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Faber 8th December 2003 08:55 PM

Question on D1 DAC
Hi all!
I'm really a dumb on digital tech, but I'm also curios and I wish to know if somebody can explain me how it works the output part of the Pass' DAC, whose schematic is on the d1-srv-man.pdf file on Passlabs site.

Well, I'm concerning on the U12 and U17 PCM chips.
It seems that the input data are the same, but the output signals follows different path, in fact they are worked out as hey are opposite in phase, obtaining a balanced output.

Ok, where is the trick? In some way the logical port before the PCMs did the work? how?

Thank you in advance,

BobEllis 8th December 2003 09:19 PM

One side is inverted digitally - check the data sheets for the DAC chip, pretty simple to do.

stefanobilliani 8th December 2003 09:27 PM

I believe that the "data" is logically inverted in one of the 2 PCM 63 s.

Anyway a starting point is to read the datasheet of the logic SN74HCB6N and see if they do the invertion that way.


Faber 8th December 2003 09:43 PM

Do you mean that transform 1 -> 0 and 0 -> 1 it's all the work to do to have an exactly inverted waveform?

I'm having some troubles finding the SN74HC86N datasheet but I'm confident to find out something on TI's website


stefanobilliani 8th December 2003 09:53 PM


Originally posted by Faber
Do you mean that transform 1 -> 0 and 0 -> 1 it's all the work to do to have an exactly inverted waveform?


That 1 and 0 are a square wave form that is inverted "logically" speaking . I am confidential with Philips I2S signals where the DATA is inverted just to have an inverted signal(or waveform) at the audio output.


stefanobilliani 8th December 2003 10:00 PM

...or the logic before the PCM63, splits the serial DATA into L channel and R channel ...

more probably ...:)


dggs 8th December 2003 10:49 PM

74HC86 are exclusive OR gates. Data at U9-A is XORed with 0, so it's pass through (1 -> 1 and 0 ->0). Data at U9-B is XORed with 1, so it's inverted (1 -> 0 and 0 ->1). If you use PCM-1704, these inverting logic is built-in. The shcematics only showed left channel. The right channel duplicates the left but uses DOR data.

wayne 8th December 2003 10:52 PM

The inversion is done on the data with one of the 74HC86 gates.
The other gate is set as noninverting. They also act as buffers and impedance matchers.

guido 9th December 2003 12:37 PM

Using just the inverter gives the inverted signal, however not 100%: the correct procedure for inverting would be:

invert all bits except the last bit (LSB).

So doing it this way means that you are one bit 'off'.

Not audible i guess, but another 'useless fact'


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