Another Kind of Tone Control?

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 company designers had reduced the band count to keep the price affordable. The frequencies of the bands are expressed in a way that today's audiophiles 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.

The manual for the equalizer carries the copyright year of 1975. The op amp ICs available at that time weren't even close to having good enough performance for use in an audio device. Heath engineers built high gain amplifiers with discrete components. The circuits of these amplifiers will not be shown.

This circuit works much the same as the one that is used today. A series resonant circuit either reduces the amount of negative feedback around the right hand amplifier causing its gain to be increased at that frequency resulting in an increase in output. Or, the circuit reduces the amplitude of the input signal at its resonant frequency causing a decrease in output at that frequency.

Let's start by looking at the circuit when all of the pots are set to center which causes the wipers to be at ground potential. The tuned circuits will have no effect because both ends of each circuit are grounded. The input signal passes through a unity gain amplifier whose purpose is to provide a low and constant output impedance to the equalizer circuit. All of the pots are nothing more than resistors in parallel from the noninverting input of the op amp on the right to ground. These are 50 k ohm pots and the grounded tap is in the exact center. That means 25 k ohms of resistance from each end of each pot to ground. When five 25 k ohm resistors are connected in parallel the combined resistance is 5 k ohms. These are the values in parentheses in figure 4. The right hand amplifier which we will refer to as amplifier 2 is a noninverting amplifier with gain greater than unity. An equivalent circuit is shown in figure 4 below.

Figure 4 Equivalent circuit with all pots set to midrange.

For a verbal description click here.

So what we see is a resistive attenuator which is virtually unloaded by the near infinite input impedance of a noninverting amplifier. Then the noninverting amplifier has the same amount of gain as was lost in the attenuator giving an overall gain of unity or 0 dB. The attenuation of the resistive attenuator is given by

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

Converting to dB gives

20 Log (0.806452) = -1.868 dB.

The gain of the amplifier is given by

A2 = 1 + Rf / R1 = 1 + 1200 / 5000 = 1.24.

Converting to dB gives,

20 Log (1.24) = +1.868 dB.

Gain values multiply and dB values add. The values of A1 and A2 multiply out to unity and the dB values add to 0. If you operate more by intuition then look at it this way. The first law of op amps states that if the amplifier is not in saturation and feedback is properly connected, the voltages at the two inputs will be identical. There is a voltage divider coming off of the output of OA1 and another identical voltage divider coming off of the output of OA2. If the voltages at the outputs of A1 and A2 are identical then the voltages at the outputs of each voltage divider will be identical. These two voltages just happen to be the two inputs of OA2. If the output voltages of OA1 and OA2 are equal then the combined gain of the first voltage divider and the amplifier containing OA2 is unity.

If we move one of the pots downward off of center ground the tuned circuit associated with that pot is connected from the noninverting input of OA2 to ground as shown in figure 5. The series resistor is that part of the slide pot from the wiper to the bottom end. As the pot is moved downward the resistor grows smaller which increases its effect on the signal. Also as an unwanted side effect the Q of the tuned circuit grows higher. A series resonant circuit has its lowest impedance at resonance which means that at this frequency the effective resistance of R2 is decreased causing the input voltage to be reduced at and near resonance.

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

For a verbal description click here.

Figure 6 shows the effect when one of the pots is pushed upward. This causes the resonant circuit to reduce the amount of voltage that is fed back to the inverting input. This causes an increase of gain at and near resonance. The behavior is the same as for attenuation above, it is applied to the other attenuator circuit.

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

For a verbal description click here.

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.

For a verbal description click here.

Upon seeing this I said to myself "This isn't right. I can't show this. They'll laugh me off the web." My next step was to get out the actual equalizer. I found it didn't work so I did what you would expect. I found a piece of melted solder wedged in under a vertical mount capacitor that was shorting out the signal. "How did that affect both channels." I wondered. While I was at it I repaired the LED power indicator light which had not worked for many years. Back together and setup for testing.

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

For a verbal description click here.

You can sort of see it on the computer screen but here is a proper view of the four settings used in the Spice simulations shown above.

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

For a verbal description click here.

If you look back at the schematic above you will see that the value of the series resistor is quite different for different frequency bands. Ideally the value of this resistor would be the same in every band. The total resistance in each tuned circuit is made up of the discrete resistor plus the resistance of the inductor. In the Spice simulation the inductor is ideal so the resistor is the only resistance in the circuit. The value of all 5 resistors is 330 ohms. I found this value by setting it for 12 dB of boost and cut at the peaks of the curves for pots all the way up or down. The close similarity between the simulated curves and the actual device is remarkable.

