When a silicon diode is forward biased over its normal operating current range, how does the voltage across the diode behave as current changes?

Difficulty: Easy

Correct Answer: remains approximately the same

Explanation:


Introduction / Context:
In circuit design, diodes are often treated as near-constant-voltage drops when forward biased. This approximation simplifies analysis of limiters, clippers, rectifiers, and reference circuits, though the true I-V curve is exponential.



Given Data / Assumptions:

  • Silicon junction diode operating in forward conduction (tens of microamps to hundreds of milliamps).
  • Temperature is not changing rapidly during the observation.
  • We are discussing qualitative behavior, not an exact exponential equation.


Concept / Approach:
The diode equation i ≈ I_s * (e^(V_d/(nV_T)) − 1) implies that a large change in current corresponds to a relatively modest change in V_d. Practically, V_d hovers around 0.6–0.8 V for silicon across a wide current range, enabling the “constant-voltage” model.



Step-by-Step Solution:
Increase forward current from I1 to I2 by a factor (e.g., 10×).V_d increases only slightly (on the order of ~60–90 mV per decade for silicon at room temperature).Therefore, across normal operating ranges, the forward voltage remains approximately constant.Designers exploit this by modeling the diode as a ~0.7 V drop for quick calculations.


Verification / Alternative check:
Datasheets show forward voltage vs current curves with a relatively flat region. Temperature coefficients (about −2 mV/°C) also shift V_d, but not enough to change the qualitative statement.



Why Other Options Are Wrong:
Direct or inverse proportionality would suggest linear behavior, which is not how diodes operate.

Proportional to source voltage is irrelevant; the diode drop depends on junction physics, not directly on the source value.

Random oscillation is not a correct description; temperature effects are predictable.



Common Pitfalls:
Overusing the constant-voltage model at very low or very high currents where the approximation breaks down. Always consult I-V curves for precision work.



Final Answer:
remains approximately the same

More Questions from Diodes and Applications

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion