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I'm leaning towards the 4x10 for it's 4 selectable settings that can be changed via remote control.

I would be running one sub line (unbalanced) out from my Pioneer AVR into the mini. Then using the appropriate (4 out) 48 KHz plugin for the best <10 Hz optimization.

What concerns me is the fact that the MiniDsp specs are wonky at best. In reality all I need is the 2x4 (1 in 4 out) but the input voltage is limiting. Very close to clipping once a person adds 15+DB of boost down low.

I was lucky and got the help of a knowledgeable person. Here is his assessment after reading the spec sheets on the 2x8 / 4x10 and their voltage ratings.

The conclusion was it can be either 4 volts or 8 volts, but not both.

I've gone round and round with MiniDsp and they can't answer the question with clarity.

Does anyone know if the Mini-Dsp 2x8/4x10 has an input limit of 4 or 8 volts?

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I just looked at the input voltage spec of the 2x8 and it is contradictory. I'll explain why below.

A balanced line-level input uses a circuit called a differential amplifier (diff amp), usually implemented with an IC op-amp and four resistors. I'm going to do a little math on this thing, because it's the only way I know to explain it without being ambiguous myself. In typical operation, the diff amp takes two input voltages that vary with time, neither of which is zero. Let''s name these voltages as follows:

Let input voltage #1 be called v1(t)

Let input voltage #2 be called v2(t)

The output voltage of the diff amp is a single ground-referenced voltage that is applied to the input of the A/D converter chip. Let this voltage be named as follows:

Let the diff amp output voltage (= input voltage to the A/D) be called vO(t)

Now let the gain from input 1 of the diff amp to the diff amp output, with input 2 grounded be called A1. Also, let the gain from input 2 of the diff amp to the diff amp output, with input 1 grounded be called A2. In linear operation, superpositon applies, which means vO(t) is determined by a simple formula:

vO(t) = A1 * v1(t) + A2 * v2(t)

We're almost there. To get a formula that has a better relationship to reality, we must make a change of variables. We define the difference-mode voltage vDM(t) as the difference between v1(t) and v2(t), and we define the common-mode voltage vCM(t) as the average of v1(t) and v2(t). In formulas, it looks like this:

vDM(t) = v1(t) - v2(t)

vCM(t) = 0.5 * (v1(t) + v2(t))

Now we can calculate vO(t) in terms of the difference-mode and common-mode input voltages. I'll skip the math in between and go straight to the result:

vO(t) = (A1 + A2) * vCM(t) + 0.5 * (A1 - A2) * vDM(t)

Now, here is the trick: If we design A2 so it's equal to the negative of A1, then by the formula above, the signal vO(t) applied to the A/D depends only on the difference between the two input voltages (the difference-mode voltage vDM(t)) and all dependency on the average of the input voltages (the common-mode voltage vCM(t)) disappears! This is called common-mode rejection. You may have heard of that term.

In a typical input diff amp of a balanced system, A1 = 1 and A2 = -1, giving:

vO(t) = vDM(t)

This follows from the longer formula above, with A1 = 1 and A2 = -1.

So, how does this relate to the miniDSP input voltage spec? (finally!). Well, the 2x8 spec says the maximum balanced input voltage is 8 Volts RMS (= 20 dBu). But remember, a properly-designed diff amp responds only to the difference-mode voltage at the inputs, and rejects the common-mode voltage. So this 8 Volt RMS balanced input voltage in a typical balanced system is achieved by having two equal-amplitude 4 Volt RMS signals at the input that are 180 degrees out of phase with each other. But there's another way, and that is to make one of the input voltages zero, and the other 8 Volts RMS. That also gives the same 8 Volt RMS difference-mode input voltage.

The conclusion? If the 8 Volt RMS balanced maximum input voltage is correct, then that means the maximum unbalanced input voltage is also 8 Volts RMS (because the diff amp only responds to the difference of the two input voltages, and within limits, it doesn't matter how that difference is achieved). But the spec says 4 Volts RMS!

Now there is one caveat, which deals with an unspecified parameter that relates to nonlinear operation: the maximum common-mode input voltage. From the math above, the common-mode input voltage is the average of the voltages at the two inputs, For a fully balanced system, the average of the input voltages is always zero for all time, because the two input voltages are negatives of each other. For a single-ended input signal of 8 Volts RMS, the common-mode input voltage is 4 Volts RMS (the average of 8 and zero). In a proper design, with a typical op-amp, the circuit will be safe with this common-mode input voltage.

So if all is well, it looks like the 2x8 has 4x the input voltage capability of the 2x4 (8 Volts RMS vs. 2 Volts RMS), assuming the 8 Volt RMS balanced maximum input voltage spec is correct.

Sorry for all the math, but if you take the time to understand the above, you will know more than miniDSP tech support and the people who wrote the miniDSP specs.