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R2R Dac tech question..

Eoin

pfm Member
So if a DAC is of the resistor ladder type, how does it cope with 16 and 24 bit..

Either 16 or 24 rungs on the ladder, Shirley..

And while we are here, DSD.. how’s that done..

Thanks, experts. 👍
 
Ha great question, Hans Beekhuysen has a review of i think the ares II in which he explains how the magic works
 
So if a DAC is of the resistor ladder type, how does it cope with 16 and 24 bit..

Either 16 or 24 rungs on the ladder, Shirley..
I have never dealt with ladder DACs in practice so this is theoretical. Beware that I may not have learned about better modern techniques and I am open to correction.

As resolution requirements increase, ladder DACs soon end up needing really extreme matching of their resistors. Automated laser trimming of each using a test set can work up to some point.

Beyond that, one of the ways to make a high-resolution ladder DAC is to segment it. In effect there are two DACs. One handling the most significant bits and one handling the least significant bits. Each DAC has a reduced resistor matching requirement. Then take the LSB DAC output, attenuate it by the resolution of the MSB section and add it to the MSB DAC output.

Theoretically simple but practically difficult. I have seen such DACs show monotonicity error at the stitching point but I have also seen some startlingly good ladder DACs recently so people must have found good techniques that I don't know about.
 
I think my question is stupider than that, don’t we need a different ladder with 24 rungs vs a ladder with 16 rungs. The steps would never match up..

Or are they making a 24 bit ladder and resampling a 16 bit signal onto a 24 bit input to the ladder
 
OK I think I get the question (well, I hope so).

If you had a 24-bit R-2R ladder DAC and wanted to convert 16 bit audio data you apply the 16 bits to the 16 most significant inputs to the DAC and put zeros (or digital noise) onto the least significant inputs. This isn't re-sampling (a change to the sample rate), but yes it is re-quantization (a change to the number of bits). You don't need a separate R-2R ladder for 16 bits if you have one that copes with 24.

If you had a 16-bit R-2R ladder DAC and wanted to convert 24 bit audio data (at reduced resolution) you add digital noise to the least significant bits of the audio data and apply the most significant 16 bits of the result to the DAC inputs.

This applies to any technology of DAC, not just R-2R ladders. Digital noise ("dither") is a technique that eliminates "quantization distortion" which happens when you change resolution (16->24 or 24->16 bits) with no precautions. Dither replaces quantization distortion with more benign random noise.

Have I still missed the point? If so sorry.
 
That makes perfect sense.

Never underestimate the magnitude of the stupid. 👍

Thanks!

I should have paid more attention in Jim Audiomisc’s lectures.

📚 🍻
 
I don't know for certain, but guess that if it is MSB first, then a 16 bit word will just activate from bit 24 (MSB) to bit 9 (LSB) for 16-bit audio. The remaining 8 bits will remain unused.
Ooh, I see @John Phillips has already replied as such.
Does raise the Q though...even 16-bit audio (2^16 = 65536. 1/65536 = 0.0000153 or 15 parts-per-million) would therefore require 15ppm accuracy resistors at MSB. Even then, that is +/- 1LSB so I suspect you'd need 10ppm at worst. So, unless you convert it as 2-off 12-bit converters (or 16-bit and 8 subsidiary bits say) and somehow sum the results (with scaling as indicated) you'd need 10 parts-per-billion for 24 bit.... (I think!) 😦
 
In case of any experimentation interest, I've got one of these boards plugged into my rasp pi stack at the moment. Not expensive, and sounds pretty damn good, although you get an audible click at start/stop/sample rate change (somewhat helped by muting control in fifopi, but not completely eradicated).
 
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A ladder DAC requires only R and 2R resistor values, that's the point of the laddering. If the DAC was implemented without the ladder, it would need R, R/2, R/4, R/8 etc, which leads to the problems you have mentioned with laser trimming etc.

Have a look here: https://en.wikipedia.org/wiki/Resistor_ladder

To answer the original question, yes, you load the bits from the MSB, leaving the lower unused bits grounded.
 


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