CMOS digital logic loading Which one of the following factors does NOT influence the loading (fan-out/speed) of a CMOS gate that is driving several other CMOS inputs?

Introduction / Context:
CMOS loading determines how many inputs a single CMOS output can reliably drive at the desired speed. It is primarily a capacitive loading problem because CMOS inputs behave like small capacitors that must be charged and discharged through finite output resistances.

Given Data / Assumptions:

• Driving device: one CMOS gate output.
• Loads: multiple CMOS gate inputs on the same net.
• We consider rise/fall timing and fan-out constraints.

Concept / Approach:
Propagation delay and edge rates in CMOS are dominated by the RC time constant formed by the driver’s output resistance (Rout) and the total load capacitance (Cload). Cload is the sum of the input capacitances of all receiving gates plus parasitics from traces and packages. Load gates do not present “output capacitance” to the net—only their inputs do.

Step-by-Step Solution:
Identify charging effect: charging time depends on Rout_high * Cload.Identify discharging effect: discharging time depends on Rout_low * Cload.Cload is primarily the sum of input capacitances of the receiving gates and interconnect capacitance.“Output capacitance of the load gates” is not connected to the driven net; therefore it does not affect the driver’s loading.

Verification / Alternative check:
CMOS data sheets specify Cin (input capacitance) and IOH/IOL with corresponding Rout or drive strength; edge rate calculations use Cin, not the output capacitance of other devices.

Why Other Options Are Wrong:
Charging time via driver output resistance: directly impacts rising edges.Discharging time via driver output resistance: directly impacts falling edges.Input capacitance of loads: the dominant component of Cload.

Common Pitfalls:
Confusing input and output capacitances, and forgetting PCB trace capacitance which also adds to Cload.

Final Answer:
Output capacitance of the load gates