Slew rate calculation — check the claim: It takes 4 µs for an op-amp output to transition from −14 V to +14 V (a 28 V step). Is the slew rate 3.5 V/µs as stated, or a different value?

Introduction / Context:Slew rate (SR) is the maximum rate of change of an op-amp’s output voltage and sets a hard limit on how fast the output can swing. It is crucial for large-signal performance, preventing distortion in fast waveforms and step responses. This problem checks a numeric SR claim using simple arithmetic.

Given Data / Assumptions:

• Voltage change ΔV = +14 − (−14) = 28 V.
• Transition time Δt = 4 µs.
• Definition: SR = ΔV / Δt (in V/µs).

Concept / Approach:Compute SR = 28 V / 4 µs = 7 V/µs. The statement “3.5 V/µs” is exactly half of the correct value and therefore incorrect. No RMS versus peak distinction applies here; slew rate is a time-derivative limit on the instantaneous output voltage, determined by internal current limiting and compensation capacitances.

Step-by-Step Solution:

Verification / Alternative check:If the amplifier had a 3.5 V/µs limit, the same 28 V step would require 8 µs, not 4 µs. This cross-check confirms that 4 µs implies 7 V/µs.

Why Other Options Are Wrong:

• Correct: Not correct, as the computed value is 7 V/µs.
• Only correct for peak values, not RMS: Slew rate is not an RMS concept.
• Only correct with capacitive loading: External load may influence behavior but does not alter the basic ΔV/Δt calculation given.

Common Pitfalls:Confusing full-swing step with half-swing; mixing bandwidth limits (small-signal) with slew-rate (large-signal) limitations.

Final Answer:Incorrect