Binary Ladder

The problem

Build a 3-bit R-2R ladder DAC from resistors only. The test bench drives the three bits on the nets “b2” (most significant), “b1” and “b0” with 0 or 1.8 V. The ladder has three nodes: “out” (the output, at the most significant end), “n1” and “n0”. A 2R resistor joins each bit to its node (“b2” to “out”, “b1” to “n1”, “b0” to “n0”); a series resistor R joins “out” to “n1” and “n1” to “n0”; and a 2R resistor from “n0” to ground terminates the ladder.

At every node, looking toward the end, the resistance is 2R: each step splits the current in two, so each bit contributes half as much to the output as the bit above it: Vout = VDD·(4·b2 + 2·b1 + b0)/8. Use R = 10 kΩ and 2R = 20 kΩ.

The test bench

• Supply: not needed.
• The test bench applies the codes 000 to 111 to “b2”, “b1”, “b0” one after another (as separate runs) and measures the voltage of “out” for each. It reports the worst deviation from the ideal staircase in units of one least significant bit (LSB = 225 mV).

Numbers to hit

• ◆Worst deviation from the ideal staircase at most 0.1 LSB