# Bandpass filter based on the GIC circuit

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#### v9260019

##### Member level 2
Hello all

"As shown, the center-frequency gain is 2 but may be reduced to 1 by "splitting" the feed-in resistor QR" ~~~ I wonder to know how to do it ??

thanks a lot

#### Euler's Identity

I'll take a stab at it...

You can reduce the gain by dividing the input signal by two, which, technically, doesn't reduce the gain but, rather, reduces the input signal, which, in turn, reduces the output.

However, the input resistance is QR; therefore, your divider must look like QR in the end.

Per Thevenin, you open I and short V.

Shorting your input signal to ground puts the upper resistor of your divider in parallel with the lower, giving Rin = R(upper)||R(lower).

Hence, you'll need to set Rin = QR.

The way I see it, QR is the same as the source resistance, so unless you're driving the BPF with an op amp, a low Z source, you'll need to add it to R(upper), first.

Incidentally, when I think of a BPF, the circuit you gave is not what I think of; I'm wondering about your chosen circuit's significance.

#### Euler's Identity

I derived the circuit and, yep, it's as it says. But what an odd circuit!

Who came up with that? How was it done?

I'm wondering if I've seen it but drawn another way.

Why use two op amps? There has to be some advantage to this thing...

Anyone? Has anyone at least seen it before??

Duh...

One glaring advantage is the fact that the Q is independent!

Who came up with this thing? I'd like to learn to think like this guy, design a circuit like this.

I did a nodal analysis on it and it does work, but seeing a circuit like this just shows me how non-creative I am. Op amps injecting the proper currents and Q easily adjusted, independent of fo...

#### moltar

##### Newbie level 2
Posted with understanding that this thread is older than dirt
and the images are no longer available:

(or dual amplifier- DABP) bandpass filter. Since there still
I thought I'd add my \$.02

When you remove the input resistor and its shunt capacitor, you
are left with the classic GIC topology.
This design can be made even more interesting by transforming the
Z1, Z2 and Z3 R's to C's and the C's to R's (including the input
R & C as well).
Q is now determined by the value of the input capacitor (formerly
an R). The center frequency can be changed by varying the value of
Z2 (formerly a cap) with no change in amplitude, and best of all,
the gain is adjustable by changing the ratio of Z4, Z5.
By way of example, here's some values to play with:
Cin = 2nf
Rshunt = 15K
Z1 = 1.5nf
Z2 = 100K pot, rev audio taper wired as a rheostat in series w/3.6K
Z3 = 10nf
Z4 = 49.9K
Z5 = 6.8K

You will end up with a bandpass filter that emulates a "Cry Baby" wah
pedal - only it's inductorless.
Range: ~400 Hz --> ~2.2 KHz
Gain: ~18 dB
Same 'Q'.

#### LvW

Perhaps it helps to give some light on the background of the described circuit:

It is a simple R-L-C band pass applied to the BRUTON impedance transformation.
This results in an C-R-D bandpass, whereby D is an active element called "Frequency dependent negative resistor (FDNR)". This element is realized by a GIC structure with two capacitors. By the way: Two other capacitor arrangements are possible within the GIC block.

#### moltar

##### Newbie level 2
It's good to see that this thread still holds some interest.
And you sir, are EXACTLY right.
that the transformation results in a BP filter which is no longer
of the constant bandwidth type.
Incidentally, I personally shy away from the use of the other
capacitor arrangements since they fail to easily provide a proper
return path for the amplifier bias currents.