You are right.  I was pondering this further after I sent my first
message and realized that you could set the second DAC to 1/256 the
total output of the first DAC then add them together.  That does require
extra active components and board space though.  I was on this train of
thought mainly because you cannot do the same thing with ADCs nearly as
easily.

-Adam

Olin Lathrop wrote:

>>Strictly speaking, two 8 bit DACs would be equivilant to one 9 bit DAC.
>> Each DAC can only represent 256 different levels, so two of them can
>>only represent 512 levels.
>>
>
>Strictly speaking, two 8 bit DACs could be equivalent to one 16 bit DAC
>theoretically.  Each DAC can create 256 different levels, so two of them can
>create 256 x 256 = 65535 different levels.
>
>If the two D/As were perfect, you could scale one by 1/256 of the other and
>get a 16 bit D/A.  In reality, errors in the high D/A will cause gaps and
>overlaps in the final output range.  You can, however, scale the low D/A by
>a larger number, like 1/64 for example.  It doesn't need to be a power of
>two, but does need to be large enough that the range of the low D/A can
>cover the worst case range between adjacent LSBs of the high D/A.  That
>would guarantee you can cover the full dynamic range with 14 bit resolution,
>within a 16 bit number space.  Due to errors in the low D/A and crossover
>problems between different values of the high D/A, you won't actually get 14
>bit accuracy, but 12 and maybe 13 bits are achievable.  Another way of
>looking at this is that you can send 65536 different numbers to this
>circuit, of which 4096 of them can be chosen to emulate a 12 bit D/A.  There
>is no way to know which 4096 of the 65536 numbers are the right ones without
>measuring.  You could measure all 65536 voltages, then pick the best 4096 of
>them and store those 16 bit values in a table.  You would now have a quite
>serviceable 12 bit D/A as long as things don't drift much.
>
>I did something like this once where I made a D/A using the standard R / 2R
>ladder, but the ratio wasn't exactly 2.  Each lower bit was worth a little
>more than 1/2 the previous, enough to cover the additional error range of
>the previous resistor.  In this case I strung together 16 bits like this,
>did all the measurements with a computer controlled voltmeter, and
>demonstrated an accurate 10 bit D/A.  This method has other problems, and
>this was only a proof that you don't *need* resistors accurate to 1 part in
>1000 to make a 10 bit D/A with +- 1/2 LSB error.
>
>
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