Comment:

This feedback loop assumes that if the amplification factor before positive feedback is even slightly less than (1.0), the resulting infinite series will converge, on some finite amplification factor. However, as the reader can see, ‘VR1′ and ‘VR2′ can be turned all the way to the right, thus connecting the feedback summations directly to the Source of the two MOSFETs.

I am assuming that the gain there is slightly less than unity, as following from changes in Gate voltage. However, because there are tolerances in all the resistors, ~~the feedback gain may reach unity~~ in theory. (:1) If this should happen, the user of the appliance should have a tremendous experience and learn not to do that again.

Explanation:

The resistor-capacitor combinations of (R11,C4) and (R12,C5) produce a rudimentary first-order, low pass and high pass filter each. The MOSFETs achieve that their Source voltages will follow their Gate voltages simply, while small changes to their Drain voltage should be the inverse of those to Source voltage, times (9), as long as the current flowing through ‘R2′ and ‘R3′ can be neglected.

*Two resistors were removed from an earlier version of this circuit, that were meant as an enhancement, but that would have prevented the circuit from working*.

The way the inverse-logarithmic-taper Variable Resistors (=Potentiometers) work is, that with their wiper at centre-position, its resistance to the right electrode should be 10kΩ, while its resistance to the left electrode should be 90kΩ. This leads to a design barrier because the resistances of ‘R8′ and ‘R10′ automatically need to be 9x ‘R7′ and ‘R9′, so that the inverted signals cancel out on the wipers.

*An added complication to this arrangement is that, due to the phase-shifts of the first-order filters, (α) is some multiple of a complex number, that multiple being determined by the setting of the Bass and Treble controls, the complex number determined by the frequency, which will appear normally in the denominator, and thus result in an attenuation, which my notes above are only a gross simplification of. However, well below and above their corner-frequencies, the low-pass and high-pass filters will exhibit little phase-shifting, so that the final result below is still correct*:

This also achieves that the maximum attenuation can become:

1 / (( (200kΩ / 220kΩ) * (90kΩ | 220kΩ) / 10kΩ) + 1)

In other words, with the wiper of each potentiometer turned all the way to the left, ‘R2′ and ‘R3′ will act as though in parallel with ‘R8′ and ‘R10′, thus diminishing the achievable inverted gain somewhat.

‘R1′ … ‘R6′, together with the operational amplifier, form a summation circuit, with the input to *this* circuit as one of its inputs.

(Updated 10/21/2019, 19h35 … )

(As of 10/19/2019, 12h40 : )

**1:)**

Because the issue may be non-trivial, of the open-loop gain achieving (+1), and therefore, of the circuit becoming an oscillator, I have updated the schematic above, to reduce the feedback of the low-pass and high-pass filters by approximately 10%. This should have negligible impact on performance, and avoid unstable behaviour, as long as the tolerances of ‘R1′ … ‘R6′ are within 2%.

(Update 10/21/2019, 19h35 : )

Also, I would consider adding a parallel capacitor of 80pF to the non-inverting input of the operational amplifier, which should act as a low-pass filter with ‘R1′ … ‘R4′, with a corner-frequency of 39.8kHz, just because the open-loop gain of simple stages can become unpredictable at high frequencies. This value should respect the fact, that some Humans can hear frequencies *at least* as high as 20kHz.

(Update 10/21/2019, 14h00 : )

There is another issue with this circuit that I should point out. Depending on how far the principle is taken, the voltage-swings at the Drain of the MOSFETs can become wider than the application allows, since the application may call for high input signal amplitudes, and low supply voltages. If the MOSFETs are biased ideally, then ‘R8′ and ‘R10′ would bisect the supply voltage. Yet, when the wipers of the potentiometers are adjusted all the way to the right, then ‘R2′ and ‘R3′ will act as though connected in parallel with ‘R7′ and ‘R9′. This will already drive the inverted voltage swings at the Drain of the MOSFETs to approximately 10x those at the Source.

The best way to solve this problem may be, *to decrease* the overall sensitivity of the circuit to ‘Vin’, while preserving the gain of the feedback loops. Fortunately this can be achieved, by *increasing* ‘R1′. However, if the reader wishes to do so, he must not forget to recompute ‘R4′, as I’ve done before, so that acting in parallel, ‘R1′ … ‘R4′ form a virtual 50kΩ resistor.

As shown, the circuit should be ‘reasonably well-behaved’ with an input signal-voltage of 50mV peak, and a supply voltage of 12V. But even then, if the wipers are turned more than 4/5 to the right, distortion can result.

(Update 10/21/2019, 17h20 : )

```
Vcc = 12V, Vin = ±1V ->
R1 = 4MΩ
C1 = 2nF
R4 = 95kΩ
Vout = ±50mV
```

(Update 10/21/2019, 17h20 : )

The MOSFET bias circuit incorporates the somewhat arbitrary assumption, that the inputs to the operational amplifier draw an additional 2.5μA of current. Since ‘R8′ and ‘R10′ of the bass and treble circuit are only expected to draw approximately 89μA at 8V, and since the voltage across ‘R1′ in the bias-determining circuit would have ~9.3VDC applied to it, such an odd value results. And *yes*, this means that only ~58μA of bias current will be flowing through the transistor.

Also, in the bass and treble circuit, together with ‘R2′ and ‘R3′, the isolation capacitors ‘C2′ and ‘C3′ form rudimentary high-pass filters with a cutoff frequency of ~18Hz, thereby preventing changes in the settings of ‘VR1′ or ‘VR2′ from affecting the amount of bias current flowing through the MOSFETs. A type of argument that could come to mind would be, ‘The treble feedback loop does not need to operate below 400Hz, therefore C3 could be replaced with a 2nF capacitor, that would also take up less space on the circuit board.’ However, this argument would not be 100% true, because ‘R2′ and ‘R3′ also define the gain of the summation-circuit at all the frequencies where ‘C2′ and ‘C3′ conduct. Therefore, changing those capacitor values will also affect how the entire circuit behaves over frequencies spanning down to 18Hz. In order to prevent non-uniformities in the frequency response which a listener will be able to hear, ‘C2′ and ‘C3′ should therefore be kept at 40nF. Hypothetically, if ‘R3′ was simply removed from the circuit, then the summation sub-circuit would end up with greater than unit gain, *including the other feedback loop*. And, adjusting ‘C3′ so that it becomes non-conductive in an audible part of the spectrum will have the same effect.

(Update 10/21/2019, 17h20 : )

There is a weakness to the design which I should mention. In the bass and treble circuit, in my initial explanation, I pretended that when the potentiometer wipers are in centre position, ‘R2′ and ‘R3′ behave as though they formed a path between the non-inverting operational amplifier input and zero. The fact is that, when those wipers are in centre position, the potentiometers exhibit a kind of output impedance of 10kΩ. They do not feed any signal to ‘R2′ or ‘R3′, but for the purpose of predicting what would happen, when there is a voltage-swing at the non-inverting operational amplifier input, ‘R2′ and ‘R3′ would have another, 10kΩ resistor in series.

However, with *one* potentiometer-wiper turned *all the way to the left*, this added, series resistance would effectively increase to 47kΩ, which means that overall gain would increase very slightly, in a way that does not favour any audible frequency-ranges. The amount of parallel resistance connected to the non-inverting operational amplifier input would become 52kΩ instead of 50kΩ, which means that overall gain would increase by ~4%.

In the case where ‘R1′ has been replaced with 4MΩ…, the parallel resistance would only increase to 51kΩ, thereby increasing voltage gain by 2%.

Dirk