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A question To Jim Thompson

J

Joerg

Jan 1, 1970
0
Habib said:
Jim,

Could you please explain me how this :
http://www.analog-innovations.com/SED/GyratorFilterNotch.pdf
actually works ?

i'm tired doing maths but i launch LTSpice on your reference and ... how
could i put it ... it's seems not to be a Notch Folter to me.

Regards, H.


You can just attach your LTSpice file here, or copy it into your post.
Then people can try to massage it and re-post the corrected version, one
that (hopefully) runs.
 
J

Jamie

Jan 1, 1970
0
Don't forget to connect power to the OpAmps!!

TomC
What do you need that for? I thought everything was wireless these days?

Jamie
 
M

miso

Jan 1, 1970
0
In chip design, you minimize the number of "floating" nodes, where in
this vernacular floating means not being driven by a low impedance
circuit, i.e. op amp. The parasitics will ruin the circuit response much
more than the phase effects (say Q enhancement) of the opamps.

Generally you have one op amp (or more!) per pole in integrated circuit
designs. And leapfrog design rules!
 
J

John Smith

Jan 1, 1970
0
Jim Thompson said:
...>

To the right of C2 is a gyrator-created inductance. And Tim is
correct, check your nodes.

For no reason other than curiosity, what is the maximum frequency that a
circuit like this could sensibly be used at? Assuming suitable OpAmps and
layout.
 
G

George Herold

Jan 1, 1970
0
"Jim Thompson" <[email protected]> wrote in
message news:[email protected]...



For no reason other than curiosity, what is the maximum frequency that a
circuit like this could sensibly be used at? Assuming suitable OpAmps and

[snip]



It's a function of the gain-bandwidth product (GBW) of the OpAmps

used. I'd simulate to make sure, but I'd guess a reasonable

rule-of-thumb would be around GBW/10.

Hi Jim, I've never done a notch thing, But I assume it's like a bandpass turned upside down. In which case the GBW requirements are going to be something like
Fmax * Q ~ GBW/10

George H.
 
G

George Herold

Jan 1, 1970
0
If each op-amp has a voltage gain k/s, the impedance looking into the

positive input of U1 is



V- (2s + k) k C1 R3 R4 s
--- = -----------------------------------------
I 2 C1 R4 s^3 + (3 k C1 R4) s^2 + k s + k^2
This looks right dimensionally, and it goes to V/I = C1 R3 R4 s as the GBW

product (GBW = k / (2 * pi)) goes to infinity. So I think I have my head
screwed on straight.
Dinner was recent and rich, and intuition is coming slowly to me, but if

you assume that C1 R4 << k, then the first thing that will show up as

frequency increases is the
s + k/2

----------

s + k



part. That's going to make the impedance lead even more than just an

inductance, which is oddball. I'm pretty sure that'll appear as an

inductor with some negative resistance, which will artificially narrow up

any tank circuit you throw the gyrator into (or make it oscillate, if the

Q and frequency are high enough). I don't think that the dynamics of C1

and R4 will contribute much unless you let C1 R4 approach k -- and dangit,

I'm not going to figure out if doing so makes things better or worse;

that's your job!



It looks like it'll have a seriously underdamped pole pair with a damping

rate of k/8 and a natural frequency that roughly depends on sqrt(k/

(2*C1*R4)). Whether this causes problems or not -- I dunno. As long as

the amplifier gain is strictly equal to k/s then it's unconditionally

stable, and I think that a single additional pole (A = k / (s * (s + a)))

or other equivalent lag would actually make it more stable, not less. I

wouldn't bet on that stability unless I actually did some paying work on

it (and besides, you can do that part).



(BTW: that last paragraph came completely from sketching a root locus

diagram and calculating the intersection of the poles-to-infinity locus

with the real line -- I think that's the first time I've actually done

the intersection thing to predict the behavior of a real live system, and

not just as a homework exercise for a class. So -- 28 years after I took

the class, I actually used that tidbit for more than getting a grade).

Hi Tim, well I'm not going to try and write down the loop equations. I assume you did them correctly. But my 'simple minded' understanding is that everything to the right of C2 looks like an inductor. (Maybe I have this wrong?) So it's a series RCL. Now if I was to take the output from the otherside of C2, then it would be a bandpass filter... And that's where I got the factor of Q in the max frequency estimate. Hey, if this type of gyratorcircuit can give me a bandpass response without the Q factor that would benice to know.

George H.
 
G

George Herold

Jan 1, 1970
0

Hey that's neat Jim! So with one circuit you change the input 'tap' and get all three outputs from the same point!
"scratch, scratch".. Scribbling in the back of my notebook.

So for the figure on page five where you add a series resistor to the C1 branch are the colors for the same GBW of the previous graphs.
Green = 100kHz
Red = 200kHz
Blue = 500kHz.
(I'm even more amazed if you can make a 100kHz GBW opamp give a BP responsewith a Q of 10 at 20kHz.)

A recent project had a 80 kHz Q=10 filter (state variable). In the firstgo-round I had 8MHz GBW opamps in it and the Q-enhancement was something over 30% (I can't recall the exact numbers.) They were replaced with AD825's (26MHz.)

I guess the difference might be that in the SV filter a Q of 10 is also a gain of 10, whereas your filter looks to have a gain of one.

Thanks again,

George H.
(now waiting on the next high Q filter need)
 
G

George Herold

Jan 1, 1970
0
You have more faith in me that me!

Grin... Well for sure I have more faith in you than in me.
When there's op-amps in the mix I generally do the stuff with voltage

dividers and expressions for the op-amp outputs. If you assume zero

input current to the op amps and zero output impedance then there really

aren't any complicated nodes.
I guess I could do it, and maybe get it right. My big problem is that making sense of the resulting complicated algebra expression would take me even longer.

It's hard for me to look at it and see where the important bits are.

George H.
 
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