Parametric EQ: What am I looking at?

[chuckfalcon]

Hi everyone!  I am interested in adding a Parametric EQ to a pedal that I have been building.  Nothing fancy, just something simple and effective.  I came across this schematic and it got me puzzled:



Here are the questions that I am having:

1.) +Vb?  The area that is represented at the bottom.  +9V and ground is already represented.

2.)  Which IC do I use for something like this?  I have only used standard NPN and PNP transistors in the past and have not dove into IC's yet.  How many legs does this IC have?  Is it 1 or 2 IC's?

I know this is basic stuff, I am sure but if any one could help me out with this one I would be most appreciative.

Thanks a bunch

[chuckd666]

+Vb is a connection from the 10k/10k/47uf network at the bottom left. So the bits that say +Vb connect to this.

You could probably use a TL074 quad IC for this seeing as it has more than two sections. Someone else can probably shine more light on this one than me!

[Kipper4]

Probably worth a look here

http://www.diystompboxes.com/smfforum/index.php?topic=113018.0

[blackieNYC]

Yeah, go read that thread.
And read "simple parametric" at geofex. That's where your drawing is from. It tells, if I have it right,how that op amp circuit is imitating an inductor so you can change frequencies with a simple variable resistor instead of a variable cap or inductor. The circuit from C1 downward is a short at your dialed-in frequency. A short leading to a DC supply is much like ground for AC signals. Notice the 10k pot pans this resonant circuit between the + or - input of the last op amp 1b. One way, your filter  frequency is added, the other way it is subtracted using that 10k pot.
PS    Don't use a TL0 for the gyrator. Use a 5532. Read thread.

[PBE6]

I have a question about this circuit too. I'm just trying to do some paper analysis on it to get a better idea of the boost/cut in dB depending on the component values, and the algebra gets a bit unwieldy pretty quickly if I assume there is an independent voltage at the pot wiper. However, both ends of the pot are at the same voltage because they're connected to the opamp inputs. Is it possible that the wiper is also at that same voltage? That would make the math a lot easier if it's true.

[blackieNYC]

Just starting to get a handle on these.  I'm not sure but at resonant frequency the wiper must appear as the DC supply, attenuating the AC. I don't know if you'll actually see 4.5 v there.
I'll spoil your math's ending - I get 7dB of boost. Not sure about cut right now. But tell us what you find. My 7dB is probably low, given the tolerances.

[PRR]

> Is it 1 or 2 IC's?

1-1/2 ICs. (We mostly use dual opamps. Singles are not cheaper, and quads are too many wires in too little space.) This has three opamps, so 1.5 packages. However you can add any reasonable number of frequency bands, with one extra opamp each! So a 2-band makes much sense: use-up two whole packages. But as a trial-build, stick with one band. "Ignoring" the un-used opamp often works; to be *sure* it won't make trouble wire it unity-gain and tie the input to the bias source.)

> both ends of the pot are at the same voltage because they're connected to the opamp inputs.

Yes.

> Is it possible that the wiper is also at that same voltage?

Seems logical.

So how can it do anything at all??

Follow the *current*. There's little or no cross-pot current because both ends are at the same voltage. So the current in each side can be summed around the node mostly ignoring the pot.

And set the pot to extremes.

And assume the "resonance" pot is ZERO. That part is not in most implementations (look around). It IS a useful add-on for hardly any cost and one more knob for your panel, but is not essential.

With pot in center it reduces to a unity-gain follower.

Assume C1 is perfect cap, gyrator is perfect coil... but wait, it can't be perfect because there is a 470r resistor on the output. The simple assumption is a perfect L-C series network, zero impedance at resonance, plus a 470r resistor. So at resonance the L-C-R is just 470r resistive.

Let us assume R.G. is no dummy (or worked from well-tried plans) and that it does appropriate things each side of resonance.

At resonance, pot at left end, we have a 2.7K+470r divider, or 6.7 loss, unity-gain after that. This is 16dB.

At resonance, pot at right end, we have unity-gain then a 470r+2.7K gain network, or 6.7 gain. This is 16dB.

If you find the resonant frequency documented (however this form is adjustable), you can use the known value of C1 to work out the apparent value of the "L" which the gyrator is faking. You can plot these reactance lines on a Reactance Chart and work out (by hand!!) the approximate gain at off-resonance points, and with the minimum 470r or the maximum 10.47K series. (You should also allow for the shunt resistance of 51K+1Meg, but that's too much for hand-calcs when we have computers.)

