TPA3251: Inductors -- several questions

Hi Adam,

A couple of follow-up questions:

> The inductors we use are actually linear within the current range that we use them

But that's not the non-linearity I was referring to --- the magnetic core exhibits hysteresis, which makes the current-voltage function non-linear. This is the part that I find troublesome. In the document, they talk about losses/inefficiencies due to this effect, but nothing is mentioned about the non-linearity introduced. Am I missing something and the hysteresis has no effect due to some aspect I'm overlooking?

> In there you will find information about the few output filter configurations you listed.

I don't understand why the type 1 filter (which is "option 3" in my diagram) cannot be used for AD modulation. In the document, they talk about common mode as the determining factor. However, in BTL mode, we ensure that one input is the (additive) inverse of the other one; the outputs are always V1 and PVDD-V1. There is symmetry in the filtered output, so why would type 1 filter not be suitable.

Can someone shed some light here? What am I missing?

> In PCB inductors cannot handle the power we need.

Thanks --- I had suspected that power requirements might be the show-stopping-number here (I guess technically, we can design it regardless, but I gather that the traces are going to be thick enough to make the size of the inductor ridiculously impractical?)

Thanks,
Carlos
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Carlos,

We haven't had issues with these and Hysteresis is not usually an issue for us. An air-core inductor that would satisfy our needs would be Giant, so we don't use them.

The document mentions that type 1 is for AD only. You are mentioning the opposite.

A PCB inductor is not at all practical for us.

Regards,

-Adam

Thanks Adam,

Clarifying:

> The document mentions that type 1 is for AD only. You are mentioning the opposite.

D'oh! Yes, I did say it backwards --- I meant to say "I don't understand why type 1 cannot be used for BD" (or equivalently, why would it only be suitable for one type of modulation, namely AD).

As I understand it, the limitation is that type 1 filters can only be used when the two output pins (the switching/"carrier" signals) are the complement of each other --- and this is clearly not the case with the TPA3251 in BTL mode, since the two audio signals are in principle independent (and if anything, when we ensure that the inputs are the complement of each other, then it will boil down to BD modulation). Correct?

I can understand the distinction between the "carrier" signals being the complement of each other vs. the audio (the low-pass-filtered) signals being the complement of each other. But I can't visualize why this would make any difference on the output when measured differentially as in the BTL configuration. If someone can shed some light, that would be appreciated.

EDIT for some additional thoughts:

The document's argument w.r.t. this issue is a bit questionable.  For type 2 analysis, it states that "From inspection, the type 2 filter can be split into an equivalent common-mode filter" (shown in Figure 13).  But this transformation is only valid because V1 = -V2.  (V1 and V2 being the voltages at the output pins).  But then, if V1 = -V2, then this is AD modulation. 

There's an easy way to solve the circuits and see that the output is proportional to (V1-V2) in both cases.  Consider type 1 configuration using inductances L/2 and capacitance C, and type 2 configuration using inductance L and capacitances C.  Let ZL = sL, ZC = 1 / sC, and Zo = R || ZC  (the capacitor in parallel with the speaker).  For type 1, the current flowing through the loop is:

i = (V1 - V2) / (ZL/2 + Zo + ZL/2).  The voltage at the output is:  Vo = i · Zo =  (V1 - V2) Zo / (Zo + ZL)

For type 2, we can use linear superposition.  Set V2 = 0 and solve for Vo_1.  In the resulting circuit, now the ZL/2 item appears in parallel with R + (ZL/2 || ZC).  The circuit is symmetric, so the expressions will be exactly the same for Vo_1 and Vo_2.  In the end, the output voltage is  Vo = (V1 - V2)  (ZL || ( ··· ) ) / (ZL/2 + (ZC || ( ··· ))) --- I didn't work the equations any further to verify that if V1 = -V2 then the expression reduces to the one for type 1;  but anyway, it is obvious by symmetry that the output has to be proportional to (V1-V2).

For low frequencies, replacing the inductor with a wire and the capacitors with open circuits, we obtain exactly the same output voltage for both filter types.  The question is whether the behaviour for high frequencies is different, and in particular, which filter configuration produces lower contents of high frequencies (if any).

The irony is:  by looking at these equations, the intuition tells me that for BD modulation (when the voltages at the output pins are not constrained to be the complement of each other), the type 2 filter is the one that should not be used --- in any case, it looks like an odd complication that the resistor (the load) "leaks" current from one output pin to the other one when the path through the capacitors avoids that.

Sorry that I may be overcomplicating things;  but I would certainly appreciate any comments on this!

Thanks,
Carlos
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Huh --- there seems to be one "show-stopping number" to my argument ....

The whole point I've been after with the type 1 filter is to use two L/2 inductors instead of two L inductors... As far as the 1/sqrt(LC) formula goes, that's what the math says. However, since the value of R affects the selection of L (provided that one wants critical damping), then with type 1 configuration, now the resistance in parallel with C will be R, and not R/2; thus, for a given cutoff frequency and a given choice of Q factor, the value of the inductance doubles with type 1 configuration, bringing it back to two inductors with inductance L :-(

I'm leaving the previous post (instead of editing to remove this morning's edit) since I still stand by my comment regarding the document's questionable argument in the analysis of type 1 filters.

Thanks,
Carlos
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