I think duplication of this circuit would not be practical. First, center tapped slide pots are somewhat rare. Second even if you can get them they will most likely be linear. The pots used by Heath have a pronounced nonlinear taper. The resistance from either end to center is 25 k ohms. If the wiper is set half way between the center tap and either end the resistance from the wiper to the end is 1.1 k ohms. That's a shame because the circuit could easily be adapted to tubes and work rather well. Since I no longer have a use for the Heathkit eq I might just gut it except for the controls and rebuild it with tubes. Don't hold me to that.

Figure 10 Showing no Difference From Figure 7.

For a verbal description click here.

Going back to the simulation used to make figure 7 I removed the ground from only one band, the one in the middle. If it did make any difference it would be immediately obvious in figure 10. Adding the ground does make the circuit easier to understand but that appears to be its only benefit.

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.

For a verbal description click here.

This is not going to be easy because all intuition fails us. I have gone so far as to modify my pot symbol to show the wiper in the exact center. Let's start assuming nothing connected to the wiper. That would be the case if we injected DC into an equalizer circuit because the capacitors in the series tuned circuits would prevent the passage of direct current. If we apply 1 VDC to the terminal on the left, the amplifier will pass the voltage with unity gain. To remind you of the first law of op amps, both inputs will always be at the same potential. That means that both ends of the pot will be at the same potential. There isn't any current in the wipers because of the capacitors. If there is no voltage across a resistor there is no current flowing in it by the law of Mister Ohm. Current does not flow into or out of the inputs of the op amp because it is an F E T input amp. This forces us to the conclusion that there is no current in either Ri or Rf. Consequently the output of OA1, both inputs of OA2, and the output of OA2 are at the same potential, namely 1 volt DC.

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.

For a verbal description click here.

The impedance of a series resonant circuit is minimum and usually quite low at resonance. Moving this low impedance from the center to the left end of the pot will cause an increase in current through Ri. The first and second laws work together to make both ends of the pot have the same voltage. There is more resistance from the right end of the pot to the wiper than there is from the left end to the wiper. It takes less current from the right than it does from the left to equalize the voltages at the ends of the pot. There will be less current through Rf, than through Ri. The output voltage of OA2 will be reduced to satisfy the second and first laws and the amplitude will be reduced at and near the resonant frequency of the tuned circuit. Thus moving the pot wiper to the left causes a cut in amplitude at and near the resonant frequency.

In figure 13 the pot has been moved to the right.

Figure 13 Equivalent Circuit With Pot Set to Boost Side.

For a verbal description click here.

Now the situation is reversed. More resistance from the left end to the wiper means less current through that part. This also means less current through Ri than through Rf. More current through Rf means a greater voltage drop and more output from OA2. Moving the pot wiper to the right causes the amplitude to be boosted at and near the resonant frequency. The output voltage from OA1 remains constant for the discussion of both figures 12 and 13.

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.

For a verbal description click here.

Look at the top four bands in figure 14 and compare to figure 4. Except for component values they are identical. These bands use inductors that appear similar to transistor radio IF transformers complete with adjustment slug. The bottom 6 bands use gyrators to replace real inductors. A real inductor for the 31 Hz band would have to be 5.6 Henrys. Apparently the performance of gyrators at frequencies of 1 kHz and lower is close enough to that of real inductors to make using them feasible. No doubt a high performance equalizer could be made with real inductors for all bands but its price would place it firmly in the professional audio market.

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

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

For a verbal description click here.

The gyrator consists of C1, R1, R2, and the amplifier. C2 is the capacitor which tunes the inductance of the gyrator to a specific frequency.

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. Multiplying the f0 equation through by 2π gives,

ω = 1 / (sqrt(R1R2C1C2)).

Squaring gives,

ω2 = 1 / (R1R2C1C2).

Multiplying through by C2 gives,

ω2C2 = 1 / (R1R2C1).

We can now write,

ω ω C2 = 1 / (R1R2C1).


ωC2 = 1 / Xc2.

Making the substitution,

ω / Xc2 = 1 / (R1R2C1).

Inverting both sides gives,

Xc2 / ω = (R1R2C1).

These calculations are being done at the resonant frequency, therefore,

XL = XC. Therefore,

XLG / ω = (R1R2C1).

Where XLG is the inductance of the gyrator.


XLG = ωLG, or, LG = XLG / ω

Therefore the inductance of the gyrator is,

LG = (R1R2C1).

Derivation not shown but equation included for completeness,

RG ≈ R2.

Where RG is the resistive component of the inductance of the gyrator.

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 checked 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 is clear that the BSR unit does not have these special taper pots installed.

I have now returned to working on my webpages after a break. I'll explain, or offer excuses, on my daily log page. Meanwhile I still have parametric equalizers to go. And then there is doing it with tubes. I have corrected some mistakes in the equations so you may want to take another look at the derivation above.

Watch this space.

This page last updated Sunday, March 07, 2021.