[PBE6]

Awesome! Thanks Paul, that sheds a lot of light on this for me. I was mostly interested in working out the maximum cut/boost, with a passing interest in figuring out why the 10k pot is an ideal choice here. I'll probably still work out the formula for the in-between points later, but for now my main question has been answered.

Thanks!!

[PRR]

> why the 10k pot is an ideal choice

It's not "ideal". Very much a compromise.

Low values, especially with multiple bands, raise the Noise Gain.

High values "damp" the resonator so not much happens until the end-stop.

Plagiarize. R.G.'s values look good for general stage use. RANE's guru studied this intensively and has both product-schematics and White Papers on the site.

There are other topologies with clearer interaction, but they use more opamps. Parametric EQ didn't happen (much) until opamps got affordable. Even then most designers tried to minimize the number of opamps, for cost, hiss, and because you can really go crazy servoing every darn aspect of a filter, then the PCB is bigger than the knob-panel. (Today the wise designer dumps it in a DSP and gives you a iPhone app to control it....)

[chuckfalcon]

Thanks for the quick replies, everyone.  I'm still trying to wrap my head around all of this

> Is it 1 or 2 IC's?

So I could use 2 MN5532 op amps?  On one, one side would be "U1a" and the other side would be "U1b", and on the second IC, one side would be "U2a"?  Would the OTHER side of the 2nd IC be unconnected?

OK, I think something just clicked, while typing this, and looking at the pinout chart of the MN5532.  The V+ and the V- pins, are the same pins represented in both the "U1a" and "U1b".  Is this correct?

[blackieNYC]

The V+ and V- pins supply both of the two op amps "halves" in the one chip. Zat what you mean? Or all four opamp circuits in a TL074.
Don't leave any unused op amps completely unconnected.  Connect the inverted input to the output, and connect the non-inverting input to Vbias thru a 100k resistor. Or find something fun to do with it.  Or make your Vbias supply with it.

[chuckfalcon]

The V+ and V- pins supply both of the two op amps "halves" in the one chip. Zat what you mean? Or all four opamp circuits in a TL074.
Don't leave any unused op amps completely unconnected.  Connect the inverted input to the output, and connect the non-inverting input to Vbias thru a 100k resistor. Or find something fun to do with it.  Or make your Vbias supply with it.

AH HA!  yes!  That is what I was talking about with the V+ and V-.  What is "Vbias"?  in this situation, non-inverting input > 100K resistor > Vbias.  What is the Bbias?

[blackieNYC]

Vbias is generally the halfway point between V+ and V-, in this case +9 and ground. 4.5v is made by the voltage divider (or some other means) of the two equal value resistors. This is the bias point that the AC voltage will swing around, the idea being to avoid hitting the rails, among other things.
Someone else should jump in here.
If several points need to be tied to the bias, my understanding is that you don't want to short them all together, so a large value resistor is used to distribute the voltage to each point. I think you and I both need to read some.   Geofex.

[PBE6]

I finally worked out the formula for the parametric EQ:

Vout/Vin = {[AB + A(Rf+Rl) + BRl] + [(A+B)(Leff*w-1/(wC))]*j} / {[AB + B(Rin+Rl) + ARl] + [(A+B)(Leff*w-1/(wC))]*j}

Where:

Rin = input resistor (2.7k in RG's article)
Rf = feedback resistor (2.7k in RG's article)
Rl = load resistor (470 ohm in RG's article)
A = Rpot*x
B = Rpot*(1-x)
Leff = effective inductance of gyrator = Rl*Rground*C2 (or R1*R2*C2 in RG's article)
C = capacitor that is NOT part of the Leff calculation (or C1 in RG's article)
w = 2*pi*freq
j = SQRT(-1)

You can use the above equation to determine the magnitude of the response at a given frequency. Some fun things I noticed when playing around with it:

1. Setting Rin and Rf to different values makes the peak boost/cut levels different, although it will also move the "centre" location (I.e. flat response) away from the middle pot rotation.

As noted by PRR above, the max values can be found by rotating the pot to the extremes and simplifying the circuit. Peak boost is 20*log(1+Rf/Rl), max cut is 20*log(Rl/(Rl+Rin)).

2. Also as noted above, larger pot values makes the control relatively inactive around the middle position, pushing the bulk of the response to the extremes of rotation. RG indeed picked some good values (as expected!).

3. The model can be expanded to include Rres from RG's article by adding it into the Rl value. This drops the maximum boost/cut available, but reduces the Q for a wider EQ effect.

Other than that, it really just tells you how goofy the response looks when you deviate significantly from RG's values for the pot and Rin/Rf. Sometimes it just ain't broke and don't need fixin'!