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

[Al Davis <ad82@xxxxxxxxxxxxxxxx>]

As you might have guessed by my hint, it is the inductor part of 
a series tuned RLC circuit in a graphic equalizer.

On Thursday 26 January 2006 22:00, Dan McMahill wrote:
> 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).

In fairness.. "over some frequency range" applies to ordinary 
inductors too.  A real 10 Henry inductor would exhibit similar 
high frequency issues.

> 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.

This is why high-Q active filters use a different design, 
usually with 3 op-amps per section.  All single op-amp filters 
perform well only with low Q.    I wouldn't use this circuit 
for higher than about 2.  The filter in the equalizer has an 
unloaded Q of 1.3.  Other loading lowers the Q to 1.0, which 
was the design goal, at full boost and cut.  At lower boost or 
cut, the adjustment pot puts more resistance in series, 
lowering the Q to about 0.1 at small amounts of boost or cut, 
hence the claim of "proportional Q".

> 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.

Even with a Q of 1, at high audio frequencies, this is 
significant.  I designed it, then we built it.  At high 
frequencies, boost and cut were not as much as predicted, due 
to finite op-amp bandwidth.  Based on measurements, I 
recalculated to fix the boost/cut and Q, to meet the design 
goal of 1.0 loaded and +/- 12 dB.  It took a few iterations to 
get it right.

Another issue that was hard to calculate (impossible with the 
resources available at the time) was the loading effect of the 
adjacent bands.  Build it, try it, design again.  That's how 
the unloaded Q ended up at 1.3.  Actually, the controls are in 
2 sets, with alternate bands on each, so the adjacent band 
loading was two octaves away, where it had only a minor effect.  
I first tried it with all 10 bands on one bus, but the loading 
interactions were too much.

I considered a hybrid design, using gyrators for low frequencies 
and ordinary inductors at high frequencies, in this case the 8 
kHz and 16 kHz bands, but ended up using the gyrator circuit 
for all 10 bands.

The Q spec of 1.0 came from the desire to have no visible ripple 
in the frequency response when adjacent bands are adjusted 
together, and at lower amounts of adjustment to really have no 
ripple.