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Old 9th July 2011, 12:58 AM   #11
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Well RIN is still equal to RFB assuming a gain of 1. When I say 'the input impedance is zero' I mean that the impedance of everything BEFORE RIN is zero. Imagine the two RIN's are connected together, which is obviously the case when the volume pot is turned all the way down.
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Old 9th July 2011, 07:54 AM   #12
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Quote:
Originally Posted by Rescue Toaster View Post
Well RIN is still equal to RFB assuming a gain of 1.
I agree that this would lead to a local FDA gain of one, but seeing that the differential voltage divider presents an attenuator even at maximum pot resistance, the overall gain would be below one in case of Rf/Rin, no?

What am I missing? Am I wrong in assuming that the differential divider imposes an attenuation at full volume? I don't see it...

Quote:
When I say 'the input impedance is zero' I mean that the impedance of everything BEFORE RIN is zero. Imagine the two RIN's are connected together, which is obviously the case when the volume pot is turned all the way down.
Woah, that's something entirely different than I took from your first description.

I totally agree that in front of the two Rin everything appears as zero with the pot turned down.
But that's not news, as this would be the case with a regular opamp and an unbalanced volume attenuator, too. Of course I assumed the pot to be shunting everything in front of Rin (and thus, in case of the differential circuit, connecting RinA and RinB) and the gain formula is still a=Rf/Rin.

Basically, when I'm looking at local FDA gain/feedback in the above circuit I have to account for the effective input impedance not being constant.

I now understand that in (the easier) case of the volume pot at zero the gain is simply Rf/Rin, whereas in case of the open pot the total effective input impedance seen by the FDA would resolve to

Rtot = Rin + (2Rout+Rpot)/2 (as explained in chapter 11 of the app note above).

Right?
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Old 9th July 2011, 02:44 PM   #13
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Originally Posted by sek View Post

Rtot = Rin + (2Rout+Rpot)/2 (as explained in chapter 11 of the app note above).

Right?
Input impedance in front of Rin should be (2*Rout || Rpot), and then you would add 2* Rin before doing gain calcs, in terms of differential impedance.

I guess my overall point was as long as Rin & Rfb are reasonably larger than Rout, the gain error shouldn't be too bad. Unless you really *must* know the exact gain of the circuit at every setting of Rpot, who cares? If the gain goes from (for example) 1 at minimum volume to 0.8 or something at max volume, is that such a big deal?
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Old 10th July 2011, 12:00 AM   #14
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Quote:
Originally Posted by sek
Rtot = Rin + (2Rout+Rpot)/2 (as explained in chapter 11 of the app note above)
Sorry, typo! I meant to write:

Rtot = Rin + (2Rout || Rpot)/2

as described in the source I referred it from.

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Originally Posted by Rescue Toaster
Input impedance in front of Rin should be (2*Rout || Rpot) and then you would add 2* Rin before doing gain calcs, in terms of differential impedance.
I agree to your description of the differential input impedance, but incorporated into the gain function a=Rf/Rtot is this impedance split in half.

As of my understanding, gain is calculated from Rf/Rtot (with Rtot being the impedane in front of either FDA input, respectively). According to the literature, differential gain doesn't depend on whether both differential inputs participate in signal contribution (i.e. if one of the inputs is tied to ground).

Although, you could jot it down as: a = 2Rf / 2Rtot

The author of some of TI' app notes on the topic also explains it from a different angle in an article on EE Times (last diagrams and formulae all the way down on the page).

Also related: AN slyt310 (Input impedance matching with fully differential amplifiers) and AN slyt326 (Output impedance matching with fully differential operational amplifiers).

Unfortunately I wasn't able to dig up any literature on variable differential voltage dividers in fully differential amp circuits.

Quote:
I guess my overall point was as long as Rin & Rfb are reasonably larger than Rout, the gain error shouldn't be too bad.
Agreed, although that doesn't help noise performance, nor stability...

Quote:
If the gain goes from (for example) 1 at minimum volume to 0.8 or something at max volume
I'd say that at maximum volume the gain should ideally be one or whatever the final design requires.
It may then rise slightly with reduced volume setting, as the level reduction by the pot will dominate the gain change.

That should be alright.

Quote:
is that such a big deal?
We're safe to say that gain variation can be minimized by increasing FDA loop impedance level.
But the practical FDAs like THS4130 or OPA1632 don't seem to get a manufacturer's recommendation for higher levels of feedback resistor values. Looking at fig.2, p.6 in the THS4130 datasheet, stability/ringing (from high Q) seems to be an issue here. And recommendet values in fig. 38, p.18 barely surpass 4kOhm even for a gain of 10.

This is of serious concern for me.

Maybe I shoud think about trying a circuit topology having a differential attenuator embedded between two differential buffers (see attachment), similar to what user alexw88 did with a dual stepped attenuator in his Balanced Preamp.

Cheers,
Sebastian.
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