A transistor amplifier shows an input current varying from 20 µA to 40 µA, while the output collector current varies from 0.5 mA to 1.5 mA over the same range. What is the small-signal (ac) current gain β_ac = ΔIc / ΔIb?

Introduction / Context:
Small-signal (ac) current gain β_ac captures how changes in base current translate to changes in collector current around a bias point. This is crucial for incremental analysis, gain prediction, and linearized modeling of transistor stages.

Given Data / Assumptions:

• Input current ranges from 20 µA to 40 µA → ΔIb = 20 µA.
• Output current ranges from 0.5 mA to 1.5 mA → ΔIc = 1.0 mA.
• Operation in active region; linear approximation valid over the range.

Concept / Approach:
By definition, β_ac = ΔIc / ΔIb. Convert units consistently, then divide. Keep in mind that β_ac can differ slightly from the DC β, but for many educational problems they are close in value.

Step-by-Step Solution:
1) Compute ΔIb = 40 µA − 20 µA = 20 µA.2) Compute ΔIc = 1.5 mA − 0.5 mA = 1.0 mA.3) Convert ΔIc to µA: 1.0 mA = 1000 µA.4) β_ac = 1000 µA / 20 µA = 50.

Verification / Alternative check:
Back-of-envelope: a 2× increase in Ib (20 → 40 µA) yields a 3× increase in Ic (0.5 → 1.5 mA). The absolute slope from endpoints gives β_ac = 50, consistent with these changes.

Why Other Options Are Wrong:
0.05 and 5: off by orders of magnitude due to unit mistakes.

20: corresponds to ΔIc = 0.4 mA, not 1.0 mA.

500: would require ΔIc = 10 mA for the given ΔIb.

Common Pitfalls:
Unit mix-ups (mA vs µA) and confusing endpoint ratio (1.5/0.5 = 3) with slope ratio (ΔIc/ΔIb). Always compute the incremental change.

Final Answer:
50