Digital sound - DIGITAL SOUND: DAC: dynamic channel separation method.

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 DAC: dynamic
 channel separation
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(c)VTS'98-99

    I don’t know what THIS is called scientifically-
mu, and is it even called ;). And I will
call this method the Dynamic Method
Channel Separation (MDRC), fortunately invented
I did it myself.

 So, now I will tell you how to
Pull out more channels for the DAC.

 ┌────┐ ┌────┐ ┌────┐
 │RG >│ │DAC>│ │ │ │ ─┼──── A
 D ═╡ ╞════╡ ├─┤ >─┐│ │ ─┼──── B
0-7═╡8bit╞════╡8bit│ │/ ││ │ ─┼──── C
 │ │ │ │ ││ ┼ ─┼──── D
 A ├────┤ └────┘ └┼─┘ │
0-1═╡2bit╞═══╗            ├────┤
    └─┬──┘   ╚════════════╡ADR │
WR    │        2-5M       └────┘
ЦАП───┘       ┌─══─┐
              │    │         ──┐
         2-5M ││  │           ┤ и так
A(BCD)──┬─══──┴o >─┴──>A(BCD)  ┤
        ╪     ┌┤/ высокооммый  ┤ 4 раза
        ┴     ┴   усилитель    ┤
    запоминающий               ┤(A,B,C,D)
    конденсатор              ──┘


    Итак,  обычно  на  каждом канале стоит
свой  запоминающий цифровой регистр, далее
ЦАП и выход канала. Но можно сделать иначе
-  перенести запоминающий элемент в анало-
говую  часть,  где он представляет из себя
простойcapacitor. In this case you will need
only one register, one DAC, analog
key  (analogous to the strobe output of the re-
hyster)  on each channel and the condensation itself
tor. But this has its own problems. Firstly,
discharge will occur due to leakage
capacitor.  Therefore he must have some
high-quality dielectric and connect to vy-
juice amplifier.  In addition, you need
regeneration. Secondly, the inertia of the con-
of the densator appear not only when discharging
de, but also when charging. Those. need to be provided
min charge circuit resistance - first
This also applies to analog keys. A
to increase the charging time, we will use
do not use the duration of the recording strobe in
port (iorq,wr), as for digital register
ra, and the entire interval between the OUT commands
into the channel. For these purposes, it is necessary to enter
where to latch the data for the DAC and the address of the ca-
the person to whom this data is intended.

 And now let's go down to the sinful earth and
let's do a little math.

    So, the resistance of the charge circuit is
row 250 Ohm (switch supply 10-15v), min
charging time t1. Resistanceleaks during
mem 5 MOhm, discharge time t2. Container capacity
capacitor C. Voltage setting error
marriages - 3% (actually - max possible
noise level from the DMRK method).

    Capacitor charge formula:

 U = E*(1-e^(-t/t)) (1)

    Discharge formula:

 U = Uo*e^(-t/t) (2)

 where U is the voltage across the capacitor,
 E - EMF of the current source,
 Uo - initial voltage across the condenser
tore,
 e=2.718... - the basis of natural lo-
garithms,
 t=RC - time constant,
 t - charge (discharge) time - argument
functions.

    From here we have the following:

    To charge:

 t = t/(-ln Q) (3)

    For rank:

 t = t/(-ln(1-Q)) (4)

    where t is the charge (discharge) time,
    Q - installation (saving) error
voltage.

    we have:

 t1 = R1*C = 250*C  ;charge
 t2 = R2*C = 5e+6 *C  ;bit
 t2/t1 = R2*C/R1*C = R2/R1 = 20000

 Q=0.03

 -ln Q = -lnO.03 = 3.5
 -ln(1-Q) = -lnO.97 = 0.03

    you can also write:

 t2/t1 = (t2/0.03)/(t1/3.5) =
 = (t2/t1)*117

 but t2/t1 = 20000:

 t2/t1 = 20000/117 = 171

    i.e. for normal operation with the specified
error and specified resistances
charge/discharge, the discharge time should be
171 times longer charging time.

 Let's set specific values.
So, the min charging time is equal to the time between
two consecutive OUT commands, i.e.
11 CPU cycles.  In the turbo 7 MHz, at worst
case, with a short IORQ, this is the timeequals
11*0.14 = 1.5Chmks. Then max discharge time
yes, i.e.  the regeneration period will be equal to
1.54*171 = 263 µs, hence the min frequency of the
generation 3.8 kHz.  In case of digital sound
this is the min sampling frequency.
    In reality, the charging time will be longer (due to
why put OUTs in a row?), and the period of re-
generation - less (for samples at 8 kHz -
12Smkc).  This makes it possible to reduce Q, i.e.
improve quality.

 To calculate the capacity you need to calculate t1
or t2.

    For example, from formula (3):

 t1 = t1/(-ln Q) = 1.54/3.5 = 0.HHmks,

i.e. 4.4e-7 sec. From here:

 C = t1/R1 = 4.4e-7/250 = 1.76e-9 F,

i.e. 1760 pF...

    p.s.: There is, however, a way to reduce noise
We have practically no use for this method. I
I won’t tell you yet, maybe someone will
guess ;) There will be a diagram, of course
more complicated, but still easier to classify
ches.