A question To Jim Thompson

You have more faith in me that 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 was just analyzing the circuit to the right of C2. It looks like an inductor to the extent that the op-amp gains look like infinity. At frequencies where this approximation does not hold, the gyrator is going to stop looking like a pure inductance.

Which pretty much verifies that GBW does matter.

The resistance in series with C1 would give the virtual inductor a virtual parallel conductance, if I'm not mistaken.

I'm not sure what the best way to verify this would be if you were really paranoid about stability issues. Probably you'd want to put a voltage source in series with an op-amp output, and suck out the open-loop transfer function.

Hey that's neat Jim! So with one circuit you change the input 'tap' and ge t 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 br anch 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 response with a Q of 10 at 20kHz.)

A recent project had a 80 kHz Q=10 filter (state variable). In the first go-round I had 8MHz GBW opamps in it and the Q-enhancement was something o ver 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 g ain 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)

Grin... Well for sure I have more faith in you than in me.

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.

Here it is (1mH design value) versus frequency and various OpAmp GBW values (10, 20, 50 & 100MHz)...

What caught me by surprise was the real part going negative. ...Jim Thompson

It figures that would happen, from the root-locus of the impedance.

Pretty pictures, but man those thin little lines aren't showing up well on my screen. You're challenging my eyesight here.

The negative R with GBW is probably why putting the resistance in series with the inductance-setting cap makes it less responsive to GBW variation

-- that's probably taking out some of the negative resistance effects.

I use WxMaxima for actually grinding through the algebra. It's got a nice symbolic solver, and it's free.

That took some head-scratching over here, too. With op-amps, if you aren't making the "V+ == V-" assumption (which implicitly assumes infinite gain), then you get your big nasty transfer function and take the limit as the gain goes to infinity. This is usually pretty easy (you just scratch out anything that's not multiplied by the gain, then you scratch out the gain) and provides a nice double-check against the solution you got from the zero input voltage assumption.

Admittedly, going from there gets complicated. Fortunately, you rarely need to go into that much detail.

In the case of this circuit, I set the op-amp gain to A = k/s; i.e., I made them perfect integrators with a set GBW product. Then I head- scratched for a bit, and saw how I could make a root locus, and did.

I just did lens replacement on the left eye... right was done about 10 years ago... my eyesight is 20:20, but I still need readers. ...Jim Thompson