4-bit inverted R-2R DAC and why better

An inverted (current-mode) R-2R ladder DAC applies Vref to the ladder input. The bits switch each 2R leg's current either to the op-amp's virtual ground (bit 1) or to true ground (bit 0). The op-amp converts the total current into a voltage.

Circuit (4-bit)

 Vref
  |  I     I/2       I/4       I/8
  +---R----+----R----+----R----+
  |        |         |         |
  2R       2R        2R        2R      2R
  |I/2     |I/4      |I/8      |I/16    |
  S3       S2        S1        S0      GND
 (b3)     (b2)      (b1)      (b0)
  1 -> to Iout line   0 -> to GND

  Iout line --> (-) op-amp, Rf from output to (-)
                (+) to GND        Vo = −Iout·Rf

Derivation

Both switch positions are at 0 V (virtual ground or ground), so the ladder always sees the same load. The resistance looking into the ladder from Vref is R, so:

I = Vref / R

At each node the current divides equally (2R down, 2R to the right), so the branch currents are I/2, I/4, I/8, I/16. The current reaching the op-amp:

Iout = (Vref/R)(b3/2 + b2/4 + b1/8 + b0/16)
     = (Vref/R)(D/16)

The op-amp output:

Vo = −Rf · Iout
   = −(Rf/R) · Vref (b3/2 + b2/4 + b1/8 + b0/16)
   = −(Rf/R) · Vref · D / 2⁴

With Rf = R: Vo = −Vref·D/16. For example, 1000 gives −Vref/2.

Why the inverted R-2R ladder is better

• Node voltages stay constant whatever the code, because each switch only moves current between two 0 V points. Stray capacitances do not charge and discharge, so the DAC settles fast with fewer glitches. In the non-inverted ladder the node voltages change with every bit.
• The reference source sees a constant load (R), so it stays steady.
• The switches always work at ground potential, so simple MOS switches give accurate, equal behaviour.