot.

11. Dec 16, 2012

### xortan

Hello,

So I did as you suggested, the phase looks completely off and the magnitude is either giving me wrong numbers or the same numbers back as the ideal op-amp. I even tried coming up with the equation for the G(s) and substituted into the above formula, simplified and plotted in matlab. It looks like my cut-off frequency shifted by a decade, also, the magnitude plot goes up then starts going down around 10 KHz. Am I missing something here?

12. Dec 16, 2012

### Staff: Mentor

What expression did you use for G(s)?

13. Dec 16, 2012

### xortan

I thought it might be Avo / (s + 100), with Avo being the open-loop gain at low frequency. I substituted that and my transfer function for the op-amp into that feedback equation, and tried simplifying and got a plot that could make sense I think. Tried doing it this way because reading the magnitudes and phases from the plots gave me numbers that weren't making sense to me.

14. Dec 16, 2012

### Staff: Mentor

When s=0 that gives the low-frequency (near DC) gain as Avo/100, and a corresponding corner frequency (-3dB) of ω=100 rad/sec. How do these compare with the open loop gain of that op-amp in the datasheet?

15. Dec 16, 2012

### xortan

The one in the datasheet are in Hertz so I'm off by a factor of 2 pi. It looks like the gain is 10^4.5 at DC. So should my equation become Avo / ( s / 628 +1)? The corner frequency on the datasheet is at 100 Hz.

Once I get the equation for that graph do I sub that and my transfer function for the op-amp into that feedback equation, simplify and plot? Or can I just pull the info I need off these graphs and I'm just doing it wrong?

16. Dec 16, 2012

### Staff: Mentor

You read the DC gain as 10⁴⋅⁵? But you can't linearly interpolate off a log plot. I read it as around 2x10⁵. At 100Hz the graph shows gain has dropped to 6x10³. That's a lot more than a 3dB drop! You should be looking for a gain of 0.7x2x10⁵.

On the graph you reference in your initial post, I'd estimate the -3dB corner frequency to be around 2Hz.

But are you sure that's the plot you should be using? You have read the full datasheet, including how you select its gain-bandwidth using one of three BIAS MODES. Retrieve the datasheet using the link I gave in my first reply.

Indeed, have you indicated the frequency range over you wish to use the filter? How precise does the response need to be, in comparison with the ideal?

17. Dec 16, 2012

### Staff: Mentor

You are modelling the op-amp as an ideal amplifier with a constant gain, followed by a low-pass filter. That's how the op-amp performs in real life, approximately.

18. Dec 16, 2012

### xortan

Looking at this attachment I see it says the load is 100 k-ohm..I have been looking at the proper one when doing the calculations tho they are the same. I see how you got the DC gain but I don't get how you get it at 100 Hz, it looks like they are at the same height..

The frequency range is from 10 Hz to 1 Mhz and the response does not need to be that precise, I am interested in how the frequency response given in the datasheet will affect the frequency response of my filter.

19. Dec 16, 2012

### Staff: Mentor

At 100Hz the gain is < 10⁴, placing it more than 20dB down.

It seems (based on my brief reading) that the BIAS MODE is largely directed towards reducing the differential offset towards the ideal. This hardly seems important to most filter apps, so I think you'd be choosing the bias that maximizes gain around 1MHz. Low bias will stretch GB to 1.7MHz

20. Dec 17, 2012

### Staff: Mentor

Here's some semilog graph paper showing how the logarithmic cycle is divided into 10 equal steps. http://www.science-projects.com/SemiLog0.GIF [Broken]

Revising my estimates: gain is about 3x10⁵ at 1Hz, and 7x10⁴ at 100Hz.

The GB product changes to some extent with supply voltage VDD and BIAS. The graph you provide is for VDD =5V. So that's the supply voltage you will be using? (See Fig 27 of the datasheet I referred to.)

Last edited by a moderator: May 6, 2017