answers (Re: gEDA-user: ideal op amps in gschem)

[Dan McMahill <dan@xxxxxxxxxxxx>]

since no one offered a complete set of answers, here is one version

1.  It's an inductor over some frequency range.  At very low frequencies
the cap looks like an open and you just see R2 and you have 100 ohms
looking in.  Once you get past f=1/(2*pi*R1*C1) (or about 1.6 Hz), the
op-amp starts to make the voltage on the right side of R2 track what you
drive in on the left and you see a higher and higher impedance as the
gain through C1, R1, and the op-amp gets closer to 1.  A little math
says that I_R1 = 1/R2 * (Vin - Vin*R1*C1*s/(R1*C1*s + 1)) which can
easily be solved for Zin = R2 * (R1*C1s + 1) which is a 10 Henry
inductor above a few Hz.  This inductive behaviour will continue until
the impedance level is high enough that you can no longer ignore the
parallel impedance of the R1-C1 network which basically looks like just
R1 at higher frequencies.  2*pi*f * 10Henries = 1 Meg gives 16 kHz.
So ideally this circuit is a 10 Henry inductor from 1.6 Hz to 16 kHz
(noting that right at those 2 frequencies its not such a hot inductor).


2.  If the op-amp gain is 100, then the "unity" gain buffer only has a
gain of 100/101 or about 0.99.  This means that even if you took out the
C1-R1 circuit, and connected the input signal straight to the op-amp
+input you'd get I_R1 = (1/R1)*(Vin - 0.99 Vin) or you'd get 10k ohms or
so.  A 10 Henry inductor has an impedance of 10k at 160 Hz so the effect
of the op-amp having a gain of 100 is to kill the bandwidth by a factor
of 100!  One moral of the story is you can't just look at the op-amp
with unity feedback and say you don't need piles of gain.  In this case
you need a gain of at least 10,000 (1Meg / 100 ohms) to not have the 
op-amp be the primary limitation on bandwidth.

3.  If the op-amp is ideal except for a 1 MHz gain-bandwidth product
then the op-amp open loop gain magnitude at 10 kHz is 100, at 1 kHz its 
1000, etc.  This gets back to the issue in question #2.  So solve for 
the frequency where 2*pi*f* 10 Henries = 100 ohms * 1 MHz/f (the right 
side is the impedance magnitude you see without the R1-C1 network).  In 
this case you get 1.2 kHz.  So the 1 Mhz op-amp hooked up as a buffer 
actually limited the circuit bandwidth to 1.2 kHz in this case!  In 
fact, to not limit the circuit bandwidth, the op-amp needs a gain of 
10,000 at 16 kHz or a 160 MHz gain-bandwidth product (assuming a fixed 
single pole rolloff).  Of course op-amps with external compensation pins 
can do better by having a 2 pole rolloff for some time followed by a 1 
pole region near crossover.

4.  If you draw 1.5 uA of input current, that multiplied by 1 Meg will
give you a 1.5V DC input to the op-amp.  At best you've eaten in to the
signal swing at the op-amp output by 1.5 volts (3 if you're looking at
peak-to-peak sinusoids), at worst, you have connected 1.5 volts DC
through 100 ohms to the circuit driving this.  That may or may not be a
problem.

The place I saw this circuit was an audio high pass filter.  Looked
wonderful on the low end, but suprise suprise, it started to roll off
around a couple of kHz (It was a 2-3 MHz op-amp).

One point of this circuit though is that you must be very careful with
how you model your "ideal" op-amps.  It would be easy for a beginner to
really blow the modeling of this circuit.  It would be easy to blow the
design too (as evidenced by the real circuit I saw).

-Dan