Equalizers first came on the consumer electronics scene in the early 1970s. They had been used by professional sound engineers for many years before that. The earliest ones must have been built around vacuum tubes although I have never seen a tube circuit. The first ones to hit the market were too expensive for my income.## My First Equalizer.

Avid builders of Heathkits in those years will remember a line on the order form asking "What new kit would you like us to make?" On a couple of orders I wrote the words "Graphic equalizer" in this space. I must not have been the only one because shortly thereafter the Heathkit AD1305 was announced. Naturally I bought one and put it together. Assembly was easy and it worked right off.

**Figure 1** Photo of AD-1305.

The knob on the right almost looks like it belongs but it is a level control which I added shortly after putting into use. It should have been a slider but they were hard to find in those days and anyway I doubt if I could have cut the slot in the panel for it.

I wished it had been a 10 band but I supposed that Heat designers had reduced the band count to keep the price affordable. The frequencies of the bands are expressed in a way that today's audio files are not accustomed to. Instead of giving the center frequency of each band the lower and upper limits of each band are given. More than likely the customary band specifications we know had not yet been firmly established.

**Figure 2** Close up of one channel showing frequencies.

Even though it is inadequate to today's demands it will serve to illustrate the principle on which equalizers operate. The circuit is slightly atypical because the pots have a center tap which is grounded. I have not seen any other EQ circuit in which this is done.

**Figure 3** Simplified Schematic of one channel of the AD-1305.

For a verbal description click here.

**Figure 4** Equivalent circuit with all pots set to midrange.

A1 = R2 / (R1 + R2) = 1200 / 6200 = 0.806452.

20 Log (0.806452) = -1.868 dB.

The gain of the amplifier is given by,

**Figure 5** Equivalent circuit showing a pot moved downward from center.

**Figure 6** Equivalent circuit showing a pot moved upward from center.

## The Spice Simulation.

Simulating a pot in spice requires two resistors that are stepped in value using commands in spice that are made for that. A pot with a center tap is double trouble. It takes 4 resistors and only half of the pots rotation can be simulated at one time. A connection must be changed in the schematic to do the other half. My simulation looked like this for the boost half of the control. The three graph lines are Green full boost (+12 dB) or cut (-12 dB) on all bands, blue (+6 dB or -6 dB) boost or cut, and red zero no boost or cut (center).

**Figure 7** Spice Simulation of the Upper and Lower Halves of the Slider Pots.

**Figure 8** Photograph showing the equalizer, Analog Discovery, and Computer.

**Figure 9** Four part picture showing +12, +6, -6, and -12 dB settings on the real equalizer..

**Figure 10** Showing no Difference From Figure 7.

## It Still Works Without the Ground, It's Just A Lot Harder To Explain.

Below you will find the equivalent circuit of equalizer with all pots combined in parallel.

**Figure 11** Equivalent Circuit With Ground Removed and Pot Centered.

## But What About AC?

Now we'll apply an alternating test voltage to the input at the left. Whatever the test frequency it's going to be close enough to one of the tuned circuit resonances for current to flow through it. That's going to pull the center of the pot closer to ground. Current will now flow through the pot. By the first law of op amps the two ends of the pot must be at the same voltage. The only way for the center of the pot to be at a voltage different from the ends while the ends are at the same voltage is for current to flow in (or out) of both ends of the pot. This requires that the currents in Ri and Rf must be equal and in opposite directions.Now the second law of op amps comes into play. It states that "the output of the op amp will assume the voltage needed to prevent the first law from being violated." Let's take a snapshot of voltages and currents when the input wave is positive. The output of OA1 is positive and current is flowing from it and to the right through Ri. Then through the left hand half of the pot and to ground through one or more of the tuned circuits. Meanwhile current is flowing out of OA2 to the left through Rf. Then through the right hand half of the pot through one or more of the tuned circuits to ground.

Because the two halves of the pot have the same resistance the voltage across them will be equal. This satisfies the first law of op amps. This also says that the currents in each half of the pot are equal. Because Rf and Ri, have the same resistance values the voltage drop across them will be equal. Therefore the output voltages of OA1 and OA2 are equal. QED. With all pots set to center (flat) the gain of the system is unity for all audio frequencies.

Now for the case of one or more pots set to the left as shown in figure 12 below.

**Figure 12** Equivalent With Pot Set to Cut Side.

**Figure 13** Equivalent Circuit With Pot Set to Boost Side.

## A More Modern Circuit.

Below is a simplified version of the BSR EQ-110X equalizer. What are all those op amps doing there? They are obeying the laws of op amps.

**Figure 14** Simplified Schematic of BSR EQ-110X.

Here is the circuit of a gyrator and its resonating capacitor C2.

**Figure 15** Circuit of a gyrator and its resonating capacitor C2.

## Here Is How It Works.

The capacitor that is closest to the pot wiper is what tunes the circuit to resonance at the center frequency of the band in question. The next capacitor in line combines with the resistor to ground to form a phase lead network. This phase leading signal is fed to the noninverting input of the op amp which has been strapped for unity gain. This phase leading signal comes back to the junction of the two capacitors through a relatively low value resistor. The current that is fed back to the node is phase leading. In an inductor the current leads the applied voltage and thus the node where the two capacitors and resistor connect looks like one end of an inductor which has the other end grounded. The frequency of resonance is given by,f0 = 1 / (2π*sqrt(R1R2C1C2))

Q = sqrt(R1R2C1C2) / (R2(C1 + C2))

The only equation that is missing is one that gives the inductance value. It can be obtained from the resonant frequency equation by removing C2 by division as follows.

First let's put the equation in terms of ω so we won't have to lug 2π around. Dividing the f0 equation through by 2π gives,

ω = 1 / (sqrt(R1R2C1C2)).

Squaring gives,

ω

^{2}= 1 / (R1R2C1C2).Multiplying through by C2 gives,

ω

^{2}C2 = 1 / (R1R2C1).We can now write,

ω ω C2 = 1 / (R1R2C1).

But,

ωC2 = 1 / Xc

_{2}.Making the substitution,

ω / Xc

_{2}= 1 / (R1R2C1).Inverting both sides gives,

Xc

_{2}/ ω = (R1R2C1).These calculations are being done at the resonant frequency, therefore,

X

_{L}= X_{C}. Therefore,X

_{LG}/ ω = (R1R2C1). But,X

_{L}= &omega ;LG, or, LG = X_{LG}/ ωTherefore the inductance of the gyrator is,

LG = (R1R2C1).

Derivation not shown but equation included for completeness,

RG ≈ R2.

## Conclusions On Graphic Equalizers.

Although a special center tapped pot is not necessary for a graphic equalizer it is clear from the simulations that a special taper is required. I haven't check in detail but it may be that an audio taper is needed. The only catch is that it needs to taper on each end. That is, as the wiper is moved in from the end, either end the resistance changes slowly at first and more rapidly as the center point is approached. It also seems as if the BSR unit does not have these special taper pots installed. I need to make more tests before I can make a definitive statement.Meanwhile I have been called away to finish another project. And with this and parametric equalizers to go. And then there is doing it with tubes. I'll get back to this as soon as I can. Meanwhile, please be patient.

## To Be Continued.