Zener diode small-signal behavior: a zener has breakdown voltage 10 V and zener (dynamic) resistance r_z = 8.5 Ω. When the current is 20 mA, what is the additional voltage drop contributed by r_z (beyond the nominal 10 V breakdown)?

Introduction / Context:
Real zener diodes do not exhibit a perfectly flat I–V curve in breakdown. The slope is represented by a small dynamic resistance r_z, which adds an incremental voltage proportional to current changes. Designers must account for this when creating references or clamps.

Given Data / Assumptions:

• Nominal breakdown V_z0 = 10 V (at or near a test current).
• Dynamic resistance r_z = 8.5 Ω.
• Operating current I = 20 mA = 0.02 A.

Concept / Approach:
The additional voltage due to dynamic resistance is ΔV = I * r_z for the operating point (relative to the intercept). Therefore compute ΔV directly from given I and r_z. If the total zener voltage were requested, it would be V_total ≈ V_z0 + ΔV.

Step-by-Step Solution:
Step 1: Convert current: 20 mA = 0.02 A.Step 2: Compute ΔV = I * r_z = 0.02 * 8.5 = 0.17 V.Step 3: Because the question asks for the additional voltage, report 0.17 V (not the sum).

Verification / Alternative check:
If asked for the total, V_total ≈ 10 V + 0.17 V = 10.17 V; this corresponds to option 10.17 V, but the stem explicitly requests the additional voltage only.

Why Other Options Are Wrong:

• 10.17 V: This is the total voltage, not the increment.
• 20 V and 18.8 V: Unrelated numerically to r_z * I.
• None of the above: Incorrect because 0.17 V is correct.

Common Pitfalls:
Mixing up “incremental” with “total” voltage leads to choosing 10.17 V. Always read whether the problem asks for ΔV or V_total.

Final Answer:
0.17 V.