Silicon diode conduction check: If the external polarity matches the diode orientation and the measured potential difference across a silicon diode exceeds about 0.7 V, the diode is considered:
-
Anot working
-
Bforward biased
-
Creverse biased
-
Dan open switch
Answer
Correct Answer: forward biased
Explanation
Introduction / Context:Technicians often assess diode orientation and bias using quick multimeter checks. For silicon PN diodes, forward conduction typically begins around 0.6–0.7 V at room temperature. Recognizing this threshold allows rapid fault isolation in rectifiers, clamp circuits, and logic protection paths.
Given Data / Assumptions:
- Silicon diode (not Schottky or germanium).
- Polarity applied in the forward direction.
- Observed potential difference (PD) > 0.7 V.
Concept / Approach:Forward bias reduces the depletion region, allowing majority carriers to cross the junction. The diode’s I–V curve shows an exponential rise in current beyond approximately 0.7 V (silicon), with exact value depending on temperature and current level. Therefore, a PD above ~0.7 V with matching polarity indicates forward bias and conduction.
Step-by-Step Solution:Check polarity: anode at higher potential than cathode.Measure PD: if around or above ~0.7 V, carriers are injected; current flows.Conclude operational state: diode is forward biased.Expect some temperature dependence: hotter junction → slightly lower forward voltage.
Verification / Alternative check:Most DMMs have a diode-test mode; a good silicon junction typically reads 0.6–0.7 V in forward direction, and OL (open) in reverse for low test currents.
Why Other Options Are Wrong:Not working / open switch: contradicts evidence of forward conduction.Reverse biased: would show near-zero current and high resistance at low voltage.
Common Pitfalls:Confusing Schottky diodes (forward drop ~0.2–0.4 V) and germanium (~0.3 V) with silicon. Always consider device type and temperature.
Final Answer:forward biased