Low-pass filter with unideal op-amp

In summary, the TLC271 op-amp has a fixed 90-degree phase shift over its useful frequency range, which can be determined by the bias level. To determine the frequency-dependent gain and phase shifting of the TLC271, a Pspice model can be used.
  • #1
xortan
78
1
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
xortan said:
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).
Do you mean that when built around the ordinary ideal inverting op-amp, that expression would give the gain?
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)
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]
How can I determine the frequency dependent gain and phase shifting of TLC271 on the frequency response of the circuit?
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
 
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  • #3
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
Enthalpy said:
Every op amp exhibits 90° phase lag over its useful frequency range. It's the effect of the compensating capacitor.
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
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 :)
 
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  • #6
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.
 
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  • #7
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
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
Yes. In real life, transfer functions (like G here) are often measured data, not defined by a formula.
 
  • #10
Thank you so much! This helps me a lot.
 
  • #11
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
What expression did you use for G(s)?
 
  • #13
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
xortan said:
I thought it might be Avo / (s + 100), with Avo being the open-loop gain at low frequency.
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
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
xortan said:
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?

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
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
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
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
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.)
 
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  • #21
Yeah that is the supply voltage I am using, I understand how to read log scale, I was looking at a different graph then the one I post (sorry). Even if I change the bias mode I still have no idea how to look at the graph provided in the datasheet and modify the ideal response I plotted in matlab. That's all I want to know.
 
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  • #22
Excuse me in making a comment that is not related to the question here.

As I practice design engineer, this is not a LPF I would design.
1) If I want a 1st order LPF with gain of 1 only, I use a simple RC, not with opamp.
2) You are having pole frequency of something like 10KHz from the graph, at frequency where the opamp unity GBWP of below 1MHz, you are only -30dB down. You are going to see the effect of the opamp.
3)If you really want to do this, I question the opamp choice. Look at the bode plot, the second pole of the opamp kicks in before unity gain, you have extra phase shift that likely cause a peak close to cross over. The situation might be made worst because you are working in the gain of << 1 because you only start at gain of 1! At least find an opamp that has higher GBWP that absolutely have a single pole cross over.

I know this is a practice circuit, but you still want to think about real world design. Look at a simple second order active filter:

http://en.wikipedia.org/wiki/Sallen%E2%80%93Key_topology

In the diagram, C2 is at the input of the opamp, this will roll off the high frequency before even going into the opamp. This will mitigates the peaking effect for the opamp and produce better result.

I am not saying this won't work, I did not do simulation, but there is potential problem just by looking at the circuit.
 
  • #23
xortan said:
Even if I change the bias mode I still have no idea how to look at the graph provided in the datasheet and modify the ideal response I plotted in matlab. That's all I want to know.
Are you giving MATLAB the circuit, or the transfer function? (I don't know matlab.)

If you want to simulate the circuit, in place of the ideal differential amplifier use that ideal differential amplifier followed by a low-pass filter (R and C) followed by an ideal non-inverting buffer amplifier (perhaps having an output impedance). Then around that as your amplifier, you arrange your filter's passive components.
 
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  • #24
NascentOxygen said:
Are you giving MATLAB the circuit, or the transfer function? (I don't know matlab.)

If you want to simulate the circuit, in place of the ideal differential amplifier use that ideal differential amplifier followed by a low-pass filter (R and C) followed by an ideal non-inverting buffer amplifier (perhaps having an output impedance). Then around that as your amplifier, you arrange your filter's passive components.

Matlab deals with transfer functions. It has no schematic capture.

A better tool for exploring circuits is LTspice, and it's free! http://www.linear.com/designtools/software/#LTspice
 

1. What is a low-pass filter with an unideal op-amp?

A low-pass filter with an unideal op-amp is a circuit that uses an operational amplifier (op-amp) to attenuate or filter out high-frequency signals from a given input signal. However, due to the limitations of the op-amp's performance, the output signal may not perfectly match the desired output.

2. How does an unideal op-amp affect the performance of a low-pass filter?

An unideal op-amp can affect the performance of a low-pass filter in several ways. It can introduce noise, distortion, and offset in the output signal, which can result in a deviation from the desired output. It can also have limitations in its frequency response and gain, which can affect the filter's cutoff frequency and overall attenuation.

3. What are the common causes of unideal behavior in op-amps?

The common causes of unideal behavior in op-amps include non-linearities in the input-output relationship, finite gain and bandwidth, input offset voltage, input bias current, and input impedance. These factors can lead to deviations from the ideal op-amp model, resulting in less than perfect performance in low-pass filter applications.

4. How can we mitigate the effects of unideal op-amps in low-pass filters?

One way to mitigate the effects of unideal op-amps in low-pass filters is by using design techniques such as feedback and compensation. Feedback can help reduce distortion and noise, while compensation can improve the op-amp's frequency response and gain. Other techniques include using op-amps with better specifications, selecting appropriate component values, and implementing higher-order filters.

5. What are the limitations of using unideal op-amps in low-pass filters?

The limitations of using unideal op-amps in low-pass filters include a reduction in signal-to-noise ratio, distortion of the output signal, and a decrease in the filter's overall performance. Additionally, as the complexity of the filter increases, the effects of unideal op-amps become more prominent, making it challenging to achieve the desired output. Therefore, it is crucial to carefully consider the op-amp's specifications and design techniques to mitigate these limitations.

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