## Analysis for Challenge 9

The derivation of CX proceeds as follows:  We first ask, for Figure 1, what total capacitance is “seen” looking to the left of 16C (MSB-3)?  The answer is 16C.  For equivalency in Figure 2, the total capacitance looking into (and from the right of) CX must be 1C.  CX is in series with 16C, so we can write

1/CX + 1/16C  = 1/1C

1/CX               = 1/1C – 1/16C

1/CX               = (16/16 – 1/16)/C

1/CX               =  15/16C

CX               = (16/15)C

which is precisely the value used by Culurciello and Andreou
(see Figure 2, p. 859, at https://engineering.purdue.edu/elab/data/papers/TCASIIadc06.pdf).  Also, please notice that the total required C is nearly a factor of 8 less when compared with the Σ=256C version.

Another Reference

Motorola M68HC11 Reference Manual.  You will notice (Figure 12-3, p. 471) that CX has taken on a slightly different value than what was derived in this circuit challenge.  This is because of the introduction of a half-unit-value (C/2) capacitor.  The use of C/2 is optional; the Motorola write-up (beginning on page 467) explains their reason for including it.  The derivation of CX proceeds in the same manner as above, and will (when C/2 is introduced) yield CX=1.1C.

Is There a Limit to the Number of Bridge Capacitors?

Actually, there can be a bridge capacitor placed between each “bit”.  The result is a C-2C ladder (analogous to the R-2R ladder).  This writer has determined the circuit would look like the figure below (without the switches, comparator circuitry, etc.).