Consider a hypothetical scenario where mobility of PMOS (holes) = mobility of NMOS (electrons)

To achieve equal rise and fall time for an inverter built with these transistors:

Question

Consider a hypothetical scenario where mobility of PMOS (holes) = mobility of NMOS (electrons)

To achieve equal rise and fall time for an inverter built with these transistors:

BlackTom AI · Accepted Answer

The question presents a scenario where the mobility of PMOS (holes) is equal to the mobility of NMOS (electrons), and asks what is needed to achieve equal rise and fall times for an inverter built with these transistors.
Option analysis:
Option 1: 'The widths of of PMOS and NMOS must be equal' — this statement claims that using identical channel widths for both transistor types will yield the same drive strength when their mobilities are the same. If mobility is equal, the transistor with the same width will produce similar current for a given overdrive, assuming other factors (thresholds, oxide thickness, equally sized overdrives) are aligned. In short, equal mobility removes the typical asymmetry that requires compensating width ratios, so equality of widths is a reasonable approach to equalize the pull-up and pull-down strengths and, hence, the rise and fall times.
Other potential considerations (not present as separate options here) would include drive strength balance through W/L ratios or process-dependent thresholds. Without differences in mobility, you do not need extra width scaling to compensate for mobility differences, so equal widths is consistent with the stated condition.
Since there are no alternative options provided to critique, there is nothing to refute within the given set. The conclusion follows logically from the assumption of equal mobility: matching widths yields matching drive currents and, therefore, similar rise/fall times.
To reinforce the intuition: in a CMOS inverter, the pull-up is PMOS and the pull-down is NMOS. When their mobilities are equal, the transistors can pass similar current levels for the same overdrive. Matching widths ensures those currents—and thus the charging and discharging rates of the load capacitance—are similar, producing equal rise and fall times.

FA25 ECE 555 001 Q2- Layout/Inverter

Consider a hypothetical scenario where mobility of PMOS (holes) = mobility of NMOS (electrons) To achieve equal rise and fall time for an inverter built with these transistors:

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