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Lowpass filter with unideal opamp 
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#1
Dec1412, 05:58 PM

P: 59

Hello,
I am designing an opamp to be used as a lowpass 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! 


#2
Dec1412, 07:51 PM

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http://www.ti.com/product/tlc271a?CMP=AFCconv_SF_SEP#doctype1 


#3
Dec1412, 08:01 PM

P: 661

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 gainbandwidth 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
Dec1412, 08:37 PM

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


#5
Dec1512, 01:45 AM

P: 59

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



#6
Dec1512, 03:35 PM

P: 59

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. 


#7
Dec1512, 04:36 PM

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See http://en.wikipedia.org/wiki/Closed...nsfer_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
Dec1512, 04:49 PM

P: 59

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
Dec1512, 05:47 PM

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



#10
Dec1612, 01:30 PM

P: 59

Thank you so much! This helps me a lot.



#11
Dec1612, 05:43 PM

P: 59

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 opamp. 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 cutoff frequency shifted by a decade, also, the magnitude plot goes up then starts going down around 10 KHz. Am I missing something here? 


#12
Dec1612, 08:29 PM

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What expression did you use for G(s)?



#13
Dec1612, 08:35 PM

P: 59

I thought it might be Avo / (s + 100), with Avo being the openloop gain at low frequency. I substituted that and my transfer function for the opamp 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
Dec1612, 08:55 PM

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#15
Dec1612, 09:03 PM

P: 59

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 opamp 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
Dec1612, 09:40 PM

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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 gainbandwidth 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
Dec1612, 09:45 PM

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You are modelling the opamp as an ideal amplifier with a constant gain, followed by a lowpass filter. That's how the opamp performs in real life, approximately.



#18
Dec1612, 09:55 PM

P: 59

Looking at this attachment I see it says the load is 100 kohm..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. 


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