As per the book on Digital Integrated Circuits by Rabaey Et al. in the static CMOS implementation of Full adder circuit with Co= carry out and S= Sum enter image description here As mentioned in the marked point of the attached image, the intrinsic load capacitance of the Co signal is 2 diffusion and 6 gate capacitances, plus the wiring capacitance.

  • Can anyone please explain how are these 2 diffusion and 6 gate capacitance are occuring?
  • What are the transistor sizes for this circuit?

1 Answer 1


The claim the circuit is slow contains a mention of ripple-carry adder:

  • it is assumed that \$C_o\$ is routed to \$C_i\$ of exactly one identical adder.
    I think it easy to count six gates labelled \$C_i\$.
  • lamenting the capacitive load on \$C_o\$ suggests to make the transistors driving it wider.

Figure 11-4 doesn't literally implement \$C_o = AB + BC_i + AC_i\$, but an odd mix of
\$C_o = (A + B)C_i + AB\$ (N, pull-down) and \$C_o = (A + B)(AB + C_i)\$ (P, pull-up)

You can reduce the number of gates driven by \$C_i\$ by one or two duplicating one or both of the "internal \$\neg C_o\$ (X) transistors".

  • \$\begingroup\$ (I fail to see 2 diffusion capacitances - yet.) (Going to try and reduce #gate on \$C_i\$.) \$\endgroup\$
    – greybeard
    Commented Apr 7, 2023 at 6:11
  • \$\begingroup\$ The diffusion cap is the inverter driving Co. Yeah, I would not consider this load per se, but on the other hand, it's capacitive, and if it were a complex gate with a single device trying to swing a heavily-dotted output, then it does matter. Like the complex gates shown here tend to have the parallel devices pushed away from the output towards the supplies to reduce "load" on X and S-bar. \$\endgroup\$
    – stevesliva
    Commented May 16, 2023 at 2:09

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