[MOVED] Third order active butter worth filter acting strange

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BartlebyScrivener

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I have tried to design a third order active filter with a cut off freq of 20kHz. Using 10 kOhm resistors, I make that the capacitors should be 795pF using 2*pi*f = 1/CR

I made a folded cascode amplifier, with an output buffer stage



then added a first order filter to the output



then added the first order to the input, and the cap to the output.



But I seem to get a first order response when I do an ac sweep? Also, the -3dB point is around 1kHz instead of 20kHz.



What am I doing wrong?! Thanks.
 

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Your schematics are almost impossible to see since they are negatives with a black background and are covered with chicken-pox dots.
Ain't CADENCE lovely?

Why didn't you use a simple opamp filter instead?
 

Sorry, I can't see how the various bits of circuit fit together. Can you show it as one schematic?
 

Thanks for looking, I don't know how to get the pictures any clearer I am afraid. I am using this circuit as I am following a worksheet from uni, I have to design a third order filter using the opamp I made. Seems simple enough, until I try and simulate it!

Here is a picture showing the entire schematic

 

You should first check the amplifier gain or the transfer characteristic of the buffer, with and without 10k load.
 

Thanks FVM, I tried that, the amplifier seems to work fine, I have a gain of ~80 dB with a unity gain frequency of ~20 MHz

Without 10 kOhm



With 10 kOhm

 

Yes, that is the circuit I am trying to recreate, but with the folded cascode amplifier. I have changed my pics to show node voltages. And added some pictures of the opamp performance in the previous post.

Even with changing 3 dB, I can not understand why the roll off is only 20 dB per decade?

My problem was that I had a DC voltage of 2.5 V. Schoolboy error. I am now getting a corner frequency of 10 kHz.



But .... why am I not getting a 60 dB per decade roll off? And why is the passband gain -6 dB and not 0?

Thanks everyone.
 
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But .... why am I not getting a 60 dB per decade roll off? And why is the passband gain -6 dB and not 0?
Click to expand...
It won't happen with a correctly working buffer amplifier. It should be mentioned that the designed filter (presuming a suitable buffer amplifier) is a third order Sallen-Key filter, but not corresponding to the Butterworth prototype.
 

A Butterworth filter has a flat bandpass then a sharp corner. Its slope is steep.
The filter you are using is sub-Bessel which is very droopy. It begins dropping its output at frequencies far from its corner and takes many octaves to reach maximum steepness.
 

Audioguru said:
A Butterworth filter has a flat bandpass then a sharp corner. Its slope is steep.
Click to expand...

The steepness of a filter (roll-off behaviour) depends on the filter order ONLY (n*20 dB/dec, n=order) - independent on the approximation used (Butterworth, Bessel, Chebyshev).
 

You final output stage looks suspect. How much quiescent current is there in the final stage? A fixed voltage to the gate of the transistor is a poor way of doing things even for a proof of principle. A current source or current mirror would be better.

Keith
 

In a more informative version of the schematic, we would see the transistor voltages and currents.

The steepness of a filter (roll-off behaviour) depends on the filter order ONLY (n*20 dB/dec, n=order) - independent on the approximation used (Butterworth, Bessel, Chebyshev).
Click to expand...

To illustrate the behaviour of different third order filters, see below a butterworth characteristic (blue) versus the equal RC design shown in post #1 and #7 (yellow).

 

LvW said:
The steepness of a filter (roll-off behaviour) depends on the filter order ONLY (n*20 dB/dec, n=order) - independent on the approximation used (Butterworth, Bessel, Chebyshev).
Click to expand...
This filter has such a low Q (it is very droopy) that it "takes all day" to reach its 60dB per octave slope (it takes a few octaves before reaching a proper 60dB/oct slope).
Look at where it begins its gradual rolloff. Look at its blurred corner if you even can find it.
 

it takes a few octaves before reaching a proper 60dB/oct slope
Click to expand...
Less than 3, see post #13. Obviously, the curve in post #8 can't be related to the designed (in deed very poor) filter third order filter. Apart from the curve shape, the 6 dB gain drop already indicates it clearly.
 

I based my comments about the droopy filter having a gradual slope on THE FIRST graph, not the 13th graph.
60dB per decade is 18dB per octave. (Corrected terms)
 

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I based my comments about the droopy filter having a gradual slope on THE FIRST graph, not the 13th graph.
60dB per decade is 18dB per octave. (Corrected terms)
Click to expand...
I see. But a said 10 kHz filter with the third pole taking effect at about 100 MHz won't be seriously discussed as "droopy" third order filter. It's just defective.
 

Audioguru said:
I based my comments about the droopy filter having a gradual slope on THE FIRST graph...
Click to expand...
Well that's a bit pointless. The graph you're referring to shows the response of a circuit that's known to be defective. It's blindingly obvious to everyone (including the OP) that that is not the desired response. That's why he started this thread in the first place.

Do you have any idea what the problem is with his opamp design, or how to fix it?
 

I wonder why he calls the droopy filter (with a low Q) a "Butterworth" filter? A Butterworth filter has a much higher Q. Then it will have a flat low frequency passband, a sharp corner then immediately cut high frequencies.


The opamp has much less voltage gain than purchased opamps. I think its compensation capacitor value is too high which cuts high frequencies (and cuts positive feedback at the corner) and reduces the slew rate.
 
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