Steve Hardy wrote:

> Now a question for you gurus: given that each of 4 outputs can be in
> one of 3 possible states, one could conjecture that it is possible to
> make a DAC with 3^4 = 729 possible output levels.  Is this possible in
                        81
> practice?  Ignore the imperfections of the outputs - assume they are
> exactly 0V or 5V or infinite impedance, and your resistors and opamp
> are perfect.  Is it possible to make a linear (not just monotonic)
> DAC?


Consider the op-amp circuit shown below:


                  +5V
                  ^
                  \    I0                   Rf
             2R0  /   --->         +------/\/\/\----+
           R0     \    R0          |   |\           |
    P0 --/\/\/\---+--/\/\/\--+-----+---|-\          |
                             |         |  \         |
                             |         |   \        |
                             |         |U1 /--------+--- Vout
                +5V          |         |  /
                 ^           |    0V --|+/
                 \     I1    |         |/
             2R1 /    --->   |
          R1     \   R1      |
   P1 --/\/\/\---+--/\/\/\---+
                             |
                 +5V         |
                 ^           |
                 \           |
            2R2  /     I2    |
                 \   --->    |
          R2     |    R2     |
   P2 --/\/\/\---+--/\/\/\---+
                             |
                 +5V         |
                 ^           |
                 \           |
            2R3  /      I3   |
                 \     --->  |
          R3     |    R3     |
   P3 --/\/\/\---+--/\/\/\---+

This circuit has an output voltage equal to:

Vout = - (I0 + I1 + I2 + I3)*Rf


Where Ix =

   Ix = 1*5V/(8Rx)   when Px=0V
      = 2*5V/(8Rx)   when Px=open circuit
      = 3*5V/(8Rx)   when Px=5V

Suppose we pick Rf = R3 = 3R2 = 9R1 = 27R0. Then the
output voltage is

Vout = -5/8*((1/3^3)*S0 + 1/(3^2)*S1 + (1/3^1)*S2 + 1*S3)
  Where Si = 1,2,3 for Px = 0,open,5

Here's a little table illustrating the first few values:
    P0    P1    P2    P3   Vout
---------------------------------
     0     0     0     0   -0.9259
    open   0     0     0   -0.9491
     5     0     0     0   -0.9722
     0    open   0     0   -0.9954
    open  open   0     0   -1.0185
     5    open   0     0   -1.0417

etc
     5     5     5     5   -2.7778

The annoying -0.9259V offset voltage is easily removable.
But the series is linear and monotonic, each step being
-23.15mV.

Perhaps someone else would be kind enough to check the
arithmetic...

Scott