# Help me analyze this EQ circuit

#### rpschultz

Joined Nov 23, 2022
416
The below circuit is from the Fishman Loudbox Mini acoustic guitar amplifier, the tone controls portion. The low and mid are fixed frequency peaking EQ and the High is a shelf. I understand the low and mid but can't figure out the high:

Low:
f = 1/(2 PI sqrt(R14 R15 C11 C12)) = 53 Hz

Mid:
f = 1/(2 PI sqrt(R16 R17 C13 C14)) = 766 Hz
(I'm reasonably sure this is right since Fishman has told me the Mid to be 750 hz).

High:
I thought it was f = 1/(2 PI C15 R18) = 4200 Hz, but that seems too low (maybe not). Is U2A involved with the High, or is that an output buffer?

Thanks.

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#### Audioguru again

Joined Oct 21, 2019
6,690
The opamp and R18 seem to be missing a connection to ground.

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#### Jony130

Joined Feb 17, 2009
5,488
The U2C and U2D are nothing more than a simulated inductor circuit.
L1 = 3.3kΩ*3.3kΩ*820nF = 8.9H and L2 = 3.3kΩ*3.3kΩ*120nF = 1.3Hz.
Thus, together with a series capacitor (C11 and C13) creates a series resonant tank circuit. And resonance frequency of F = 1/( 2Π√(LC) ).
The maximum "boost" will be Av_max = 1 + 20kΩ/( 3.3kΩ+ 3.3kΩ) = 4 V/V = 12dB.

But for the high frequency, we have a simple RC circuit. And if you set the high to max gain it will be around 1+20Ωk/5.6kΩ = 4.6 V/V = 13dB for the signal frequency larger than F = 0.16/(6.8nF * 5.6kΩ) = 4.2KHz

#### rpschultz

Joined Nov 23, 2022
416
Huh, OK. Seemed like 4200 Hz was low, but I guess not. That's a fair number (3-1/2) octaves above high E, so probably a good spot for it. Thanks for the confirmation.

#### rpschultz

Joined Nov 23, 2022
416
So in this circuit, with the switch is set to SHELVING (C2 bypassed). If I want a fixed frequency by removing VR1, do I need U1? Because in the original post, the HIGH only had 1 op amp instead of 2, it was a simple RC circuit with only gain and a fixed frequency of 4200... but I think that is a LPF (not shelving). So if I want a shelving Low knob, then it needs an op amp as seen below. Right? You can have passive LPF and HPF, but shelving filters need an op amp?

#### Ian0

Joined Aug 7, 2020
9,808
What you have there is a gyrator circuit, which simulates an inductor.
Inductor and capacitor make a bandpass filter.
All the component values are in the LM833 datasheet. (page 14)

#### Audioguru again

Joined Oct 21, 2019
6,690
The circuit (in this post) from the datasheet of the LM833 dual audio opamp is completely different to the circuit in post #5.

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#### Ian0

Joined Aug 7, 2020
9,808
The circuit (in this post) from the datasheet of the LM833 dual audio opamp is completely different to the circuit in post #5.
What's different about it? (apart from the shelving switch)
Although it is different from the circuit in post #1

#### Audioguru again

Joined Oct 21, 2019
6,690
Yeah, they are the same.

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#### rpschultz

Joined Nov 23, 2022
416
I'm trying to figure out the Q formula. Here's a nice calculator
Guitar Pedals: Gyrator Filter Calculator (muzique.com)

I understand how Q and bandwidth are related. I can simulate this in Eagle and get the same frequency. Anyone know the formula for the Q based on R1, R2, C1, C2?

#### Ian0

Joined Aug 7, 2020
9,808
L=R1.R2.C1
it‘s in the Wikipedia article
then f=1/(2π√(LC))

#### Jony130

Joined Feb 17, 2009
5,488
I understand how Q and bandwidth are related. I can simulate this in Eagle and get the same frequency. Anyone know the formula for the Q based on R1, R2, C1, C2?
Your simulated inductor circuit will have this equivalent circuit.

From this equivalent circuit we can see why we want to have R1 >> R2.
As for the quality factor of a series resonance circuit in will be Q ≈ 1/R1 * √(L/C1)

#### rpschultz

Joined Nov 23, 2022
416
OK, I think I figured it out. Based on the below diagram and corresponding AMZ calculator.

fo = 1/2/Pi/sqrt(R1 R2 C1 C2)
Q = 2 Pi fo R1 C2
BW (Hz) = fo/Q -> bandwidth

Then this link helps figure out the BW in Octaves.

Simplest one: N (BW in octaves) = 2/Ln(2)*asinh(1/2/Q)
The bandwidth (BW) is based on -3db from the center frequency fo, f1 and f2 on either side of the center. It's interesting because it's not equal frequencies on either side due to the nature of how octaves work, although it's close.

(I solved this equation in Mathematic)
f1 = (sqrt(BW^2+4*fo^2)+BW)/2
f2 = (sqrt(BW^2+4*fo^2)-BW)/2

So if you pull up the default values for that AMZ, you get:
R1 = 68k
R2 = 1k2
C1 = 560n
C2 = 56n
fo = 99.54 Hz
Q = 2.38

Then from further calcs:
N = .6017 (octaves)
f1 = 80.765 Hz
f2 = 122.56 Hz
f average = 101.663 -> which is a little higher than fo.

#### Audioguru again

Joined Oct 21, 2019
6,690
The circuit in post #13 is a Sallen-Key highpass filter with a flat high frequency response:

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#### rpschultz

Joined Nov 23, 2022
416
The schematic in #13 is vague on the ground. Here's how I modeled it to get a peak band EQ:

The cursor is at -3db from fo, 130 Hz is close to the AMZ calculation.

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#### rpschultz

Joined Nov 23, 2022
416
The circuit in post #13 is a Sallen-Key highpass filter with a flat high frequency response:
What program are you using to simulate this? Eagle has its' limitations, maybe I need to learn something else... especially since it has an EOL date of June 2026.

#### Audioguru again

Joined Oct 21, 2019
6,690
You are wrongly mixing the biasing of opamps that usually have a power supply with the usual V+, 0V and V- with a single positive supply and 0V. Then the output of the opamps are rectified positive pulses instead of linear AC audio.

Most of us use the free simulation program LTspice.

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#### rpschultz

Joined Nov 23, 2022
416
So the voltage source on the right is DC. It says DC 9 AC 0.

The voltage source on the left is AC, says DC 0 AC 1. This represents the frequency input for simulation.

#### rpschultz

Joined Nov 23, 2022
416
Here's another example I've been looking at from Elliot Sound.

#### Audioguru again

Joined Oct 21, 2019
6,690
The circuit in post #5 and post #13 here from Elliot Sound have the opamps biased at 0V and are powered with V+, 0V and V-. To use a single positive supply voltage, the opamps need to be biased at half the supply voltage, not 0V.

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